DSP Lab Report

DSP Lab Report

Kurian Abraham B070027EC

Shanas P. Shoukath B070059EC

T. Venkateswarlu B070427EC

V. A. Amarnath B070032EC

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Kurian Abraham, Shanas P. ShoukathT. Venkateswarlu and V. A. Amarnath is licensed under a

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Contents

I Lab Report 1 1

1 z Plane and Pole - Zero Plots 11.1 Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 z-Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Pole-Zero Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 FIR lters 52.1 Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Responses of FIR lters . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 First Degree Transfer Function . . . . . . . . . . . . . . . 62.2.2 Linear Phase Filters . . . . . . . . . . . . . . . . . . . . . 82.2.3 Minimum Phase Filters . . . . . . . . . . . . . . . . . . . 9

3 Linear Convolution 113.1 Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Illustrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Exercise 15

II Lab Report 2 16

1 Overlap Save Method 161.1 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2 Overlap Save Routine . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Overlap Add Method 182.1 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Overlap Add Routine . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Inference 20

III Lab Report 3 21

1 Low Pass Filter Design using Window Method 211.1 Window method for FIR lter design . . . . . . . . . . . . . . . . 211.2 Window Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 231.3 Designing a Lowpass Filter . . . . . . . . . . . . . . . . . . . . . 23

1.3.1 Specications . . . . . . . . . . . . . . . . . . . . . . . . . 231.3.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.3.3 Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.3.4 LPF Frequency Response . . . . . . . . . . . . . . . . . . 251.3.5 LPF in action . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.4 Designing a Bandpass Filter . . . . . . . . . . . . . . . . . . . . . 301.4.1 Specications . . . . . . . . . . . . . . . . . . . . . . . . . 301.4.2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.4.3 Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.4.4 BPF Frequency Response . . . . . . . . . . . . . . . . . . 321.4.5 BPF in action . . . . . . . . . . . . . . . . . . . . . . . . . 33

IV Lab Report 4 37

1 Fast Fourier Transform 37

2 Decimation In Time FFT Algorithm 38

3 Decimation In Frequency FFT Algorithm 39

4 Spectrum Analysis 41

5 FIR Filter Design using Frequency Sampling Method 55

V Lab Report 5 61

1 IIR Filters 61

2 IIR Design through Analog Filters 622.1 Butterworth Approximation . . . . . . . . . . . . . . . . . . . . . 622.2 Chebyshev Approximation . . . . . . . . . . . . . . . . . . . . . . 64

3 Transforming Analog Filter to Digital Filter 683.1 Impulse Invariant Technique . . . . . . . . . . . . . . . . . . . . . 683.2 Bilinear Transformation . . . . . . . . . . . . . . . . . . . . . . . 69

4 Design of Filters 724.1 Butterworth Filter Design . . . . . . . . . . . . . . . . . . . . . . 72

4.1.1 Specications . . . . . . . . . . . . . . . . . . . . . . . . 724.1.2 Design using Butterworth approximation . . . . . . . . . 724.1.3 IIT Method . . . . . . . . . . . . . . . . . . . . . . . . . . 734.1.4 BLT Method . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.2 Chebyshev Filter Design . . . . . . . . . . . . . . . . . . . . . . . 754.2.1 Specications . . . . . . . . . . . . . . . . . . . . . . . . 754.2.2 Design using Chebyshev approximation . . . . . . . . . . 754.2.3 IIT Method . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2.4 BLT Method . . . . . . . . . . . . . . . . . . . . . . . . . 77

4

4.3 Integrator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.3.1 IIT Method . . . . . . . . . . . . . . . . . . . . . . . . . . 784.3.2 BLT Method . . . . . . . . . . . . . . . . . . . . . . . . . 78

4.4 Comb Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5 Observations 805.1 Butterworth Filter . . . . . . . . . . . . . . . . . . . . . . . . . . 805.2 Chebyshev Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.3 Integrator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.4 Comb Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

6 Code 966.1 Butterworth Filter . . . . . . . . . . . . . . . . . . . . . . . . . . 966.2 Chebyshev Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.3 Comb Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

VI Lab Report 6 100

1 Lattice Realization of FIR Filters 1001.1 Filter of order 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1001.2 Filter of order 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011.3 Filter of order M . . . . . . . . . . . . . . . . . . . . . . . . . . . 1021.4 Design Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . 1031.5 Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1041.6 Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1051.7 Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

VII Lab Report 7 111

1 Introduction 111

2 Levinson Durbin Recursion Algorithm 111

3 Inverse Filter 117

4 Observations 118

5 Code 126

6 Inference 130

VIII Lab Report 8 131

1 Introduction 131

5

2 Quantisation Technique 1312.1 Uniform Quantization . . . . . . . . . . . . . . . . . . . . . . . . 1312.2 Non Uniform Quantization . . . . . . . . . . . . . . . . . . . . . 1322.3 PDF Optimized Non Uniform Quantizer . . . . . . . . . . . . . . 132

2.3.1 Determination of optimum ∆k . . . . . . . . . . . . . . . 133

3 Logarithmic Companding 1353.1 A-law Companding . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.2 µ-law Companding . . . . . . . . . . . . . . . . . . . . . . . . . . 137

4 Digital Companding:Segmentation of Companding Characteristics 1374.1 Segmentation of A-law Companding Characteristics . . . . . . . 1384.2 Segmentation of µ-law Companding Characteristics . . . . . . . . 138

5 Observations 140

6 Code 143

Part I

Lab Report 1

Date: 31st December, 2009

1 z Plane and Pole - Zero Plots

1.1 Concepts

IntroductionLaplace transform is utilized for analysis of continuous time signals whereas

z-transform is used to analyse discrete time signals. z-transform is dened as

X(z) =

∞∑n=−∞

x(n)z−n (Eq. I.1.1)

where z is a complex variable.

Laplace domain to z domain

X(s) =

∞

−∞

x(t)e−stdt (Eq. I.1.2)

where s = σ + jω.When x(t)is discretized to x(n),

t = nT

where T is the sampling period.

z = esT (Eq. I.1.3)

⇒ X(z) =

∞∑n=−∞

x(n)z−n

Mapping from s-plane to z-planeFrom (Eq. I.1.3),

z = eσejω

= rejw (Eq. I.1.4)

When s-plane is mapped to z-plane, the jω axis maps onto unit circle. Thepoints on LHP of s-plane maps onto interior of this circle and the points onRHP maps onto exterior.

1

Pole-Zero PlotLet

H(z) =N(z)

D(z)

Solving N(z) = 0 and D(z) = 0 we obtain the zeros and poles of H(z).

Stability Criterion All poles of H(z) should lie within the unit circle.

1.2 z-Plane

Code

clf;

w = linspace(0,2*pi,100);

r = 1; %plot z plane

z = r*exp(1j*w);

plot (z);

axis equal;

hold on;

n = [1 2 3];

d = [1 0 0];

ze = roots(n); %finding zeros and poles

po = roots(d);

plot (real(ze), imag(ze), òr');

hold on;

plot (real(po), imag(po), `xr');

xlabel(`real z');

ylabel(ìmag z');

title(`z-Plane');

grid on;

2

Figure

Figure 1: z-Plane

1.3 Pole-Zero Plot

Code

clf;

w = linspace(-pi,pi,1000);

r = 1;

z = r*exp(1j*w);

plot (z); %plotting z plane

axis equal;

hold on;

z = exp(1j*w);

n = [1, 2, 3, 4];

d = [1, 0, 0, 0];

ze = roots(n);

po = roots(d);

plot (real(ze), imag(ze), òr'); %plot zeros and poles

hold on;


xlabel(`real z');

ylabel(ìmag z');

title(`Pole Zero Plot');

grid on;

3

Figure

Figure 2: Pole Zero Plot

1.4 Inference

For the given transfer function, the pole-zero plot is obtained(see gure 2). Here,we observe that all three zeros are outside the unit circle and we have multiplepoles at origin. Hence, the given transfer function is stable whereas the inversetransfer function is not stable.

4

2 FIR lters

2.1 Concepts

IntroductionFilters with nite impulse responses are called FIR lters. A lter with

transfer function H(z) is stable if all its poles lie within the unit circle.A causal FIR lter can be represented as

H(z) =

N−1∑n=0

h(n)z−n (Eq. I.2.1)

where h(n) is its impulse response. There exists (N − 1) zeros for H(z) and(N − 1) poles which lies on z = 0. Therefore, FIR lters are stable.

Linear Phase FiltersLinear phase lters do not alter the shape of the original signal given to it.

It simply introduces a phase shift to the signal. So, linear phase lters do notcause distortion to the input signal.

Condition for linear phase

h(n) should be finite durationN (Eq. I.2.2)

h(n) = h(N − 1− n) (Eq. I.2.3)

Solving,

H(ejω) =

e−jω (N−1)

2

[2∑N

2 −1n=0 h(n) cos

(ω(n− N−1

2 ))]

N is even

e−jω(N−1)

2

[h(N−1

2 ) + 2∑N−3

2n=0 h(n) cos

(ω(n− N−1

2 ))]

N is odd

(Eq. I.2.4)From (Eq. I.2.4), the underlined term represents the phase

φ = −ω (N − 1)

2

Hence, any lter with transfer function satisfying (Eq. I.2.2) and (Eq. I.2.3)is linear phase.

Minimum Phase FiltersA lter is called minimum phase lter when its transfer function and inverse

transfer functions are stable and causal. A stable lter has all its poles withinthe unit circle whereas its zeros can be anywhere on z-plane. Thus, for minimumphase lter all its zeros and poles lie within the unit circle.

5

2.2 Responses of FIR lters

2.2.1 First Degree Transfer Function

H(z) = 1 + z−1

CodeHere we have plotted real and imaginary part of H(ejw) versus ω.

clf;

w = linspace(-3*pi,3*pi,1000);

r = 1;

z = r*exp(1j*w);

figure(1);

hold on;

plot (real(z), imag(z)); %plotting z plane

axis ([-2,2,-2,2]);

axis equal;

z = exp(1j*w);

n = [1, 1];

d = [1, 0];

ze = roots(n);

po = roots(d);

plot (real(ze), imag(ze), òr'); %pole zero plot

hold on;


xlabel(`real z');

ylabel(ìmag z');

title(`Pole Zero Plot');

grid on;

figure(2);

h = 1 + 1*z.^(-1); %finding DTFT

subplot (2,1,1);

plot (w,real(h));

grid on;

title (`magnitude response'); %magnitude response

xlabel (`w');

ylabel (`magnitude');

subplot (2,1,2);

plot (w,imag(h));

grid on;

title (`phase response'); %phase response

xlabel (`w');

ylabel (`phase');

6

Figures


Figure 4: Real and Imaginary

Inference

H(ejω) = 1 + e−jω

Real part of H(ejω) is cos(ω)with a DC oset 1 and imaginary part −sin(ω).

7

2.2.2 Linear Phase Filters

H(z) = 1 + 2z−1 − 3z−2 + 2z−3 + z−4

CodeUsing code on on page 6, and replacingn = [1, 2, -3, 2, 1];

d = [1, 0, 0, 0, 0];

And then plotting magnitude and phase responses.

Figures


8

Figure 6: Magnitude and Phase Response

InferenceFrequency response is obtained by taking DTFT of the transfer function.

That is evaluating the transfer function along the unit circle once. Zeros try tobring the magnitude response to zero and poles try to make it innity. When ω isvaried from -π to π, whenever the product of the Eucledian distances from zerosis minimum, magnitude reponse will encounter a local minimum. Similarly, inthe case of poles response will have a local maxima. Zeros lying on unit circlemakes the response zero and poles on unit circle makes the response innite atthat ω.

2.2.3 Minimum Phase Filters

H(z) = 1 + z−1 +z−2

4

CodeUsing code on on page 6, and replacingn = [1, 1, .25];

d = [1, 0, 0];

And then plotting magnitude and phase responses.

9

Figures


Figure 8: Magnitude and Phases Response

10

3 Linear Convolution

3.1 Concepts

y(n) =

∞∑k=−∞

x(n− k)h(k) =

∞∑k=−∞

x(k)h(n− k)

where x(n) is the input to the LTI system with impulse reponse h(n) andy(n) is its output.

y(n) = x(n) ∗ h(n) (Eq. I.3.1)

In z-domain,Y (z) = X(z)H(z) (Eq. I.3.2)

3.2 Convolution

Code

x = [1 2 3];

y = [3 2 1];

xn = length (x);

yn = length (y);

zn = xn+yn-1;

z = zeros(1, zn);

x1 = [x, zeros(1, yn-1)]; %zero padding

y1 = [y, zeros(1, xn-1)];

for i = 1:zn %convolving

for j = 1:i

z(i) = z(i) + x1(j)*y1(i-j+1);

end

end

z

Resultz = 3 8 14 8 3

3.3 Illustrations

Unit impulse response (rect(n) ∗ δ(n))

Code

x = 0:1:22;

y = [1,zeros(1,20)];

H = [ 1, 1, 1 ];

xn = length (H); %convolution

yn = length (y);

11

zn = xn+yn-1;

z = zeros(1, zn);

x1 = [H, zeros(1, yn-1)];

y1 = [y, zeros(1, xn-1)];

for m = 1:zn

for n = 1:m

z(m) = z(m) + x1(n)*y1(m-n+1);

end

end

clf;

plot(x,z);

axis([0,10,0,4]);

grid on;

Figure

Figure 9: Impulse Response

12

Unit step response (rect(n) ∗ u(n))

Using code from on page 11 withy = ones(1,21);

Figure

Figure 10: Step response

Exponential Input (rect(n) ∗ exp(n))

Using code from on page 11 witha = 0:1:7;

y = exp(a);

Figure

Figure 11: rect(n) ∗ exp(n)

13

Square input


y = square(a);

Figure

Figure 12: Square input

Sinusoidal input


y = sin(a);

Figure

Figure 13: Sinusoidal input

14

4 Exercise

Are all linear phase lters, minimum phase?

Consider a linear phase lter with transfer function,

H(z) = a0 + a1z−1 + a2z

−2 + ...+ anz−n (Eq. I.4.1)

where a0 = 1 for FIR lters. Let zeros of Eq. I.4.1 be ri, where i = 0, 1, 2, ...n.Hence, we have

∏i

ri =

−1 n is odd

1 n is even(Eq. I.4.2)

Assuming here n is even, we may have 2k zeros which are complex conjugatesand (n− 2k) real zeros. Now, consider dierent cases of absolute value of thesezeros and their corresponding locations on z-plane.

1. absolute value of each zeros is one⇐⇒ all of these lie on unit circle. Thisviolates condition specied in 2.1 for minimum phase lters.

2. absolute value of each complex zeros is one and real zeros exist in α and1/α form where α < 1 ⇐⇒ zero with absolute value 1/α lies outside theunit circle violating the minimum phase condition.

3. absolute value of each real zeros is one and complex zeros have absolutevalue greater than one ⇐⇒That is, both the complex zeros lies outsidethe unit circle violating the minimum phase condition.

4. absolute value of all zeros are not equal to one, is a combination of abovementioned cases 3 and 4 ⇐⇒ This one also violates the minimum phasecondition.

Corollary: For a minimum phase lter to be linear phase the transfer functionshould be symmetric and should satisfy Eq. I.4.2. From above case 1 - 4, thisis never satised.

15

Part II

Lab Report 2

Date: 7th January, 2010

1 Overlap Save Method

1.1 Concept

The sequence x(n) of length n is the input to the LTI system with impulseresponse h(n) of length m( n). x(n) is partitioned into blocks of length l withrst(m−1) terms are taken from previous block. Product of DFT of blocks andof h(n) is computed. IDFT of the result gives the circularly convolved result.First (m−1) terms of each resultant block is discarded and the remaining termsgives the result.

Figure 14: Overlap Save

16

1.2 Overlap Save Routine

Code

%routine for h*x

h=[1,1,1];

x=[3,-1,0,1,3,2,0,1,2,1];

l=3;

%linear convolution result

conv(x,h)

m=length(h);

L=length(x);

%creating blocks of length `l'

y=[zeros(1,m-1),x];

r=rem(length(y),(l-(m-1)));

y=[y,zeros(1,r)];

if r==0

y=[y,zeros(1,m-1)];

end

res = zeros(1,L+m-1);

k=1;

pos=0;

H=fft(h,l);

while k <= length(res)

temp=zeros(1,l);

for p=0:l-1

temp(p+1)=y(k+p);

end

temp=fft(temp,l);

temp=temp.*H;

temp=ifft(temp,l);

p=1;

while p <= l-m+1

res(p+pos)=temp(p+m-1);

p=p+1;

end

pos=pos+l-m+1;

k=k+l-m+1;

end

res

Output

ans =

3 2 2 0 4 6 5 3 3 4 3 1

res =

3 2 2 0 4 6 5 3 3 4 3 1

17

2 Overlap Add Method

2.1 Concept

The sequence x(n) of length n is the input to the LTI system with impulseresponse h(n) of length m( n). x(n) is partitioned into blocks of nite length,l. These blocks are linearly convolved with h(n) one by one. The result ofprevious step gives sequences of length (l +m− 1). This result contains errorsin rst and last (m − 1) terms due to original sequence x(n) being partitionedinto blocks. The last (m−1) terms of a block is added to the rst (m−1) termsof succeeding block with all other terms retained as such.

Figure 15: Overlap Add

18

2.2 Overlap Add Routine

Code

%routine for h*x

h=[1,1,1];

x=[3,-1,0,1,3,2,0,1,2,1];

l=3;

%linear convolution result

conv(x,h)

m=length(h);

n=length(x);

%creating blocks of length `l'

x=[x,zeros(1,l-rem(n,l))];

xn = length(x);

k=1;

res=zeros(1,n+m-1);

pos=1;

while k <= xn

temp = zeros(1,l+m-1);

for p=1:l

temp(p)=x(p+k-1);

end

temp=conv(temp,h);

k=k+l;

for p=1:l+m-1

res(pos)=res(pos)+temp(p);

if (pos>=n+m-1)

break;

end;

pos=pos+1;

end;

pos=pos-m+1;

end;

res

Output

ans =

3 2 2 0 4 6 5 3 3 4 3 1

res =

3 2 2 0 4 6 5 3 3 4 3 1

19

3 Inference

Both overlap save and overlap add methods are ecient in computing linearconvolution of long sequences. These techniques reduce computational com-plexity.

References

[1] Emmanuel C. Ifeachor and Barrie W. Jervis, Digital Signal Processing: A

Practical Approach, 2nd Ed., Pearson Education

[2] Bill Rison, EE 451: Digital Signal Processing, http://www.ee.nmt.edu/~rison/ee451_fall99/

20

Part III

Lab Report 3


1 Low Pass Filter Design using Window Method

1.1 Window method for FIR lter design

As we know, the impulse response hd(n) and the frequency response of a lterHd(ω) is related as,

hd(n) =1

2π

π

−π

Hd(ω)ejωndω, −∞ < n <∞ (Eq. III.1.1)

−5 0 5−0.5

0

0.5

1

n →

Ideal lowpass filter impulse response

−5 0 50

0.5

1

1.5

2

w →

Ideal lowpass filter frequency response

Figure 16: Ideal Low Pass Filter

21

Evaluating hd(n) from ideal low pass lter Hd(ω) as shown in gure 16,

hd(n) =

2fcsin(nωc)

nωc, n 6= 0,−∞ < n <∞

2fc, n = 0(Eq. III.1.2)

where fc is the cuto frequency.The impulse response plotted in reveals that hd(n) is symmetrical about

n = 0. Hence, the lter will have a linear phase reponse. But, the practicalproblem is that hd(n) extends from −∞ < n <∞, i.e. the lter is not FIR.

Eq. III.1.2 derived above is a non causal lter. If we take the frequencyresponse, Hd(ω)

Hd(ω) =

e−jωα |ω| ≤ ωc0 otherwise

where −α is the slope of phase response Hd(ω). That is in time domainevery frequency component undergoes a delay of α.

From Eq. III.1.1 we have, hd(n)

hd(n) =

sin(ωc(n−α))

π(n−α) , n 6= αωcα , n = α

(Eq. III.1.3)

The causal FIR is obtained by windowing Eq. III.1.3 with w(n)(exists from0 to N-1).

h(n) = hd(n) ∗ w(n)

where hd(n) is symmetric about α and w(n) is symmetrical about (N−1)/2.Hence, for h(n) to be symmetric,

α =(N − 1)

2

Using a rectangular window introduces ripples and overshoots near the tran-sition region in the frequency response. This phenomenon is called Gibb's phe-nomenon.

To minimize the ripples we need to use windows with smoother transitionand to get a close approximation to Hd(ω) we need to retain as many coecientsof hd(n) as possible. Several window functions available to obtain a FIR lterfrom hd(n). The selection of the window is based on the stopband attenuationrequired and the order of the lter.

22

1.2 Window Functions

Name Time domain sequence [0 ≤ n < N ] k Min. stopband att.

Bartlett 1-2|n−N−1

2 |N−1 4 -25 dB

Hanning 12

(1− cos

(2πnN−1

))4 -44 dB

Hamming 0.54− 0.46 cos(

2πnN−1

)4 -53 dB

Blackman 0.42− 0.5 cos(

2πnN−1

)+ 0.08 cos

(4πnN−1

)6 -74 dB

1.3 Designing a Lowpass Filter

1.3.1 Specications

Cuto frequency, fc = 1000HzStopband edge frequency, fs = 1500HzPassband attenuation, Ap ≈ 0 dBStopband attenuation, As ≥ 50 dBSampling frequency, fsam = 10000Hz

For a stop band attenuation of 50 dB, Hamming window can be used. Therequired order of lter, N is given by

N =4fs

transitionwidth

Substituting values, we obtain N = 81. (Here we have assumed fc = fp).

Note: h(n) is designed using ωc = ωp. Verify if h(n) meets the designspecications. If passband attenuation is not satised choose a dierent ωcvalue. and repeat the design steps. Also try to minimize N maintaining thespecications for ease of implementation.

1.3.2 Design

Hamming widow is given by,

w(n) = 0.54− 0.46 cos

(2πn

N

), 0 ≤ n < N (Eq. III.1.4)

The windowed impulse response is given by,

h(n) = hd(n) ∗ w(n), 0 ≤ n < N

23

1.3.3 Code

%Low Pass Filter Design

sam = 10000;

fp = 1000/sam;

fs = 1500/sam;

att = -50;

fc = fp;

%selecting Hamming window

k = 4;

N = ceil(k/(fs-fp));

a = (N-1)/2;

%windowing

n = 1:1:81;

hd = (sin(2*pi*fc.*(n-1-a))./(pi*(n-1-a))).*(.54-(.46*cos(2*pi*(n-1)./80)));

hd(a+1) = 2*fc;

%Frequency Transformation

H=fft(hd,1800);

x = linspace(-pi,pi,1800);

%plotting magnitude response

plot(x,fftshift(20*log(abs(H))));

grid on;

xlabel('Frequency (\omega) \rightarrow');

ylabel('|H(e^j\omega) (db) \rightarrow');

title('Low Pass Filter Response');

%impulse as input

x=[zeros(1,1000),1,zeros(1,1000)];

l=100;

m=length(hd);

n=length(x);


subplot(2,1,1);

plot(x);

title('Unit Impulse Input');

xlabel('Time\rightarrow');

ylabel('Amp\rightarrow');

%convolution through overlap add method on page 18

xn = length(x);

k=1;

res=zeros(1,n+m-1);

pos=1;

while k <= xn

temp = zeros(1,l);

for p=1:l

temp(p)=x(p+k-1);

24

end

temp=conv(temp,hd);

k=k+l;

for p=1:l+m-1


if (pos>=n+m-1)

break;

end;

pos=pos+1;

end;

pos=pos-m+1;

end;

%plotting

subplot(2,1,2);

plot(res)

title('Unit Impulse Response');

xlabel('Time\rightarrow');

ylabel('Amp\rightarrow');

axis([950,1150,-.1,.3]);

1.3.4 LPF Frequency Response

−4 −3 −2 −1 0 1 2 3 4−300

−200

−100

0

100

Frequency (ω) →

|H(e

jω)|

(d

b)

→

Low Pass Filter Response

−1 −0.5 0 0.5 1−4

−2

0

2

4

Frequency (ω) →

an

gle

(H(e

jω)

→

Low Pass Filter Response

Figure 17: Frequency Response

25

1.3.5 LPF in action

0 500 1000 1500 2000 25000

0.2

0.4

0.6

0.8

1Unit Impulse Input

Time→

Am

p→

950 1000 1050 1100 1150−0.1

0

0.1

0.2

0.3Unit Impulse Response

Time→

Am

p→


0 100 200 300 400 5000

0.5

1

1.5

2Unit Step Input

Time→

Am

p→

0 100 200 300 400 500−1

0

1

2Step Response

Time→

Am

p→

Figure 19: Step Response

26

0 20 40 60 80 100 1200

0.2

0.4

0.6

0.8

1Exponential Input

Time→

Am

p→

0 50 100 150 200−0.1

0

0.1

0.2

0.3Exponential Response

Time→

Am

p→

Figure 20: Exponential Response

0 50 100 150 200−1

0

1

2Square Input

Time→

Am

p→

0 50 100 150 200−2

−1

0

1

2Square Response

Time→

Am

p→

Figure 21: Square Response

From fourier series of square wave, we know that it has its natural frequencyand its odd harmonics as frequency. So when we pass it through a low passlter, all frequency components beyond the cut o freequency is attenuated and

27

so we obtain a distorted square wave as output.

0 20 40 60 80 100−2

−1

0

1

2Sinusoidal Input

Time→

Am

p→

fc = 500 Hz

0 20 40 60 80 100−1

−0.5

0

0.5

1Sinusoidal Response

Time→

Am

p→

(a) fc in pass band

0 20 40 60 80 100−2

−1

0

1

2Sinusoidal Input

Time→

Am

p→

fc = 1.8 kHz

0 20 40 60 80 100−1

−0.5

0

0.5


Time→

Am

p→

(b) fc in stop band

Figure 22: Sinusoidal Input

28

0 20 40 60 80 100−4

−2

0

2

4Sinusoidal Input

Time→

Am

p→

fc1

= 300 Hz, fc2

= 2000 Hz

0 2000 4000 6000 8000 10000 120000

1000

2000

3000

4000Frequency Response

Frequency (ω) →

Figure 23: Composite Sinusoidal Input

Lets give a composite sinusoid as in gure 23. According to our design, cuto frequency = 1000 Hz.

x(t) = sin(600πt) + sin(4000πt)

x(t)have frequency components of 300Hz and 2kHz. When we pass x(t)through the designed LPF, we expect it to remove 2kHz component. Theoutput and its frequency response is shown in gure 23. We see that we haveonly 300Hz component in the output as expected.

We observe that all the outputs are delayed by 40 as α = 40.

29

1.4 Designing a Bandpass Filter

1.4.1 Specications

Passband: −1 ≤|H(ejω)| ≤ 1 and 0.42π ≤ ω ≤ 0.61πStopband: |H(ejω)| < −60dB and 0 ≤ ω ≤ 0.16π, 0.87π ≤ ω ≤ π

1.4.2 Design

The Bandpass lter is obtained as the dierence of two low pass lters havingcuto frequencies ωc1and ωc2. ωc1and ωc2 are selected in such a way that ripplesin the passband is as specied. The window function is selected depending onthe stopband attenuation.

ωc1 = 0.3πωc2 = 0.7πωs1 = 0.16πωs2 = 0.87πωp1 = 0.42πωp2 = 0.61π

Stopband Attenuation ≤ −60dB =⇒ k = 6

N = max(2πk

ωs2 − ωp2,

2πk

ωp1 − ωs1) = 47

α =N − 1

2= 23

∴ The window selected is Blackman window.

w(n) = 0.42− 0.5 cos

(2πn

N − 1

)+ 0.08 cos

(4πn

N − 1

)

hd(n) =sin(ωc2(n− α))

π(n− α)

∴ Required response is

h(n) = hd(n)w(n)

1.4.3 Code

%band pass filter design

%specifications

wc2=.61*pi;

wc1=.42*pi;

ws1=.16*pi;

ws2=.87*pi;

30

N=min((wc1-ws1),(ws2-wc2));

N=ceil(20*pi/N);

if rem(N,2)~=1

N=N+1;

end;

a=(N-1)/2;

%selecting Blackman Window

k=1:N;

hd=(sin(wc2*(k-1-a))./(pi*(k-1-a))-sin(wc1*(k-1-a))./(pi*(k-1-a)))

W=(.42-(.5*cos(2*pi*(k-1)./(N-1)))+(.08*cos(4*pi*(k-1)./(N-1))));

h(k)=hd.*W;

h((N+1)/2)=(wc2-wc1)/pi;

x=linspace(-pi,pi,1800);

H=fft(h,1800);

%plotting

plot(x,fftshift(20*log(abs(H))));

grid on;

xlabel('Frequency \rightarrow');

ylabel('|H(e^j\omega)| \rightarrow');

title ('Band Pass Filter Response');

%unit impulse input

x=[1,zeros(1,100)];

%plotting input

subplot(2,1,1);

plot(x)

axis([-100,100,0,2])

xlabel('Time \rightarrow')

ylabel('Amp \rightarrow')

Title('Unit Impulse Input')

%convolution through overlap add method given 2

l=100;

m=length(h);

n=length(x);


xn = length(x);

k=1;

res=zeros(1,n+m-1);

pos=1;

while k <= xn

temp = zeros(1,l);

for p=1:l

temp(p)=x(p+k-1);

end

temp=conv(temp,h);

k=k+l;

for p=1:l+m-1

31


if (pos>=n+m-1)

break;

end;

pos=pos+1;

end;

pos=pos-m+1;

end;

%plotting output

subplot(2,1,2);

kl = linspace(-pi,pi,n+m-1);

plot(res);

xlabel('Time \rightarrow');

ylabel('Amp \rightarrow');

Title('Unit Impulse Response');

1.4.4 BPF Frequency Response

−4 −3 −2 −1 0 1 2 3 4−350

−300

−250

−200

−150

−100

−50

0

50

Frequency →

|H(e

jω)|

→

BandPass Filter Response


32

1.4.5 BPF in action

−100 −50 0 50 1000

0.5

1

1.5

2

Time →

Am

p →

Unit Impulse Input

0 50 100 150−0.4

−0.2

0

0.2

0.4

Time →

Am

p →

Unit Impulse Response


−100 −50 0 50 1000

0.5

1

1.5

2

Time →

Am

p →

Step Input

0 50 100 150−0.4

−0.2

0

0.2

0.4

Time →

Am

p →

Step Response

Figure 26: Step Response

For step input to a band pass lter, we observe that dc component of input isremoved as low frequency components are eliminated.

33

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1

Time →

Am

p →

Exponential Input

0 50 100 150−0.4

−0.2

0

0.2

0.4

Time →

Am

p →

Exponential Respnse

Figure 27: Exponential Response

0 50 100 150 200−2

−1

0

1

2

Time →

Am

p →

Square Input

fc=.45pi

0 50 100 150 200−2

−1

0

1

2

Time →

Am

p →

Square Response

Figure 28: Square Response

34

0 20 40 60 80 100−2

−1

0

1

2

Time →

Am

p →

Sinusoidal Input

sin(.45pi.x)

0 20 40 60 80 100−2

−1

0

1

2

Time →

Am

p →

Sinusoidal Response

(a) in passband

0 20 40 60 80 100−2

−1

0

1

2

Time →

Am

p →

Sinusoidal Input

sin(.91pi.x)

0 2000 4000 6000 8000 10000 12000−0.1

−0.05

0

0.05

0.1

Time →

Am

p →

Sinusoidal Response

(b) in stop band

Figure 29: Sinusoidal Response

35

0 20 40 60 80 100−2

−1

0

1

2

Time →

Am

p →

Sinusoidal Input

sin(.91pi.x)+sin(.45*pi.x)

0 2000 4000 6000 8000 10000 120000

2000

4000

6000

Frequency(ω) →

Sinusoidal Response

Figure 30: Composite Sinusoidal Input

When a composite sinusoid is given as input, we observe that only sinusoidsof frequency fp1 < f < fp2 are not attenuated.

We observe that all the outputs are delayed by 23 as α = 23.

References


Practical Approach, 2nd Ed., Pearson Education

[2] John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:

Principles, Algorithms, and Applications, 3rd Ed., Prentice Hall Interna-tional, Inc

36

Part IV

Lab Report 4


1 Fast Fourier Transform

Concepts

Circular convolution of two sequences can be done by calculating the discretefourier transforms (DFT) of two sequences and multiplying those two DFTs andtaking inverse discrete fourier transform of the resulted signal will give the re-sult. But when the sequences are of very long length, nding the DFTs of suchsequences takes more time and computations, because it needs to perform anumber of multiplications proportional to the square of the length of the se-quences say N , after padding them with sucient number of zeroes. So tocompute discrete fourier transform faster than this, the need to have an algo-rithm which reduces the number of multiplications and one of those algorithmsis fast fourier transform (FFT).

In this algorithm, the periodicity and symmetry properties of the twiddlefactor, ωknN = e−j(2π/N)kn is exploited which reduces the total number of com-putations.

Symmetry property

ωN (k + N/2) = −ωN (k)

Periodicity property

ωN (k +N) = ωN (k)

In fast fourier transform, the sequence is split into two subsequences of lengthN/2 and calculate the N/2 point DFT of those two sequences and adding thosetwo DFTs to calculate the DFT of original sequence. This process is repeatedby making this N/2 point DFTs into two subsequences and so on until 2 pointDFTs is reached. In this process the total number of multiplications reduces toa value equal to N ∗ log2N from a value which was proportional to N2.

Types of Fast Fourier Transforms

There are two types of fast fourier transforms depending upon whether samplingis done in the time domain or in frequency domain.

1. Decimation In Time FFT

2. Decimation In Frequency FFT

37

2 Decimation In Time FFT Algorithm

Here the original signal is sampled into two subsequences depending upon theirposition. All the samples present in odd places are grouped into one sequenceand all the samples present in even places are grouped into another sequence.Then nd the N/2 point DFTs of both sequences. Let g(n) represent evennumbered samples sequence and h(n)represent odd numbered sample sequence.These sequence are added using symmetry and periodicity of twiddle factor,ωN . In this way a 8 point DFT is reduced to two 4 point DFTs and the numberof multiplications can still be reduced by computing the two 2 point DFTs tocalculate each of these 4 point DFTs.

Properties

1. Input is in bit reversal order

2. Output is in normal order

3. In place computation is possible

Figure 31: DIT FFT

38

X[k] =

N−1∑n=0

x[n]ωknN , k = 0, 1, ..., N − 1

then separating even and odd terms, and substituting n = 2r for n even andn = 2r + 1 for n odd.

X[k] =

(N/2)−1∑r=0

x[2r](ω2N

)kr+ ωkN

(N/2)−1∑r=0

x[2r + 1](ω2N

)kras ω2

N = ωN/2

X[k] =

(N/2)−1∑r=0

x[2r]ωkrN/2 + ωkN

(N/2)−1∑r=0

x[2r + 1]ωkrN/2

= G[k] + ωkNH[k], k = 0, 1, ..., N − 1

First term is N/2 point DFT of even terms and second term is N/2 point DFTof odd terms. This procedure can be repeated till two point DFTs are reached.Then making use of symmetry and periodicity of twiddle factor, the requiredN point DFT is obtained. The ow graph of 8 point DFT using DIT FFT isshown in gure 31 on the previous page.

3 Decimation In Frequency FFT Algorithm

Here the sequence of length N is split into two sequences of length N/2 by takingthe rst half of total number of samples. A sequence, g(n) is made by addingthe element in the rst sequence to the corresponding element in the secondsequence. Similarly, another sequence, h(n) is made by subtracting an elementof second sequence from the corresponding element in rst sequence. Then theN/2 point DFTs of g(n) and h(n) ∗ ωnN to get N point DFT of the originalsequence, where ωN is twiddle factor. These sequences can be further split intosubsequences in the same manner as mentioned above and apply the procedure,so that total number of multiplications reduces.

Properties

1. Input is in normal order

2. Output is in bit reversal order

3. In place computation is possible

39

Figure 32: DIF FFT

X[k] =

N−1∑n=0

x[n]ωknN , k = 0, 1, ..., N − 1

for even numbered samples

X[2r] =

N−1∑n=0

x[n](ω2N

)nrr = 0, 1, ..., (N/2)− 1

=

(N/2)−1∑n=0

x[n]ω2nrN +

N−1∑n=N/2

x[n]ω2nrN

=

(N/2)−1∑n=0

x[n]ω2nrN +

(N/2)−1∑n=0

x[n+ N/2]ω2(n+N/2)rN

using periodicity of twiddle factor

X[2r] =

(N/2)−1∑n=0

(x[n] + x[n+ N/2])ωnrN/2, r = 0, 1, ..., (N/2)−1 (Eq. IV.3.1)

similarly for odd numbered samples

X[2r + 1] =

(N/2)−1∑n=0

(x[n]− x[n+ N/2])ωnNωnrN/2, r = 0, 1, ..., (N/2)− 1

(Eq. IV.3.2)

40

Eq. IV.3.1 is the N/2 point DFT of the N/2 point sequence obtained byadding the rst half and the last half of the input sequence. Eq. IV.3.2 is theN/2 point DFT of the sequence obtained by substracting the second half of theinput sequence from the rst half and multiplying the resulting sequence by ωnN .

Thus from Eq. IV.3.1 and Eq. IV.3.2,

g[n] = x[n] + x[n+ N/2]

h[n] = x[n]− x[n+ N/2]

DFT is computed by rst forming g[n] and h[n], then computing ωnNh[n] andnally computing the N/2 point DFT of these sequences to obtain even samplesand odd samples respectively. The ow graph of 8 point DFT using DIF FFTis illustrated in gure 32 on the preceding page.

4 Spectrum Analysis

Sinusoidal Signal

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Magnitude →

Sinusoidal Input

f=500Hz

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

Frequency(ω) →

Magnitude →

Sinusoidal Response

ω=π/10

Figure 33: Sinusoidal Signal

The DFT of a sine or cosine wave will give only two spikes at conjugate fre-quencies and there will not be any other frequencies. This is because pure sineor cosine functions contains only one frequency at all times.

Observations From gure 33, we observe that the frequency components onboth sides of origin at the frequencies equal to 2πf

fsamwhere f is the analog

frequency of the given signal and fsam is the sampling frequency and these two

41

spikes will have an amplitude which is equal to half the amplitude of originalsignal.

Pulse Input

0 100 200 300 400 500 6000

0.2

0.4

0.6

0.8

1

Time(n) →

Mag

nitu

de

→Pulse Input

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

Frequency(ω) →

Magnitude →

Response

Figure 34: Pulse Input

Observations The DFT of a step function will give a sinc function centredat the origin, which is of amplitude same as that of step function. If the widthof the pulse is small, the sinc would be more spead out and if the pulse is oflonger duration, the sinc would be sharper.

42

Square Input

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Ma

gnitud

e →

SquareWave Input

f=500Hz

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

Frequency(ω) →

Magnitude →

Response

Figure 35: Square Input

Concepts When a square wave is applied as input, the frequency spectrumindicates that the frequency components are present on both sides of origin atfrequencies equal to the digital frequency of the square wave applied and all itsodd harmonics.

Consider a square wave of period 2L, and in [0, 2L] it can be represented as,

f(x) = 2 u(x)− u(x− L) − 1 where x ε [0, 2L]

Fourier series representation of f(x),

f(x) =4

π

∞∑n=1,3,5...

1

nsin(nπxL

)From the fourier series representation of square wave, we can observe that

it has coecients of nonzero value for fundamental frequency and for its oddharmonics only.

Observations From gure 35, we observe that the amplitude of the frequencycomponents present at fundamental frequency on either side of the origin willhave an amplitude equal to half the amplitude of square pulse and the sidecomponents amplitude decreases as we move farther from the centre.

Frequency Modulated Signals

Concepts Frequency modulation is the angle modulation in which the instan-taneous frequency of the signal is modulated in proportional with the amplitude

43

of the message signal.If fc is frequency of carrier andm(t)is the message signal then the modulated

signal instantaneous frequency is

f(t) = fc + km(t) (Eq. IV.4.1)

where k is the frequency sensitivity of the signal.Angle θ of FM signal is,

θ = 2π

tˆ

0

f(τ)dτ (Eq. IV.4.2)

Linear FM (Chirp Signal)For chirp function, m(t) = t, then using (Eq. IV.4.1)

f(t) = fc + k t

from (Eq. IV.4.2),

⇒ θ = 2πfct+ 2πkt2

2

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Magnitude →

Linear FM

Slope=.4, f=200Hz, fs=500Hz

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

Frequency(ω) →

Magnitude →

Response

Figure 36: Chirp Signal (smaller slope)

44

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Ma

gn

itud

e →

Linear FM

Slope=24, f=200Hz, fs=500Hz

−4 −3 −2 −1 0 1 2 3 40

0.01

0.02

0.03

0.04

0.05

0.06

Frequency(ω) →

Mag

nitu

de

→

Response

Figure 37: Chirp Signal (larger slope)

Observations From gure (36), we observe that as k is a small value we havelesser frequency deviation and hence frequency spectrum gives lesser spread.As k increases deviation from fc increases (see gure (37)). The high frequencycomponents become much dominant supressing the lower frequencies. Since theenergy of the signal is constant the amplitude of the spectrum decreases as thespectrum spreads out.

Sinusoidal ModulationFor a sinusoidal modulating signal,m(t) = Acos(2πfmt) then from (Eq. IV.4.1),

f = fc + kAcos(2πfmt)

= fc +∇fcos(2πfmt)

where ∇f is frequency deviation.Modulation index, β

β =∇ffc

Modulated wave is

s(t) = cos(2πfct+ βsin(2πfmt)

When β << 1r,

s(t) = cos(2πfct)− βsin(2πfct)sin(2πfmt)

which resembles amplitude modulated signal. This is narrowband FM.

45

For other β,

s(t) = Re [exp (j2πfct+ jβsin(2πfmt))] (Eq. IV.4.3)

and using complex fourier series,

s(t) =

∞∑n=−∞

Jn(β)cos [2π(fc + nfm)t] (Eq. IV.4.4)

where Jn(β) is nth order Bessel Function of the rst kind.

Jn(β) =1

2π

π

−π

exp (j(βsinx− nx)) dx

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Ma

gn

itud

e →

Sine Modulated FM

β=.2,fc=50Hz,f

m=5

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

Frequency(ω) →

Magnitude →

Response

Figure 38: Narrowband FM (β=0.2)

Observations From (Eq. IV.4.3), we expect frequency components at ±ω.

ω = 2πfcfsam

and is veried (see gure 38). Here β will be less and so the frequencydeviation. So in the frequency spectrum we will have a component at fc only.

46

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Magnitude →

Sine modulated FM

β=2, fc=50Hz, f

m=3, f

s=200Hz

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

Frequency(ω) →

Magnitude →

Response

(a) fm=3

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

Time(n) →

Magnitude →

Sine modulated FM

β=2, fc=50Hz, f

m=5, f

s=200Hz

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

Frequency(ω) →

Magnitude →

Response

(b) fm=5

Figure 39: Wideband FM (β=2)

As expected from (Eq. IV.4.4), we observe frequency components at f c ±nfm. The components are centred at fc and the separation between adjacentcomponents is fm. In this case β will be high. So there will be a spread offrequency components innitely located on both sides of the carrier frequencypresent at a distances of fc, 2fc, 3fc etc. Here the amplitude of the carrier wavewill not be constant for all modulation indices unlike in an AM and narrowband FM. The amplitude of carrier component varies with β as J0(β). Thephysical explaination for this is that the envelop of FM signal is constant. Sothe average power developed across load resistor is constant. Here when carrieris modulated to generate FM signal, power in the side frequencies may appearonly at the expense of power originally generated in the carrier, thereby makingamplitude of carrier wave dependent on β. When fm is increased from 3 to 5we observe that the spectrum spreads out more (see gure 39).

47

Kernel of DFT

0 100 200 300 400 500 6000

0.2

0.4

0.6

0.8

1

Time(n) →

Magnitude →

0 200 400 600 800 1000 12000

0.2

0.4

0.6

0.8

1

Frequency(ω) →

Magnitude →

0 200 400 600 800 1000 12000

0.5

1

1.5

Time(n) →

Magnitude →

0 200 400 600 800 1000 12000

0.2

0.4

0.6

0.8

1

Frequency(ω) →

Magnitude →

0 200 400 600 800 1000 12000

0.2

0.4

0.6

0.8

1

Time(n) →

Magnitude →

Figure 40

Consider that DFT of signal x(t) is X(ω). If you nd the DFT of X(ω) andnormalize, the resultant signal is same as x(−t). Once more taking DFT ofx(−t) will give X(−ω) and on applying DFT fourth time and normalizing willgive back x(t). This is because while we are nding the DFT, we are justrotating the samples on a unit circle. So when DFT is performed twice, 180o

out of phase signal is obtained and performing DFT four times will give youroriginal signal back. So here when a pulse function is applied as input and DFTof this function gives a sinc. On applying DFT once again and normalizing givesa mirror image of original pulse signal. On applying DFT two more times andnormalizing gives back your original rectangular signal (see gure 40).

Inference Taking DFT each time may be visualised as rotating the sequenceanti-clockwise by π/2 in time frequency plane.

So in this case eigen values of DFT is j, −1, −j, 1 and kernel is j.

F (x(t)) = X(ω)

F (X(ω)) = F2 (x(t)) = x(−t)

F3 (x(t)) = X(−ω)

F4 (x(t)) = x(t)

48

Code

DIT FFT

clc;

x=[1 2 1 1 1 2 3 1 4 5];

n = length(x);

N=2^(ceil(log2(n)));

z=[x, zeros(1,N-n)];

fft(z,N)

y=bitrevorder(z);

r=log2(N);

for p=1:r

pow=2^p;

wn=exp(-j*2*pi/pow);

o=1;

while(o<=N)

for q=1:pow/2

y(o+pow/2+(q-1))=y(o+pow/2+(q-1))*wn^(q-1);

end

for q=1:pow/2

g=y(o+(q-1))+y(o+pow/2+(q-1));

h=y(o+(q-1))-y(o+pow/2+(q-1));

y(o+(q-1))=g;

y(o+pow/2+(q-1))=h;

end

o=o+pow;

end

end

y

DIF FFT1

function [y]=jan28_2(x)

clc;

n = length(x);

N=2^(ceil(log2(n)));

z=[x, zeros(1,N-n)];

a = log2(N);

y = z;

for p = 1:a

temp=[];

w = exp(-j*2*pi/(2^(a-p+1)));

for q = 1:(2^(p-1))

for m = 1:2^(a-p)

1jan28_2() used in computation of DFT in other programs

49

temp = [temp y((2^(a-p))*(q-1)*2+m)+y((2^(a-p))*(q-1)*2+m+2^(a-p))];

end

for m = 1:2^(a-p)

temp = [temp (w^(m-1))*(y((2^(a-p))*(q-1)*2+m)-y((2^(a-p))*(q-1)*2+m+2^(a-p)))];

end

end

y = temp;

end

y=bitrevorder(y);

end

Spectrum Analysis

% sine

figure(1);

pt = 2048;

n = linspace(0,pt-1,pt);

y = sin(2*pi*500*n/10000);

u=length(y)

subplot(2,1,1);

plot(y);

xlabel('Time(n) \rightarrow')

ylabel('Magnitude \rightarrow')

title('Sinusoidal Input')

text(80,1.5,'f=500Hz')

axis([0,100,-2,2]);

res = jan28_2(y)./pt;

subplot(2,1,2);

w=linspace(-pi,pi,pt);

plot(w,fftshift(abs(res)));

xlabel('Frequency(\omega) \rightarrow')


title('Sinusoidal Response')

text(2.7,.35,'\omega=\pi/10')

figure(2);

%pulse

pt = 1024;


y = [ones(1,100),zeros(1,500)];

subplot(2,1,1);

plot(y);



title('Pulse Input')

res = jan28_2(y)./(100);

subplot(2,1,2);

50


plot(w,fftshift(abs(res)))



title('Response')

figure(3);

% square

pt = 2048;


y = square(2*pi*500*n/10000);

subplot(2,1,1);

plot(y);



title('SquareWave Input')

text(80,1.5,'f=500Hz')

axis([0,100,-2,2]);

res = jan28_2(y)./(pt);

subplot(2,1,2);


plot(w,fftshift(abs(res)))



title('Response')

% linear fm

figure(4);

pt = 1024;

n = linspace(0,2,1000);

y = sin(2*pi*(200*n+12*n.^2));

subplot(2,1,1);

plot(y);

axis([0,100,-2,2]);



title('Linear FM')

text(60,1.5,'Slope=24, f=200Hz, f_s=500Hz')

res = jan28_2(y)./(pt);

subplot(2,1,2);





title('Response')

y = sin(2*pi*(200*n+.2*n.^2));

res = jan28_2(y)./(pt);

figure(5)

51

subplot(2,1,1);

plot(y);

axis([0,100,-2,2]);



title('Linear FM')

text(60,1.5,'Slope=.4, f=200Hz, f_s=500Hz')

subplot(2,1,2);




title('Response')

figure(6);

% narrow fm

pt = 2048;

n = linspace(0,10,2000);

y = cos(2*pi*50*n+.2*sin(2*pi*5*n));

subplot(2,1,1);

plot(y);

axis([0,100,-2,2]);



title('Sine Modulated FM')

text(60,1.5,'\beta=.2,f_c=50Hz,f_m=5')

res = jan28_2(y)./(pt);

subplot(2,1,2);





title('Response')

% wideband fm

figure(7);

pt = 2048;

n = linspace(0,10,2000);

y = cos(2*pi*50*n+2*sin(2*pi*5*n));

subplot(2,1,1);

plot(y);

axis([0,100,-2,2]);



title('Sine modulated FM')

text(60,1.5,'\beta=2, f_c=50Hz, f_m=5, f_s=200Hz')

res = jan28_2(y)./(pt);

subplot(2,1,2);


52




title('Response')

% wideband fm

figure(8);

pt = 2048;

n = linspace(0,10,2000);

y = cos(2*pi*50*n+2*sin(2*pi*3*n));

subplot(2,1,1);

plot(y);

axis([0,100,-2,2]);



title('Sine modulated FM')

text(60,1.5,'\beta=2, f_c=50Hz, f_m=3, f_s=200Hz')

res = jan28_2(y)./(pt);

subplot(2,1,2);





title('Response')

% fft

figure(9);

pt = 1024;


y = [ones(1,100),zeros(1,500)];

subplot(3,2,1);

plot(y);



res = jan28_2(y);

res1 = jan28_2(res)./pt;

res2 = jan28_2(res1);

res3 = jan28_2(res2)./pt;

subplot(3,2,2);

plot(n,fftshift(abs(res/100)));



subplot(3,2,3);

plot(n,abs(res1));



subplot(3,2,4);

plot(n,fftshift(abs(res2/100)));

53



subplot(3,2,5);

plot(n,abs(res3));



Normalisation of FFT

Lets consider a rectangular pulse of length N in time domain.

x[n] =

1 0 ≤ n < N

0 otherwise

The DTFT of x[n] gives,

X(ω) =

∞∑n=0

x[n]e−jωn

=

N−1∑n=0

e−jωn

=1− e−jωN

1− e−jωusing L' Hospital's Rule to nd X(0)

X(0) =jNe−jωN

je−jω

∣∣∣∣ω=0

= N

From Parseval's Theorem, energy of any signal, x[n] in time domain andfrequency domain are equal.

N−1∑n=0

x2[n] =1

N

N−1∑k=0

|X[k]|2

Hence, after taking Fourier Transform, the result is normalised with N .

References

[1] Alan V. Oppenheim, Ronald W. Schafer and John R. Buck, Discrete-Time

Signal Processing, 2nd Ed., Pearson Education


Principles, Algorithms, and Applications, 3rd Ed., Prentice Hall Interna-tional, Inc.

[3] Simon Haykin, Communications Systems, 4th Ed., Wiley India

54

Date: 4th February 2010

5 FIR Filter Design using Frequency Sampling

Method

Concepts

In frequency sampling method, the ideal frequency response of the requiredlter is given, say H(ω). N samples of H(ω) is taken by sampling at kFsam/N,k = 0, 1, 2, ..., N − 1. The lter coecients h(n) is obtained by taking inverseDFT of the sampled values,

h(n) =1

N

N−1∑k=0

H(k)ej(2πnkN ) (Eq. IV.5.1)

where H(k), k = 0, 1, 2, ..., N − 1, are samples of ideal frequency response.From (Eq. IV.5.1),

h(n) =1

N

N−1∑k=0

H(k)ej(2πnkN )

=1

N

N−1∑k=0

|H(k)| e−j(2παkN )ej(

2πnkN )

=1

N

N−1∑k=0

|H(k)| ej(2π(n−α)k

N )

=1

N

N−1∑k=0

|H(k)|[cos

(2π(n− α)k

N

)+ j sin

(2π(n− α)k

N

)]

=1

N

N−1∑k=0

|H(k)| cos

(2π(n− α)k

N

)since h(n) is real, imaginary part of h(n) is zero. And for h(n) to be linear

phase lter, h(n) needs to be symmetrical. For n even,

h(n) =2

N

N/2−1∑k=0

|H(k)| cos

(2π(n− α)k

N

)+H(0) (Eq. IV.5.2)

and for n odd, the upper limit of sum is (N−1)/2.(Eq. IV.5.2) is implemented to obtain the FIR lter coecients.

55

Ideal Low Pass Filter

Here samples are taken from the ideal low pass lter as shown in gure 41. Aswe observe the designed FIR lter have ripples in the response, where ever thereis as sudden transition in the frequency reponse in ideal lter. This is becausetheir is a sudden transition in frequency domain. And so we have taken less orno samples (as in this case) in the transition band of the response.

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

frequency (ω) →

mag

nitu

de

→

Ideal Low Pass Filter

−4 −3 −2 −1 0 1 2 3 40

0.5

1

1.5

frequency (ω) →

mag

nitude

→

Figure 41: Ideal Low Pass Filter with less number of samples in transition band

Now let's compare the resultant FIR lter with the ideal lter by placingthe responses one over the other as in gure 42 on the next page.

56

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

1.2

1.4

frequency (ω) →

magnitude →

Figure 42: Comparison of ideal versus designed FIR lters

Let's now redesign the FIR lter by including more samples from transitionband of the response. As we observe in gure 43 on the following page, thedesigned lter have smoother transition band and less ripples.

57

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1

frequency (ω) →

mag

nitud

e →

Low Pass Filter (smooth transition band)

−4 −3 −2 −1 0 1 2 3 40

0.5

1

1.5

frequency (ω) →

ma

gnitud

e →

Figure 43: Ideal Low Pass Filter with more number of samples in transitionband

Arbitary Frequency Response

For an arbitary frequency response as in gure 44 on the next page, the lterobtained is show below.

58

−4 −3 −2 −1 0 1 2 3 40

0.5

1

1.5

2

frequency (ω) →

mag

nitud

e →

Arbitrary Frequency Response

−4 −3 −2 −1 0 1 2 3 40

0.5

1

1.5

2

2.5

frequency (ω) →

ma

gnitu

de →

Figure 44: Arbitary Frequency Response

Frequency sampling method can be used to obtain FIR lter from any ar-bitary frequency response satisfying the symmetricity property.

Code

N = 255;

alpha = (N-1)/2;

H = [ones(1, 50), zeros(1,156), ones(1,50)];

figure(1);

subplot(2,1,1);

w = linspace(-pi, pi, N+1);

plot(w,H);

xlabel('frequency (\omega) \rightarrow');

ylabel('magnitude \rightarrow');

title('Ideal Low Pass Filter');

h = zeros(1, N);

for k = 0 : alpha

59

for m = 1 : alpha

h(k+1) = h(k+1) + 2*H(m+1)*cos(2*pi*m*(k-alpha)/N);

end;

h(k+1) = (h(k+1)+H(1))/N;

h(N-k) = h(k+1);

end;

pt = 1024;

subplot(2,1,2);

w = linspace(-pi, pi, pt);

z = exp(j*w);

freq = 0;

for k = 0 : N-1

freq = freq + h(k+1)*z.^(-k);

end

plot(w, fftshift(abs(freq)));

xlabel('frequency (\omega) \rightarrow');

ylabel('magnitude \rightarrow');

60

Part V

Lab Report 5

Date: 4th & 11th February, 2010

1 IIR Filters

Basic Features

IIR digital lters which can be realized is characterised by the recursuive equa-tion,

y[n] =

∞∑k=0

h[k]x[n− k] (Eq. V.1.1)

=

N∑k=0

bkx[n− k]−M∑k=1

aky[n− k]

where h[k] is the impulse response of the lter, bk and ak are the ltercoecients of the lter, and x[n] and y[n] are the input and output to the lter.

The transfer function for the IIR lter is,

H(z) =b0 + b1z

−1 + . . .+ bNz−N

1 + a1z−1 + . . .+ aMz−M(Eq. V.1.2)

=

∑Nk=0 bkz

−k

1 +∑Mk=1 akz

−k

From (Eq. V.1.1), it can be noted that output, y[n] depends on the pastoutputs, y[n−k] and present as well as past input samples, x[n−k], that is IIRlter is a feedback system and hence the name Innite Impulse Response (IIR)lter.

The transfer function of the IIR lterm, H(z), given by (Eq. V.1.2) can berepresented as,

H(z) =K(z − z1)(z − z2). . .(z − zN )

(z − p1)(z − p2). . .(z − pM )

where z1, z2, . . .are the zeros and p1, p2, . . .are the poles of the transfer func-tion, H(z).

61

2 IIR Design through Analog Filters

IIR lters are generally designed by rst designing its analog counterparts andthen transforming it to digital lters. This method is preferred as analog lterdesign procedures are highly advanced and the classical lters like Butterworth,Chebychev and elliptic are readily available.

In case of IIR lter design ,the most common practice is to convert thedigital lter specications into analog low pass prototype lter specications,to determine the analog low pass lter transfer function H(s) meeting thesespecications, then to transform it into the desired digital lter transfer function.This approach is widely used because

1. Analog approximations are highly advanced.

2. They usually yield closed form solutions.

3. Extensive tables are available for analog lter design.

2.1 Butterworth Approximation

The low pass Butterworth lter is characteristics by the following magnitudesquared frequency response,

|H(jΩ)|2 =1

1 +(

ΩΩC

)2N(Eq. V.2.1)

where N is the order of the lter and ΩC is the 3dB cut o frequency of thelow pass lter.

Figure 45: Classical Butterworth Low Pass Filter Response

The response of typical Butterworth low pass lter is depicted in Figure 45on page 62. It can be seen that the response is monotonic in pass band and stop

62

band regions. The response is also said to be maximally at due to the initialatness with a slope of almost zero at DC.

Normalising (Eq. V.2.1),

|H(jΩ)|2 =1

1 + Ω2N

⇒ H(jΩ)H(−jΩ) = H(s)H(−s)

=1

1 +(sj

)2N

H(s)H(−s) =1

1 + (−1)Ns2N

=1

1 + (−s2)N

Now, poles of transfer function are obtained by solving,

1 + (−s2)N = 0

solving, the poles sk are obtained as

s =

ej2πk/2N , for N odd

ej(2k−1)π/2N , for N even(Eq. V.2.2)

Here, the poles are placed on the circumference of the unit circle (as thefunction is normalised ).

F =f

fs(Eq. V.2.3)

where F is the digital frequency,f is the analog frequency and fs is the samplingfrequency.

To nd n

α = 10log

(1 +

(Ω

Ωc

)2n)dB (Eq. V.2.4)

The attenuation in the passband should not exceed αmax and that in thestopband should be atleast αmin.

(Eq. V.2.4)

⇒ Ωc =Ω(

10α10 − 1

) 12n

(Eq. V.2.5)

Solving for n in (Eq. V.2.4) by substituting for α and Ω in the passbandand stopband,

n =1

2

log(

100.1αmin−1100.1αmax−1

)log(

ΩsΩp

) (Eq. V.2.6)

63

Maximal Flatness All derivatives of the |H(jΩ)|2 up to but not includingthe 2N th derivative are zero, resulting in "maximal atness". If the requirementto be monotonic is limited to the passband only and ripples are allowed in thestopband, then it is possible to design a lter of the same order that is atterin the passband than the "maximally at" Butterworth. Such a lter is inverseChebyshev lter or Type II Chebyshev lter.

2.2 Chebyshev Approximation

The Butterworth approximation, which is monotonic in both the pass band andstop band, gives us a magnitude response which exceeds the specications forthe pass band ripple. The problem is that the order of the lter will be high.Usually, we can tolerate some amount of ripple, and it would be advantageous tohave a trade-o between the ripple and the lter order, so that we can obtain alter which meets the specications and has a lower order than its Butterworthcounterpart. For this we use the Chebychev approximation. The characteristicof Chebyshev approximation is that it minimises the error between the idealisedand the actual lter over the range of lter allowing some ripples in pass band.

Type I, with equal ripple in the pass band, monotonic in the stop band.

Type II, with equal ripple in the stop band, monotonic in the pass band.

Type I Chebyshev Filter is characterised by the magnitude squared responsegiven by,

|H(Ω)|2 =1

1 + ε2C2N

(ΩΩp

)where CN

(ΩΩp

)is a Chebyshev polynomial which exhibits equal ripple in

the pass band, N is the order of the polynomial as well as that of the lter, andε determines the pass band ripple, which in decibels is given by,

passband ripple ≤ 10 log10(1 + ε2)

Chebyshev polynomial is given by,

CN (x) =

cos(N cos−1(x)) , |x| ≤ 1

cosh(N cosh−1(x)) , |x| > 1

64

Figure 46: Classical Chebyshev Low Pass Filter Response

A typical response of a type I Chebyshev characteristic is shown in Figure46 on page 65.

Here

α = 10log(1 + ε2C2

n (Ω))

(Eq. V.2.7)

Pole Calculation

1 + ε2C2n (Ω) = 0 (Eq. V.2.8)

Substituting for Cnand solving (Eq. V.2.8) gives

cos (nu) cosh (nv)− sin (nu) sinh (nv) = ± ε

(Eq. V.2.9)

where ω = u+ v = cos−1(s

)(Eq. V.2.9) gives

cos (nu) cosh (nv) = 0

⇒ cos (nu) = 0

∴ u =

(2m+1)π2n ,m = 0, 1, 2....n− 1

sin (nu) sinh (nv) = ±1

ε

⇒ v = ± 1

nsinh−1

(1

ε

)= ±a

65

Then,

s = cosω

Replace s→ sωp

⇒ s =ωp cos

(1n

[(2m+ 1) π2 ± sinh−1

(1ε

)]),m = 0, 1, 2....n− 1

(Eq. V.2.10)

Guillemin's Algorithm

Guillemin's algorithm is used to transform a Butterworth lter design to aChebyshev design in a simple and easier way. The steps to obtain Chebyshevpoles from Butterworth poles is given below.

The Butterworth angles are,

φk =2k + 1

n

π

2, k = 0, 1, 2, ...(2n− 1)

From (Eq. V.2.10),

−uk = sinh a sinφk

±vk = cosh a cosφk

The complementary angle to φk is ψk.

ψk =π

2− 2k + 1

n

π

2

From the above diagram,

66

sinφk = cosψk

cosφk = sinψk

Hence,

−uk = sinh a cosψk

±vk = cosh a sinψk

67

3 Transforming Analog Filter to Digital Filter

After analog lter design is completed as per the specications, the design needsto be converted into digital domain. The commonly used techniques are

1. Impulse Invariance

2. Bilinear Transformation

3.1 Impulse Invariant Technique

Impulse invariant technique is used to obatin the digital lter whose impulseresponse is as close as possible to analog lter's impulse response. In the impulseinvariant method, starting with the analog lter transfer function, H(s), theimpulse response, h(t) is obtained using the Laplace transform. h(t) is thensuitably sampled to produce h(nT ), and the desired transfer function H(z) isthen obtained by z-transforming h(nT ), where T is the sampling interval.

The analog lter impulse response, ha(t) is given by

ha(t) = L−1 (H(s))

Now, sampling h(t) at t = nT ,

h[n] = ha(nT )

and then now H(z) by taking z-transform of h[n].Consider the case when no poles have multiplicity more than one, H(z) can

be obtained directly from H(s).

H(s) =

N∑k=1

Aks− pk

(Eq. V.3.1)

ha(t) = L−1 (H(s))

⇒ ha(t) =

∑Nk=1Ake

pkt , t ≥ 0

0 , t < 0

h[n] = Tha(nT ), n ≥ 0

⇒ h[n] =

N∑k=1

AkTepknT (Eq. V.3.2)

⇒ H(z) =

N∑k=1

AkT

1− epkT z−1

=

N∑k=1

AkTz

z − epkT(Eq. V.3.3)

68

Ignoring zero at z = 0 for (Eq. V.3.3) causes the transfer function to bedelayed.

H(z) =

N∑k=1

AkT

z − epkT(Eq. V.3.4)

From (Eq. V.3.4), it can be infered that to transform analog transfer func-tion, H(s) to digital transfer function, H(z) by replacing pk in (Eq. V.3.1) withepkT . The transformations are given below:

s → z

H(s) → TH(z)

pk → epkT

Relation between Analog and Digital Frequencies Ω is the analogfrequency and ω is the digital frequency.

s = jΩ , z = ejω

jΩT = jω

⇒ Ω =ω

T(Eq. V.3.5)

Hence, from (Eq. V.3.5) it can be seen that analog frequency and digitalfrequency are linearly related.

3.2 Bilinear Transformation

Bilinear transformation is much more important and useful than IIT method.The method is derived by approximating the rst order dierential equation witha dierence equation. This transformation maps the analog transfer function,H(s) from the s-plane onto the discrete transfer function, H(z) in the z-plane.The entire jω axis in the s-plane is mapped onto the unit circle , and the lefthalf s-plane is mapped inside the unit circle and the right half s-plane is mappedoutside the unit circle in the z-plane.

69

z = esT

⇒ s =1

Tln(z)

=2

T

[(z − 1

z + 1

)1

+1

3

(z − 1

z + 1

)3

+1

5

(z − 1

z + 1

)5

+ . . .

]

⇒ s ≈ 2

T

(z − 1

z + 1

)∴ s =

2

T

(1− z−1

1 + z−1

)(Eq. V.3.6)

⇒ z =

(1 + sT/2

1− sT/2

)Hence,

H(s) → H(z)

s → 2

T

(1− z−1

1 + z−1

)Relation between Analog and Digital Frequencies Ω is the analogfrequency and ω is the digital frequency.

s = jΩ , z = ejω

s =2

T

(1− z−1

1 + z−1

)jΩ =

2

T

(1− e−jω

1 + e−jω

)⇒ Ω =

2

Ttan

(ω2

)(Eq. V.3.7)

⇒ ω = 2 tan−1

(TΩ

2

)(Eq. V.3.8)

Hence, from (Eq. V.3.5) it can be seen that analog frequency and digitalfrequency are not linearly related (see Figure 47 on page 71). This deviation ofΩ− ω from linear relation is known as frequency warping.

To compensate for this eect, the analog frequencies are prewarped beforeapplying bilinear transformation using (Eq. V.3.8).

70

Figure 47: Ω− ω plot

71

4 Design of Filters

4.1 Butterworth Filter Design

4.1.1 Specications

αmax = 0.5 dBαmin = 20 dBΩp = 1000 rad/secΩs = 2000 rad/secfsampling = 10000Hz

4.1.2 Design using Butterworth approximation

Now Ωc is calculated using α and Ω in the passband and stopband and n from(Eq. V.2.5). From the two Ωc value,the one giving better attenuation is selected.

(Eq. V.2.6)⇒ n = 4.83, so n is taken as 5.

(Eq. V.2.5)⇒ Ωc =

1234 , whenΩ = Ωs, α = αmin

1263 , whenΩ = Ωp, α = αmax

Ωc = 1234 =⇒ α = 21dB. That is αmin increased by 5.

Ωc = 1263 =⇒ α = 0.4dB. That is αmax decreased by 20.

So Ωc is chosen as 1263 rad/sec.

As n is odd, (Eq. V.2.2) gives the poles as

sk =exp

(2πk

5

), k = 1, 2...10

From these poles, the ones lying on the left half plane is chosen for stability.The selected poles as per this criterion are,

s3 = −0.309 + 0.951

s4 = −0.809 + 0.587

s5 = −1

s6 = −0.809− 0.587

s7 = −0.309− 0.951

The analog transfer function is given by,

H(s) =1

(s− s3) (s− s4) (s− s5) (s− s6) (s− s7)

=1

(s+ 1) (s2 + 0.618s+ 1) (s2 + 1.62s+ 1)

72

Replacing s → sΩc

= s1263 ,

H(s) =12635

(s+ 1263) (s2 + 783s+ 12632) (s2 + 2046s+ 12632)

Considering it as a cascade of one rst order and two second order systemsgives,

H(s) = H1(s)H2(s)H3(s)

where

H1(s) =1263

s+ 1263

H2(s) =12632

s2 + 783s+ 12632

H3(s) =12632

s2 + 2046s+ 12632

4.1.3 IIT Method

H1 (s) =1263

s+ 1263

∴ H1 (z) =0.1263

1− 0.8814z−1(Eq. V.4.1)

H2 (s) =(1263)

2

s2 + 783s+ (1263)2

=664

s+ 390 + 1201− 664

s+ 390− 1201

∴ H2 (z) =0.664

1− (0.955− 0.115) z−1− 0.664

1− (0.955 + 0.115) z−1

=0.1527z−1

1− 1.9z−1 + 0.925z−2(Eq. V.4.2)

H3 (s) =(1263)

2

s2 + 2046s+ (1263)2

=1076.9

s+ 1023 + 740− 1076.9

s+ 1023 + 740

∴ H3 (z) =0.108

1− (0.9− 0.667) z−1− 0.108

1− (0.9 + 0.667) z−1

=0.0144z−1

1− 1.804z−1 + 0.815z−2(Eq. V.4.3)

H (z) = H1 (z)H2 (z)H3 (z)

73

where

H1 (z) =Y1 (z)

X (z)(Eq. V.4.4)

H2 (z) =Y2 (z)

Y1 (z)(Eq. V.4.5)

H3 (z) =Y (z)

Y2 (z)(Eq. V.4.6)

(Eq. V.4.1) and (Eq. V.4.4) gives

y1[n] = 0.1263x[n] + 0.88y1[n− 1]


y2[n] = 0.152y1[n] + 1.9y2[n− 1]− 0.925y2[n− 2]


y[n] = 0.0144y2[n] + 1.804y[n− 1]− 0.815y[n− 2]

Figure 48: Signal Flow Graph

4.1.4 BLT Method

Prewarping the frequencies gives

Ωp = 1000.83 rad/sec

Ωs = 2006.69 rad/sec

H1 (z) =1 + z−1

16.83− 14.83z−1(Eq. V.4.7)

H2 (z) =1 + 2z−1 + z−2

242z−2 − 500z−1 + 262(Eq. V.4.8)

H3 (z) =1 + 2z−1 + z−2

226z−2 − 500z−1 + 277(Eq. V.4.9)

74


y1[n] =1

16.83(x[n] + x[n− 1] + 14.83y1[n− 1])


y2[n] =1

262(y1[n] + 2y1[n− 1] + y1[n− 2] + 500y2[n− 1]− 242y2[n− 2])


y[n] =1

277(y2[n] + 2y12[n− 1] + y2[n− 2] + 500y[n− 1]− 226y[n− 2])


4.2 Chebyshev Filter Design

4.2.1 Specications

αmax = 0.5 dBαmin = 20 dBΩp = 1000 rad/secΩs = 2000 rad/secfsampling = 10000Hz

4.2.2 Design using Chebyshev approximation

To nd ε, put α = αmax and ω = ωp in (Eq. V.2.7)

ε = 0.3493

∵Cn = 1, whenω = ωpTo nd n put α = αmin and ω = ωs in (Eq. V.2.7)

n = 4

75

∴The poles are obtained as

⇒ s =ωp cos

((2m+ 1) π8 ± 0.4435

),m = 0, 1, 2, 3

s1 = −175.338 + 1016.238

s2 = −423.305 + 419.464

s3 = −423.305− 419.464

s4 = −175.338− 1016.238

H1 (s) =(1031.25)

2

s2 + 350.676s+ (1031.25)2

H2 (s) =(596.93)

2

s2 + 846.61s+ (595.93)2

−1.5 −1 −0.5 0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.5

Pole Zero Plot


76

4.2.3 IIT Method

H1 (s) =523.249

s+ 175.338 + 1016.238+

−523.249

s+ 175.338− 1016.238

H1 (z) =0.05232

1− e−(0.01753+0.10162)z−1+

−0.05232

1− e(−0.01753+0.10162)z−1

=0.01043z−1

1− 1.9549z−1 + 0.9655z−2

⇒ y1[n] = 0.01043x[n− 1] + 1.9549y1[n− 1]− 0.9655y1[n− 2]

H2 (s) =423.303

s+ 423.303 + 420.9+

−423.303

s+ 423.303− 420.9

H2 (z) =0.04233

1− e−(0.04233+0.04209)z−1+

−0.04233

1− e(−0.04233+0.04209)z−1

=0.034z−1

1− 1.915z−1 + 0.9655z−2

⇒ y[n] = 0.000321y1[n− 1] + 1.915y[n− 1]− 0.9187y[n− 2]

4.2.4 BLT Method

Prewarping the frequencies gives

Ωp = 1000.83 rad/sec

Ωs = 2006.67 rad/sec

∴ Poles are

s1 = −175.484 + 1017.073

s2 = −423.654 + 421.28

s3 = −423.654− 421.28

s4 = −175.484− 1017.073

H1 (s) =(1032.1)

2

s2 + 351.49s+ (1032.1)2

∴ H1 (z) =0.00251

(1 + 2z−1 + z−2

)1.0202− 1.99468z−1 + 0.985z−2

⇒ y1[n] = 0.00246x[n] + 2x[n− 1] + x[n− 2] + 1.9552y1[n− 1]− 0.9654y1[n− 2]

H2 (s) =(597.44)

2

s2 + 847.35s+ (597.44)2

∴ H2 (z) =0.00089

(1 + 2z−1 + z−2

)1.043− 1.998z−1 + 0.9585z−2

⇒ y[n] = 0.000855 (y1[n] + 2y1[n− 1] + y1[n− 2]) + 1.9156y1[n− 1]− 0.919y1[n− 2]

77

4.3 Integrator

The analog transfer function of integrator is given by,

H(s) =1

s

4.3.1 IIT Method

H (z) =1

1− z−1

⇒ y (n) = x (n) + y (n− 1)

Here the digital lter obtained has a zero at z = 0. So to ignore this a delayis given to the input. So

H (z) =z−1

1− z−1

⇒ y[n] = x[n− 1] + y[n− 1]

4.3.2 BLT Method

Taking T = 1

H (z) =

(1 + z−1

)2 (1− z−1)

⇒ y[n] = 0.5x[n] + 0.5x[n− 1] + y[n− 1]

4.4 Comb Filter

The poles and zeros are selected in such a manner that, the frequency responseof the lter looks like a comb. Here, the zeros are placed on the jω axis ons-plane at equally spaced locations and poles are placed on left half plane notfar away from jω axis as shown in the gure. Poles tries to take the response toinnite value where as the zeros placed at regular intervals cause the responseto have dips. So the analog transfer function, H(s) is chosen as

H (s) =s (s+ a) (s− a)

(s+ b+ c) (s+ b− c) (s+ b+ d) (s+ b− d)

Taking a = 2, b = 1, c = 1, d = 3

78

H (s) =s(s2 + 4

)(s2 + 2s+ 2) (s2 + 2s+ 10)

H1 (s) =s

(s2 + 2s+ 2)

H2 (s) =

(s2 + 4

)(s2 + 2s+ 10)

using BLT Method where T = 1

H1 (z) =1− z−2

5− 2z−1 + z−2

H2 (z) =4(1 + z−2

)9 + 6z−1 + 5z−2

y1[n] =1

5(x[n]− x[n− 2] + 2y1[n− 1]− y1[n− 2])

y[n] =1

9(4y1[n]− 4y1[n− 1]− 6y[n− 1]− 5y[n− 2])


79

5 Observations

5.1 Butterworth Filter

IIT Method

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude

→

Frequency Response

−4 −3 −2 −1 0 1 2 3 4−300

−250

−200

−150

−100

−50

0

n →

Magnitude →

−4 −3 −2 −1 0 1 2 3 4−3

−2

−1

0

1

2

3

ω →

Phase →

Figure 52: Filter Response

80

BLT Method

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude →

Frequency Response

−4 −3 −2 −1 0 1 2 3 4−4

−2

0

2

4

X: 1.29Y: −1.845

ω →

Phase →

−4 −3 −2 −1 0 1 2 3 4−150

−100

−50

0

ω →

Magnitude →


81

−4 −3 −2 −1 0 1 2 3 4−120

−100

−80

−60

−40

−20

0

ω →

Magnitude(d

b)

→

Frequency Response →


0 2000 4000 6000 8000 10000 120000

0.2

0.4

0.6

0.8

1

n

Am

plit

ude →

Impulse Response →

0 100 200 300 400 500 600 700−0.02

0

0.02

0.04

0.06

n →

Am

plit

ude →


82

0 100 200 300 400 500 600 700−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

n →

Am

plit

ude →

Pulse Response

0 100 200 300 400 500 600 700 800−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

ω →

Magnitude →

0 100 200 300 400 500 600 700−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

n →

Am

plit

ude →

Pulse Response

0 100 200 300 400 500 600 700 800−0.2

0

0.2

0.4

0.6

0.8

1

1.2

ω →

Magnitude →

Figure 56: Pulse Response

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude →

Sinusoidal Response

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude →

Figure 57: Sinusoidal Response (pass band frequency)

83

0 100 200 300 400 500 600 700−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

n →

Am

plit

ude →

Sinusoidal Response

0 100 200 300 400 500 600 700−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

n →

Am

plit

ude →

Figure 58: Sinusoidal Response (stop band frequency)

84

−4 −3 −2 −1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

X: 0.0764Y: 0.4865

ω →

Magnitude →

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude →

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

n →

Am

plit

ude

→

Sinusoidal Response

Figure 59: Sinusoidal Response (mixed sinusoids)

85

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Pulse Input

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Figure 60: Pulse Response (including all poles)

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Pulse Input(Dominant pole alone)

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Figure 61: Pulse Response (excluding all poles other than the dominant pole)

86

5.2 Chebyshev Filter

IIT Method

−4 −3 −2 −1 0 1 2 3 4−120

−100

−80

−60

−40

−20

0

Frequency(rad/sam) →

Gain

in d

B

Frequency response


−0.1 −0.05 0 0.05 0.1

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0


Gain

in d

B

Frequency response

Figure 63: Frequency Response (zoomed view of pass band)

87

BLT Method

−4 −3 −2 −1 0 1 2 3 4−160

−140

−120

−100

−80

−60

−40

−20

0

20


Am

plit

ude

Frequency Response


−0.1 −0.05 0 0.05 0.1

−4

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

1


Am

plit

ude

Frequency Response

Figure 65: Frequency Response (zoomed view of pass band)


88

0 100 200 300 400 500 600 7000

0.5

1

1.5

Time(n) →

Am

plit

ud

e

Pulse Input

0 100 200 300 400 500 600 700−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude

Figure 67: Pulse Response (including all poles)

The rising and falling edges of the pulse excites the system causing a dampedsinusoid as output as seen in the case of the impulse response of the system.The frequency of the sinusoid is expected to be that of dominant pole frequency.Here, from the dominant pole frequency is 1016rad/sec. And from the Figure67 on page 89, frequency of sinusoid, f

f =

(166− 107

10000

)−1

× 2π = 1064.9rad/sec

Here, we observe that there is a variation of frequency from what is expected.This is because of the eect of non dominant poles. So, let us take the lterwith only dominant pole and other poles are removed. The response of such asystem is shown in Figure 68 on page 90.

89

0 100 200 300 400 500 600 7000

0.5

1

1.5

Time(n) →

Am

plit

ude

→

Pulse input(Dominant pole alone)

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

1.5

2

Time(n) →

Am

plit

ude →

Figure 68: Pulse Response (excluding all poles except the dominant pole)

With the system with only dominant pole, the frequency of sinusoid f ,

f =

(142− 80

10000

)−1

× 2π = 1014rad/sec

And as expected, we obtain f equal to dominant pole frequency.

0 100 200 300 400 500 600 700−2

−1

0

1

2

Time(n) →

Am

plit

ude

Sinusoidal Input(Passband)

f=0.075 rad/sam

0 100 200 300 400 500 600 700−1

−0.5

0

0.5

1

Time(n) →

Am

plit

ude

Figure 69: Sinusoidal Response (pass band frequency)

90

0 100 200 300 400 500 600 700−1.5

−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Sinusoidal input(stopband)

f=0.75 rad/sam

0 100 200 300 400 500 600 700−1.5

−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Figure 70: Sinusoidal Response (stop band frequency)

0 100 200 300 400 500 600 700−2

−1

0

1

2

Time(n) →

Am

plit

ude →

Sinusoidal input(Mixed sine)

0 100 200 300 400 500 600 700−1.5

−1

−0.5

0

0.5

1

1.5

Time(n) →

Am

plit

ude →

Figure 71: Sinusoidal Response (mixed sinusoids)

91

0 100 200 300 400 500 600 700−1

0

1

Time(n) →

Am

plit

ude

→

Preprocessor

0 100 200 300 400 500 600 700−1

0

1

Time(n) →

Am

plit

ude →

0 100 200 300 400 500 600 700 800−0.05

0

0.05

Time(n) →

Am

plit

ude →

Figure 72: Preprocessing of input signal

If we have a preprocessing block to which the input signal is fed before givingto our system which has a zero at z=1. This causes the eect of poles of pulseinput to be compensated or nullied.

z-transform of input signal

X(z) =1− z−N

1− z−1

and transfer function of preprocessing system is

P (z) = 1− z−1

So the output of preprocessing unit is

Y1(z) = 1− z−N

y1[n] = δ[n]− δ[n−N ]

So, we can observe two impulses as in Figure 72 on page 92. And once thisis fed into our Chebyshev lter, we obtain two exponentially decaying sinusoidswhich is excited by these impulses.

92

5.3 Integrator

IIT Method

0 200 400 600 800 1000 12000

0.2

0.4

0.6

0.8

1Integrator Impulse Response

Time(n) →

Am

p →

−4 −3 −2 −1 0 1 2 3 4−1

−0.5

0

0.5

1


Am

p →


0 200 400 600 800 1000 12000

0.2

0.4

0.6

0.8

1Integrator Impulse Response(Delayed input)

Time(n) →

Am

p →

−4 −3 −2 −1 0 1 2 3 4−60

−40

−20

0


Am

p →

Figure 74: Delayed Impulse Input

93

BLT Method

0 100 200 300 400 500 600 700 800 900 10000

0.2

0.4

0.6

0.8

1Integrator Impulse Response

Time(n) →

Am

p →

−4 −3 −2 −1 0 1 2 3 40

0.2

0.4

0.6

0.8

1


Am

p →


0 100 200 300 400 500 600 700 800 900 1000−1

−0.5

0

0.5

1Integrator(Sinusoidal Input)

Time(n) →

Am

p →

0 100 200 300 400 500 600 700 800 900 1000−10

0

10

20

30

40

Time(n) →

Am

p →


94

0 100 200 300 400 500 600 700 800 900 1000−1

−0.5

0

0.5

1Integrator(Square wave Input)

Time(n) →

Am

p →

0 100 200 300 400 500 600 700 800 900 10000

10

20

30

40

50

Time(n) →

Am

p →

Figure 77: Square Input

0 100 200 300 400 500 600 700 800 900 10000

2

4

6

8

10Integrator(Ramp Input)

Time(n) →

Am

p →

0 100 200 300 400 500 600 700 800 900 10000

1000

2000

3000

4000

5000

Time(n) →

Am

p →

Figure 78: Ramp Input

95

5.4 Comb Filter

−4 −3 −2 −1 0 1 2 3 4−400

−300

−200

−100

0Comb Filter


Magnitude

−4 −3 −2 −1 0 1 2 3 4−3

−2

−1

0

1

2

3


Phase


6 Code

6.1 Butterworth Filter

clc;

x=[1,zeros(1,10000)];

%x=[zeros(1,50),ones(1,10),zeros(1,1000)];

%t=linspace(0,999,1000);

%x=sin(2*pi*1.2*t/100)+sin(2*pi*12*t/100);

figure(5);

subplot(3,1,1)

plot(x);

xlabel('n \rightarrow')

ylabel('Amplitude \rightarrow')

title('Frequency Response ')

axis([0,700,-1,1])

y=[];

pt=700;

for i=1:pt

if(i==1)

y(i)=.1263*x(1);

96

else

y(i)=.88*y(i-1)+.1263*x(i);

end

end

z(1)=0;

for i=2:pt

if(i==2)

z(i)=.0152*y(1);

else

z(i)=1.908*z(i-1)-.925*z(i-2)+.0152*y(i-1);

end

end

w=[];

w(1)=0;

for i=2:pt

if(i==2)

w(i)=.01436*z(1);

else

w(i)=1.804*w(i-1)-.818*w(i-2)+.01436*z(i-1);

end

end

w1=-pi:2*pi/(pt-1):pi;

l1=length(w);

n=0:l1-1;

subplot(3,1,2)

plot(w1,(fftshift(20*log((abs(fft(w)))))))

%plot(n,(w))

xlabel('n \rightarrow')


%figure(3);

subplot(3,1,3)

plot(w1,(fftshift(angle(fft(w)))))

xlabel('\omega \rightarrow')

ylabel('Phase \rightarrow')

%xlabel('\omega \rightarrow')

%ylabel('Magnitude(db) \rightarrow')

%title('Frequency Response \rightarrow')

6.2 Chebyshev Filter

clc

x=[1 zeros(1,999)];

%x=[zeros(1,50),ones(1,300),zeros(1,1000)];

%t=linspace(0,999,1000);

%x=sin(2*pi*1.2*t/100)+sin(2*pi*12*t/100);

figure(1);

97

pt=700;

y(1)=0;

y(2)=0.01043*0.944*x(1)+1.9549*y(1);

for i=3:pt

y(i)=1.9549*y(i-1)-0.9655*y(i-2)+0.01043*0.944*x(i-1);

end

w(1)=0;

w(2)=(3.4154*10^(-3))*y(1)+1.915*w(1);

for i=3:pt

w(i)=1.915*w(i-1)-0.9187*w(i-2)+(3.4154*10^(-3))*y(i-1);

end


l1=length(w);

n=0:l1-1;

%plot(n,(w))

%axis([0,700,-1,1.5])

plot(w1,20*log10(fftshift(abs(fft(w,pt)))))

grid on

xlabel('Frequency in radians per sample')

ylabel('Gain in dB')

title('Frequency response')

6.3 Comb Filter

clc;

x=[1,zeros(1,10000)];

figure(5);

%subplot(3,1,1)

%plot(x);

%axis([0,700,-1,1])

y=[];

pt=1024;

for i=1:pt

if(i==1)

y(i)=4*x(i)/17;

elseif(i==2)

y(i)=(-2*y(i-1)+4*x(i))/17;

else

y(i)=(-2*y(i-1)-1*y(i-2)+4*x(i)-4*x(i-2))/17;

end

end

z=[];

for i=1:pt

if(i==1)

98

z(i)=(8*y(i))/25;

elseif(i==2)

z(i)=(-18*z(i-1)+8*y(i))/25;

else

z(i)=(-18*z(i-1)-9*z(i-2)+8*y(i)+8*y(i-2))/25;

end

end

w=z;


l1=length(w);

n=0:l1-1;

subplot(3,1,2)

plot(w1,(fftshift(20*log10(abs(fft(w))))))

subplot(3,1,1)

plot(w1,abs(fft(w)))

%plot(n,(w))

%figure(3);

subplot(3,1,3)

plot(w1,(fftshift(angle(fft(w)))))

References

[1] Alan V. Oppenheim, Ronald W. Schafer and John R. Buck, Discrete-Time

Signal Processing, 2nd Ed., Pearson Education, 2002


Practical Approach, 2nd Ed., Pearson Education, 2003

99

Part VI

Lab Report 6

Date: 18th February, 2010

1 Lattice Realization of FIR Filters

Consider a FIR lter with system function, Hm(z)

Hm(z) = Am(z) m = 0, 1, 2 ..., M − 1

where Am(z) is the polynomial

Am(z) = 1 +

m∑k=0

αm(k)z−k m ≥ 1

with A0(z) = 1. The unit impulse response of the mth order lter ishm(0) = 1 and hm(k) = αm(k), k = 1, 2, ..., m. m is the degree of the polyno-mial Am(z). Let αm(0) = 1.

If x(n) is the input sequence to the lter Am(z) and y(n) is the outputsequence,

y(n) = x(n) +

m∑k=1

αm(k)x(n− k) (Eq. VI.1.1)

Figure 80: Direct form realization of (Eq. VI.1.1)

1.1 Filter of order 1

Suppose m = 1

y(n) = x(n) + α1(1)x(n− 1) (Eq. VI.1.2)

This lter can be realized using lattice structure as shown in Figure 81 onpage 101. Here both the inputs are x(n) and output is taken from upper branch.The output is same as (Eq. VI.1.2) if K1 = α1(1) where K1is lattice parametercalled reection coecient.

100

Figure 81: First order Lattice Filter

1.2 Filter of order 2

For m = 2,

y(n) = x(n) + α2(1)x(n− 1) + α2(2)x(n− 2) (Eq. VI.1.3)

By cascading two lattice stages (Eq. VI.1.3) can be realized. The output ofrst stage is

f1(n) = x(n) +K1x(n− 1) (Eq. VI.1.4)

g1(n) = K1x(n) + x(n− 1)

and output of second stage is

f2(n) = f1(n) +K2g1(n− 1) (Eq. VI.1.5)

g2(n) = K2f1(n) + g1(n− 1)

From (Eq. VI.1.4) and (Eq. VI.1.5),

y(n) = f2(n) = x(n) +K1x(n− 1) +K2 [K1x(n− 1) + x(n− 2)]

y(n) = x(n) +K1 (1 +K2)x(n− 1) +K2x(n− 2)(Eq. VI.1.6)

Comparing (Eq. VI.1.3) and (Eq. VI.1.6),

α2(2) = K2 α2(1) = K1(1 +K2)

or

K2 = α2(2) K1 =α2(1)

1 + α2(2)

101

1.3 Filter of order M

As order 2 lter was developed in 1.2, a lter of any order, say M can berealized by cascading the lattice stages. Such a lter is described by

f0(n) = g0(n) = x(n)

fm(n) = fm−1(n) +Kmgm−1(n− 1) (Eq. VI.1.7)

gm(n) = Kmfm−1(n) + gm−1(n− 1)

where m = 1, 2, ..., M − 1. The recursive equations in z-domain are

F0(z) = G0(z) = X(z)

Fm(z) = Fm−1(z) +Kmz−1Gm−1(z) (Eq. VI.1.8)

Gm(z) = KmFm−1(z) + z−1Gm−1(z)

And output of (M − 1)th stage lter is output of the (M − 1)th order FIRlter,

y(n) = fM−1(n)

As mth order FIR lter can be expressed using output of m stage latticefm(n),

fm(n) =

m∑k=0

αm(k)x(n− k), αm(0) = 1 (Eq. VI.1.9)

z-transform of (Eq. VI.1.9) gives

Fm(z) = Am(z)X(z)

⇒ Am(z) =Fm(z)

F0(z)

Similarly for gm(n) can also be expressed like fm(n) as in (Eq. VI.1.9). Butit is seen that the coecients are in the reverse order to that given in (Eq.VI.1.9).

gm(n) =

m∑k=0

βm(k)x(n− k)

where

βm(k) = αm(m− k), k = 0, 1, 2, ..., m & βm(m) = 1

or in z domain,

102

G(z) = Bm(z)X(z)

⇒ Bm(z) =Gm(z)

X(z)

Now lets see the relation between Am(z) and Bm(z)

Bm(z) =

m∑k=0

βm(k)z−k

=

m∑k=0

αm(m− k)z−k

= z−mm∑l=0

αm(l)zl

Bm(z) = z−mAm(z−1)

From the above relation, the zeros of FIR lter with system function Am(z)are the reciprocals of the zeros of Bm(z). Thus, Bm(z) is called the reversepolynomial of Am(z).

From (Eq. VI.1.8), and dividing each equation with X(z),

A0(z) = B0(z) = 1

Am(z) = Am−1(z) +Kmz−1Bm−1(z) (Eq. VI.1.10)

Bm(z) = KmAm−1(z) + z−1Bm−1(z)

wherem = 1, 2, ..., M−1. Therefore, a stage of lattice lter can be expressedas [

Am(z)Bm(z)

]=

[1 Km

Km 1

] [Am−1(z)

z−1Bm−1(z)

][

Am(z)z−1Bm(z)

]=

1

1−K2m

[1 −Km

−Km 1

] [Am(z)Bm(z)

]1.4 Design Procedure

Given the FIR lter coecients for direct form realization, the correspondinglattice lter parameters, Ki can be determined. For a lter of m stages, Km =αm and to compute Km−1, Am−1 polynomial is required. From (Eq. VI.1.10),

Am−1(z) =Am(z)−KmBm(z)

1−K2m

, m = M − 1, M − 2, ..., 1

Hence all Ki are determined by stepping down from m = M − 1 to m = 1.

103

1.5 Code

c l c ;p=[2 3 4 5 ] ;a = [ ] ;b= [ ] ;H=[1 ,p ] ;l=length (H) ;k=p( l −1);c=[k ] ;

% determining the K va lue sf o r n=1: l−2

l=length (H) ;a=H;f o r m=1: l

b (m)=a ( l−m+1);endb=b ( 1 : l ) ;H=(a−k*b)/(1−k ^ ( 2 ) ) ;H=H( 1 : l −1);k=H( l −1);c=[k , c ] ;

endck=c ;

%ex c i t i n g the l a t t i c e s t r u c tu r e with an impulsel =100;x=ze ro s (1 , l ) ;f 1=ze ro s (1 , l ) ;g1=ze ro s (1 , l ) ;x (1)=1;f 0=x ;g0=x ;n=1;f o r n=1:4

f o r i =1: li f i==1

f1 ( i )= f0 ( i ) ;g1 ( i )=k (n)* f 0 ( i ) ;

e l s ef 1 ( i )=k (n)* g0 ( i−1)+f0 ( i ) ;g1 ( i )=k (n)* f 0 ( i )+g0 ( i −1);

endend

104

n=n+1;f 0=f1 ;g0=g1 ;

endstem ( f1 ) ;

1.6 Observations

Impulse Response

0 5 10 15 200

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5Impulse Response

n→

am

plit

ude

→


For the direct lter coecients [1 2 3 4 5], the corresponding lattice parame-ters are computed and is

k =

1 0.50000 0.33333 0.25000 5.00000

When the lattice lter is excited with impulse input, the output obtained isequal to the direct lter coecients as expected.

105

Frequency Response

−4 −3 −2 −1 0 1 2 3 42

4

6

8

10

12

14

16Frequency Response

ω (rad/sec) →

mag

nitu

de

→


Pulse Response

0 10 20 30 40 500

5

10

15Pulse Response

n→

am

plit

ude

→

Figure 84: Pulse Response

106

Sinusoidal Response

0 200 400 600 800 1000−20

−10

0

10

20f=1Hz

−4 −3 −2 −1 0 1 2 3 40

2000

4000

6000


n→

am

plit

ude

→


White Gaussian Noise is given as input to lattice structure, it is obserevedthat, lattice structure introduces correlation to the signal. Therefore, latticestructure is similar to a correlator. Here, when WGN is given as input, latticebeing a FIR lter causes truncation of the signal and thereby producing a morecorrelated output.

When output is observed after each lattice stage, the signal is getting moreand more correlated. Each stage of lattice contributes to introduction of corre-lation into input WGN.

0 20 40 60 80 100 120−40

−20

0

20

40

60

80

100

Figure 86: Output after stage 1 of lattice lter

107

0 20 40 60 80 100 120−40

−20

0

20

40

60

80

100

120

Figure 87: Output of stage 2 of lattice lter

0 20 40 60 80 100 120−40

−20

0

20

40

60

80

100

120

140


108

0 20 40 60 80 100 120−50

0

50

100

150

200


0 20 40 60 80 100 120−2000

−1000

0

1000

2000

3000

4000

5000

6000


1.7 Inference

Lattice structures are easy and simple to implement. Addition or deletionof a part of lattice lter does not aect the other stages of the lter. So, latticelters are used in linear prediction applications. For a mth order lter, thereare only m reection coecients and it is possible to use this lter as a lterof any order from 1 to m, without the need of change of previous coecients.But, if we had used a direct form realization of mth order lter, we would havem(m+1)

2 lter coecients to implement a lter of order between 1 and m.

109

References



110

Part VII

Lab Report 7

Date: 25th February & 11th March, 2010

1 Introduction

At the transmitting side, the messge signal is passed through a decorrelatorto give a signal which is approximately white. Then, only the coecients of thelter needs to be transmitted along with the compressed error signal.

y(n) −→[A(z)

]−→ w(n)

By transmitting the lter coecients, A(z) and the error signal, w(n) themessage signal can be reconstructed. It is also possible to reconstruct the mes-sage signal without transmitting the error signal, w(n) as the error signal isapproximately white, it is sucient if a white noise of same power as error isgenerated locally.

Receiver side,

w(n) −→[

1/A(z)]−→ y(n)

where y(n) is the predicted message signal.

2 Levinson Durbin Recursion Algorithm

Consider ideal sampled signal, y(n). Each sample of the signal, y(n) canbe represented as linear combination of previous samples. Lets represent nthsample of y(n),

y(n) = a1yn−1 + a2yn−2 + . . .+ amyn−m

en = y(n)− y(n) (Eq. VII.2.1)

(n−m) depends on the extend of correlation of y(n), and m gives the orderof the lter and en is the prediction error.

Determining the order of the lterLets assume the order of the lter, m = 1. That is,

y(n) = −a1yn−1

en = yn + a1yn−1 (Eq. VII.2.2)

111

Now to obtain more accurate prediction, the error should be minimized, i.e.minimize

∑e2n. Take z-transform of (Eq. VII.2.1),

En(z) = Y (z) + a1z−1Y (z)

En(z)

Y (z)= A(z) = 1 + a1z

−1

Error energy is to be minimized. Error energy, ξ

ξ = E[e2n

](Eq. VII.2.3)

minimizing (Eq. VII.2.3),

⇒ ∂ξ

∂a1= 0

This can be intepreted in vector space as projection of a vector on to 1D.Error is minimum when projected perpendicularly.

ξ = E[e2n

]⇒ ∂ξ

∂a1= 2E

[en∂en∂a1

]= 0

now from (Eq. VII.2.2),

∂en∂a1

= yn−1

⇒ E [enyn−1] = 0 (Eq. VII.2.4)

E [(yn + a1yn−1) yn−1] = 0

E [ynyn−1] + a1E [yn−1yn−1] = 0

⇒ RY Y (1) + a1RY Y (0) = 0 (Eq. VII.2.5)

a1 = −RY Y (1)

RY Y (0)

As RY Y (0) is the maximum value, a1 < 1⇒ zeros are within the unit circleand hence the lter is minimum phase and is invertible. (Eq. VII.2.4) is knownas the orthogonality relation.

yn −→[A(z)

]−→ en

112

Figure 91: Direct Form Realization

Lets compute the energy left in the error sequence, en, when the error isminimum

ξ(a1)min = E[e2n

]= E [en (yn + a1yn−1)]

= E [enyn] + a1E [enyn−1]

= E [(yn + a1yn−1) yn]

ξ(a1)min = RY Y (0) + a1RY Y (1) (Eq. VII.2.6)

= RY Y (0)[1− a2

1

]From (Eq. VII.2.5) and (Eq. VII.2.6),

[RY Y (0) RY Y (1)RY Y (1) RY Y (0)

]︸︷︷︸

[1a1

]︸︷︷︸ =

[ξ0

]︸︷︷︸

R a ξ

Ra = ξ

The matrix R is symmetricBut the error, en obtained in one dimensional space need not be the min-

imial projection error possible. May be two dimensional or some other higherdimensional results in much smaller error, i.e. more uncorrelation.

Let m = 2,

yn = − (a′1yn−1 + a′2yn−2)

en = yn − yn

following similar procedure as described above,

113

∂ξ(a′1, a′2)

∂a′1= 0

∂ξ(a′1, a′2)

∂a′2= 0

Here, R is a 3× 3 matrix and RY Y (0) RY Y (1) RY Y (2)RY Y (1) RY Y (0) RY Y (1)RY Y (2) RY Y (1) RY Y (0)

1a′1a′2

=

ξ00

Hence, en also decreases. en can be further minimized by adding more

dimensions. It is required to obtain a completely white error signal so thatsignal need not be transmitted.

en = yn + a′1yn−1 + a′2yn−2

A(z) = 1 + a′1z−1 + a′2z

−2

Figure 92: Direct Form Realization

Filter A(z) is generating the prediction error en and so A(z) is called theprediction error lter. Drawback of direct form realization is that wheneverlter order is changed or updated, all coecients computed earlier are also tobe redened. So it is computationally intensive.

So we require a lter which does not change coecient values when thedimension of the lter changes except that new coecients are introduced. Gapfunction method is utilized to make such a lter.

114

Let g(1) = E [enyn−1], and for rst order, g(1) = 0

g(1) = E

[(1∑

m=0

amyn−m

)yn−1

]

=

1∑m=0

amE [yn−myn−1]

=

1∑m=0

amRY Y (1−m)

Now for kth order,

g(k) =

k∑m=0

amRY Y (k −m)

= ak ∗RY Y (k)

The lter is similar to convolution operation.

RY Y (k) −→[1, a1, a2, . . .

]−→ g(k)

g(k) = E [enyn−k] = 0

For rst order lter we have a zero at 1, and for a second order lter weshould have zeros at 1, 2. Lets assume that our prediction error lter is of ordertwo, i.e. we have a gap length of two. Lets assume we are extending a alreadydesigned rst order lter two second order without aecting the existing ltercoecients.

g(k) = ak ∗R(k) (Eq. VII.2.7)

g(1) = E [enyn−1] = 0

Optimal rst order predictor, y(n) = yn+a1yn−1. Lets dene a new param-eter called reection coecients. For rst order predictor, γ1 = −a1. To extendto second order lter, we create a gap function of second order.

g′(k) = g(k) + γ2g(2− k) (Eq. VII.2.8)

g′(k) = a′(k) ∗R(k) (Eq. VII.2.9)

already we have, g′(1) = 0

weneed, g′(2) = 0

⇒ 0 = g(2) + γ2g(0)

∴ γ2 = −g(2)

g(0)

115

Relation between γ in terms of R

g(0) = E [enyn] = E[e2n

]= εmin

and

g(2) = a0R(2) + a1R(0)

g(0) = a0R(0) + a1R(1)

∴ γ2 = −(a0R(2) + a1R(1)

a0R(0) + a1R(1)

)= −

(R(2)− γ1R(1)

εmin

)similarly

γ3 = −(a0R(3) + a1R(2) + a2R(1)

a2R(2) + a1R(1) + a0R(0)

)in general

γm+1 = −(amR(1) + am−1R(2) + . . . + a0R(m+ 1)

amR(m) + am−1R(m− 1) + . . . + a0R(0)

)z Domain relationships Take z-transform of (Eq. VII.2.7) and (Eq. VII.2.8)to give

G′(z) = G(z) + γ2z−2G(z−1)

and

G(z) = A(z)SY Y (z)

G′(z) = A′(z)SY Y (z)

from the above,

A′(z)SY Y (z) = A(z)SY Y (z) + γ2z−2A(z−1)SY Y (z−1)

⇒ A′(z) = A(z) + γ2z−2A(z−1)

⇒ A′(z) = A(z) + γ2z−1AR(z) (Eq. VII.2.10)

where AR(z) represents the reversed sequence of z transform of ak,

AR(z) = z−1A(z−1)

also,

A′(z−1) = A(z−1) + γ2zAR(z−1)

⇒ A′R(z) = z−1AR(z) + γ2A(z) (Eq. VII.2.11)

116

Representing (Eq. VII.2.10) and (Eq. VII.2.11) as matrix form,[A′(z)A′R(z)

]=

[1 γ2z

−1

γ2 z−1

] [A(z)AR(z)

]In general,

[AM+1(z)ARM+1(z)

]=

[1 γM+1z

−1

γM+1 z−1

] [AM (z)ARM (z)

](Eq. VII.2.12)

(Eq. VII.2.12) is called the forward Levinson Durban Recursion and canbe easily implemented using Lattice Structures. The order of the lter canbe determined by prediction error analysis. Error decreases when the order ofthe lter is increased, but the lter order is xed by xing a threshold valueof error. When the threshold is satised, the order of the lter need not beincreased further.

3 Inverse Filter

Figure 93: Inverse Filter

For transmitting a given message signal, it is passed through a decorrelator oran analyser. It computes the reection coecients and lter coecients for op-timum error energy. These lter coecients are transmitted over the channel tothe receiver. The receiver uses this set of coecients for predicting the messagesignal from a locally generated WGN of energy of the error signal generated atthe transmitter. At the receiver, the lter is called synthesiser (inverse lter).Linear prediction can be implemented by Residual Excited Linear Prediction orby Code Excited Linear Prediction.

117

4 Observations

Gap Function

0 5 10 15 20 25 30−20

−10

0

10

20

30

40

50

0 5 10 15 20 25 30−10

0

10

20

30

40

50

0 5 10 15 20 25 30−10

0

10

20

30

40

50

0 5 10 15 20 25 30 35−10

−5

0

5

10

15

20

25

30

35

40

0 5 10 15 20 25 30 35 40−5

0

5

10

15

20

25

30

35

40

0 5 10 15 20 25 30 35 40−5

0

5

10

15

20

25

30

35

40

Figure 94: Gap Function

Here while a gap is added, actually a new new dimension is added to thelter. The new component in orthogonal to the rest, so a zero component isobtained.

Linear Prediction

When x=[1 2 3 4 5 3 2 3 1 5 6 6 3]; is given as input to the predictionerror lter, the reection coeents and lter coeents are found as

118

Figure 95

And the following gures illustrates decorrelator in action,

0 5 10 15 20 25 30−20

−10

0

10

20

30

40

50

0 5 10 15 20 25 30 35−10

0

10

20

30

40

50

0 5 10 15 20 25 30 35 40−5

0

5

10

15

20

25

30

35

40

0 5 10 15 20 25 30 35 40 45−5

0

5

10

15

20

25

30

35

40

Figure 96: Output after each stage of decorrelator

119

0 10 20 30 40 50 60−5

0

5

10

15

20

25

30

35

40

Figure 97: Output of decorrelator

Predicted signal

0 5 10 150

1

2

3

4

5

6

7Linear predictor

input⇒ red; predictedwith error signal⇒ blue; predictedwithwgn⇒ green

Figure 98: Predicted Signal

Speech Signal Processing

When a speech signal is to be transmitted, the signal is passed through thedecorrelator and at receiver either error signal or WGN is used to excite theinverse lter. Input sequence is given as

x = [-120 186 -348 -517 555 -5 -434 17 595 -225 -473 189 237 -3

-188 117 -249 12 617 -932 -92 970 -497 -592 873 -175 -314 178 -249

149 481 -884 -166 1087 -497 -991 1034 606 -1205 523 644 -706 205 320

-642 512 294 -892 3 688 -525 -298 412 -9 -331 9 -79 11 80 -505 -200

120

449 -196 -538 144 363 -341 -203 373 106 -98 86 227 54 88 37 -136 337

359 -281 -4 772 317 -34 427 610 296 318 90 -192 -17 -583 -1108 -863

-946 -1505 -1381 -899 -1012 -637 -73 -1 594 1112 867 1369 1585 968

1164 1420 709 403 402 181 105 220 540 882 693 538 503 304 -296 -1051

-1640 -2066 -2706 -3267 -3150 -3013 -2460 -2001 -1129 -5 1023 1795

2596 3491 3482 3114 2508 2256 1581 231 -612 -666 -1222 -1628 -1490

-980 -747 -402 284 827 1530 2143 2147 2117 1361 1078 418 -603 -2202

-3100 -3740 -4572 -5140 -4530 -3996 -3005 -1731 -308 1260 2898 4169

5079 5698 5376 4730 3589 2403 760 -658 -1734 -2531 -3103 -3249 -2651

-1758 -870 -17 659 1860 2621 2790 3551 3473 1997 837 380 -1169 -3620

-4224 -5309 -6060 -5987 -5841 -4288 -3035 -1301 703 2439 4205 5564

6520 6693 5896 4825 3262 1413 -429 -1887 -2974 -3409 -3690 -3565 -2645

-1439 -79 1097 1682 2373 3352 3431 3091 3414 2477 450 -891 -1903 -3887

-5615 -5664 -6152 -6432 -5684 -4636 -2713 -569 1284 3138 4577 5609

6757 7069 6248];

The signal is segmented into eight segments with 32 samples each. Theneach of them are processed separately.

0 50 100 150 200 250 300−8000

−6000

−4000

−2000

0

2000

4000

6000

8000Speech signal

Figure 99: Input Sequence

121

0 5 10 15 20 25 30 35−1000

−800

−600

−400

−200

0

200

400

600

800

1000

(a) Predicted using error signal

0 5 10 15 20 25 30 35−1000

−800

−600

−400

−200

0

200

400

600

800

1000

(b) Predicted using WGN

Figure 100: Speech Segment 1

0 5 10 15 20 25 30 35−1500

−1000

−500

0

500

1000

1500


0 5 10 15 20 25 30 35−1500

−1000

−500

0

500

1000

1500



122

0 5 10 15 20 25 30 35−2000

−1500

−1000

−500

0

500

1000


0 5 10 15 20 25 30 35−2000

−1500

−1000

−500

0

500

1000



0 5 10 15 20 25 30 35−4000

−3000

−2000

−1000

0

1000

2000


0 5 10 15 20 25 30 35−4000

−3000

−2000

−1000

0

1000

2000



123

0 5 10 15 20 25 30 35−4000

−3000

−2000

−1000

0

1000

2000

3000

4000


0 5 10 15 20 25 30 35−4000

−3000

−2000

−1000

0

1000

2000

3000

4000



0 5 10 15 20 25 30 35−6000

−4000

−2000

0

2000

4000

6000


0 5 10 15 20 25 30 35−6000

−4000

−2000

0

2000

4000

6000

8000



124

0 5 10 15 20 25 30 35−8000

−6000

−4000

−2000

0

2000

4000

6000

8000


0 5 10 15 20 25 30 35−8000

−6000

−4000

−2000

0

2000

4000

6000

8000



0 5 10 15 20 25 30 35−8000

−6000

−4000

−2000

0

2000

4000

6000

8000


0 5 10 15 20 25 30 35−8000

−6000

−4000

−2000

0

2000

4000

6000

8000



125

5 Code

Decorrelator

%deco r r e l a t o r ( l e v i n s on s r e cu r s i on )

c l e a rc l cx=[1 2 3 4 5 3 2 3 1 5 6 6 3 ] ;%c a l c u l a t i n g r e f l e c t i o n c o e f f i e n t sa1=length (x ) ;% g1=a1 ;r1 = [ ] ;p1=xcorr (x , x ) ;ord = 6 ;f o r i=a1 : l ength ( p1 )

r1 ( i−a1+1)=p1 ( i ) ;ende = [ ] ;a = [ ] ;ar = [ ] ;gam= [ ] ;g = [ ] ;a (1)=1;a(2)=− r1 (2)/ r1 ( 1 ) ;gam(1)=a ( 2 ) ;

ar (1)=a ( 2 ) ;ar (2)=a ( 1 ) ;e (1 ) = p1 ( ( l ength ( p1 )+1)/2) ;

f o r i =2: ordg=conv (a , p1 ) ;f i g u r e ( i )stem ( g ) ;gam( i ) = −g ( a1+i )/ g ( a1 ) ;b=a ;a=[a 0 ] + gam( i )* [ 0 ar ] ;ar=gam( i ) * [ b 0 ]+[0 ar ] ;e ( i )=e ( i −1)*(1−gam( i −1)^2);

endi=i +1;e ( i )=e ( i −1)*(1−gam( i −1)^2);gama

126

efn = x ;gn = x ;

%l e v i n s on s r e cu r s i onf o r i = 1 : ord

f t = fn ;fn = [ f t 0]+gam( i )* [ 0 gn ] ;gn = [0 gn]+gam( i ) * [ f t 0 ] ;

end

Speech Processing

%speech p ro c e s s i ng

c l cc l fc l e a r ;x = [−120 186 −348 −517 555 −5 −434 17 595 −225 −473 189 237 −3 −188 117 −249 12 617 −932 −92 970 −497 −592 873 −175 −314 178 −249 149 481 −884 −166 1087 −497 −991 1034 606 −1205 523 644 −706 205 320 −642 512 294 −892 3 688 −525 −298 412 −9 −331 9 −79 11 80 −505 −200 449 −196 −538 144 363 −341 −203 373 106 −98 86 227 54 88 37 −136 337 359 −281 −4 772 317 −34 427 610 296 318 90 −192 −17 −583 −1108 −863 −946 −1505 −1381 −899 −1012 −637 −73 −1 594 1112 867 1369 1585 968 1164 1420 709 403 402 181 105 220 540 882 693 538 503 304 −296 −1051 −1640 −2066 −2706 −3267 −3150 −3013 −2460 −2001 −1129 −5 1023 1795 2596 3491 3482 3114 2508 2256 1581 231 −612 −666 −1222 −1628 −1490 −980 −747 −402 284 827 1530 2143 2147 2117 1361 1078 418 −603 −2202 −3100 −3740 −4572 −5140 −4530 −3996 −3005 −1731 −308 1260 2898 4169 5079 5698 5376 4730 3589 2403 760 −658 −1734 −2531 −3103 −3249 −2651 −1758 −870 −17 659 1860 2621 2790 3551 3473 1997 837 380 −1169 −3620 −4224 −5309 −6060 −5987 −5841 −4288 −3035 −1301 703 2439 4205 5564 6520 6693 5896 4825 3262 1413 −429 −1887 −2974 −3409 −3690 −3565 −2645 −1439 −79 1097 1682 2373 3352 3431 3091 3414 2477 450 −891 −1903 −3887 −5615 −5664 −6152 −6432 −5684 −4636 −2713 −569 1284 3138 4577 5609 6757 7069 62 48 ] ;f i g u r e (1 )p l o t ( x )g r id ont i t l e ( ' Speech s i gna l ' )sgmt = ze ro s ( 8 , 3 2 ) ;

f o r i = 1 :8

f o r j = 1 :32

sgmt ( i , j ) = x ( j + ( i −1)*32);

end

end

%f i l t e r c o e f f i c i e n t s computation

ab = x ;

l = length ( ab ) ;ord = 50 ;

127

ryy1 = xcorr ( ab ) ;ryy = [ ] ;

f o r i = l : l ength ( ryy1 )

ryy ( i−l +1) = ryy1 ( i ) ;

end

a = [ ] ;arev = [ ] ;

a (1 ) = 1 ;a (2 ) = −ryy (2)/ ryy ( 1 ) ;

arev (2 ) = a ( 1 ) ;arev (1 ) = a ( 2 ) ;gamma = [ ] ;g = [ ] ;

gamma(1) = a ( 2 ) ;

f o r j = 2 : ord

g = conv ( ryy1 , a ) ;

gamma( j ) = −g ( l+j )/ g ( l ) ;

b = a ;

a = cat (2 , a , z e r o s ( 1 , 1 ) ) + gamma( j )* cat (2 , z e r o s ( 1 , 1 ) , arev ) ;arev = gamma( j )* cat (2 , b , z e r o s ( 1 , 1 ) ) + cat (2 , z e r o s ( 1 , 1 ) , arev ) ;

end

gamma

f0 = ab ;g0 = ab ;

gamma(1) = gamma( 1 ) ;

f o r j = 1 : ord

128

f o r i = 1 : l

i f ( i == 1)

f1 ( i ) = f0 ( i ) ;g1 ( i ) = gamma( j )* f 0 ( i ) ;

e l s e

f 1 ( i ) = f0 ( i ) + gamma( j )* g0 ( i −1);g1 ( i ) = gamma( j )* f 0 ( i ) + g0 ( i −1);

end

end

f0 = f1 ;g0 = g1 ;

end

%pr ed i c t i on us ing s i g n a l e here us ing e r r o r s i g n a l

var in = var ( f 1 ) ;e = f1 ;

o = ord ;

z = [ z e r o s ( o+1, l ) ] ;t = [ z e r o s ( o+1, l ) ] ;kr = f l i p l r (gamma) ;

f o r i = 1 : lz (1 , i ) = e ( i ) ;

end

f o r i = 1 : li f i == 1

f o r j = 1 : oz ( j +1, i ) = z ( j , i ) ;t ( j , i ) = kr ( j )* z ( j +1, i ) ;

endt ( o+1, i ) = z ( o+1, i ) ;

129

e l s ef o r j = 1 : o

z ( j +1, i ) = z ( j , i ) − kr ( j )* t ( j +1, i −1);t ( j , i ) = kr ( j )* z ( j +1, i ) + t ( j +1, i −1);

endt ( o+1, i ) = z ( o+1, i ) ;

endendf o r j = 1 : l

pred ( j ) = z ( o+1, j ) ;end

f i g u r e (1 )c l fp l o t ( ab , ' b ' )g r id onhold onp lo t ( pred , ' ro ' )g r id on

6 Inference

In speech transmission, Levinson Durbin Algorithm helps in reducing thebandwidth and power requirement for transmission. Here, the speech is analysedand the lter coecients along with the error variance is transmitted insteadof the entire signal being transmitted. At receiver, the inverse lter synthesisesthe original signal using these lter coecients.

This algorithm can be used for various applications like vocal tract modellingand speech synthesis.

References



130

Part VIII

Lab Report 8

Date: 18th March, 2010

1 Introduction

Most of the real world signals are continuous amplitude, time varying signals.So whenever there is a need for signal processing, analog to digital conversion isessential. The signal is sampled initially and then passed onto a quantizer. Theoutput of a sampler can be visualised as weighted impulse train whose weightsare determined by the signal amplitude at that specic instant. Quantizer helpsin amplitude quantization. Quantizer maps the output sample amplitudes tothe set of quantization values.

2 Quantisation Technique

The signal amplitude range is segmented into l segments, and each of theseis represented by a amplitude value say Ik.

Ik = x : xk ≤ x ≤ xk+1 ∀ 1 ≤ k ≤ l

Then each of these segments Ik is assigned a representative amplitude valuesay yk. Then either the amplitude levels are encoded and transmitted or elsethe index k itself can be transmitted. At receiver, the encoded signal is decodedand transformed to the respective amplitude level. This process inevitably in-troduces error due to approximating the message signal to various nite quan-tization levels. This error is called quantization error. Thus for ecient signalreconstruction, quantization error should be minimized.

2.1 Uniform Quantization

Uniform quantization has uniform step size. The dierence between thequantization levels are xed. The step size for kth interval is given as

∆k = Ik+1 − IkHere, in uniform quantization

∆1 = ∆2 = . . . = ∆l = ∆

131

2.2 Non Uniform Quantization

Uniform quantizers are not used in speech quantization even though theirdesign is simple and easy to implement. Speech signals have higher probabiliyof low amplitude components than larger amplitude components. So, uniformquantizers would introduce a lot more of quantization error.

So non uniform quantizers are used in such cases. Non uniform quantizeruses dierent step sizes in dierent amplitude regions, therey minimizing error.In small amplitude region, step size is much smaller than in high amplituderegions.

2.3 PDF Optimized Non Uniform Quantizer

Assumptions for a PDF optimized non uniform quantizer are

The number of quantization levels ,L is large as a esult of high bit rate, sothat the step size ∆k will be small for all intervals. So, it can be assumedthat pX(x) is uniform in each interval Ik for k = 0, 1, 2, . . . , L.

The input signal is upper bounded at |xmax|. The input signal often showsdecaying pdf and probability of tail is lower. Hence, overload error is lower.

The pdf is symmetric.

Error variance,

σ2Qe =

ˆq2pQ(x)dq

=

L∑k=1

ˆ xk+1

xk

(x− yk)2pX(x)dx (Eq. VIII.2.1)

In the interval Ik(xk ≤ x ≤ xk+1), pX(x) is assumed to be uniform. pX(x) =pX(yk) where yk is in between xk and xk+1. The probability of x in Ik is

Pk = P (XεIk)

= pX(yk)∆k (Eq. VIII.2.2)

Here ∆k is the length of interval Ik, i.e., ∆k = xk+1− xk. Substituting (Eq.VIII.2.2) in (Eq. VIII.2.1),

σ2Qe =

L∑k=1

pk∆k

ˆ xk+1

xk

(x− yk)2dx (Eq. VIII.2.3)

To nd the optimum reconstruction level and reduce the error variance.Hence,

132

∂σ2Qe

∂yk= 0, k = 1, 2, . . . , L

−2

ˆ xk+1

xk

(x− yk) dx = 0[x2

2

]xk+1

xk

− yk (xk+1 − xk) = 0

⇒ yk =1

2(xk + xk+1) (Eq. VIII.2.4)

Hence, it can be concluded that the reconstruction levels should be in themiddle of Ik for the minimum error variance. Since, PX(x) is a constant in Ik,this translates to yk being the centroid of Ik.

Now substitute (Eq. VIII.2.4) in (Eq. VIII.2.3),

σ2Qe(min) =

L∑k=1

Pk∆k

ˆ xk+1

xk

(x− yk)2dx

=

L∑k=1

Pk∆k

(x− yk)3

3

∣∣∣∣∣xk+1

xk

=

L∑k=1

(∆2k

12

)Pk (Eq. VIII.2.5)

Here,∆2k

12 is the error variance condition in the interval Ik. That is, theerror variance is dependent only on the length of the interval Ik. Thus, it is the

parameter ∆k which is to be optimized. For uniform quantizer, σ2Qe(min) = ∆2

12 ,where ∆ = ∆k.

2.3.1 Determination of optimum ∆k

We have

σ2Qe(min) =

1

12

L∑k=1

pX(yk)∆3k & Pk = pX(yk)∆k

Lets dene, αk = (pX(yk))13 ∆k, then

σ2Qe(min) =

1

12

L∑k=1

α3k

with the constraint∑Lk=1 αk =constant. That is,

133

L∑k=1

αk =

L∑k=1

(pX(yk))13 ∆k =

ˆ xk+1

xk

(pX(x))13 dx

We need to nd the optimum value of αk under the above constraint. La-grange Multiplier technique is used to incorporate the constaint.

∂

∂αk

[σ2Qe + λ

L∑k=1

αk

]= 0 for k = 1, 2, . . , L

where λ is the Lagrange Mmultiplier

∂

∂αk

[1

12

L∑k=1

α3k + λ

L∑k=1

αk

]= 0

1

4

L∑k=1

α2k + λL = 0

⇒ λ = − 1

4L

L∑k=1

α2k

Thus,

σ2Qe + λ

L∑k=1

αk = σ2Qe −

1

4L

L∑k=1

α2k

L∑j=1

αj

= σ2

Qe −1

4L

L∑k=1

(pX(yk))23 ∆2

k

L∑j=1

(pX(yj))13 ∆k

= σ2Qe −

1

4L

L∑k=1

pX(yk)∆3k

as pX(yk)pX(yi) = 0 for j 6= k. Now

L∑k=1

αk =

L∑k=1

pX(yk)13 ∆k

'ˆ xmax

−xmaxpX(x)

13 dx

' constant

⇒ pX(yk)∆3k = constant

⇒ Pk∆2k = constant

134

From the above result, we can interpret that the interval of higher proba-bility of occurence must have a smaller step size. This agrees with our earlierassumption that the smaller amplitudes require smaller step sizes. We have,

σ2Qe =

1

12

L∑k=1

Pk∆2k and Pk∆2

k = constant

Hence, all intervals produce identical quantization error variance contribu-tions. Therefore we need to use a non uniform quantizer which practicallydicult to implement. So, we go for companding technique which in generaltransforms an arbitrary pdf to uniform pdf. The uniform pdf can be easilyquantized using the uniform quantizers. The expander helps in retriving theoriginal pdf from unifom pdf.

3 Logarithmic Companding

The companding techniques implemented for dierent signal distributionsare dierent. Speech signals being exponentially distributed, logarithmic com-pression leads to a uniform distribution. To maintain the transmission quality ofloud and soft sounds, voice signals require relatively constant signal to quantiza-tion noise ratio over a wide range of amplitude levels. This requires logarithmiccompression to be implemented.

We have,

σ2Qe =

x2max

3L2

ˆ xmax

−xmaxpX(x)

(dC(x)

dx

)−2

dx

⇒ SNR =σ2X

σ2Qe

=3L2

x2max

´ xmax−xmax x

2pX(x)dx

´∞−∞ pX(x)

(dC(x)dx

)−2

dx

We require constant SNR, i.e. dC(x)

dx should be always proportional to x

dC(x)

dx= (kx)−1

where k is a constant. Thus,

SNR =3L2

k2x2max

⇒ C(x) =1

klnx+ d ;x > 0

135

where d is constant of integration. Thus, logarithmic companding results inconstant SNR. Using the condition,

C(xmax) = xmax

C(x) =

(1

kln|x|xmax

+ xmax

)sgn(x)

but C(x) → −∞ as x → 0. This is not practically implementable. So, werequire C(x) → 0 for x → 0. Two companding laws which incorporate thiscorrection are

1. A-law companding

2. µ-law companding

3.1 A-law Companding

The companding characteristic is linear near origin, i.e. |x| ≤ xmax/A andlogarithmic for large x. Logarithmic curve is expressed as

C(x) = k′(

1 + lnA(x)

xmax

)and using the condition, C(xmax) = xmax,

k′ =xmax

1 + lnA

A-law companding characteristic is given by

C(x) =

A(x)1+lnAsgn(x) ; 0 ≤ |x|

xmax≤ 1/A

xmax1+ln( A(x)

xmax)

1+lnA sgn(x) ; 1/A ≤ |x|xmax

≤ 1

Note: The parameter À' controls the degree of compression and may bechosen so that large amplitude changes in input is not reected at the output.The ratio of maximum to minimum slope is A. The practical value of A is 87.56

136

3.2 µ-law Companding

The µ-law characteristic is neither strictly logarithmic nor strictly linear any-where. It is approximately linear at low levels i.e. µ|x| << xmax or x <<µ−1xmax and approximately logarithmic at high levels i.e. x >> µ

−1xmax.

C(x) = k'log

[1 +

µ|x|xmax

]Apply the condition, C(xmax) = xmax,

k′

=xmax

log(1 + µ)

Hence, the µ-law characteristic is given by

C(x) =xmaxloge(1 + µ|x|

xmax)

log(1 + µ)sgn(x)

C(x) =

xmaxloge(

µ|x|xmax

)

log(1+µ) sgn(x) ;µ |x| xmaxµ|x|

log(1+µ) ;µ |x| xmax

4 Digital Companding:

Segmentation of Companding Characteristics

The logarithmic form of companding laws is slow to calculate. In order toimplement companding in a digital transmission system, we need a digital formof the companding laws. Digital companding involves sampling an analog signal,converting it to a PCM code and digitally compressing the code. At the receiver

137

end, the compressed PCM code is received, expanded and decoded. Digitalcompanding is obtained by approximating the companding characteristics bylinear segments. This process is called segmentation.

4.1 Segmentation of A-law Companding Characteristics

In A-law companding we compress a 13-bit code to an 8-bit code. We againapproximate the curve using 8 straight line segments. The slope of each succes-sive segment is half that of the previous segment, except segment 0 and 1 whichhave the same slope:

Segment 0 1 2 3 4 5 6 7Step size 32 32 64 128 256 512 1024 2048

Algorithm for A-law companding

Encoding:

1. Determine the bit position of the most signicant 1-bit among bits 5through 11 of the input.

2. If such a 1-bit is found, the segment code becomes that position minus 4.Otherwise, the segment code is 0.

3. The 4 bits following this 1-bit position determine the quantization code.If segment code is 0, the quantization code is half the input value.

4. The sign bit is same as the input sign bit.

Decoding:

1. If the segment code is non-zero, multiply quantization code by 2 and add33. Multiply result by 2 raised to the power of the segment code minus 1.

2. If the segment code is zero, multiply quantization code by 2 and add 1.

3. Append sign bit.

4.2 Segmentation of µ-law Companding Characteristics

In µ-law encoding, we compress a 14-bit code to an 8-bit code. The curveis segmented and approximated as straight lines. The slope of each successivesegment is half that of the previous segment. This is done so that smalleramplitudes need smaller step size. If the rst segment has a step size of 32, wehave

138

Segment 0 1 2 3 4 5 6 7Step size 32 64 128 256 512 1024 2048 4096

The output levels are equispaced. Hence, output step size is 128/8 = 16.This leads to segmentation of the curve as shown in the gure. The sameprocedure can be applied to the negative half of the characteristic. In order tomake the segment end points multiples of 2, we add a bias of 33. The motivationbehind such a step is ease in computation. Thus, the maximum allowable inputamplitude reduces to 214 33 = 8159.

Each segment is further sub-divided into 16 equally spaced quantizationintervals (of step size 1 each). In this way, we form an 8-bit compressed codeconsisting of sign bit, a 3-bit segment identier and a 4-bit quantization intervalidentier. The 4-bit quantization code identies the quantization interval withinthe segment. Sinse each segment change implies a change of 16, the format ofthe 8-bit code is

P S2 S1 S0 Q3 Q2 Q1 Q0

where P is sign bit, S2 − S0is segment code, Q3 −Q0 is quantization.Consider any one segment, say, segment 1 (64-128) long with its 16 quanti-

zation levels. Any 12-bit number falling in this segment can be reperesented as00 0000 01ab cdef. Each quantization interval has an input step size 4 and anoutput step size 1. Alternatively, a change in the input by 4 elicits a change inthe output by 1. Since 4 = 22 this leads us to the conclusion that changes in thelower 2 bits bits 1 and 0 are ignored by the compressor. Thus, the compressedversion of 00 0000 01ab cdef is 0 001 abcd (after ignoring bits 1 and 0). As thesegment number increases, the number of lower bits ignored increases.

Algorithm for µ-law companding:

Encoding:

1. Add 33 to absolute value of input.

2. Determine bit position of the most signicant 1-bit among bits 5 through12of the input.

3. Subtract 5 from the position. The resulting number gives the segmentcode.

4. The 4 bits following the bit-position determined in step2 form the 4-bitquantization code.

5. The sign bit is the same as the sign bit of the input.

139

Decoding:

1. Shift left by 1 position and add 33.

2. Shift left by the amount of segment code.

3. Decrement the result by 33.

4. Append sign bit.

5 Observations

A-law Companding

PDF Transformation

0 100 200 300 400 5000

200

400

600

800

1000

1200

1400

1600

1800Histogram of input

Amplitude→

Fre

qu

en

cy→

(a) Input Histogram

0 100 200 300 400 5000

100

200

300

400

500

600

700

800

900Histogram of companded input

Amplitude→

Fre

qu

en

cy→

(b) Companded Histogram

Figure 108: A-law Companding

140

Companding

0 50 100 150 200 250 300−8000

−6000

−4000

−2000

0

2000

4000

6000

8000Companded output−Alaw

t→

Am

plit

ude

→

Figure 109: Companded Speech Signal

µ-law Companding

PDF Transformation

0 100 200 300 400 500 600 7000

100

200

300

400

500

600

700Histogram of input

Amplitude→

Fre

que

ncy→

(a) Input Histogram

0 100 200 300 400 500 600 7000

20

40

60

80

100

120

140

160

180

200Histogram of companded input

Amplitude→

Fre

que

ncy→

(b) Companded Histogram

Figure 110: µ-law Companding

141

Companding

0 50 100 150 200 250 300−8000

−6000

−4000

−2000

0

2000

4000

6000

8000Companded output−ulaw

t→

Am

plit

ude

→

Figure 111: Companded Speech Signal

In both these companded outputs, we observe that the low amplitude com-ponents are amplied more than higher amplitude components. So, we havealmost a uniform PDF as per requirement.

ReconstructionHere we have taken a mixed sinusoid as input to the compander (black

colour) and it is companded by a µ-law compander (red colour). The compandedoutput is used to reconstruct the original input and this expanded signal is shownwith blue colour.

142

0 50 100 150 200 250 300 350 400 450−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Companding And Reconstruction

t→

Am

plit

ud

e→

Figure 112: Compression and Expansion

6 Code

In MATLAB, it is possible ot create a uniform PDF but here we have aspeech sequence which is exponentially distributed. So, we need to transformto uniform PDF.

A-law Companding

%Alaw speech compandingc l cA=87.6x=[−120 186 −348 −517 555 −5 −434 17 595 −225 −473189 237 −3 −188 117 −249 12 617 −932 −92 970 −497−592 873 −175 −314 178 −249 149 481 −884 −166 1087−497 −991 1034 606 −1205 523 644 −706 205 320 −642512 294 −892 3 688 −525 −298 412 −9 −331 9 −79 11 80−505 −200 449 −196 −538 144 363 −341 −203 373 106 −9886 227 54 88 37 −136 337 359 −281 −4 772 317 −34 427610 296 318 90 −192 −17 −583 −1108 −863 −946 −1505−1381 −899 −1012 −637 −73 −1 594 1112 867 1369 1585968 1164 1420 709 403 402 181 105 220 540 882 693 538

143

503 304 −296 −1051 −1640 −2066 −2706 −3267 −3150−3013 −2460 −2001 −1129 −5 1023 1795 2596 3491 34823114 2508 2256 1581 231 −612 −666 −1222 −1628 −1490−980 −747 −402 284 827 1530 2143 2147 2117 1361 1078418 −603 −2202 −3100 −3740 −4572 −5140 −4530 −3996−3005 −1731 −308 1260 2898 4169 5079 5698 5376 47303589 2403 760 −658 −1734 −2531 −3103 −3249 −2651 −1758−870 −17 659 1860 2621 2790 3551 3473 1997 837 380 −1169−3620 −4224 −5309 −6060 −5987 −5841 −4288 −3035 −1301703 2439 4205 5564 6520 6693 5896 4825 3262 1413 −429−1887 −2974 −3409 −3690 −3565 −2645 −1439 −79 1097 16822373 3352 3431 3091 3414 2477 450 −891 −1903 −3887 −5615−5664 −6152 −6432 −5684 −4636 −2713 −569 1284 3138 4577

5609 6757 7069 62 48 ] ;ymax=max(x ) ;

f i g u r e (4 )y=x ;n=length (y ) ;i =1:n ;p l o t ( i , y ) ;g r i d on ;hold on ;C=0;f o r l =1: l ength (y )

i f abs ( y ( l ))<(ymax/A)C( l )=A.* abs (y ( l ) ) ./ (1+ log (A) ) . * s i gn (y ( l ) ) ;

e l s eC( l )=ymax.*((1+ log ( (A.* y ( l ) )/ymax))/(1+ log (A) ) ) . * s i gn (y ( l ) ) ;

endend

n=length (C) ;i =1:n ;p l o t ( i ,C, ' r ' ) ;g r i d on ;t i t l e ( 'Companded output−Alaw ' )x l ab e l ( ' t \ r ightarrow ' )y l ab e l ( ' Amplitude\ r ightarrow ' )

144

µ-law Companding

%ulaw speech compandingc l cc l fu=255;y=[−120 186 −348 −517 555 −5 −434 17 595 −225 −473189 237 −3 −188 117 −249 12 617 −932 −92 970 −497−592 873 −175 −314 178 −249 149 481 −884 −166 1087−497 −991 1034 606 −1205 523 644 −706 205 320 −642512 294 −892 3 688 −525 −298 412 −9 −331 9 −79 11

80 −505 −200 449 −196 −538 144 363 −341 −203 373 106−98 86 227 54 88 37 −136 337 359 −281 −4 772 317−34 427 610 296 318 90 −192 −17 −583 −1108 −863 −946−1505 −1381 −899 −1012 −637 −73 −1 594 1112 867 13691585 968 1164 1420 709 403 402 181 105 220 540 882

693 538 503 304 −296 −1051 −1640 −2066 −2706 −3267−3150 −3013 −2460 −2001 −1129 −5 1023 1795 2596 34913482 3114 2508 2256 1581 231 −612 −666 −1222 −1628−1490 −980 −747 −402 284 827 1530 2143 2147 2117 13611078 418 −603 −2202 −3100 −3740 −4572 −5140 −4530−3996 −3005 −1731 −308 1260 2898 4169 5079 5698 53764730 3589 2403 760 −658 −1734 −2531 −3103 −3249 −2651−1758 −870 −17 659 1860 2621 2790 3551 3473 1997 837380 −1169 −3620 −4224 −5309 −6060 −5987 −5841 −4288−3035 −1301 703 2439 4205 5564 6520 6693 5896 48253262 1413 −429 −1887 −2974 −3409 −3690 −3565 −2645−1439 −79 1097 1682 2373 3352 3431 3091 3414 2477 450−891 −1903 −3887 −5615 −5664 −6152 −6432 −5684 −4636−2713 −569 1284 3138 4577 5609 6757 7069 62 48 ] ;ymax=max(y ) ;

f i g u r e (1 )n=length (y )i =1:np l o t ( i , y , ' b ' )hold onC=ymax* ( ( l og (1+(u . * ( abs ( y ) . / ymax ) ) ) ) . / l og (1+u ) ) . * s i gn (y ) ;

n=length (C)i =1:np l o t ( i ,C, ' r ' )g r i d ont i t l e ( 'Companded output−ulaw ' )x l ab e l ( ' t \ r ightarrow ' )y l ab e l ( ' Amplitude\ r ightarrow ' )

145

Reconstruction

t =0 : 0 . 0 1 : 4 ;x=8158*( s i n (2* pi *2 .* t )+ s i n (2* pi *3* t ) ) . / 2 ;x=x+33;l en=length (x ) ;x1=x . /max(x ) ;p l o t ( x1 , ' black ' )y=ze ro s ( len , 1 4 ) ;

f o r i =1: l en ;i f x ( i ) >=0

x ( i )= c e i l ( x ( i ) ) ;s=0y ( i , : )= [ de2bi ( x ( i ) , 13 ) s ] ;

e l s e i f x ( i )<0s=1;x ( i )= f l o o r ( x ( i ) ) ;y ( i , : )= [ de2bi (−x ( i ) , 13 ) s ] ;

endendd=y ( : , 1 4 ) ;t r i a l 1 =0;a=ze ro s ( len , 4 ) ;b=ze ro s ( len , 3 ) ;z=ze ro s ( len , 8 ) ;

f o r j =1: l ent r i a l 1 =0;

f o r i =0:13

i f y ( j ,13− i )==1t r i a l 1=13−i−1break ;

endend

t r i a l 2=t r i a l 1 −5;

i f t r i a l 2==8break ;

146

end

b( j , : )= de2bi ( t r i a l 2 , 3 ) ;

a ( j , : )= y ( j ,13− i −3:13− i ) ;

endz=[a b d ] ;

f i n a l=ze ro s (1 , l en ) ;t r i a l 3=ze ro s (1 , 7 )

f o r k=1: l eni f z (k,8)==1

t r i a l 3=z (k , 1 : 7 ) ;f i n a l ( k)=−1*bi2de ( t r i a l 3 ) ;

e l s e i f z (k,8)==0t r i a l 3=z (k , 1 : 7 ) ;f i n a l ( k)=bi2de ( t r i a l 3 ) ;

end

end

hold onp lo t ( f i n a l /max( f i n a l ) , ' r ' )hold o f fs i gn=z ( : , 8 ) ;segment=z ( : , 5 : 7 ) ;q=z ( : , 1 : 4 ) ;q=[ z e ro s ( len , 1 ) q ] ;p=bi2de (q )p=p+33;segment=bi2de ( segment ) ;p=p .*2 .^ segmentp=p−33;new=de2bi (p , 1 3 ) ;new=[new s i gn ] ;t r i a l 4=ze ro s (1 , l en )t r i a l 5=ze ro s ( 1 , 1 4 ) ;

f o r i =1: l eni f new( i ,14)==0

t r i a l 5=new( i , 1 : 1 3 ) ;t r i a l 4 ( i )=bi2de ( t r i a l 5 ) ;

147

e l s e i f new( i ,14)==1t r i a l 5=new( i , 1 : 1 3 ) ;t r i a l 4 ( i )=−1.*bi2de ( t r i a l 5 ) ;

end

endop=t r i a l 4 /max( t r i a l 4 ) ;hold onp lo t ( op , ' b ' )t i t l e ( ' Companding And Reconstruct ion ' )x l ab e l ( ' t \ r ightarrow ' )y l ab e l ( ' Amplitude\ r ightarrow ' )hold o f f

References

[1] Fathima Shifa Akbar, Geetu George, Kala K. and Sneha Sara Suresh,Mini Project Report on A-Law and µ-Law Companding: Principles,

Formulations and Implementation, NIT Calicut 2006

148

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