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DSP NOTES PREPAREDBY 1Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldDIGITAL SIGNAL PROCESSING

A signal is defined as any physical quantity that varies with time, space or another

independent variable.A system is defined as a physical device that performs an operation on a signal.System is characterized by the type of operation that performs on the signal. Such

operations are referred to as signal processing.

Advantages of DSP1. A digital programmable system allows flexibility in reconfiguring the digital

signal processing operations by changing the program. In analog redesign of hardware is

required.2. In digital accuracy depends on word length, floating Vs fixed point arithmetic etc.

In analog depends on components.3. Can be stored on disk.

4. It is very difficult to perform precise mathematical operations on signals in analogform but these operations can be routinely implemented on a digital computer using

software.5. Cheaper to implement.6. Small size.

7. Several filters need several boards in analog, whereas in digital same DSP

processor is used for many filters.Disadvantages of DSP

1. When analog signal is changing very fast, it is difficult to convert digital form.(beyond 100KHz range)2. w=1/2 Sampling rate.3. Finite word length problems.

4. When the signal is weak, within a few tenths of millivolts, we cannot amplify thesignal after it is digitized.

5. DSP hardware is more expensive than general purpose microprocessors & micro

controllers.2Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World6. Dedicated DSP can do better than general purpose DSP.Applications of DSP1. Filtering.

2. Speech synthesis in which white noise (all frequency components present to thesame level) is filtered on a selective frequency basis in order to get an audio signal.

3. Speech compression and expansion for use in radio voice communication.

4. Speech recognition.5. Signal analysis.6. Image processing: filtering, edge effects, enhancement.7. PCM used in telephone communication.

8. High speed MODEM data communication using pulse modulation systems such as

FSK, QAM etc. MODEM transmits high speed (1200-19200 bits per second) over a

band limited (3-4 KHz) analog telephone wire line.

9. Wave form generation.Classification of SignalsI.Based on Variables:

f(t)=5t : single variable1.

2.

3.

4.f(x,y)=2x+3y : two variablesS1= A Sin(wt) : real valued signal

S2 = A ejwt : A Cos(wt)+j A Sin(wt) : Complex valued signal

S1(t) 5.S4(t)=: Multichannel signalS2(t)S3(t)Ex: due to earth quake, ground acceleration recorderIr(x, y,t)6.

I(x,y,t)=multidimensionalIg(x, y,t)Ib(x, y,t)II.Based on Representation:3Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldIII.Based on duration.1.

2.

3.

4.

5.right sided: x(n)=0 for n<N

left sided :x(n)=0 for n>N

causal : x(n)=0 for n<0Anti causal : x(n)=0 for n0

Non causal : x(n)=0 for n >NIV.Based on the Shape.

n 01. (n)=0=1n=02. u (n) =1

=0n0n<0Arbitrary sequence can be represented as a sum of scaled, delayed impulses.4Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldP (n) = a-3 (n+3) +a1 (u-1) +a2 (u-2) +a7 (u-7)Or

x(n) =x(k) (n k)kn

u(n) = (k) = (n) + (n-1)+ (n-2)…..k

= (n k)k0

3.Discrete pulse signals.Rect (n/2N) =1

= 0n Nelse where.5.Tri (n/N) = 1- n /N n N

= 0 else where.1. Sinc (n/N)= Sa(n/N) = Sin(n/N) / (n/N), Sinc(0)=1

Sinc (n/N) =0 at n=kN, k= 1, 2…

Sinc (n) = (n) for N=1; (Sin (n) / n=1= (n))

6.Exponential Sequencex (n) = A n

If A & are real numbers, then the sequence is real. If 0< <1 and A is +ve, thensequence values are +ve and decreases with increasing n.

For -1< <0, the sequence values alternate in sign but again decreases in magnitude

with increasing n. If >1, then the sequences grows in magnitude as n increases.7.Sinusoidal Sequencex(n) = A Cos(won+ ) for all n8.Complex exponential sequence5Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldIf = ejwo

A = A ejx(n) = A ej nejwon

= A n Cos(won+ ) + j A n Sin(won+ )If >1, the sequence oscillates with exponentially growing envelope.If <1, the sequence oscillates with exponentially decreasing envelope.

So when discussing complex exponential signals of the form x(n)= A ejwon or real

sinusoidal signals of the form x(n)= A Cos(won+ ) , we need only consider frequencies

in a frequency internal of length 2 such as < Wo < or 0 Wo<2 .V. Deterministic (x (t) = t x (t) = A Sin(wt))

& Non-deterministic Signals. (Ex: Thermal noise.)VI.Periodic & non periodic based on repetition.Power & Energy SignalsVII.Energy signal: E = finite, P=0 Signal with finite energy is called energy signal.

Energy signal have zero signal power, since averaging finite energy over infinitetime. All time limited signals of finite amplitude are energy signals.Ex: one sided or two sided decaying. Damped exponentials, damped sinusoidal.1 x(t) is an energy signal if it is finite valued and x

2

(t) decays to zero fasten thantas t .

Power signal: E =, P 0, P Ex: All periodic waveformsNeither energy nor power: E=, P=0 Ex: 1/ t t 1 E=, P=, Ex: tn

VIII.Based on Symmetry1.2.

3.EvenOddx(n)=xe(n)+xo(n)x(-n)=xe(-n)+xo(-n)

x(-n)=xe(n)-xo(n)Hidden4.Half-wave symmetry.xe(n)= 1 [x(n)+x(-n)]2

6Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldxo(n)= 1 [x(n)-x(-n)]2Signal Classification by duration & Area.a. Finite duration: time limited.b. Semi-infinite extent: right sided, if they are zero for t < where = finitec. Left sided: zero for t >Piecewise continuous: possess different expressions over different intervals.

Continuous: defined by single expressions for all time. x(t) = sin(t)

Periodic: xp (t) = xp (t nT)T

1

For periodic signals P = x(t) 2 dtT0

X rms = PFor non periodicT

1

P = Lt x(t) 2 dtTo0To

Xavg = Lt x(t)dt0

x(t) = A cos( 2 fo t + ) P=0.5 A2

x(t) = A e j( 2 fo t + )

P=A27Downloaded From JNTU World (http://JNTU-World.com)


E= A2 b

Q.E = 1 A2 bE = 1 A2 b23

e- t dt = 10

Q.Ex = 1 A2 0.5T + 1 (-A)2 0.5T = 0.5 A2 T228Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldPx = 0.5 A2

Q.Ey = [ 1 A2 0.5T] 2 = 1 A2 T33Py = 1 A23 x(t) = A e

jwt

is periodicT

1

x(t) 2 dt = A2

Px =T0

x(2t -6 ): compressed by 2 and shifted right by 3 OR shifted by 6 and compressed

by 2. x(1-t): fold x(t) & shift right by 1 OR shift right and fold. x(0.5t +0.5) Advance by 0.5 & stretched by 2 OR stretched by 2 & advance by 1.y (t) = 2 x [- (t 2) ] = 2 x[ ]t22 x( t + ) ; 5 + =-1; - + =1 => = -1/3333; = 2/3Area of symmetric signals over symmetric limits (- , )

Odd symmetry: x0 (t) dt =0

Even symmetry: xe (t) dt = 2 xe (t) dt0

Xe (t) +Ye (t): even symmetry.9Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldXe (t) Ye (t): even symmetry.Xo (t) +Yo (t): odd symmetry.Xo (t) Xo (t): even symmetry.

Xe (t) +Yo (t): no symmetry.

Xe (t) Yo (t): odd symmetry.Xe(n)= 1 [x(n)+x(-n)]2Xo (n) = 1 [x (n)-x (-n)]2 Area of half-wave symmetry signal always zero.

Half wave symmetry applicable only for periodic signal.

F0 = GCD ( f1,f2)T = LCM (T1, T2) Y(t) = x1(t) + x2(t)Py= Px1+Px2

Y(t)rms = Py U(0) = 0.5 is called as Heaviside unit step.

X(t) = Sin(t) Sin( t)= 0.5 cos (1- )t – 0.5 cos (1+ ) tW1=1-W2=1+ almost periodic OR non periodic.

Px = 0.5[0.52 +0.52] =0.25 WArea of any sinc or Sinc 2 equals area of triangle ABC inscribed within the main lobe.10Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldEven though the sinc function is square integrable ( an energy signal) , it is not1absolutely integrable( because it does not decay to zero faster than )t (t) = 0 t 0= t=0 ()d = 1An impulse is a tall narrow spike with finite area and infinite energy.The area of impulse A (t) equals A and is called its strength. How ever its hight at

t=0 is .= 2 (t) – 2e-t u(t)2 e-t (t) = 2 (t) [ [t- ]] = 1 (t )

2

I 2 = cos(2t)(2t 1)dt = cos(2t)0.5(t 0.5)dt = 0.5 cos(2 t) at t=-0.5 = -0.52

44

x1(t) = x(t) (t-kts ) =x(kts) (t-kts)kk11Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldx1(t) is not periodic.The doublet ’(t) =0t 0

'(t)dt 0= undefinedt=0 ’ (-t) = - ’ (t) then Odd function. [ [t- ]] = 1 (t )Differentiating on both sides1 ’ [ [t- ]] = '(t )With =-1 ’ (-t) = - ’ (t)d [x(t) (t )]= x’ (t) (t- ) + x (t) ’ (t- )dt= x’ ( ) (t- ) + x (t) ’ (t- )-----------1Ord [x(t) (t )] = d [x() (t )]= x ( ) ’ (t- ) -----------2dtdt1 = 2x’ ( ) (t- ) + x (t) ’ (t- ) = x ( ) ’ (t- )12Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World x (t) ’ (t- ) = x ( ) ’ (t- ) - x’ ( ) (t- )

x (t) ’ (t- ) dt =x ( ) ’ (t- ) dt -x’ ( ) (t- ) dt= 0- x’ ( ) = - x’ ( )

Higher derivatives of (t) obey n(t) = (-1)n n(t) are alternately odd and even,and possess zero area. All are eliminating forms of the same sequence that generate

impulses, provided their ordinary derivatives exits. None are absolutely integrable.

The impulse is unique in being the only absolutely integrable function from among

all its derivatives and integrals (step, ramp etc)What does the signal x(t) = e-t ’(t) describe?x(t) = ’ (t) – (-1) (t) = ’ (t) + (t)2

I = [(t 3)(2t 2)] 8cos(t)'(t 0.5)]dt2

= 0.5 (t-3) t 1 - 8 dtd [cost]t 0.5= 23.1327 Answer.Operation on Signals:1. Shifting.x(n) shift right or delay = x(n-m)x(n) shift left or advance = x(n+m)

2. Time reversal or fold.

x(-n+2) is x(-n) delayed by two samples.

x(-n-2) is x(-n) advanced by two samples.

Orx(n) is right shift x(n-2), then fold x(-n-2)x(n) fold x(-n) shift left x(-(n+2)) = x(-n-2)

Ex:x(n) = 2, 3 , 4 , 5, 6, 7 .Find 1. y(n)=x(n-3) 2. x(n+2) 3. x(-n) 4. x(-n+1) 5. x(-n-2)1.y(n)= x(n-3) = 0 ,2,3,4,5,6,7 shift x(n) right 3 units.13Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World2.

3.

4.

5.x(n+2) = 2,3,4,5,6 ,7 shift x(n) left 2 units.x(-n) = 7,6,5, 4 ,3,2 fold x(n) about n=0.x(-n+1) = 7,6,5,4,3,2 fold x(n), delay by 1.x(-n-2) = 7,6,5,4,3,2 fold x(n), advanced by 2.3. a. Decimation.

Suppose x(n) corresponds to an analog signal x(t) sampled at intervals Ts. The signal

y(n) = x(2n) then corresponds to the compressed signal x(2t) sampled at Ts and contains

only alternate samples of x(n)( corresponding to x(0), x(2), x(4)…). We can also obtain

directly from x(t) (not in compressed version). If we sample it at intervals 2Ts (or at a

1sampling rate Fs =). This means a two fold reduction in the sampling rate.2TsDecimation by a factor N is equivalent to sampling x(t) at intervals NTs and implies anN-fold reduction in the sampling rate.b. Interpolation.

y(n) = x(n/2) corresponds to x(t) sampled at Ts/2 and has twice the length of x(n)

with one new sample between adjacent samples of x(n).

The new sample value as ‘0’ for Zero interpolation.The new sample constant = previous value for step interpolation.The new sample average of adjacent samples for linear interpolation.

Interpolation by a factor of N is equivalent to sampling x(t) at intervals Ts/N and

implies an N-fold increase in both the sampling rate and the signal length.

Ex:Decimation

Step interpolation

nn/2

, 2, 6, 4, 8 , 6, 8 , 1, 6, 6, 8, 8111n2n

Step interpolationDecimation , 2, 6, 4, 8 , 1,2,2,6, 6,4,4,8, 81

nn/21

, 2, 6, 4, 8

1n2n

Since Decimation is indeed the inverse of interpolation, but the converse is notnecessarily true. First Interpolation & Decimation.Ex:x(n) = 1 1, 2, 5, -114Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldx(n/3) = 1,0,0, 2 2,0,0,5,0,0,-1,0,0 Zero interpolation.= 1,1,1, 2 ,2,2,5,5,5,-1,-1,-1 Step interpolation.= 1, , 5 , 2 , 3,4,5,3,1,-1, - ,- 1 Linear interpolation.4233334. Fractional Delays.

It requires interpolation (N), shift (M) and Decimation (n): x (n - M ) = x ( (Nn M ) )NNx(n) = 2, 4, 6 , 8, find y(n)=x(n-0.5) = x ( 2n 1)2g(n) = x (n/2) = 2, 2, 4, 4, 6 , 6, 8,8 for step interpolation.n 1h(n) =g(n-1) = x() = 2, 2, 4, 4 , 6, 6,8,82y(n) = h(2n) = x(n-0.5) = x( 2n 1) = 2, 4 , 6, 82ORg(n) = x(n/2) = 2,3,4,5, 6 ,7,8,4 linear interpolation.h(n) = g(n-1) = 2,3,4,5, 6, 7,8,4g (n) = h(2n)=3,5,7,4Classification of Systems1. a. Static systems or memory less system. (Non Linear / Stable)Ex. y(n) = a x (n)= n x(n) + b x3(n)= [x(n)]2 = a(n-1) x(n)

y(n) = [x(n), n]If its o/p at every value of ‘n’ depends only on the input x(n) at the same value of ‘n’

Do not include delay elements. Similarly to combinational circuits.

b. Dynamic systems or memory.If its o/p at every value of ‘n’ depends on the o/p till (n-1) and i/p at the same value of

‘n’ or previous value of ‘n’.Ex. y(n) = x(n) + 3 x(n-1)15Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World= 2 x(n) - 10 x(n-2) + 15 y(n-1)Similar to sequential circuit.2. Ideal delay system. (Stable, linear, memory less if nd=0)Ex. y (n) = x(n-nd)nd is fixed = +ve integer.3. Moving average system. (LTIV ,Stable)m2

y(n) = 1/ (m1+m2+1)x(n k)km1

This system computes the nth sample of the o/p sequence as the average of (m1+m2+1)samples of input sequence around the nth sample.If M1=0; M2=55

y(7) = 1/6 [x(7 k)]k0

= 1/6 [x(7) + x(6) + x(5) + x(4) + x(3) + x(2)]y(8) = 1/6 [x(8) + x(7) + x(6) + x(5) + x(4) + x(3)]So to compute y (8), both dotted lines would move one sample to right.4. Accumulator. ( Linear , Unstable )n

y(n) = x(k)kn1

=x(k) + x(n)k= y(n-1) + x(n)16Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldx(n) = …0,3,2,1,0,1,2,3,0,….y(n) = …0,3,5,6,6,7,9,12,12…O/p at the nth sample depends on the i/p’s till nth sampleEx:x(n) = n u(n) ; given y(-1)=0. i.e. initially relaxed.1n

x(k)

y(n) =+x(k)kk0nn

n(n 1)

= y(-1) +x(k) = 0 + n =2k0k0

5. Linear Systems.If y1(n) & y2(n) are the responses of a system when x1(n) & x2(n) are the respective

inputs, then the system is linear if and only if

[x1(n) x2(n)]= [x1(n)] + [x2(n)]= y1(n) + y2(n) (Additive property)[ax(n)] = a [x(n)] = a y(n) (Scaling or Homogeneity)

The two properties can be combined into principle of superposition stated as

[ax1(n)bx2(n)] = a [x1(n)] + b [x2(n)]Otherwise non linear system.6. Time invariant system.

Is one for which a time shift or delay of input sequence causes a corresponding shift

in the o/p sequence.y(n-k) = [x(n k)]TIV

TV7. Causality.

A system is causal if for every choice of no the o/p sequence value at index n= no

depends only on the input sequence values for n no.

y(n) = x(n) + x(n-1) causal.y(n) = x(n) + x(n+2) + x(n-4) non causal.

8. Stability.17Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldFor every bounded input x(n) Bx < for all n, there exists a fixed +ve finite valueBy such that y(n) By <.PROPERTIES OF LTI SYSTEM.

1. x(n) =x(k)(n k)k

y(n) = [ x(k)(n k)] for lineark

x(k) [ (n-k)] for time invariantk

x(k)h(n k)= x(n) * h(n)kTherefore o/p of any LTI system is convolution of i/p and impulse response.

y(no) =h(k)x(no k)k1

=h(k)x(no k)+ h(k)x(no k)kk0

= h(-1) x(n0+1) + h(-2) x(n0+2)……….+h(0) x(n0) + h(1) x(n0-1) + ….y(n) is causal sequence if h(n) =0n<0y(n) is anti causal sequence if h(n) =0 n0y(n) is non causal sequence if h(n) =0 |n|>N

Therefore causal system y(n) =h(k)x(n k)k0n

If i/p is also causal y(n) =h(k)x(n k)k0

2. Convolution operation is commutative.

x(n) * h(n) = h(n) *x(n)3. Convolution operation is distributive over additive.x(n) * [h1(n) + h2(n)] = x(n) * h1(n) + x(n) * h2(n)

4. Convolution property is associative.

x(n) * h1(n) * h2(n) = [x(n) * h1(n)] * h2(n)18Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World5 y(n) = h2 * w(n) = h2(n)*h1(n)*x(n) = h3(n)*x(n)6

h (n) = h1(n) + h2(n)

7 LTI systems are stable if and only if impulse response is absolutely summable.

y(n) =h(k)x(n k) h(k) x(n k)kkSince x (n) is bounded x(n) bx<

y(n) Bxh(k)k

S=h(k) is necessary & sufficient condition for stability.k8 (n) *x(n) = x(n)9 Convolution yields the zero state response of an LTI system.10 The response of LTI system to periodic signals is also periodic with identicalperiod.y(n) = h (n) * x(n)19Downloaded From JNTU World (http://JNTU-World.com)


=h(k)x(n k)k

y (n+N) =h(k)x(n k N)kput n-k = m

=h(n m)x(m N)m

=

=

h(n m)x(m)mm=k

h(n k)x(k) = y(n) (Ans)kQ. y (n)-0.4 y(n-1) =x (n). Find causal impulse response? h(n)=0 n<0.

h(n) = 0.4 h(n-1) + (n)h(0) = 0.4 h(-1) + (0)=1h(1) = 0.4 h(0) = 0.4h(2) = 0.42

h(n) = 0.4n for n0Q. y(n)-0.4 y(n-1) = x(n). find the anti-causal impulse response? h(n)=0 for n 0

h(n-1) = 2.5 [h(n)- (n)]h(-1) = 2.5 [h(0)- (0)] = -2.5

h(-2) = -2.52 . …….. h(n) = -2.5n valid for n-1Q. x(n)=1,2,3 y(n)=3,4 Obtain difference equation from i/p & o/p information

y(n) + 2 y(n-1) + 3 y(n-2) = 3 x(n) + 4 x(n-1) (Ans)Q. x(n) = 4,4,, y(n)= x(n)- 0.5x(n-1). Find the difference equation of the inverse

system. Sketch the realization of each system and find the output of each system.

Solution:The original system is y(n)=x(n)-0.5 x(n-1)The inverse system is x(n)= y(n)-0.5 y(n-1)

y (n) = x (n) – 0.5 x(n-1)Y (z) = X (z) [1-0.5Z-1]20

Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldY(z)=1-0.5 Z-1

SystemX (z)Inverse Systemy (n) – 0.5 y(n-1) =x(n)Y (z) [1-0.5 Z-1] = X (z)Y(z)

[1-0.5 Z-1] -1X (z)g (n) = 4 (n) - 2 (n-1) + 4 (n-1) - 2 (n-2) = 4 (n) + 2 (n-1) - 2 (n-2)y (n) = 0.5 y(n-1) + 4 (n) + 2 (n-1) – 2 (n-2)

y (0) = 0.5y(-1) + 4 (0) = 4y(1) = 4y(2) = 0.5 y(1) - 2 (0) = 0y(n) = 4, 4 same as i/p.Non Recursive filtersRecursive filtersNN

y(n) =ak x(n-k)y(n) =ak x(n-k) –bk y(n-k)kk0k1

for causal systemPresent response is a function of the

present and past N values of the

excitation as well as the past N values

of response. It gives IIR o/p but not

=ak x(n-k)k0

For causal i/p sequence21Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldalways.N

y(n) =ak x(n-k)k0

y(n) – y(n-1) = x(n) – x(n-3)Present response depends only on present

i/p & previous i/ps but not future i/ps. It gives

FIR o/p.Q. y(n) = 1 [x (n+1) + x (n) + x (n-1)] Find the given system is stable or not?3Let x(n) = (n)h(n) = 1 [ (n+1) + (n) + (n-1)]3

1h(0) =31h(-1) =3

1h(1) =3

S=h(n)< therefore Stable.Q. y(n) = a y(n-1) + x(n) given y(-1) = 0

Let x(n) = (n)h(n) = y(n) = a y(n-1) + (n)h(0) = a y(-1) + (0) = 1 = y(0)

h(1) = a y(0) + (1) = ah(2) = a y(1) + (2) = a2 . . . . . . . h(n) = an u(n) stable if a<1.y(n-1) = 1 [ y(n) – x(n)]a1y(n) = [ y(n+1) – x(n+1)]a22Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World1y(-1) = [ y(0) – x(0)]=0ay(-2) = 0

1n 1Q. y(n) =y(n-1) + x(n)for n0= 0otherwise. Find whether given system is time variant or not?Let x(n) = (n)h (0) = 1 y(-1) + (0) = 1

h(1) = ½ y(0) + (1) = ½

h(2) = 1/6h(3) = 1/24if x(n) = (n-1)

y(n) = h(n-1)1

n 1h(n-1) = y(n) =h(n-2) + (n-1)n=0n=1

n=2h(-1) = y(0) = 1 x 0+0 =0h(0) = y(1) = ½ x 0 + (0)= 1h(1) = y(2) = 1/3 x 1 + 0 = 1/3h(2) = 1/12h (n, 0) h(n,1) TVQ. y (n) = 2n x(n) Time varyingQ. y (n) = 1 [x (n+1) + x (n) + x (n-1)] Linear3Q. y (n) = 12 x (n-1) + 11 x(n-2) TIVQ. y (n) = 7 x2(n-1) non linear

Q. y (n) = x2(n) non linearQ. y (n) = n2 x (n+2) linearQ. y (n) = x (n2) linearQ. y (n) = ex(n) non linearQ. y (n) = 2x(n) x (n) non linear, TIV23Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World(If the roots of characteristics equation are a magnitude less than unity. It is anecessary & sufficient condition)Non recursive system, or FIR filter are always stable.

Q. y (n) + 2 y2(n) = 2 x(n) – x(n-1) non linear, TIV

Q. y (n) - 2 y (n-1) = 2x(n) x (n) non linear, TIV

Q. y (n) + 4 y (n) y (2n) = x (n) non linear, TIV

Q. y (n+1) – y (n) = x (n+1) is causal

Q. y (n) - 2 y (n-2) = x (n) causalQ. y (n) - 2 y (n-2) = x (n+1) non causalQ. y (n+1) – y (n) = x (n+2) non causal

Q. y (n-2) = 3 x (n-2) is static or Instantaneous.

Q. y (n) = 3 x (n-2) dynamicQ. y (n+4) + y (n+3) = x (n+2) causal & dynamicQ. y (n) = 2 x (n )If =1 causal, static <1 causal, dynamic >1 non causal, dynamic 1 TVQ. y (n) = 2(n+1) x (n) is causal & static but TV.Q. y (n) = x (-n) TVSolution of linear constant-co-efficient difference equationQ. y(n)-3 y (n-1) – 4 y(n-2) = 0 determine zero-input response of the system;

Given y(-2) =0 & y(-1) =5Let solution to the homogeneous equation be

yh (n) = n

n - 3 n-1 - 4 n-2

=0 n-2[ 2 - 3 - 4] =0 = -1, 4

n

yh (n) = C1 1n + C2 2 = C1(-1)n + C2 4n

y(0) = 3y(-1) +4 y(-2) = 1524Downloaded From JNTU World (http://JNTU-World.com)


C1+ C2 =15

y (1) = 3y (0) +4 y (-1) = 65 -C1+4C2 = 65Solve: C1 = -1 & C2=16y(n) = (-1)n+1 + 4n+2 (Ans)If it contain multiple roots yh(n) = C1 1n + C2 n 1n + C3 n2 1n

or 1n [C1+ nC2 + n2 C3….]Q. Determine the particular solution of y(n) + a1y(n-1) =x(n)

x(n) = u(n)Let yp (n) = k u(n)k u(n) + a1 k u(n-1) =u(n)To determine the value of k, we must evaluate this equation for any n 1k + a1 k =1

1k =1 a1

11 a1yp (n) =u(n) Ansx(n)yp(n)1. AK2. Amn

3. Anm

Kmn

Ko nm + K1nm-1 + …. Km

K1 Coswon + K2 Sinwon4. A Coswon or A SinwonQ. y(n) = y(n-1) - 1 y(n-2) + x(n)5x(n) = 2n n 066Let yp (n) = K2n

K2n u(n) = K 2n-1 u(n-1) - 1 K 2n-2 u(n-2) + 2n u(n)566For n 24K = (2K) - 1 K +45Solve for K=8/566yp (n) = 8 2n

Ans5Q. y(n) – 3 y(n-1) - 4 y(n-2) = x(n) + 2x(n-1) Find the h(n) for recursive system.


JNTU-World.comJNTU WorldWe know that yh (n) = C1 (-1)n + C2 4n

yp (n) =0 when x(n) = (n)for n=0y(0) - 3y(-1) - 4 y(-2) = (0) + 2 (-1)y(0) =1y(1) = 3 y(0) +2 = 5C1 + C2 =1-C1 + C2 =5 Solving C1 = 1 ; C2 =655 h(n) = [ 1 (-1)n + 4n ] u(n) Ans655ORh(n) – 3 h(n-1) -4 h(n-2) = (n) + 2 (n-1)

h(0) = 1h(1) =3 h(0) + 2 = 5plot for h(n) in both the methods are same.

Q. y(n) – 0.5 y(n-1) = 5 cos 0.5n n0 with y(-1) = 4

yh(n) = n

n – 0.5 n-1 =0 n-1 [ -0.5] =0 =0.5 yh(n) = C (0.5)n

yp(n) = K1 cos 0.5n + K2 sin 0.5nyp(n-1) = K1 cos 0.5(n-1) + K2 sin 0.5(n-1)

= - K1 sin 0.5n - K2cos 0.5nyp(n) - 0.5 yp(n-1) = 5 cos 0.5 n

= (K1 + 0.5 K2) cos 0.5 n -(0.5 K1 – K2) sin 0.5nK1 + 0.5 K2 = 50.5 K1 – K2 =0 Solving we get: K1= 4 & K2=2

yp(n) = 4 cos 0.5 n + 2 sin 0.5n26Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldThe final responsey (n) = C (0.5)n + 4 cos 0.5 n + 2 sin 0.5nwith y(-1) = 44 = 2C-2i.e. C=3 y (n) = 3 (0.5)n + 4 cos 0.5 n + 2 sin 0.5n for n0

Concept of frequency in continuous-time and discrete-time.

1) xa (t) = A Cos ( t)x (nTs) = A Cos ( nTs)= A Cos (wn)w = Ts = rad / secw = rad / Sample

F = cycles / sec f = cycles / Sample

2) A Discrete- time – sinusoid is periodic only of its f is a Rational number.

x (n+N) = x (n)Cos 2 f0 (n+N) = Cos 2 f0 n2 f0N = 2 K => f0 = KNEx: A Cos ( ) n6w = = 2 f61f =N=12 Samples/Cycle ;Fs= Sampling Frequency;Ts=12Sampling Period

Q. Cos (0.5n) is not periodic27Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldQ. x (n) = 5 Sin (2n)

2 f = 2 => f = 1

Q. x (n) = 5 Cos (6 n)

2 f = 6 => f = 3Non-periodicN=1 for K=3 Periodic

6nQ. x (n) = 5 Cos

35632 f ==> f =for N=35 & K=3Periodic3535Q. x (n) = Sin (0.01 n)

2 f = 0.01 => f =

Q. x (n) = Cos (3 n)0.01

2for N=200 & K=1

for N=2PeriodicPeriodicfo = GCD (f1, f2) & T = LCM (T1, T2) ------- For Analog/digital signal[Complex exponential and sinusoidal sequences are not necessarily periodic in ‘n’with period ( 2 ) and depending on Wo, may not be periodic at all]Wo

N = fundamental period of a periodic sinusoidal.

3. The highest rate of oscillations in a discrete time sinusoid is obtained when

w = or -28Downloaded From JNTU World (http://JNTU-World.com)


Discrete-time sinusoidal signals with frequencies that are separated by an integral

multiple of 2 are Identical.FsFs4. - F 22

- Fs 2 F Fs- TsTs- Ts Therefore - w

5. Increasing the frequency of a discrete- time sinusoid does not necessarilydecrease the period of the signal.x1(n) = Cos ( 4n)N=83nx2(n) = Cos ()N=163/8 > 1/482 f = 3 /8

3=> f =16

16. If analog signal frequency = F =samples/Sec = Hz then digital frequency f = 1Ts

W = Ts

2 f = 2 F Ts

=> f =14;2 F =2 f = /429Downloaded From JNTU World (http://JNTU-World.com)


1F =8

f = 1; T = 8 ;N=887. Discrete-time sinusoids are always periodic in frequency.

Q. The signal x (t) = 2 Cos (40 t) + Sin (60 t) is sampled at 75Hz. What is the

common period of the sampled signal x (n), and how many full periods of x (t) does it

take to obtain one period of x(n)?F1 = 20HzF2 = 30Hzf1 = 20 4 K1f2 = 30 2 K2N275 15 N1755The common period is thus N=LCM (N1, N2) = LCM (15, 5) = 15The fundamental frequency Fo of x (t) is GCD (20, 30) = 10Hz1And fundamental period T =

Since N=15 0.1sFo1sample ---------- 1sec75=> 15 0.2S15 sample ----------- ?7530Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldSo it takes two full periods of x (t) to obtain one period of x (n) or GCD (K1, K2) =GCD (4, 2) = 2Frequency Domain Representation of discrete-time signals and systemsFor LTI systems we know that a representation of the input sequence as a weightedsum of delayed impulses leads to a representation of the output as a weighted sum ofdelayed responses.

Let x (n) = ejwn

y (n) = h (n) * x (n)

h(k)x(n k) =h(k) ejw (n-k)kk

= ejwn

h(k) e-jwkk

Let H (ejw) =h(k ) e-jwk is the frequency domain representation of the system.k y (n) = H (ejw) ejwn

ejwn = eigen function of the system.

H (ejw) = eigen valueQ. Find the frequency response of 1st order system y (n) = x (n) + a y (n-1)(a<1)Let x (n) = ejwn

yp (n) = C ejwn

C ejwn = ejwn + a C ejw (n-1)

C ejwn [1-ae-jw] = ejwn

1C = [1 ae jw]11Therefore H (ejw) = [1 ae jw]= 1 a(cosw jsin w)1H(e jw) =1 2acosw a2

aSinw

H (e jw) Tan1 (1 aCosw)31Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldQ. Frequency response of 2nd order system y(n) = x(n) - 12 y(n 2)x (n) = ejwn

y(n) cejwnp

- 1 ce jw(n2)

=ejwn jwn

ce2ce jwn (1+ 1 e2 jw) = e jwn

12016Cos2w5 4Cos2wc = 1 1c 2e2 jw

2c tan1 Sin2w 2Cos2w32Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldUNIT - IIContinuous Time ot = onTs = wonDiscrete TimePeriodic f (t) =2 nN 1

jK

ckeN

cke jkotPeriodicxp(n) =k0k DTFSNon periodicPeriodic Ck =T

1Ck = T j 2 nK

N 1f (t)e jKot

1dt

xp(n)eN0

Nn0 jK 2nTs

NTs1 x(n)ek=0 to N-1TT = N Ts

t = n Ts : dt = TsNon-Periodic f(t) =

Non – Periodic x(n) =211

F(w)e jtdX (w)e jwn

dw220

x(n)e jwnPeriodic X(w) =Non-Periodic F(w) =n

f (t)e jtdtX(w) = FT of DTS33Downloaded From JNTU World (http://JNTU-World.com)


Energy and Power221

2 x(n) x(n)x*(n) =x(n)X *(w)e jwndwE =nnn02

1= 2

X *(w)x(n)e jwn

dwn021

= 2 X*

(w)X (w)dw

02

X (w) dw

1= 222x(n) 1Therefore:E =

X (w) dw -------- Parsval’s Theorem

2n2

1N

Lt 2N 1x(n)P=for non periodic signalN n N2

1N

x(n)

N 1=for periodic Signaln0 j 2 nkN 1

x(n)x*(n) 1N 1N1

1

x(n) Ck *eN

= NNn0n0k0

2

N 1

1N

N 1

x(n)e j Nnk

C k*

=

k 0n 02N 1

Ckk 02N1

Ck

Therefore

Ex:P =E = Nk0Unit step34Downloaded From JNTU World (http://JNTU-World.com)


12N 1N

u2(n)

P = LtN 0n

N 11= NLt 2N 1 2Power Signal

E = x (n) = AejwonEx:2

12N 1N

LtAejwonP =N nN12N 1 A2

[11........]=

=

LtN

Lt A (2N 1) A22it is Power Signaland

E = 2N 1NEx:

Ex:x (n) = n u(n)neither energy nor power signalx (n) = 3 (0.5)n

n09

2

9(0.25)n

12J

11 ]x (n)

n

E =note: [10.25nn0n0

2n4

6

,0, 6,0 Ex:x (n) = 6 Coswhose period is N=4x (n) =

13

(n) 14[36 36] 18W2

xP = 4n0j 2n4Ex:x (n) = 6 ewhose period is N = 42

1P = 43

x(n) 14[36 36 36 36] 36Watts

n035Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldDISCRETE CONVOLUTION It is a method of finding zero input response of linear Time Invariant system.Ex: x(n) = u(n)

h(n) = u(n)

u(k)u(n k)

ky(n) =u(k) = 0u(n-k) = 0

n

k<0

k>n

n

1 = (n+1) u(n) = r(n+1)k 0u(k)u(n k) =k0Q. x(n) = an u(n) and h(n) = an u(n) a<1 find y(n)36Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldn

y(n) =ak an-k = an (n+1) u(n)k 0Q. x(n) = u(n) and h(n) = n u(n) <1 find y(n)

n

y(n) = k u(k) u(n-k) = k = (1- n+1) / (1- )k k0

The convolution of the left sided signals is also left sided and the convolution of two

right sided also right sided.Q. x(n) = rect ( n ) = 1n N2N= 0else wherenh(n) = rect ( 2 N )y(n)= x(n) * h(n)= [u (n+N) – u (n-N-1)] * [u (n+N) – u (n-N-1)]= u (n+N) * [u (n+N) – u (n-N-1)] – u (n-N-1)* [u (n+N) – u (n-N-1)]= u (n+N) * u (n+N) – 2 u (n+N)*u (n-N-1)] + u (n-N-1) *u (n-N-1)= r(n+2N+1) – 2r(n) + r(n-2N-1)n= (2N+1) Tri ( 2N 1)nNTri ( n ) = 1-for n NN= 0elsewhere.37Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldQ. x(n) = 2,-1,3h(n) = 1,2,2,3 Graphically Fold-shift-multiply-sumy(n) =1

22

42

43

62

-1

3-1

3-2

6-2

6-3

9y(n) = 2,3,5,10,3,910

,4Q. x(n) = 4, ,3h(n) = 2,5, 2

8

2

65

20

50

0

0

04

4

1

31641512y(n) = 8,22,11,31,4,12Note that convolution starts at n=-3↑38Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldQ)h(n):x(n):2451034_________________________________820205

6160401512____________________________________31y(n):8221141224Q. Convolution by sliding step method: h(n) = , 5, 0, 4 ; x(n)= , 1, 3i)

2 5 0 4

3 1 4ii)

2 5 0 4

3 1 4___________________

y(0) = 8_________________________2 20

y(1) = 2+20 = 22iii)2 5 0 43 1 4iv)2 5 0 4

3 1 4_______________________________________________6 5 0y(2) = 11

15 0 16

2 5 0 4y(3) = 31v)2 5 0 4Vi)3 1 43 1 4________________________

0 4 y(4) = 4_______________________12y(5) = 12

If we insert zeros between adjacent samples of each signal to be convolved, theirconvolution corresponding to the original convolution sequence with zeros inserted

between its adjacent samples.39Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World24X(z) = 2z3+5z2+4 ; X(z) = 4z2+z+3Q. h(n) = , 5, 0, 4 ; x(n)= , 1, 3Their product Y(z) = 8z5+22z4+11z3+31z2+4z+12

y(n) = 8 ,22,11,31,4,122h(n) = , 0, 5, 0, 0, 0, 4 ; x(n) = 4, 0, 1, 0, 3H(z) = 2z6+5z4+ 4 ; X(z) = 4z4+z2+3Y(z) = 8z10+22z6+31z4+4z2+12y(n) = 8,0,22,0,11,0,31,0,4,0,12

Q. Compute the linear convolution of h(n)=1,2,1 and x(n) = 1, -1, 2, 1 ,2, -1, 1, 3,1 using overlap-add and overlap-save method.h (n):x (n):

x1(n):

x2(n):

x3(n):11

1212

2-1

-11

12

2-1

-113113

1____________________________________________________________y1(n) = (h (n)*x1(n))1

y2(n) =113

12

440

1-1

5y3(n) =851y(n) =1146414851 1OVER LAP and SAVE methodh (n):x1(n):

x2(n):

x3(n):121201

102

2(N2=3)(N3+N2-1) = 51

1

1

-1-11

13

3100y1(n) = 1

y2(n) =14

4

6

6

5 21 4 41

1

4

4

7 3

1 5 8 5 1

8 5 1y3(n) =y(n) = 11440Downloaded From JNTU World (http://JNTU-World.com)


Discrete Fourier SeriesQ. Determine the spectra of the signalsa. x(n) = Cos 2 n

wo = 2 1fo =is not rational number2 Signal is not periodic.Its spectra content consists of the single frequencynb. x (n) = Cos 3after expansion x(n)= 1,0.5,-0.5,-1,-0.5,0.5fo = 1N=66 j 2 nk

1Ck = 65

x(n)e6k=0 to 5n0

j k

3 j2 k

3 j4 k

3

j5 k1x(0) x(1)e x(2)e x(3)e jk x(4)e x(5)e3

Ck = 6For k=0

Similarly

K=1

Co = 16x(0) x(1) x(2) x(3) x(4) x(5) = 0C1 = 0.5,C2 = 0 = C3 = C4 , C5 = 0.5Or2 nj 2 n

1+ 2j 2 kn

= Cos3 n 12 e j5

eC ke666x (n)=k041Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldj 2 nj 4 nj 6 nj10 n8 nj66666= Co+C1 e+C2e+ C3 e+C4 e+C5 eBy comparison C1= 12 j 2 n 56 j10nj2 n666Sincee= e= eC5 12c. x (n) = 1,1,0,0nk

143 j2

x(n)e4C =

kk=0, 1, 2, 3n0

2 k

1411e j 2=

c0 1 11 j;

c3 1

1 j c1

;

c 0 ;2

244Co 1&C0 = 02c 42& C1 =

& C2

1

4c2 0undefined2c3

4& C3 =442Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldPROPERTIES OF DFS1.Linearity

DFS ~x1(n) Ck1

DFS~x2(n) Ck2

a~x1(n) b~x2(n) aCk1 bCk2DFS2.Time Shifting

DFS ~x (n m ) e j 2mk C kN3.Symmetry

~(n) C *k

x*DFSCk =

j 2nk~x (n )eN

1N

N 1n 0

~

*

x (n) C *kj 2 N nk

C keN 1

~x (n) DFSk 0~~x(n) x(n) 1

Ck C*k Cke

*

DFSRe~x(n) DFS22~~ (n) 1

*

x(n) xj Im~x(n) DFS

Ck C *k Cko

2

DFS2~Ifx(n) is real then

~~ *(n)

x(n) x~xe (n) 243Downloaded From JNTU World (http://JNTU-World.com)


~~ *(n)x(n) x~xo (n) 212~xe (n) C C ReCk *kkDFS1DFS~xo (n) C C j ImCk *kk

2Periodic Convolution

N1

~ ~x (m)x2 (n m) Ck1Ck 2

1

DFS m0If x(n) is real

C k C *k

Re[Ck ] Re[Ck ]Im[Ck ] Im[Ck ]Ck Ck

Ck CkPROPERTIES OF FT (DTFT)1. Linearityy (n) = a x1 (n) + b x2 (n)

Y (e ) = a X1(e ) + b X2(e )

jw jw jw

2. PeriodicityH (e j(w2 ) ) = H (e)jw

3. For Complex Sequence44Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldh (n) = hR(n) + j hI(n)

[ h R (n) jh I (n) ][Cos(wn)- jSin(wn)]H (ejw

) =n-

[ h R (n) Cos(wn) h I(n)Sin(wn) = HR (e jw )n-

[ h I(n) Cos(wn) h R (n)Sin(wn) = HI (e jw )n-

H (e jw ) = HR(e jw) jH I (e jw

)H2R

(e jw) H I2

(e jw) H(e jw)H*

(e jw

)= H I (e jw)H(e jw) tan1H R (e jw)4. For Real Valued Sequence

H(e jw) =h(n)e jwnn

h(n)Cos(wn) j h(n)Sin(wn)=nn

= HR(e jw) jH I (e jw) -------------------- (a)

h(n)ejwn

H(e jw) =n

h(n)Cos(wn) j h(n)Sin(wn)=nn

= H R(e jw) jH I (e jw) -------------------- (b)From (a) & (b)H R(e jw) H R (e jw)45Downloaded From JNTU World (http://JNTU-World.com)


H I (e jw) H I (e jw

Real part is even function of w

Imaginary part is odd function of w)

H(e jw) H *(e jw)H (e jw ) H (e jw )H* (e jw ) H*(e-jw )H(e-jw ) H(e-jw )

Magnitude response is an even function of frequency=> tan-1HI(e-jw

HR(e-jw

)HI(ejw

HR(ejw ))H (e-jw ) tan -1

H (ejw

))Phase response is odd function.5. FT of a delayed Sequence

h(n k )e jwnFT [h (n-k)] = n

Put n-k = m

h(m )e jw (m k )=m

= H (e jw ) e jwk

= e jwk

h(m)e jwmm 6. Time Reversal

x (n) X (w)x (-n) X (-w)

x(n)e jwnF T [x (-n)] = nPut –n = m

x(m)e jwm X (w)46Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World7. Frequency Shifting

x(n) e jwon X (w-wo)

x (n) e jwon e-jwn

F T [x (n) e jwon ] =n

x (n) e j(wwo )n== X (w-wo)n8. a. Convolutionx1 (n) * x2 (n) X1(w) X2(w)

[x1 (n) * x2 (n) ] e-jwn

=[ x1 (k) x2 (n-k) ] e-jwnnn kPut n-k = m

=x1 (k)[x2(m)] e-jw (m+k)nm

=x1 (k) e-jwk

[x2(m)] e-jwmnm= X1(w) X2(w)b.21 [X1(w) *X2(w)] x1 (n) x2 (n)9. Parsevals Theorem1

2x1(n) x2*(n) =[X1(w) X2*(w)] dwndX (w)n x (n) jdw10.F T of Even Symmetric Sequence

jwH (e) =h (n) e-jwn

n47Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World1

=h (n) e-jwn + h (0) +h (n) e-jwnnn1Let n = -m

=h (-m) ejwm + h (0) +h (n) e-jwnm1n1Let h (-m) = h (m) for even

Therefore = h (0) + 2h (n) Cos (wn) is a real valued function ofn1

frequency 0

; H (e ) >0

jw

; H (e ) <0

jw

11.F T of Odd Symmetric SequenceFor odd sequence h (0) = 0

H (ejw

) =h (n) [e-jwn - ejwn ]n1

= -j 2h (n) Sin (wn) HI (ejw

)is a imaginary valued function

n1of freq. and a odd function of wi.e,H (e jw) = - H (e jw

)H (e jw

)= HI (ejw

)for HI (e

for HI (e

jw

jw

) > 0= - HI (ejw

)) < 0H(e jw) For w over which

for w over which

HI (e ) > 0jw

2 = 2HI (e jw ) < 02

1x(0) = 2X (w)dw12.Central Co-ordinates048Downloaded From JNTU World (http://JNTU-World.com)


x(n)X (0) =n13.ModulationX(w w0) X(w w0)Cos (won) x (n) 22FOURIER TRANSFORM OF DISCRETE TIME SIGNALS

X (w) =x (n) e-jwnn

x(n) F T exists ifnThe FT of h (n) is called as Transfer functionEx: h (n) = 1for -1 n 13= 0otherwise1

13

jw = 13 1 2Cos(w)

jw

e 1 eSol:H (ejw

) =e jwn = 13n1w

01132- 1


JNTU-World.comJNTU WorldEx:h (n) = an u (n)

n jwna e

H (ejw

) =n0

1= 1 ae jw

<1

(ae jw)n

=n0Q.x(n) = n n u(n)

n n u(n)

d 1 j dw

1e jw

e

= (1e jw)2jw

11e jw

jw n)(en

e jwn

Hint: n u(n) ==n0n0

x(n) = n

Q.0nNOrx(n) = n [ u(n) – u(n-N)]

= n u(n) – N n-N u(n-N) Using Shifting Property

1e jwN

jw] jw N

[X(w) = 1e1e1 (e jw N)=

1e jwAnsnQ.

Q.

x(n) = 1 two sided decaying exponentialx(n) = n u(n) + -n u(-n) - (n)using folding property12

1 jw 1

1e

jw 1= 1 e=1 2Cosw2x (n) = u (n) Since u (n) is not absolutely summable50Downloaded From JNTU World (http://JNTU-World.com)


we know that u (t) (w) jw1

Similarly X (w) = 1 jw + (w)1 e


JNTU-World.comJNTU WorldDFT (Frequency Domain Sampling)

The Fourier series describes periodic signals by discrete spectra, where as the

DTFT describes discrete signals by periodic spectra. These results are a consequence of the

fact that sampling on domain induces periodic extension in the other. As a result, signals that

are both discrete and periodic in one domain are also periodic and discrete in the other. This

is the basis for the formulation of the DFT.

x(n)e jwn

Consider aperiodic discrete time signal x (n) with FT X(w) =nSince X (w) is periodic with period 2 , sample X(w) periodically with N equidistancesamples with spacing w 2N .K = 0, 1, 2…..N-1 j 2 Kn

2k

X x(n)eN

N nThe summation can be subdivided into an infinite no. of summations, where each sumcontains j 2 KnN1 j 2 Kn

2k 1

x(n)eN

X N ............x(n)eN+n0nN j 2 Kn

2N1

x(n)e ..................NnN52Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World j 2 KnlNN1

x(n)e

N=lnlNPut n = n-lN

2 K (nlN )NN1 j

x(n lN)e= ln0 j 2 KnN 1

x(n lN)eN=n0l j 2N KnN1

eX(k) =xp(n)n0j 2 KnN1

eNWe know that xp(n) =Ck

n= 0 to N-1k0N 1

2 j N Knk=0 to N-1

1

Ck= Nexp(n)n01ThereforeCk=X(k)k=0 to N-1Nj 2 KnN1

X(k) en = 0 to N-11NIDFT ------------ xp (n) = Nk0This provides the reconstruction of periodic signal xp(n) from the samples of spectrumX(w).The spectrum of aperiodic discrete time signal with finite duration L<N, can be exactlyrecovered from its samples at frequency Wk= 2Nk.Prove: x(n) = xp (n)0nN-153Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldUsing IDFTj 2 Kn

x (n) = 1N1

X(k) eN

Nk0j 2 KnN1

1N1

X (k) eX (w) =[] e-jwnN

Nn0k02 K )N1

1N1 jn(w

=X (k) [e]N

Nk0n0

If we define p(w) = 1N1

e-jwn

Nn0

Sin wN jwN11 1 e jwN

N 1 e jw

==2w e2NSin2N1

2k

Therefore: X (w) = X (k) P(w-)Nk0

2kAt w =

And P (w-

X (w) =P (0) =1

) = 0 for all other valuesN

2kNN1N1

2k

X(k) =X()Nk0k0

Ex:The spectrum of this signal is sampled at frequency Wk= 2k . k=0, 1…..N-1, determinex(n) = an u(n)0<a<1Nreconstructed spectra for a = 0.8 and N = 5 & 50.

1X (w) = 1

ae jw1X (wk) =k=0, 1, 2… N-1 j 2 k1 aeN



xp (n) =x(n LN)

nlN

all

an

1 a= an

alN

0nN-1Nl0

Aliasing effects are negligible for N=50If we define aliased finite duration sequence x(n)xˆ(n) xp (n)0nN-1= 0otherwiseN1N1

Xˆ (w) xˆ(n)e jwn

x p (n)e jwn

=n0n0N 1

an

N e jwn

1

= 1 aN1

(ae jw)n

=1 aNn0n0

1 a Ne jwN

11 aXˆ (w) 1 a e jw N



j 2k

N

11 aN

eN

2K N ˆX 1 a N j 2K

1 a e

N

1 2K N == X j2k

1 aeN

2k Although Xˆ (w) X (w),the samples at Wk=are identical.N11Ex:X (w) =&X (k) = j 2 k

1 a e jw1 a eN

Apply IDFTj 2nk

1N1

eN

x (n) =using Taylor series expansion j 2k

Nk0

1 aeN

j 2nk

j 2kr

1= NN 1

eN

a

r0r

eN

k0

j2k (nr)

1N1

ear

N

=Nr0

k0

= 0exceptr = n+mN

nmN

= an

(a N )m

a x (n) =m0m0

an

= 1 aNThe result is not equal to x (n), although it approaches x (m) as N becomes .Ex:x (n) = 0, 1, 2, 3 find X (k) =? j 2kn3

x(n)e4X (k) =056Downloaded From JNTU World (http://JNTU-World.com)


X (0) =

X (1) =

x(n) = 0+1+2+3 = 6n0 j 2 n3

x(n)e4= -2+2jn0X (2) = -2

X (3) = -2-2jDFT as a linear transformation j 2Let WN eNN1

x(n)WN

nkX (k) =k = 0 to N-1n0

1x (n) = NN1

X (k)WNnk

n = 0, 1…N-1

X (0) k0 x(0)

X (1)x(1)LetxN = XN = X (N 1)x(N 1)11111 WN1

WN

WN

4

2

WN(N1)

WN2(N1)

(N1)(N1)

1 WN2

WN = WNN1 WN2(N1) WN

1The N point DFT may be expressed in matrix form as57Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldDFTIDFT

1 W NxN = NX N*XN = WN xN

x WN1X NN

1. WNKN WNK

WN1 N1 WN*N

2. WNK 2 W K

NEx:x (n) = 0, 1, 2, 31111 1111 1111 1 W41

W42

W43

1 W41

W42

W3

4

1 j 1j

1

DFT W4 == =1 W42 W44 W46

1 W42

W4

W40

W42

14

1111 W43

W46

W4

9

1 W432

W1j1 j06

1111 2 2 j11 j 1 j X4 = W4x4 =

1

= 1 1221 2 2 j31j1 j IDFT061111 1

2 2 j

14 = 4 1 j 1j

1

xWN

*

X N

1 =

= 2 Ans 2 2 2 j

1114 31j1 j Q.x (n) = 1,0.5h (n) = 0.5,158Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldFind y (n) = x (n) h (n) using frequency domain. Since y (n) is periodic with period 2.

Find 2-point DFT of each sequence.X (0) = 1.5

X (1) = 0.5H (0) = 1.5

H (1) = -0.5Y (K) = X (K) H (K)Y (0) = 2.25 Y (1) = -0.25Using IDFTy (0) = 1; y (1) = 1.25~y(n) h~(n) ~x(n) h(k)~x(nk)~

k~x(k)h~(n k)

=k~x(k)h~(k)

~ y(0)=k

= ~x(0)h~(0) ~x(1)h~(1)= 1 * 0.5 + 0.5 * 1 = 1x(k)h~(1 k)

~~y(1)=k 59Downloaded From JNTU World (http://JNTU-World.com)


= ~x(0)h~(1) ~x(1)h~(0)= 1 * 1 + 0.5 * 0.5 = 1.25

~y(2)x(k)h~(2 k)

~=k

= ~x(0)h~(2) ~x(1)h~(1)= 1 * 0.5 + 0.5 * 1 = 1

~y(n)= 1, 1.25, 1, 1.25…..Q. Find Linear Convolution of same problem using DFT

The linear convolution will produce a 3-sample sequence. To avoid timeSol.aliasing we convert the 2-sample input sequence into 3 sample sequence by padding withzero.For 3- point DFTX (0) = 1.5H (0) = 1.52 j 2 jX (1) = 1+0.5 e3

H (1) = 0.5+ e3 j 4 j 4X (2) = 1+0.5 e3H (2) = 0.5+ e3Y (K) = H (K) X (K)Y (0) = 2.25 j 2 j 4Y (1) = 0.5 + 1.25 e 3 + 0.5 e3 j 4 j8

Y (2) = 0.5 + 1.25eCompute IDFT

3

+ 0.5 e32kn

3

1y(n) = 32

Y (k )e jk 060Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldy(0) = 0.5y(1) =1.25y(2) =0.5y(n) = 0.5, 1.25, 0.5AnsPROPERTIS OF DFT1. LinearityIf h(n) = a h1(n) + b h2(n)

H (k) = a H1(k) + b H2(k)2. Periodicity H(k) = H (k+N)

3. h~(n) h(n mN)

m4. y(n) = x(n-n0) j 2kn 0NY (k) = X (k) e5. y (n) = h (n) * x (n)

Y (k) = H (k) X (k)

6. y (n) = h(n) x(n)

Y (k) = N1 H(k) X (k)7. For real valued sequence

2knN

N 1

H R (k) h(n)Cosn0

2knN

N 1H I (k ) h(n)Sinn0a.Complex conjugate symmetryh (n) H(k) = H*(N-k)h (-n) H(-k) = H*(k) = H(N-k)61Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldi.Produces symmetric real frequency components and anti symmetric

Nimaginary frequency components about the

2DFTNii.Only frequency components from 0 toneed to be computed in order2to define the output completely.b.Real Component is even functionHR (k) = HR (N-k)c.Imaginary component odd functionHI (k) = -HI (N-k)d.Magnitude function is even functionH(k) H(N k)e.Phase function is odd function

H(k) H(N k)f.

If h(n) = h(-n)H (k) is purely real

If h(n) = -h(-n)g.H (k) is purely imaginary8. For a complex valued sequencex*(n) X*(N-k) = X*(-k)

N 1DFT [x(n)] = X(k) =x(n)W N

nkn 0

N 1X*(k) =x*

(n )W N nkn 0

N 1

x*

(n)W N

nk

=X*(-k)X*(N-k) =n0N 1

x (n)W N

* nk= X*(N-k) proved

DFT [x*(n)] =n0Similarly DFT [x*(-n)] = X*(k)62Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World9. Central Co-ordinates1NN 1

1NN 1

X (k )

Nx ( ) =2

(1) k X (k)x (0) =N=evenk 0k0

N 1N 1

x(n)(1)

x(n)n

NX (0) =X ( ) =N=evenn 02n010.Parseval’s RelationN 1N 1

X (k ) 2k 0

2

x(n)N

n0

N 1

N x (n ) x * (n )Proof:LHSn 0(k )W N N 1

1 NN 1

x(n )X

k 0*nk

= Nm 0

N 1N 1

X*

(k ) x(n )Wn k

N

=

k 0n 0

2N 1N 1

X (k )X*

(k ) X (k) ==k 0k 011.Time Reversal of a sequence

x((n))N x(N n) X((k))N X(N k)Reversing the N-point seq in time is equivalent to reversing the DFT values. j 2k n

DFT x(N n) N 1

x( N n )eNn 0Let m=N-n

j 2k ( N m )N 1

x (m )eN=

=

m=1 to N = 0 to N-1n 0

j 2 k mN 1

x (m )eNm 063Downloaded From JNTU World (http://JNTU-World.com)


jN2m ( N k )N 1

x (m )e== X(N-k)m 012.Circular Time Shift of a sequence j 2 k l

x (n l ) N X (k )eN j 2k nN 1

DFTx(n l) N x(n l) N eNn0 j 2k nl1 j 2k n

N

N 1x(n l) N ex(n l) N eN=

=

+n0nl j 2k nN 1

j2k nl1

x(N n l)eNx(N n l) eN+nln0Put N+n-l = m j 2k (ml ) j2k (ml) N 12 N 1 l

x(m)ex(m)eNN=+mN lm NN to 2N-1-L is shifted to N 0 to N-1-LTherefore 0 to N-1 = (0 to N-1-L) to ( N-L to N-1)2kN

N 1

x(m)e j(ml )Therefore m0

2 k j le j 2Nk

mN 1

Nx(m)e=m02k l

N

= X(k) e jRHS13.Circular Frequency Shift64Downloaded From JNTU World (http://JNTU-World.com)


2l

x(n)e j n X (k l) NN2lN

n j 2k n

N 1

j 2l n

x(n)e j eN

x(n)eNDFT=n 02n = X (k l ) N(k l )N

N 1

x(n)e j=RHSn0

14.x(n) X(k)

km )x(n), x(n), x(n)…….x(n) M X ((m-fold replication)x(mn ) X (k), X (k),......X (k)(M- fold replication)2, 3, 2, 1 8, -j2, 0, j2Zero interpolated by M

2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1 24, 0, 0, -j6, 0, 0, 0, 0, 0, j6, 0, 015.Dualityx(n)X(k)0 K N 1X(n)N x(N-k)2

1x(n) = N

N 1

X ( )e j N n 0

1 N 1 X ( )e j 2( N k )N

Nx(N-k) = 02

1= N

1N

X ( )e j N k 065Downloaded From JNTU World (http://JNTU-World.com)


2

N 1

X ( )e j NkN x(N-k) = 02k n

N 1

X (n )e j N== DFT [ X(n) ] LHS provedn0116.Re[x(n)] X ep (k)X((k))NX *((k))NX (k) =

ep

2j Im[x(n)] X op (k)

xep (n) Re[X(k)]

xop (n) j Im[X(k)]

1xep(n) Even part of periodic sequence = x(n) x((n)) N 21 x(n) x((n)) N xop(n) Odd part of periodic sequence = 2

N 1

x(n)W N

nkProof: X(k) =

X(N-k) =

n 0N 1x(n)W N nk X ((k )) Nn 0N 1

x*

(n)W NnkX*(k) = n0

N 1

x*

(n)W N

nk

X*

((k )) NX*(N-k) =n066Downloaded From JNTU World (http://JNTU-World.com)


X ((k))N X2

*

((k)) N 11x(n) xN

(n)WN

nk*

2n0= DFT of [Re[x (n)]]LHS

N 1

x1 (n)x 2 (n) = 1*

N k 0

N 1

X 1 (k ) X (k )*217.n 0

x1 (n)x 2

*

(n)Let y(n) =

1Y(k) = N* (k )

X 1 (k ) X 2

1N

N 1

X 1 (l) X 2* (k l)=l 0

1 N 1

(l)

*Y(0) =1

X (l) X 2

Nl 0Using central co-ordinate theoremN 1

x1 (n)x 2 (n)

*Y(0) =n 0N 1

N 1x1(n)x 2

*

(n) = 1

X 1 (k )X 2

*

(k )ThereforeNn0k 0QUESTIONS1 Q.(i) 1,0,0,…….0 (impulse) 1,1,1…..1 (constant)(ii) 1,1,1,……1 (constant) ) N,0,0,…….0 (impulse) j2k

n

j2k

N1

1 ()

NN

1 N

N

e(iii) n j2k j 2k

k0

11 eN

N

N

2nko (k ko ) (k (N ko )

(iv) Cos 2

N (Impulse pair)67Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOr

Cos (2nf) = Cos (wn)

j 2nko

N j 2nko )

N

e eSol.x(n) =

2j 2nko

Nj 2n( N ko )

N

e e=2We know that 1 N (k)j2nKo

N

x(n)e X (K Ko)N(k ko) (k (N ko)x(n) 2I.

Inverse DFT of a constant is a unit sample.DFT of a constant is a unit sample.

2 Q. Find 10 point IDFT ofII.X(k) = 3

= 1k=01k9Sol. X(k) = 1+2 (k)

1

10 (k)= 1 +5

x(n) = 15 + (n)Ans3 Q. Suppose that we are given a program to find the DFT of a complex-valued sequencex(n). How can this program be used to find the inverse DFT of X(k)?N1

x(n)WNnkX(k) =n068Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldN1

X (k)WNnk1x(n) =N

k 0N1

X *(k)WNnkN x*(n) =k0

1. Conjugate the DFT coefficients X(k) to produce the sequence X*(k).

2. Use the program to fing DFT of a sequence X*(k).3. Conjugate the result obtained in step 2 and divide by N.14 Q.xp(n) = , 2, 3, 4, 5, 0, 0, 00(i) fp(n) = xp(n-2) = , 0, 1, 2, 3, 4, 5, 03(ii) gp(n) = xp(n+2) = , 4, 5, 0, 0, 0, 1, 2(iii) hp(n) = xp(-n) = 1, 0, 0, 0, 5, 4, 3, 25 Q.x(n) = 1, 1, 0, 0, 0, 0, 0, 0n = 0 to 7 Find DFT. jk j2k n1

x(n)e= 1 + e84X(k) =k = 0 to 7n0X(0) = 1+1 = 2 j

X(1) = 1+ e4= 1.707 - j 0.707 jX(2) = 1+ e

X(3) = 1+ e2= 1- j j34= 0.293 - j 0.707X(4) = 1-1 = 0By conjugate symmetry X(k) = X*(N-k) = X*(8-k) X(5) = X*(3) = 0.293 + j 0.70769Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldX(6) = X*(2) = 1+jX(7) = X*(1) = 1.707 + j 0.707

2X(k) = , 1.707 - j 0.707, 0.293 - j 0.707, 1-j, 0, 1+j, 0.293 + j 0.707, 1.707 + j0.707

6 Q.x(n) = 1, 2, 1, 0 N=4X(k) = 4, -j2, 0, j2(i) y(n) = x(n-2) = 1, 0, 1, 2 j 2k (no2)4Y(k) = X(k) e= 4, j2, 0, -j2(ii) X(k-1) = j2, 4, -j2, 0j 2 lnIDFT x(n) eNjn2

=

x(n) e= 1, j2, -1, 0(iii) g(n) = x(-n) = 1, 0, 1, 2

G(k) = X(-k) = X*(k) = 4, j2, 0, -j2(iv) p(n) = x*(n) = 1, 2, 1, 0

P(k) = X*(-k) = 4, j2, 0, -j2* = 4, -j2, 0, j2

(v) h(n) = x(n) x(n)= 1, 4, 1, 01414 [ 24, -j16, 0, j16] = 6, -j4, 0, j4H(k) =

X (k ) X (k ) =(vi) c(n) = x(n) x(n)

= 1, 2, 1, 0 1, 2, 1, 0 = 2,4,6,4

C(k) = X(k)X(k) = 16, -4, 0, -4(vii) s(n) = x(n) x(n) = 1, 4, 6, 4, 1, 0, 0

S(k) = X(k) X(k) = 16, -2.35- j 10.28, -2.18 + j 1.05, 0.02 + j 0.03, 0.02 - j 0.03, -2.18 -

j 1.05, -2.35 + j 10.28

(viii) x(n) 2 1 41 0 670Downloaded From JNTU World (http://JNTU-World.com)


14X(k) 2 1[16 4 4] 6471Downloaded From JNTU World (http://JNTU-World.com)


III UNIT: FFTN1

x(n)WN

nk

0 K N 1X(k) =n0N 1

nk ) + j Im(Wnk

)

= Re[x(n)] + j Im[x(n)] Re(WNNn0N 1N1

=Re[x(n)] Re(WN

nk

) -Im[x(n)] Im(WNnk ) +n0n0N 1

Im[x(n)] Re(W ) + Im(WNnk nk

)Re[x(n)]N

jn0

Direct evaluation of X(k) requires N 2 complex multiplications and N(N-1) complexadditions.4 N 2 real multiplications 4(N-1) + 2 N = N(4N-2) real additionsThe direct evaluation of DFT is basically inefficient because it does not use the symmetryNK

2 WNnkWNKN

WN

nk

& periodicity properties WN

&DITFFT:NN1122

x(2n 1)WN(2n1)k

x(2n)WN2nkX(k) =+n0n0(even)(odd)N 12N12

W+xo(n)WN2nk

xe(n)WN2nkK

N=

=

n0n0NN1122

xo (n)WNnk/ 2

nk

xe(n)WN / 2 +

WK

Nn0n0= Xe(k) + WNK Xo(k)72Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldAlthough k=0 to N-1, each of the sums are computed only for k=0 to N/2 -1, since Xe(k)& Xo(k) are periodic in k with period N/2NFor K N/2

2 = -WNKK WN

X(k) for K N/2NKX(k) = Xe(k-N/2) -WN2Xo(k-N/2)N = 8x(2n) = xe(n) ; x(2n+1) = xo(n)xe(0) = x(0)xe(1) = x(2)

xe(2) = x(4)

xe(3) = x(6)xo(0) = x(1)xo(1) = x(3)

xo(2) = x(5)

xo(3) = x(7)X(k) = Xe(k) + W8

k

Xo(k)k = 0 to 3= Xe(k-4) - W8k4Xo(k 4)

k = 4 to 7

80 Xo(0) ; X(4) = Xe(0) - 80 Xo(0)

81 Xo(1) ; X(5) = Xe(1) - WX(0) = Xe(0) + WWX(1) = Xe(1) + W

X(2) = Xe(2) + W81 Xo(1)

W82 Xo(2) ; X(6) = Xe(2) - 82 Xo(2)X(3) = Xe(3) + W83 Xo(3) ; X(7) = Xe(3) - W

83 Xo(3)X(0) & X(4) having same i/ps with opposite signs73Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldN

Npt DFT can be expressed as combination of pt DFT.

4This2k = 0 to N4-1(0 to 1)Xe(k) = Xee(k) + W 2k Xeo(k)

NN

N2(k)Xeo(k

NN

N WN= Xee(k- )-4

)k =to

to

2 -1( 2 to 3 )444k = 0 to N4-1Xo(k) = Xoe(k) + W 2k Xoo(k)NN

NN2(k)

N2 -1Xoo(k )= Xoe(k- ) - WNN4

k =444

For N=874Downloaded From JNTU World (http://JNTU-World.com)


Xe(0) = Xee(0) + W80 Xeo(0) ;

82 Xeo(1) ;

80 Xeo(0) ;

82 Xeo(1) ;xee(0) = xe(0) = x(0)

xee(1) = xe(1) = x(2)

xeo(2) = xe(2) = x(4)

xeo(3) = xe(3) = x(6)Xe(1) = Xee(1) + WXe(2) = Xee(0) - WXe(3) = Xee(1) - W

Where Xee(k) is the 2 point DFT of even no. of xe(n) & Xeo(k) is the 2 point DFT of oddno. of xe(n)Similarly, the sequence xo(n) can be divided in to even & odd numbered sequences asxoe(0) = xo(0) = x(1)

xoe(1) = xo(2) = x(5)

xoo(0) = xo(1) = x(3)

xoo(1) = xo(3) = x(7)W

Xo(0) = Xoe(0) + 80 Xoo(0) ;Xo(1) = Xoe(1) + W

82 Xoo(1) ;

Xo(2) = Xoe(0) - W80 Xoo(0) ;

Xo(3) = Xoe(1) - W82 Xoo(1) ;Xoe(k) is the 2-pt DFT of even-numbered of xo(n)Xoo(k) is the 2-pt DFT of odd-numbered of xo(n)

Xee(0) = xee(0) + xee(1) = xe(0) + xe(2) = x(0) + x(4)

Xee(1) = xee(0) - xee(1) = xe(0) - xe(2) = x(0) - x(4)75Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldXee(0) = xee(0) + xee(1) = xe(0) + xe(2) = x(0) + x(4)Xee(1) = xee(0) - xee(1) = xe(0) - xe(2) = x(0) - x(4)76Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldNo.ofNo.ofComplexSpeedNo.of points NMultiplications

Direct N2

Improvement Factor:StagesFFTN2

N2 Log2NN2 Log2N241644386412

32

80

1925.33

8416

32

64256

1024

40965

612.8

21.33For N=8No of stages given by= Log2N = Log28 = 3.( Log N -1 )

2No. of 2 i/p sets = 2

= 4NLog2NTotal No. of Complex additions using DITFFT is77Downloaded From JNTU World (http://JNTU-World.com)


= 8 * 3 =24Each stage no. of butterflies in the stage= 2m-q where q = stage no. and N=2m

Each butterfly operates on one pair of samples and involves two complex additions and

one complex multiplication. No. of butterflies in each stage N/2DITFFT: ( different representation) (u can follow any one) ( both representations arecorrect)NN1122

x(2n 1)WN(2n1)k

X(k) =x(2n)WN2nk +n0n0NN1122

xe (n)WNnk + WN

2kxo(n)Wnk

N / 2

=n0n0

4 pt DFT Xe(k) + WNK Xo(k)k= 0 to N/2 -1 = 0 to 3N

NNXe(k- ) - WN ) Xo(k- ) k = N/2 to N-1 = 4 to 7(K2

222 pt DFTXe(k) = Xee(k) + WN2K

Xeo(k)k = 0 to N/4-1 = 0 to 1N2(k)

Xeo(k-N/4)= Xee(k-N/4) - WN

k = N/4 to N/2 -1 = 2 to 3

k = 0 to N/4-1 = 0 to 14

Xo(k) = Xoe(k) + WN2K

Xoo(k)N2(k)

Xoo(k-N/4)= Xoe(k-N/4) - WN

k = N/4 to N/2 -1 = 2 to 34

W8 W4

2 1

N=8X(0) = Xe(0) + W80

Xo(0) ;

Xo(1) ;

Xo(2) ;

Xo(3) ;X(4) = Xe(0) - W80

Xo(0)

Xo(1)

Xo(2)

Xo(3)X(1) = Xe(1) + W8

1

X(5) = Xe(1) - W8

1

X(2) = Xe(2) + W8

2

X(6) = Xe(2) - W8

2

X(7) = Xe(3) - W8

X(3) = Xe(3) + W8

33

W0

Xeo(0) ;8

W0

Xeo(0)8

Xe(0) = Xee(0) +Xe(2) = Xee(0) -Xe(1) = Xee(1) + W8

2

Xeo(1) ;Xe(3) = Xee(1) - W82 Xeo(1)78Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldXo(0) = Xoe(0) + W80

Xoo(0) ;

Xoo(1) ;

Xo(2) = Xoe(0) - W80 Xoo(0)Xo(1) = Xoe(1) + W8

2

Xo(3) = Xoe(1) - W82 Xoo(1)N1

1n0 x(4n)WN4nk =x(0) + x(4) W84k4

x(4n)WN4nk =Xee(k) =0Xee(0) = x(0)+x(4)Xee(1) = x(0)-x(4)x(0)x(4)

x(2)x(0) x(0)x(2) x(1)

x(4) x(2)79Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldx(6)x(1)

x(5)

x(3)

x(7)x(6) x(3)x(1) x(4)

x(3) x(5)

x(5) x(6)

x(7) x(7)Other way of representation80Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldDIFFFT:N1N12

x(n)WNnk +x(n')WNn'kX(k)=put n’ = n+N/2n0n1N / 2N 12N1

2

x(n)WN

nk

x(n N / 2)WN(nN / 2)k

=+n0n0N 12N1N2

k

2

x(n N / 2)WN

nk

x(n)W

[x(n)nk W

N +=Nn0n0N12

nkW

N=+ (-1)k x(n+ N2 )]n0N12

N )]Wnk

N / 2X(2k)=[x(n) + x(n+2n0N12

Nn

N

Wnk

N / 2X(2k+1) =[x(n) - x(n+

)]W2n0Let f(n) = x(n) + x(n+N/2)

g(n) = x(n) – x(n+N/2) WNn81Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldN=8f(0) = x(0) + x(4)f(1) = x(1) + x(5)

f(2) = x(2) + x(6)

f(3) = x(3) + x(7)g(0) = [x(0) - x(4)] W8

g(1) = [x(1) - x(5)] W8

g(2) = [x(2) - x(6)] W8

g(3) = [x(3) - x(7)] W8012382Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldN14

Wnk

N / 4[ f (n)

N+ f(n+ )]4

X(4k)=n0N14

n

N / 2nk

N / 4

)W ]W[f (n)- f(n+ N4X(4k+2) =n0N14

Nnk

N / 4X(4k+1) =

X(4k+3) =

[g(n) + g(n+

)]W4n0N14

nk)WN / 2 ]

Wnk

N / 4

N[g(n) - g(n+4n0

X(4k) = f(0) + f(2) + [ f(1) + f(3) ] W 4k8

X(4k+2) = f(0) – f(2) + [ f(1) – f(3) ] W82

W84kX(0) = f(0) + f(2) + f(1) + f(3)83Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldX(4) = f(0) + f(2) – [ f(1) + f(3) ]

X(2) = f(0) - f(2) + [ f(1) - f(3)] W82

X(6) = f(0) - f(2) - [ f(1) - f(3)] W8

2

Find the IDFT using DIFFFTX(k) = 4, 1-j 2.414, 0, 1-j 0.414, 0, 1+j 0.414, 0, 1+j 2.414 Out put 8x*(n) is in bit reversal order x(n) = 1,1,1,1,0,0,0,084Downloaded From JNTU World (http://JNTU-World.com)


UNIT-IVDIGITAL FILTER STRUCTUREThe difference equationNP

M

by(n) =

ak x(n-k) +k y(n-k)kNFk1N

P

ak z kNpNF

(1 C k )Z 1

kN F

or = A Z N FM

M(1 d k )Z 1

k 1

H(z) = 1 bk z k

k1k1If bk= 0 non recursive or all zero filter.Direct Form – I1. Easily implemented using computer program.2. Does not make most efficient use of memory = M+Np+NF delay elements.Direct form-II85Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldSmaller no. of delay elements = Max of (M, Np) + NF

Disadvantages of D-I & D-II1. They lack hardware flexibility, in that, filters of different orders, having different no.

of multipliers and delay elements.2. Sensitivity of co-efficient to quantization effects that occur when using finite-precision

arithmetic.Cascade Combination of second-order section (CSOS)y(n) = x(n) + a1 x(n-1) + a2 x(n-2) + b1 y(n-1) + b2 y(n-2)1 a1Z a2Z 2

1b1Z b2Z 2

11

H(z) =Ex:86Downloaded From JNTU World (http://JNTU-World.com)


z 5Z 2

1 5 Z 1 1 Z3

2 5 Z 1 Zz534443 12 1212H(z) ==

1

1 Z Z2

1

1 Z Z2

2424z

1 1 Z1 1 Z

1 Z 2

1 Z 23 41 1 1 1 Z 1

Z 1

1 Zz == 3 4122

1 Z Z Z2424Ex:1 Z 2 Z 3Z Z 1 Z 2Z Z 1 Z 1 H(z) = =Z 1

1 1 1

1 Z1 Z 1 Z11

12482 4 8 =Z 0.65 0.45Z 1 Z 21.45 Z 11 1 1

1 Z

1 Z

4

1 Z

821.45 Z 1 0.65 0.45Z 1 Z 2= Z 1

Z 1

Z 2

1 Z1243287Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldParallel Combination of Second Order Section (PSOS)Ex:

5 Z 2 Z 3

Z 1

12 12

2

15z55 Z Z Z 1

3 12H(z) = 3 12 1212=12

Z21 Z42

1 Z Z12412

Z 3

Z 1 Z Z 152 Z 2 Z 1 1 51 724123 123

33

Z 2

Z 1

Z12 63___-____+____-_______7Z 2

Z 1

11212 37Z 2

7Z 1

73

126______________________5

4 Z 1 737 5 Z 11

2 Z

1 Z 1 2 3 4H(z) = Z Z 2

34


JNTU-World.comJNTU WorldEx:Z Z 1 Z 2H(z) =obtain PSOSZ 1 Z 1 Z 1

111248 11 Z 1 1 2ZAB

C

1 Z 1

Z 1 Z 1 Z 1

=1 1 1 Z1 Z111

24824 8 A = 8/3B = 10C = -35/3Jury – Stability Criterion

N(z)H(z) = D(z)N

D(z) =biZ Ni = bo ZN +b1 ZN-1 + b2 ZN-2 +….. bN-1 Z1 + bNi0ROWSCOEFFICIENTS12

3

4

5

6bo

b1 ……. bN

bN

Co

bN-1 ……. boC1 ……. CN-1

CN-1 CN-2 ……. Co

do d1 ……. dN-2

dN-2 dN-3 ……. do89Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World..

.2N-3r0 r1

r2

bo bNi

Ci = bNi = 0,1,…N-1bi

co

di = cN1

cN1i

ci

i = 0,1,…N-2i. D(1) > 0

ii. (-1)N D(-1) > 0bo bN

co cN1 do dN2

ro r2

iii.Ex:Z4

3Z 2Z Z 14 3 2

D(z) = 4ZH(z) = 4Z3Z 2Z Z 1

2431

2

3

4

543

12

21 1

3 4

1115 11 61611 15224 159 79D(1) = 4+3+2+1+1 = 11 > 0,(-1)4 D(-1) = 3 >0

Stable.

bo b4

do d2

co c3

Ex:4Z 2

4Z 2 7Z 21H(z) =

=2

Ans: Unstable1 74 Z 12 Z190Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldUNIT-VNon Recursive filters

y(n) = ak x(n-k)Recursive filtersNpM

y(n) =ak x(n-k) –bk y(n-k)kkNfk1

for causal systemfor causal systemNpM

=ak x(n-k)y(n) = ak x(n-k) –bk y(n-k)k0k1k0

For causal i/p sequenceIt gives IIR o/p but not always.N

Ex: y(n) = x(n) – x(n-3) + y(n-1)y(n) =ak x(n-k)k0N

P

ak z k

It gives FIR o/p. All zero filter.Always stable.General TF : H(z) = kNF

M

1 bk z kk1

bk = 0 for Non RecursiveNf= 0 for causal systemFIR filtersIIR filters

1. Linear phase no phase distortion.

2. Used in speech processing, data

transmission & correlation processingLinear phase, phase distortion.

Graphic equalizers for digital audio,tonegeneratorsfiltersfordigitaltelephone3. Realized non recursively.

4. stableRealized recursively.Stable or unstable.H(n) = an u(n) a<1 stable= 0a>1 unstable5. filter order is moreLess6. more co-efficient storageLess storage

7. Quantization noise due to finiteprecision arithmetic can be made

negligibleQuantization noise8. Co-efficient accuracy problem isMore91Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldless severe9. used in multirate DSP (variablesampling rate)IIR FILTER DESIGNButterworth, chebyshev & elliptic techniques.

Impulse invariance and bilinear transformation methods are used for translating s-plane singularities of analog filter to z-plane.

Frequency transformations are employed to convert LP digital filter design into HP,BP and BR digital filters.

All pass filters are employed to alter only the phase response of IIR digital filter toapproximate a linear phase response over the pass band.

The system function = H(s)The frequency transfer function = H(j ) = H(s) / s=jThe power transfer function = H( j) 2 = H(j ) H*(j ) = H(s) H(-s) / s=jTo obtain the stable system, the polse that lie in the left half of the s-plane are assigned toH(s).BUTTERWORTH FILTER DESIGN1The butterworth LP filter of order N is defined as HB(s) HB(-s) =2N

s

j1 c

Where s = jc

= 1orH B ( jc ) 2

H B ( jc ) db= -3dB ‘s2It has 2N poles2N

1

s

jc

2N

= -1s

j

c

S2N = -1 ( jc )2N92Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldjj 2N

= e j (e c )2N = c 2Neee2jj2m21N2m j 2NS2N = c

Sm = c

e1N2m j e2N0 m 2N 1Ex: for N=3j (42m)j 2j 4j5j 2j 7,e j ,ee6

= e33

,e3

,e ,e3

= 1200, 1800, 2400, 3000, 3600, 6001Vo(s) Vi(s)

11 RCSCS1R CS111c RS

s2

1 c

1 c

Poles that are let half plane are belongs to desired system function.1H B ( j) 2

=2N

c

1 For a large , magnitude response decreases as -N, indicating the LP nature of thisfilter.93Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldH B ( j) dB = 10log10 H B ( j) 2

2N

= -10 log10(1 ) c

As = -20 N log10

= -20 N dB/ Decade = -6 N dB/OctaneAs N increases, the magnitude response approaches that of ideal LP filter.

The value of N is determined by Pass & stop band specifications.Ex: Design Butterworth LPF for the following specifications.Pass band:-1< H( j) 2 dB 0Stop band:for0 1404( p = 1404 )H( j) 2 dB < -60

If the c is given

s for 8268( s = 8268 )2N

H( j s) 2

= [1 ]-1 < 10-6 (-60dB)

c

log(106

1)= N > 2log( s)cSince c is not given, a guess must be made.94Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldThe specifications call for a drop of -59dB, In the frequency range from the edge of thepass band (1404 ) to the edge of stop band (8268 ). The frequency difference is equal to82681404 log = 2.56 octaves.21 oct ----2.56 ------

=>- 6N dB?2.56 X - 6N dB = -59 dB’s59N = 2.56X 6 3.8There fore: N =42N

s = [1 Now H B ( js) 2

]-1 < 10-6

c2N

s 1

6> 10c

s 2N > 106 c 2N6

s 10 2N > c => 1470.3 > c c <1470.3Let c =1470.3At this c it should satisfy pass band specifications.

2N

]-1 > 0.794 (= -1dB) p H B( jp) 2

= [1

c

= 0.59This result is below the pass band specifications. Hence N=4 is not sufficient.Let N=5

6

c < s X 10 2N = 2076.810

]-1 = 0.981404 2076In the pass band H B ( jp) 2= [1Since N=5 c = 207695Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldS1 = -2076S = 2076 (cos (4 /5) j sin(4 /5)) = 2076 e j144

2, 3

S = 2076 (cos (3 /5) j sin(3 /5)) = 2076 e j1084, 5

20765

HB(s) = s 2076s2 3359s (2076)2s2 1283s (2076)21. Magnitude response is smooth, and decreases monotonically as increases from 0 to

2. the magnitude response is maximally flat about =0, in that all its derivatives up toorder N are equal to zero at =0

Ex: c=1, N=1H B ( j) 2 = (1+ 2)-1

The first derivatived

d 2H B ( j) 2 =

=0 at =021 2

The second derivatived2

H B ( j) 2 = -2 at =02

d96Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World N3. The phase response curve approachesfor large , where N is the no. of poles of2butterworth circle in the left side of s-plane.

Advantages:1. easiest to design2. used because of smoothness of magnitude response .Disadvantage:Relatively large transition range between the pass band and stop band.Other procedureAvoWhen c = 1Avs =2N

w 1 wo AvoH B (s) 2 =2N

s1

j If n is even S2N = 1 = e j(2k1)j(2k1) The 2N roots will be Sk= ek=1,2,….2N2N

Sk =Cos(2k 1) jSin(2k 1) 2N2N1where k = (2k 1) 2NTherefore: H B (s) 2 = T(s) =N / 2

(s2

2Cosks 1)k1

If N is oddS2n =1 = e j2k

Sk = e j2k / N

k=0,1,2….(2N-1)1where k = k T(s) =(N1) / 2

N

(s2

2Cosks 1)k1


JNTU-World.comJNTU World0 20log H( j) K1 for 120log H( j) K2for 2k1

1c

2n

10 10 1110 log

10 log

= K1 1

2N

1 c 2nk 2

2 10 10 1 c 1= K2 22N

1 c k1

1 2 10

10 110 10 1

2n

Dividing k 2

k1

10 10 1log10

k 2

10 10 1n = 1 22log10

choosing this value for n, results in two different selections for c . If we wish to satisfy

our requirement at 1 exactly and do better than our req. at 2, we use1

2

c =orc =for better req at 211

k1

k 2

2n2n

1010

11010

1EndCHEBYSHEV FILTER DESIGN98Downloaded From JNTU World (http://JNTU-World.com)


1S

jpDefined as Hc(S) Hc(-S) = 1 2

C2N

= measure of allowable deviation in the pass band.

CN(x) = Cos(NCos-1(x)) is the Nth order polynomial.

Let x = CosCN(x) = Cos(N )C0(x) = 1C1(x) = Cos =xC2(x) = Cos2 = 2 Cos2 -1 = 2x2-1C3(x) = Cos3 = 4 Cos3 -3 Cos = 4x3-3xetc..N

0

1

2

3

4CN(x)1x2x2-14x3-3x

8x4- 8x2 +1Two features of Chebyshev poly are important for the filter design1. CN (x) 1for x 11 Hc( j) 2112

for 0 pTransfer function lies in the range 1 21 Hc( j) 21for 0 p

Whereas the frequency value important for the design of the Butterworth filter was thec, the relevant frequency for the Chebyshev filter is the edge of pass band p .99Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World2. x 1, CN (n) Increases as the Nth power of x. this indicates that for >> p , themagnitude response decreases as -N, or -6N dB Octane. This is identical to Butterworthfilter.Now the ellipse is defined by major & minor axis.Define = 1 1 2

1

1

N N Minor r = p 2

1

N N1

Major R = p N = Order of filter.2SP = r Cos +j R SinEx:Pass band:-1< H( j) 2 dB 0Stop band:for

for

0 1404H( j) 2 dB < -60 8268Value of is determined from the pass band10 log1 1 > -1dB-1dB = 0.79421

< 100.1 12 = 0.508 = 0.508100Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldValue of N is determined from stop band inequality1

Hcjs 2 =1 2C s

jp2N

<10-6

12

10

1Since ps 5.9CN(5.9) >6

= 19692

EvaluatingC3(5.9) = 804C4(5.9) = 9416 therefore N = 4 is sufficient.

Since this last inequality is easily satisfied with N=4 the value of can be reduced to as

small as 0.11, to decrease pass band ripple while satisfying the stop band. The value =0.4

provides a margin in both the pass band and stop band. We proceed with the design with

=0.508 to show the 1dB ripple in the pass band.Axes of Ellipse: =0.508-1 + (1+0.508-2)1/2 = 4.17

1

4.17 4 4.17 4 702 (1.43 0.67) 14941

1404R =

r =

21404

4.17141

4.17 5124

27 , 5poles locations : 88

473 j572 742e j130

S1,2 = 512Cos 78 j1494Sin 78 = 55= 196 j1380 1394e j98

S3,4 = 512Cos j1494Sin8874213942

H (S) = [S S *2*742Cos(130) (742) S *2*1394Cos(98)1394] [S ]c2 2 2 2

Chebyshev filter poles are closer to the j axis, therefore filter response exhibits a

ripple in the pass band. There is a peak in the pass band for each pole in the filter, located

approximately at the ordinate value of the pole.101Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World Exhibits a smaller transition region to reach the desired attenuation in the stop band,when compared to Butterworth filter. Phase response is similar.

Because of proximity of Chebyshev filter poles to j axis, small errors in their

locations, caused by numerical round off in the computations, can results in significant

changes in the magnitude response. Choosing the smaller value of will provide some

margin for keeping the ripples within the pass band specification. However, too small a

value for may require an increase in the filter order. It is reasonable to expect that if relevant zeros were included in the system function, a

lower order filter can be found to satisfy the specification. These relevant zeros could serve

to achieve additional attenuation in the stop band. The elliptic filter does exactly this.

IMPULSE INVARIANCE METHOD

H(z) =h(n)Z nn0

H(z) (at z = eST ) =h(n)eSTn

re jw e j)T r = eT

jw e jT w TeLet S1 = j =>Z1 = eT

Z2 = eT jT j2 eT jT

e eejT

2S2 = j( ) =>T102Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldIf the real part is same, imaginary part is differ by integral multiple of 2 , this is theTbiggest disadvantage of Impulse Invariance method.

s as a2 b2

s aLet HA(S) == s a jb

s a jb hA(t) = eat

for t 0s1 = -a-jb

s2 = -a+jbCosbt= 0otherwiseh (nTs) = eanTsCos(bnTs)for n 01eaTsCos(bTs)Z 1

Cos(bTs)Z 1 Z 2

1eaTsCos(bTs)Z 1

H(z) = 1 2eaTs

= (1e(a jb)TsZ 1)(1e(a jb)TsZ)

1

The pole located at s=p is transformed into a pole in the Z-plane at Z = epTs

, however, thefinite zero located in the s-plane at s= -a was not converted into a zero in the z-plane at Z =eaTs , although the zero at s= was placed at z=0.Desing a Chebyshev LPF using Impulse-Invariance Method.

S1,2 = -473 j 572S3,4 = -196 j 1380[The freq response for analog filter we plotted over freq range 0 to 10000 . To set thediscrete-time freq range (0, ), therefore Ts = 10-4]TsZ1,2 = eS1,2Ts

= e0.148 j0.179= 0.862 e j10.2

Z3,4 = eS3,4Ts

= e0.061 j0.433= 0.94 e j24.8

k(1 2*0.862Cos10.2Z 1 0.8622Z 2)(1 2*0.94Cos24.8Z 1 0.942Z 2

H(z) =)103Downloaded From JNTU World (http://JNTU-World.com)


k(11.69Z 1 0.743Z 2)(11.707Z 1 0.88Z 2

=)Methods to convert analog filters into Digital filters:1. By approximation of derivatives/ t=nTs = x(nTs) x(nTsTs)dxdtTs1 Z 1

S =TsOrUsing forward-difference mapping based on first order approximation Z = esTs

1+STsZ 1

S =TsUsing backward- difference mapping is based on first order approximationZ 1 esTs 1STsZ 1 1 Z 1

S =

TsZTsd2

2x /t=nTs =d dx /t nTs

dt dt

dtx(nTs) x(nTs Ts) x(nTs Ts) x(nTs 2Ts)

TsTs

Ts=104Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldx(nTs) 2x(nTsTs) x(nTs 2Ts)

Ts=22

S 2 1 2Z1 Z 2

=1 Z 1

Ts2

Tsk

1 Z 1

Therefore Sk =T 1 Z 1

Therefore H(z) = Ha(s) /s=using backward differenceT

11 STs= 0.5 + 0.5(1 STs)Z =

=1 STs11

jTsTs2 11 jTs = 1Ts

2 22

Z - 0.5 = 0.5(1 STs)(1 STs)z 0.5 0.5is mapped into a circle of radius 0.5, centered at Z=0.5

Using Forward-differenceZ 1S=Z=1+STsTsu+jv = 1+ ( j)Tsif =0 u=1 and j axis maps to Z=1If >0, then u>1, the RHS-plane maps to right of z=1.If <0, then u<1, the LHS-plane maps to left of z=1.105Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldThe stable analog filter may be unstable digital filter.Bilinear Transformation

Provides a non linear one to one mapping of the frequency points on the jw axis in s-

plane to those on the unit circle in the z-plane. This procedure also allows us to implement digital HP filters from their analogcounter parts.2 Z 1 2 1 Z 1

S = Ts Z 1=

Ts 1 Z 1

Using trapezoidal rule y(n)=y(n-1)+0.5Ts[x(n)+x(n-1)]

H(Z)=2(Z-1) / [Ts(Z+1)] 2 Z 1

Ts Z 1To find H(z), each occurrence of S in HA(s) is replaced bySTs12

STsAnd Z =121/ 2

2

1Ts 2

j tan1 eTs

jTs22

2 1 = ejw

Ts1/ 2

2

j 1TsTs

j tan1e

2

1 2 2

2Ts

j2tan1 e-1 Ts

2ejw2w=2tan106Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldThe entire j axis in the s-plane -<j < maps exactly once onto the unit circle - w such that there is a one to one correspondence between the continuous-time anddiscrete time frequency points. It is this one to one mapping that allows analog HPF to be

implemented in digital filter form.As in the impulse invariance method, the left half of s-plane maps on to the inside of the

unit circle in the z-plane and the right half of s-plane maps onto the outside.In Inverse relationship is 2 tan wTs 2

w w3 w2 2 8 ....For smaller value of frequency 2 Sin= w2wTs CosTs 1 w2

Ts....24(B.W of higher freq pass band will tend to reduce disproportionately)107Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldThe mapping is linear for small and w. For larger freq values, the non linearcompression that occurs in the mapping of to w is more apparent. This compressioncauses the transfer function at the high freq to be highly distorted when it is translated to

the w-domain.Prewarping Procedure:

When the desired magnitude response is piece wise constant over frequency, thiscompression can be compensated by introducing a suitable prescaling or prewarping to the

freq scale. scale is converted into * scale. * = 2 tan TsTs 2 We now derive the rule by which the poles are mapped from the s-plane to the z-plane.

1S SpLet HA(s) =S=Sp

12 1 Z 1

Ts 1 Z 1

Ts(1 Z 1)H(z) == 2 SpTs1 22 SpTsSpTs Z 1 SpA pole at S=Sp in the s-plane gets mapped into a zero at z= -1 and a pole at Z = 22 SpTsSpTsEx:Chebyshev LPF design using the Bilinear TransformationPass band:-1< H( j) dB 0Stop band:for0 1404 =4411 radH( j) dB < -60for 8268 rad/sec =25975 rad/s

Let the Ts = 10-4 secPrewarping values are p* = 2 tan Ts = 2*104 tan(0.0702 ) = 4484 rad/sec

Ts 2 And s* = 2 tan Ts = 2*104 tan(0.4134 ) = 71690 rad/sec

Ts 2 The modified specifications are108Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldPass band:-1< H( j*) dB 0for

for

0 *4484 rad/s

*71690rad/sec

Stop band:H( j*) dB < -60Value of : is determined from the pass band ripple 10log 1 2 1 1dB = 0.508Value of N: is determined from

1

s*Hcjs*2

<10-6

=1 2

C2N

p* Since ps** 1610 1CN2(16) < 6

2

12

6

10 1CN(16) <= 19692

(0.508) C3(16) = 16301

N = 3 is sufficientUsing Impulse Invariance method a value of N=4 was required. =4.171

1

N N Major R = p * =4484

1

4.17 3 4.171

50013

22

1

4.17 3 4.17 3 22161

4484

2r =2Since there are three poles, the angles are &3S1 = r cos + j Rsin = -2216S2,3 = 2216 Cos 23 j 5001 Sin 23 = -1108 j 4331 = 4470 e j104.4


JNTU-World.comJNTU World4.43*1010

Hc(s) = (s 2216)(S 2

2223s 4470 )2

Pole MappingAt S=S1

In the Z-plane there is zero at Z = -1 and pole at Z = 22 ((22221166**11004

)

S2,3 = there are two zeros at Z=-14) 0.801Z = 2 (1108 j4331)*1044 0.801 j0.373 0.9e j24.5

2 (1108 j4331)*10

1 Z 1

1 0.801Z 1 11.638Z 1 0.81Z 2

1 2Z1 Z 2

H(z) = 4.29 * 10-3

Pole Mapping Rules:Hz(z) = 1-CZ-1 zero at Z=C and pole at Z = 0

11dZ Hp(z) =1 pole ar Z=d and zero at z=0C and d can be complex-valued number.Pole Mapping for Low-Pass to Low Pass Filters

Applying low pass to low pass transformation to Hz(z) we get

c 1

1 c = (1+c )Z

1

Z 1 1Z 1

HLZ(Z) = 1-c1Z 1

c The low pass zero at z=c is transformed into a zero at z=C1 where C1 = 1 c And pole at z=0 is Z=Similarly,1Z 1

HLP(Z)= d 1 d 1Z 1

1d d Pole at z=d => Z= 1d Zero at z=0 => z = 1 Z 1 2Z 2Z 112

H(z) = K 1 0.622Z 111.07Z 1 0.674Z 2110Downloaded From JNTU World (http://JNTU-World.com)


1 (1)(0.356)3

K =(1 0.801*0.356)(1 (0.819 j0.373)(0.356))(1 (0.819 j0.373)(0.356)) 0.029111Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldUNIT-VI()

Phase Delay: p Group Delay: g d()

dIf p = g =constant and independent of frequency are called as constant time delay orlinear phase filters.

() o o

p

Changes with frequency g = - =constant.Type 1 SequenceN 1Center of Symmetry M= integer value2N3

2

h(n) 2 h(n)CosT(n m) e jnT

H(w) = n0

N 1 T 2 Amplitude spectrum is even symmetric about w=0 & & both H(0) & H( ) can be nonzero.Type 2 Sequence112Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldh(n) = h(N-1-n)Center of Symmetry M= N 1 half-integer value2N

12

h(n)CosT(nm)e

jMT

2H(w) =n0

The Amplitude spectrum is even symmetric about w=0 & odd symmetric about w= &both H( ) is always zero for type 1 & 2 : Constant phase delay and group delay.

Type 3 SequenceN 1M= integer value2N3

2

h(n)SinT(M n) e jMT

2H(w) = jn0


JNTU-World.comJNTU WorldIt shows generalized linear phase ofMT and constant group delay of M. The2Amplitude spectrum is odd symmetric about w=0 & w= and H(0) & H( ) are always zero.

(Generalized means () may jump of at 0 if H(ejw) is imaginary.

Type 4 SequenceN

1

2

( ) ( )h n Sin T M n e jMT

2H(w) = j n0

Generalized linear phase and constant group delay of M. The Amplitude spectrum is oddsymmetric about w=0 & even symmetric about w= and H(0)=0 always.For N=even, even Symmetry h(n) = h(N-1-n)N1

h(n)e jnT

H(e jT ) =n0

N1N12

h(n)e jnT

=h(n)e jnT +

Nn

2n0

Let N-1-n = mN

120

h(n)e jnT +h(N 1 m)e jT (N1m)

=n0Nm 12

But h(N-1-m) = h(m)114Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldNN1122

h(n)e jnT +h(m)e jT (N1m)

==n0m0N1N N112 jT N12

h(m)e T (N1n)e jTh(n)e jnT

e2

j2=

=

+n0n0N

12

j(nTT (N1) j(T (N1n)T (N1)

N1e jT

2

e22

2h(n)e2n0

N1

N1 jT

N 1

2 cosTn 2

2

2h(n)e=n0N 1

N1 jT

2 N 1

----Magnitude

22h(n)cosTn=e 2 n0

N 1 2 T Linear PhasePoles & Zeros of linear phase sequences:

The poles of any finite-length sequence must lie at z=0. The zeros of linear phasesequence must occur in conjugate reciprocal pairs. Real zeros at z=1 or z=-1 need not be

paired (they form their own reciprocals), but all other real zeros must be paired with their

reciprocals. Complex zeros on the unit circle must be paired with their conjugate (that form

their reciprocals) and complex zeros anywhere else must occur in conjugate reciprocal

quadruples. To identify the type of sequence from its pole-zero plot, all we need to do is

check for the presence of zeros at z= and count their number. A type-2 seq must have an

odd number of zeros at z=-1, a type-3 seq must have an odd number of zeros at z=-1 and

z=1, and type-4 seq must have an odd number of zeros at z=1. The no. of other zeros if

present (at z=1 for type=1 and type-2 or z=-1 for type-1 or type-4) must be even.115Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldFIR Filters2Fs F Fs2Fourier series Method 2Fs 2F 2Fs22s s22

1. Frequency response of a discrete-time filter is a periodi function with period s(sampling freq).

2. From the F.S analysis we know that any periodic function can be expressed as a linear

combination of complex exponentials.Therefore desired freqency response of a discrete time filter can be represented by F.S as116Downloaded From JNTU World (http://JNTU-World.com)


H(e jT ) h(n)e jnT

nT = sampling periodThe F.S co-efficient or impulse response samples of filter can be obtained using

1h (n) = ss / 2 H(e jT )e jnTds / 2clearly if we wish to realize this filter with impulse response h(n), then it must have finiteno. of co-efficient, which is equivalent to truncating the infinite expansion of H(e jT ), whichleadstoapproximationofH(e jT ),whichisdenotedbym

H1(e jT ) h(n)e jnT

nm.N 1We choose M=, in order to keep ‘N’ no of samples in h(n).2M

h(n)Z nH1(z) =nMHowever, this filter can’t be physically realizable due to the presence of +ve powers of Z,means that the filter must produce an output that is advanced in time with respect to the i/p.N 1this difficulty can be overcome by introducing a delay M=samples.2M

h(n)Z n

ThereforeH(z) = Z-M H1(z) = Z-MnMH(z) = h(-M)Z0 + h(-M+1) Z-1 +…. +h(M) Z-2M

Let bi = h(i-M) i=0 to 2M2M

H(z) =biZ i be the transfer function of discrete filter that is physically realizable.i0Properties:1. N=2M+1, impulse response co-eff, bi = 0 to 2M.2. h(n) is symmetric about bM

Ex: M=4117Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World3. The duration of impulse response is Ti = 2MT4. Its magnitude and time delay function can be found in the following way

H(e jT ) e jMTH1(e jT

)H(e jT ) H1(e jT

)This implies that magnitude response of the filter we have desired approximates the

desire magnitude response. The time delay of H(ejw) is a constant M. thus sinusoids of

different frequencies are delayed by the same amount as they are processed by the filter, we

have designed. Consequently, this is a linear phase filter, which means that it does not

introduce phase distortion.Ex:Design a LPF (FIR) filter with frequency responseH(e jT ) 1for cs= 0 for c 2

1h(n) = sce jnTdc

2= scCos(nT)d0

2 SincnT= s nT118Downloaded From JNTU World (http://JNTU-World.com)


2= 2Fsn. Fs11 SincnT n SincnTbi = h(i-M)2M

biZ iH(z) =i0w=T 2sT 2Fs 1

2 FsEx:Design LPF that approximate following freq response.

H(F) = 1 0F1000Hz

= 0 else where 1000FFs/2When the sampling frequency is 8000 SPS. The impulse response duration is to belimited to 2.5ms

Ti = 2MT2.5*103

10M =N=2112*800119Downloaded From JNTU World (http://JNTU-World.com)


1h(n) = sc1.e jnTdc

1= 2Fs

1= Fs2FcFc

FcCos(2FnT)dF

1.e j2FnT 2dFe j2FnTdF FcFc

Fs0

1 Sin2FcnT n1 Sin(0.25n)= n________________________________________________________________OR1

w = T = 2 *1000*8000 4w 4Hc(w) = 1=0else where1

2= 1

n Sin(0.25n ) 1.e jwndw4

4h(0) = 0.25h(6) = -0.05305

h(7) = -0.03215

h(8) = 0h(1) = 0.22508h(2) = 0.15915

h(3) = 0.07503

h(4) = 0h(9) = 0.02501h(10) = 0.03183h(5) = -0.04502bi = h(i-10)

20

biZ i

H(z) =i0FIR HPF1c 1.e jnTds / 2e jnTdh(n) = s s / 2c120Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World1 e jnT

e jnT

s / 2

c

= s jnT s / 2

jnT c 1 e jcnT e jsnT

ejsnT

e jcnT

22

= sjnTjsnT

jsnT 2 1 e jcnT e jcnT ee2 j22

=s nT 2 j 2 Sin nT Sin snT = 2FsnT 2 c

1

n1sincnT Sinnsinc

nT==nFIR BPF

2h(n) = sEx:1

sinunT sinlnTucosnT d= nl

Desing a BPF for H(f) = 1 160 F 200Hz= 0 else where

Fs = 800SPSTi = 20 ms 20*10

2*3

M = Ti 8N = 172T1800

sin0.5n sin0.4nn

1 Sin2FunT Sin2FlnT =h(n) =n121Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldh(0) = 0.1h(4) = 0.07568h(5) = 0.06366

h(6) = -0.05046

h(7) = -0.07220h(1) = 0.01558h(2) = -0.09355h(3) = -0.04374h(8) = 0.0233816

biZ i

H(z) =i0

bi = h(i-8)h(-n) = h(n)WINDOWING

Disadvantage of F.S is abrupt truncation of FS expansion of the freq response. Thistruncation result in a poor convergence of the series.

The abrupt truncation of infinite series is equivalent to multiplying it with the rectangularsequence.WR(n) = 1n M= 0else whereh(n) h(n)WR (n)H(e jw) H(e jw)*WR(e jw

)1

= 2

) (

H(e j W e j(w )R

)d122Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldWR(ejw) => FT of Rectangular WindowwNN1

wNNSa

Sa2Sin

2

w1.e jwndw =2WR(ejw) =w = N 1Sin

2n22 4Main lobe width =& it can be reduced by increasing N, but area of side lobe willNbe constant.

For larger value of N, transition region can be reduced, but we will find overshoots &undershoots on pass band and non zero response in stop band because of larger side lobes.

So there overshoots and leakage will not change significantly when rectangular window is

used. This result is known as Gibbs Phenomenon.The desined window chts are

1. Small width of main lobe of the fre response of the window containing as much as of

the total energy as possible.2. Side lobes of the frequency response that decrease in energy as w tends to .3. even function about n=0N 14. zero in the range n 2123Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldLet us consider the effect of tapering the rectangular window sequence linearly from themiddle to the ends.Triangular Window:W (n) 1 2nn N 1T

N 12= 0else whereIn this side lobe level is smaller that that of rectangular window, being reduced from -138to -25dB to the maximum. However, the main lobe width is now. There is trade offNbetween main lobe width and side levels.

General raised cosine window is (1)Cos 2n

N 1for n W(n) = N 12= 0else whereIf =0.5 Hanning WindowIf =0.54Hamming WindowW (n) = 0.42 + 0.5 Cos 2n 4n Blackman Window N 1 0.08Cos N 1BKaiser Window2 2n Io 1

N 1 Io()N 1Wk (n)

for n 2= 0else where

is constant that specifies a freq response trade off between the peak height of the side

lobe ripples and the width or energy of main lobe and Io(x) is the zeroth order modified

Bessel function of the first kind. Io(x) can be computed from its power series expansiongiven by2

1 x k

k! 2

Io(x) = 1+k1124Downloaded From JNTU World (http://JNTU-World.com)


0.25x 23

+…..0.25x2

0.25x22

= 1 +++(1!)2

(3 !)2

(2 !)2

WindowPeak amplitudeTransition width

Minimum stop

band deviation dB

-21of side lobe dB

-13of main lobeRectangular

Triangular

Hanning4

k=1N-258-25

-44

-53

-74

-k=2

N-318

k=2NHamming

BlackMan

Kaiser-418

k=2N-5712

k=3N

variablevariable

If we let K1,W1 and K2,W2 represent cutoff (pass band) * stop band requirements for thedigital filter, we can use the following steps in design procedure.

1. Select the window type from table to be the one highest up one list such that the stop

band gain exceeds K2.2. Select no. of points in the windows function to satisfy the transition width for the typeof window used. If Wt is the transition width, we must have Wt = W2-W1 k.2Nwhere K depends on type of window used.K=1 for rectangular , k=2 triangular…..2ThereforeN K w2 w1


JNTU-World.comJNTU WorldIf analog freq are given, it must be converted in to Digital using w= TEx:

Apply the Hamming Window to improve the low pass filter magnitude response ontained

in ex1: 2n N 1N 1WH(n) = 0.54 + 0.46 Cosfor n 2= 0else whereN = 2M+1 = 21WH(0) = 1WH(6) = 0.39785WH(1) = 0.97749WH(2) = 0.91215

WH(3) = 0.81038

WH(4) = 0.68215

WH(5) = 0.54WH(7) = 0.26962WH(8) = 0.16785

WH(9) = 0.10251

WH(10) = 0.08Next these window sequence values are multipled with coefficients h(n), obtained in ex1,to ontain modified F.S Co eff h’(n).

h’(0) =0.25h’(1) =0.22h’(2) =0.14517h’(3) =0.0608h’(4) =0h’(5) =0.02431

h’(6) =0.02111

h’(7) =-0.0086725

h’(8) =0h’(9) =0.00256

h’(10) =0.00255126Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World2M

H’(z) = b'i Z ii0

bi' = h’(i-M)0i20h’(-n) = h’(n)Ex:Find a suitable window and calculate the required order the filter to design a LP digitalfilter to be used A/D-H(Z)-D/A structure that will have a -3dB cutoff of at 30 rad/sec andan attenuation of 50dB at 45 rad/sec. the system will use a sampling rate of 100 samples/secSol:The desired equivalent digital specifications are obtained as1Digital …..w1 wc cT 30 0.3 k1 3dB

k2 50dB1001w2 2T 45 0.45 100

1. to obtain a stop band attenuation of -50dB or more a Hamming window is shosensince it has the smallest transition band.

2. the approximate no. of points needed to satisfy the transition band requirement (or the

order of the filter ) can be found for w1 =0.3 rad &w2 = 0.45 rad, using Hamming

window (k=2), to be22.2N k w2 w1 0.45 0.3=26.65N = 27 is selected127Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU Worldthe attractive property of the Kaiser window is that the side lobe level and main lobewidth can be varied continuously by simple varying the parameter . Also as in other

window, the main lobe width can be adjusted by varying N.we can find out the order of Kaiser window, N and the Kaiser parameters to design

FIR filter with a pass band ripple equal to or less that Ap, a minimum stop band attenuation

equal to or greater than As, and a transition width Wt, using the following steps:

Step 1 :100.05Ap 1s 100.05As , p 100.05Ap 11p1p => As = 20logsAp = 20log10

1p As = 20log10 s1

=1

= 1100.05Ap

1100.05ApTherefore: solving above eq for , we get

100.05Ap -1 100.05Ap1Step 2:Calculate As using the shosen valuesAso= 20log128Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldStep 3:Calculate the parameter as follows for

= 0

= 0.5842(Aso -21)0.4 + 0.07886(Aso -21)

= 0.1102(Aso -8.7)for Aso 21 dB

for 21< Aso 50 dB

for Aso >50 dBStep 4:Calculate D as followsD = 0.9222 for Aso 21 dBAs 7.95=for Aso >21 dB14.36Step 5:Select the lowest odd value of N satisfying the inequalityN samD 1

tWsam : Angular Sampling frequency sam : Analog Freq t = s- p

= p- sfor LPFfor HPF129Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World= Min[( p1- s1), ( s2- p2)] for BPF= Min[( s1 - p1), ( p2- s2)] for BSF-3dB cutoff freq c can ve considered as follows c = 1p sfor LPF & HPF2tt c1 = p1 ;c2 p2

for BPF22t;c2 p2 2t c1 = p1 2for BSFEx:

Calculate the Kaiser parameter and the no. of points in Kaiser window to satisfy thefollowing lowpass specifications.Pass band ripple in the freq range 0 to 1.5 rad/sec 0.1 dBMinimum stop band attenuation in 2.5 to 5.0 rad /s 40 dB

Sampling frequency : 10 rad/sSol:The impulse response samples can be calculated using h(n) = n1 sincnTWhere c = 12 (1.5 2.5)=2rad/sAnd the no. of points required in this sequence can be found as followsStep1:s 100.05(40) 0.01p 100.05(0.1) 1 5.7564*103

100.05(0.1) 1Therefore we choose, 5.7564*103

Step 2:Aso = -20 log(5.7564*103 ) = 44.797 dB

Step 3 & 4: = 0.5842 ( 44.797 -21)0.4 + 0.07886 ( 44.797 -21) = 3.9524130Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldD = 2.566Step 5:10(2.566) 1 26.66N => N=271 2n 2

N 1 Io 1Wk (n) Io()W (0) Io() 1k

Io()2 2 Io3.9524 1 26 =Io(3.94)10.269 0.9899Wk (n) Io(3.9524)Io(3.9524) 10.3729131Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-1

1)What is the parsval’s theorem expression in DTFT :∑n=-∞|x(n)|2=(1/2π)2dwMatch the following:2) E = , P = 0a) power3) E , P = 04) E = , P 0, P b) Neither energy nor powerc) EnergyMatch the following5) e-t u(t)a) power6) u(t)b) Neither energy nor power7) 1/tc) Energy8) x (n) = 6e j 2 n / 4, what is the power of the signala) 36W b) 72W c) 18W d) noneMatch the following: For a real valued sequence, the DTFT follow the properties as9) Re [H (jw) ]10) Im[ H(jw) ]a) Real valued function of wb) even function of w11) F.T [even symmetric sequence]12) F.T [odd symmetric sequence]c) Imaginary valued function of wd) odd function of w13) x(n) = 4, 1, 3 h(n) = 2, 5, 0, 4 what is the output of the system.

a) 8, 22, 11, 31, 4, 12d) noneb) 8, 22, 11, 31, 4, 12c) 8, 22, 11, 31, 4, 1214) y(n) = x(n) * h(n) then y1 (n) = 0, 0, x(n), 0 * 0, h(n), 0 is equal to

a) 0, 0, y(n), 0 b) 0, 0, 0, y(n), 0, 0 c) [0, 0, y(n), 0 d) 0, y(n), 0, 015)If x(n) and h(n) are having N values each, to obtain linear convolution using circular

convolution, the number of zeros to be appended to each sequence isa) N – 1 b) 2N – 1 c) N d) N + 1

16)W49 = ?a) – j b) + j c) + 1 d) -1

17) DFT [ x* (-n) ] = ?a) X *(K) b) X * (-K) c) X *(N-K) d) none

12b3c4a

5c6a7b8a9b10d11a12c13c14b15a16a17a132Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-2

1) The region of convergence of the Z-transform of a unit step function isa)|Z| >1 b) |Z|<1c) (real part of Z ) >0d) (real part of Z ) <02) The Z T of the function f(nT) = anT isa) Z/(Z-aT)b) Z/(Z+aT) c) Z/(Z-a-T)d) Z/(Z+a-T)

3) The Z T of the function (n k) isk0a) (Z-1)/Zb) Z/(Z-1)2 c) Z/( Z-1)d) (Z-1)2/Z4) The Z T of a signal is given by X(Z)= Z-1(1-Z-4)/( 4(1-Z-1)2) its final value isa) ¼ b) 0 c) 1 d) infinity5) Consider the system shown in fig. The transfer function Y(Z) / X(Z) of the system isx(n)y(n)++Z-1

-ba

b) (1+bZ-1)/ ( 1+aZ-1)d) (1-bZ-1)/ ( 1+aZ-1)a) (1+aZ-1)/ ( 1+bZ-1)c) (1+aZ-1)/ ( 1-bZ-1)6) A linear discrete time system has the characteristic equation Z3-0.8 Z=0, the systema) is stableb) is marginally stablec) is un stabled) stability cannot be assessed from the given information7) The advantage of Canonic form realization isa) smaller no of delay elementsb) larger no of delay elementsc) hard ware flexibilityd) none3

5k1 b k(yn k)akx(n k)

8) y(n) =-the minimum no of delay elementsk2

needed to realize the system isa) 5 b) 10

c) 8d) 119) Expand CSOS Ans: Cascaded form of second order section.

PSOS Ans: Parallel form of Second order section10) To ensure a causal system, the total no of zeros must be less than or equal to the total

number of poles ( T / F )133Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World1a2a3c4c5a6a7a8c910T11) The poles or zeros at the origin do effect the magnitude response ( T / F)12) All poles and zeros of a minimum phase system lie inside the unit circle ( T / F)

13) To realize FIR filtera) no feedback paths and forward pathc) feedback paths and no forward pathb) no feedback paths and no forward pathd) feedback paths and forward path14) Find total no of complex multiplications using FFT for N=8: __________

15) Find total no of complex additions using FFT for N=8: __________

16) Find total no of real additions using direct DFT for N=8: __________11F12T13a141215241624017) What is Z T of 2 (3n) u (-n-1): ____(-2)/(1-3z-1_)___ or (-2z)/(z-3)__________18) (2M) Show the structure ofDirect form –II for 2nd order systemZZx(n)y(n)

a1Z -1

+b1

b2

b3

a2

a3Z-1

19) Show the structure of butterflyZ

Z-1

-1

bnp

anpZ-1

bm134Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOBJECTIVE PAPER-3State TRUE or FALSE

1) u(n) = (n k)K0

2) x(n) = cos 0.5n is periodic sequence3) Discrete-time sinusoidal signals with frequency that are separated by an integral

multiple of 2π are identical4) y(n) =x(-n) is time invariantMatch the following

5)h(k) 1 Zero input response6) Impulse response of difference equation is 2 linear7) y(n) = |x(n)|3 Stable8) y(n) = x(n2)4 Time invariantCHOOSE THE CORRECT ANSWER9) x(n) = Cos 0.125n , what is the period of the sequencea) 8 b) 16 c) 125 / 2 d) none10) y (n) = x (2n)a) Causalb) Non-Causalc) Time invariant d) none11) x(-n + 2) is obtained using following operartiona) x (-n) is delayed by two samplesc) x (n) is shifted left by two samplesb) x (-n) is advanced by two samplesd) none

12) In situations where both interpolation and decimation are to be performed insuccession, it is therefore best toa) Interpolate first, then decimatec) Any order we can performb) Decimate first and interpolated) none1T2F3T4F53617

4829B101112BAa13) The output of anti causal LTI system isn

K0a) y (n) =h(k)x(n k)b) y (n) =h(k)x(n k)

h(k)x(n k)K01

c) y (n) =h(k)x(n k)d) y (n) =135Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World14) (n-k) * x (n-k) is equal toa) x(n-2k) b) x(n-k)c) x(k)d) none

15) Given x(n) the y(n) = x(2n – 6) isa) x(n) is Compressed by 2 and shifted by 6

shifted by 3b) x(n) is Compressed by 2 and

d) nonec) x(n) is Expanded by 2 and shifted by 316) Decimation by a factor N is equivalent toa) Sampling x(t) at intervals ts / Nc) N fold increase in sampling rateb) Sampling x(t) at intervals tsNd) none17) In fractional delay, x(n-M/N), specify the order of operation.a) Decimation by N, shift by M, Interpolation by Nb) Shift by M, Decimation by N and Interpolation by Nc) Interpolation by N, Shift by M and Decimation by Nd) All are correct18) Given g(n) =1,2,3, find x(n) = g (n / 2), using linear interpolationa) 1, 0, 2, 0, 3

19)b) 1, 1, 2, 2, 3, 3 c) 1, 3/2, 2, 5/2, 3 d) noneh1(n)h3(n)+y(n)+x(n)h2(n)In the figure shown, how do you replace whole system with single blocka) [ h1(n) + h2(n) ] * h3(n)c) [ h1(n) + h2(n) ] h3(n)b) h1(n)h3(n) * h2(n)h3(n)

d) none20 The h(n) is periodic with period N, x(n) is non periodic with M samples, the output

y(n) isa) Periodic with period Nc) Periodic with period M

b) Periodic with period N+Md) none13C14A15B16

B17C18C19A20A136Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-4

1) If x(n) = -1, 0, 1, 2, 1, 0, 1, 2, 1, 0, -1 What is X(0)a) 6b) 10c) 0d) none2) If x(n) = 1,|n|≤2

0, other wiseFind DTFTa) sin(5w)/sinwb) sin(4w)/sinw c) sin(2.5w)/sin(0.5w) d) none of the above3) If x(n)=h(n)=u(n), then h(n) is equal toa) (n+1)u(n) b) r(n)c) r(n-1)d) none

d) None

4) if x ~ (n) = 1,0,1,1 and h ~(n) = 1, 2, 3,1 find y ~(n)a) 6 , 6, 5, 4b) 1, 2, 4, 4 c) 5, 4, 1, 0

5) x(n) = 4, 1, 3 h(n) = 2, 5, 0, 4 what is the output of the system.

a) 8, 22, 11, 31, 4, 12 b) 8, 22, 11, 31, 4, 12 c) 8, 22, 11, 31, 4, 12 d)none6) y(n) = x(n) * h(n) then y1 (n) = 0, 0, x(n), 0 * 0, h(n), 0 is equal to

a) 0, 0, y(n), 0 b) 0, 0, 0, y(n), 0, 0 c) [0, 0, y(n), 0 d) 0, y(n), 0, 07) If x(n) and h(n) are having N values each, to obtain linear convolution using circular

convolution, the number of zeros to be appended to each sequence isa) N – 1 b) 2N – 1 c) N d) N + 1

8)W49 = ?a) – j b) + j c) + 1 d) -1

9) DFT [ x* (-n) ] = ?a) X *(K) b) X * (-K) c) X *(N-K) d) none

10) If x(n)X(K), then IDFT [ X (K), X(K) ] = ?a) x (n / 2) b) 2x (n/2) c) ½ x (2n) d) none.11) Both discrete and periodic in one domain are also periodic and discrete in other

domain (T / F)12) If h(n)= -h(-n) then H(K) is purely real(T / F)137Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World13) Reversing the N point sequence in time is equivalent to reversing the DFT values (T /F)14) FT of non periodic discrete time sequence is non periodic(T / F)Match the following: For a real valued sequence, the DTFT follow the properties as15) Re [H (jw) ]16) Im[ H(jw) ]17) F.T [even symmetric sequence]18) F.T [odd symmetric sequence]

a) Real valued function of wb) even function of wc) Imaginary valued function of w

d) odd function of wn=N-1

19) Write DFF & IDFT formulas.X(k)=∑x(n)Wnnk

n=0N-1x(n)=(1/N)∑X(k)Wnnk

K=020) Total no of real multiplications in DFT is:1A2C3A A C B A A A A T B D A C45678910 11 12 13 14 15 16 17 18 19 20FTF4n2

OBJECTIVE PAPER-5

Choose the Correct Answers

1. The Fourier transform of a finite energy discrete time signal, x(n) is defined as []

b) X( )=

n=-a) X( )= x(n) ejn

x(n) enn=-

x(n) e-jn

c) X( )=x(n) e-jn

d) X( ) =n=-n=02. Inverse DFT (IDFT) of X(K) is x(n), where k=0,1,-----n-1. It is given as[ ]j2knj2knN1

N11

1a) x(n) =

c) x(n) =X(k) e

X(k) eb) a) x(n) =

d) a) x(n) =X(k) e

X(k) eNNNNn0n0

j2knk

j2knN

1

1

NNNn0n0

3. A N – periodic sequence x(n) and its DFT x(k) are known. Then the DFT of x(n) =(n) will be

a) e-j2nk

b) 1c) e-j2nok/N

d) e-j2nk

/N[]4. If the length of sequence x(n) is L and h(n) is M then the length of o/p sequence of thecircular convolution is[]a) L+M b) L+M-1 c) L if L>M d) 2L if L=M138Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldSTATE TRUE OR FALSE5. The DFT of a sequence is a continuous function of [

[]

]6. The DFT of even sequence is purely imaginary and DFT of oddsequence is purely real7. The circular shift of an N point sequence is equivalent to linear shift of its periodicextension[]8. The multiplication of DFT of two sequences is equal to DFT of the linear convolutionof two sequences[]Fill in the blanks9. The 4-point DFT of a sequence x(n) is ________

10.DFT of a sequence x(n) = (n-n0) is __________11.An N point sequence is called ________________ if it is antisymmetric about point

zero on the circle12.The two methods of sectioned convolution are ________________ &

_______________13.DFT of multiplication of two sequences DFT x1

_____________________(n)x2(n) =

14.DFT of even sequence is X(k)= ________________________& DFT of odd sequenceis X(k) = _______________________

15.To get the result of linear convolution with circular convolution of sequence x(n) &h(n), the sequences should extended to the length of __________________16. Match the following1 DFT [ x1(n) x2(n) ]a) X (N-K)2. DFT [ x*(n) ]b) 1 [ X1(k) X2(k)]N3. DFT [ x((-n))N ]c) X*(N-K)4. X1(k) X2*(k)d) x1(n) x2(n)

e) x1(n) x2*(-n)139Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World1=2=3=4=

17. Show that the given sequence x(n) = 1,-2,3,2,1,0 for the following conditionsusing concentric circles.a) x(-n)b) x(2-n)(2M)18. Compute 4-point DFT of a sequence x(n) = 1,2,0,2(2M)

OBJECTIVE PAPER-6MULTIPLE CHOICES

1. In Impulse invariant transformation, the mapping of analog frequency to the digitalfrequency isa)

one to one b) many to one d) none2. The digital frequency in bilinear transformation is

w = 2 tan-1(Ts/2) b) w = tan-1(Ts/2)c) w = 2 tan-1(Ts) d) w = 2 tan-1(/2)3. Which technique is useful for designing analog LPFc) one to manya)a)c) Both a and bButter worth filterb) Chebyshev filterd) none4. Which filter is more stable?a) Butter worthb) Chebyshevc) none5. As increases , the magnitude response of LPF approaches witha) –20Ndb/oct b) –6Ndb/oct c) –10Ndb/dec d) none6. Using Impulse invariant technique the pole at S= SP is mapped to Z-plane as

-Sa) Z=e PTs(S )b) Z=e PTs

Sc) Z=e P (Ts)d) None

TRUE or FALSE7. The disadvantage of Chebyeshev filter is less transition region8. The advantage of Butter worth filter is flat magnitude response.9. for the given same specifications order of the Chebyshev filter is more than

Butterworth filter10. Poles of Butterworth filter lies on circle.

1B2A3C

4A5B6B7F8T9F10TFILL IN THE BLANKS11.The Butterworth LPF of order N is defined as: 1/(1+(s/jΩc)2N)12.For N=3 what are the stable Butter worth angles :120 ,1800,24000140Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World13. –0.5db convert in to gain equivalent =0.99414.Let S1,2 = 2076e±j144° Ha(S)= k/(s2-10552.7s+(2076π)2

(2M)15.Given s = 2000; Ts = 10-4 ; = 2006*s

16.Using Bi-linear transformation, the pole at S = Sp is mapped into Z-plane using(2M)

Z=1-(2+SpTS)/(2-SpTs)17.Given allowable ripples in Pass band is –3 dB, the value of is 0.997(2M)OBJECTIVE PAPER-7Choose the correct Answer1. In impulse invariant transformation the mapping of analog frequency to digitalfrequency is[]a) one to oneb) many to onec) one to many none2. The digital frequency in Bi –linear transformation is[]a) = 2 tan-1( T /2)c) = 2 tan-1( T )b) = tan-1( T /2)d) = 2 tan-1( /2)3. Using bilenear transformation for T = 1sec the pole pk is in S- Plane is mapped to Z –plane using[]1 z 1

1 z 1

1 z 1

1 z 1

c) S = 2 11 zz1

1

Z 1Z 1a) S = 2b) S =d) S=4. The normalized magnitude response of chebyshev type – I filter has a value of____________ at cut off frequency are[]111a)b)

c)d) 1

2

12

11

2

5. For high pass analog filter the transformation used isa) SS/ b) S /S c) SS/c d) S c /S[

[]

]6. The magnitude response of Type I – chebyshev LPF is given by

11 2CN (/c )1a)H ( ) 2

=b)H () 2

=a

a

12C (/c )2N

11

2CN (/c )c)H ( ) 2

=d) H a () = 1a

12C (/c )2N


JNTU-World.comJNTU World7. The width of main lobe in rectangular window spectrum isa) 2/N b) 4/N c) 8/N d) 16/N[]8. The width of main lobe in Hamming window isa) 4/N b) 2/N c) 8/N d) 16/N9. The frequency response of rectangular window WR(w ) is[

[]

]Sinwn / 2Sinwn / 2Sinwn / 2

Sinwn / 2d)Sinwn / 2

a)b)c)Sinw / 2SinwSinwn10.In …………………. Window spectrum the width of main lobe is double that ofrectangular window for same value of N

a) Hamming window b) Kaiser window[]c) Blackman window d) noneState TRUE or FALSE11.The disadvantage of chebyshev filter is less transition region[]12.For chebyshev Type 2 filter ripples are present in pass bandand stop band[]

]

]13.The advantage of Butter worth filter is flat magnitude response.[

[14.for cheby shev Type 1 filter equi–ripples are present onlyin pass band.15.For same specifications, the order N of chebyshev filter is less compared to Butterworth filter.

16.FIR filter have non-linear phase characteristics.

17.FIR filters are non – recursive and stable filters.[

[

[]

]

]18.The design of Digital transformation H (z) of IIR filter is direct and FIR is indirect[]19.Poles of chebyshev filter lies on circle[]20.In FIR filter with constant phase delay the impulse response is symmetric[]142Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOBJECTIVE PAPER-8CHOOSE THE CORRECT ANSWER1. The DTFT of a sequence x(n) is[]

a)x(n)e jwn

b)x(n)ejwn

c) x(n)e jwndw d) x(n)e jwn

dwnn2. DTFT of ejwon x(n) is[

[

]

]

a) x[ e j(wwo )

]b) x[ e j(wwo )

]c) x[ e j(wwo )

]d) x[ e j(wwo )

]3. DTFT of x1[n] * x2[n] isa) X1[w] X2[w] b) 1 X1[w] X2[w] c) X1[w] * X2[w] d) 1 X1[w] * X2[w]NN4. The smallest value of N for which x(n +N) = x(n) holds is called[

[

[]a)Fundamental period b) Fundamental frequency c) fundamental signal d) None5. DFS of real part of periodic signal is

a) Xe(K) b) Xo (K) c) XR(K)]

]d) XIm(K)6. Expression for DFT isN1N1

N1

C) K0

N1

a)x(n)WN

Kn

b)x(n)WNKn

x(n)WN

Kn

d)x(n)WNKnn0n0n0

7. DFT of x1[n] x2[n] is[]a) 1 X1[K] * X2[K] b) 1 X1[K] + X2[K] c) X1[K] * X2[K] d) X1[K] + X2[K]NN8. If M & N are the lengths of x(n) & h(n) then length of x(n) * h(n) is [

a) M+ N –1 b) M + N +1 c) max (M,N) d) min (M,N)]

]9. Zero padding means[a) increasing length by adding zeros at the end of sequenceb) Decreasing length by removing zeros at the endc) Inserting zeros in between the samplesd) None of the aboveII STATE TRUE OR FALSE10. The F.T of discrete signal is a discrete function of [

[]

]11.In a discrete signal x(n), if x(n) =x(-n) then it is called symmetric signal12.The F.T of the product of two time domain sequence is equivalent to product

of their F.T13.The DFT of a signal can be obtained by sampling one period of FT of the signal

[[]

]14.DFS is same as DTFS[]143Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-9CHOOSE THE CORRECT ANSWER1. Power signal isa) Periodic b) aperiodicc) Continuousd) none[

[

]

]

2. WN nK

is j2K j2Kn2Kn

a) eb) e j2nK

c) ed) eNNN

3. When the sequence is circularly shifted in time domain by ‘m’ samples i.e. x((n-m))N

then on applying DFT, it is equivalent multiply sequence in frequency domain byj2Km j2Km2Km

a) eb) ec) e j2Km

d) e[]NNN

4. Multiplication of sequence in time domain, on apply DFT, it corresponds to circularconvolution in frequency domain and is given asa) x1(n) x2(n)b) x1(n) x2(n)c) x1(n) * x2(n)D

FT

X1(K) X2(K)X1(K)X2(K)

X1(K)

DFTD

FT

X2(K)

X1(K)X2(K)N1

d) x1(n) x2(n)D

FTK0

5. Linear convolution of two sequences N1 and N2 produces an output sequence of lengtha) N1 – N2 +1 b) N1 + N2 –1c) N1 + N2 +1d) 2N1 – N2 +1[]FILL IN THE BLANKS6. The basic signal flow graph for butterfly computation of DIT-FFT is7. The Fourier transform of discrete time signal is called ………………………

8. FFT’s are based on the ………………………….. of an N-point DFT intosuccessively smaller DFT’s.9. The Fourier transform of x(n)*h(n) is equal to …………………………..

10.Appending zeros to a sequence in order to increase the size or length of the sequenceis called ……………………..11.In N-point DFT using radix 2 FFT, the decimation is performed …………… times.

12.In 8-point DFT by radix 2 FFT, there are …………… stages of computations with…………………….. butterflies per stage.13.If DFT of x(n) is X(K), then DFT of WN x(n) is …………………….

ln


JNTU-World.comJNTU WorldANSWER THE FOLLOWING

14.What are the differences between linear and circular convolution?15.How many multiplications and additions are required to compute N-point DFT using

radix 2 FFT16.How many multiplications and additions are required to compute N-point DFT

17.What is the expression for N-point DFT of a sequence x(n) ?

18.What is the expression for N-point IDFT of a sequence X(K) ?19.Define Aliasing error.20.What is meant by Inplace computation.145Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-101. How we can calculate IDFT using FFT algorithm.(2M)2. Draw the basic butterfly diagram for DIF algorithm.3. Z[x(n)] = X(Z) then Zx(n-m) = …………………………………..4. Define convolution property in Z-Transform.5. Find the Z-Transform and ROC for the signal x(n) = an u(n).

6. Find the Z-Transform and ROC for the signal x(n) = - an u(-n-1).

7. Write the initial value theorem expression.

8. Z(n) = ……………………..Z

Z 19. Find inverse Z-Transform for X(z) =when ROC is Z<1146Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World10.What are the differences and similarities between DIT and DIF algorithms. (2M)11.Give the Direct form II realization for second order system.

12.Give the Direct for I realization for second order system.

13.What is the relationship between Z-Transform and Fourier transform.

STATE TRUE OR FALSE:14.ROC of a causal signal is the exterior of a circle of some radius r.[]

]

]15.ROC of a anti causal signal is the exterior of a circle of some radius r. [

16.ROC of a two sided finite duration frequency is entire Z-plane.[17.Direct form I required less no.of memory elements as compared to Canonic form.[ ]18.A linear time invariant system with a system function H(Z) is BIBO stable if and onlyif the ROC for H(Z) contains unit circle.[]147Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOBJECTIVE PAPER-11ANSWER THE FOLLOWING

1. What are the advantages of digital filter over analog filter.2. What is the relation between analog and digital radiant frequency in Impulse

Invariance design..3. What is the relation between analog and digital radiant frequency in Bilinear

transformation design.4. What are the drawbacks with Impulse Invariance method?

5. What is the disadvantage with Bilinear transformation technique.

6. What is the relation between S & Z in Bilinear transformation?

7. Mention any two techniques to design IIR Filter from analog filter.

8. What are the differences between Chebyshev type I and type II.


JNTU-World.comJNTU World9. What are the differences between Butterworth & Chebyshev filter.

10.What is the expression for magnitude squared frequency response of Butterworthanalog filter?

11.What is the expression for magnitude squared frequency response of Chebyshevanalog filter?TRUE OR FALSE12.Poles of Butterworth filter lies on circle.[

[]

]13.Poles of Chebyshev filter lies on circle.14.Transition bandwidth for Chebyshev is more as compared to Butterworth filter.[]]15.Butterworth filters are all pole filters.[16.Chebyshev, type-II are all pole filters.[]17.Chebyshev, type II filter exhibit equiripple behavior in the pass band and monotoniccharacteristic in the stopband.[]

18.Chebyshev, type I filter exhibit equiripple behavior in the pass band and monotoniccharacteristic in the stopband.[]

]19.Butterworth filter exhibit monotonic behavior both in passband and stopband.[

20.For the given specifications order of the Chebyshev filter is more as compared toButterworth filter.[]149Downloaded From JNTU World (http://JNTU-World.com)


OBJECTIVE PAPER-12

Define the following

Time variant system with an example(Equation)

Power signal with an example

1.

a.b.

c.

d.

e.

2.

3.

4.

5.

Dynamic systemRecursive SystemNon Recursive systemGive the example for FIR and IIR systems.

Give an example of Causal system

Write the condition to test the Linearity of the system

Plot y(n) = x(n-2) Give x(n) =1,2,3,5,6

6.Resolve the signal into impulse x(n)=4,5,4,4------ 2 Marks150Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World7.

8.

Give the expression for Convolution sum y(n)=Find the Convolution Sum Graphically with all the steps-------3 Marksx(n)=21h(n)=1 1-100 19.Write the properties of Convolution Sum--------2 Marks10.Write the expression for X(n) in terms of impulses11.

12.

Write the necessary condition for the stability of the systemWrite the general form of Difference equation151Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOBJECTIVE PAPER-13State TRUE or FALSE

1. In direct –form II realization the number of memory locations required is more thanthat of direct form –I realization[]2. An LTI system having system function H(z) is stable if and only if all poles of H(z)are out side the unit circle.[]3. The inverse Z – transform of z/z-a is an u(n)[

[]

]4. Digital filters are not realizable for ideal case.5. As the order of Butter worth filter increases than the response is closer to ideal filterresponse.[]Answer the following6. Find the transfer function H(z) of the given difference equationY(n) = 0.7 y(n-1) – 0.12y(n-2) + x(n-1) + x(n-2)7. Indicate the poles and zeros of the given system and also check the stability of thesystemz(z 1)H(z) = (z 0.2)(z 0.4)(z 0.5)(2M)8. Realize the given system function H(z) using direct form –IIH(z) = 3 3.6z 1 0.6z 2

(2M)1 0.1z 1 0.2z 2


JNTU-World.comJNTU World9. Realize the given system function H(z) using cascade form(2M)

1(1 0.5z 1)(1 0.5z 1

H(z) =)z10.Find the inverse z-transform of x(z) = (z 2)(z 3) using partial fractionmethod.(2M)11.Using cauchy residue method find the inverse z- transform ofzx(z) = (z 1)(z 2) for ROC :z >2(2M)12.Mention the two conditions to realize any digital filter13.Draw the Magnitude response of Low Pass Butter Worth filter.

14.The order of the Butter Worth filter is obtained by using the formula N______________________

15.The cut- off frequency c is obtained by using the formula___________________________153Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldOBJECTIVE PAPER-14Fill in the Blanks1. The expansion of FFT is _______________2. The main advantage of FFT is _____________

3. The number of multiplications needed in the calculation of DFT using FFT with 32-point sequence = ________________

4. __________________ number of additions are required to compute N – pt DFT usingradix –2 FFT.5. What is decimation in time algorithm.State TRUE or FALSE6. For DIT –FFT algorithm the input is bit reversed and the output is in natural order[ ]7. By using radix –2 DIT –FFT algorithm it is possible to calculate 6-point DFT.[]

]8. WNNK 1

9. WNN/K2 1[

[

[]

]10.In DIT –FFT, the input sequence is divided into smaller subsequences

Answer the following11.Calculate the DFT of the sequence x(n)=1,0,0,1 using DIT –FFT(2M)12.Draw the Butterfly diagram for 8-point DFT using DIT –FFT algorithm (2M)13.Find IDFT of the sequence X(k) = 10, 0, 10, 0(2M)154Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU World14. Write the steps for the calculation of IDFT using DIT –FFT(2M)15.Write the values of the followinga) W8

0

b) W8

2

c) W8

3

d) W8

5

OBJECTIVE PAPER-15CHOOSE THE CURRECT ANSWER1. y(n)=x(2n) is a ____________ system[

[

[

[]

]

]

]a) time invariantb) causal c) non causal d) none2. y(n) = nx2(n) is a ____________systema) Linear b) Non-linearc) time-invariant d) none3. y(n)= x(n) +x(n-1) is a ____________ systema) Dynamic b) Static c) time variant d) None

4. x(-n+2) is obtained by which of the following operationsa) x(-n) is shifted left by 2 samples b) x(-n) is shifted right by 2 samplesc) x(n) is shifted left by 2 samples d0 none5. The necessary and sufficient condition for causality of an LTI system is [ ]

a) h(n) =0 for n=0 b) h(n) =0 for n>0 c) h(n) =0 for n<0d) none

6. The convolution of two sequences x(n) =h(n) = 1, 2, -1a) 1,4,2,-4,1 b) 1,-4,1,2,4 c) 1,1,2,-4,4d) 4,-4,2,1,1

[

]II STATE TRUE OR FALSE7. An IIR system exhibits an impulse response for finite interval

8. If the energy of a signal is infinite then it is called energy signal

9. Static systems does not require memory[ T/F ]

[ T/F ]

[ T/F ]10. A linear system is stable if its impulse response is absolutely summable[T/F ]155Downloaded From JNTU World (http://JNTU-World.com)

JNTU-World.comJNTU WorldIII Answer the following:11. The average power of a discrete time signal with period N is given by ___________

12. The convolution sum of causal system with causal sequence is ____________13. Give the graphical representation of the following discrete signals.i)x(n) = (5-x) [ 4(x) – 4(x-3)ii)x(n) = -0.5(n+1) + 0.5(n) – 0.75 (n-2)14. x(n) = 3, -2, 1,0,-1 show for x(-n)(1M)

(2M)

15. If x(n) = 1,2,-2,-1 show for x(n-2) & X( -n+2)16. Find the convolution of u(n) * u(n-2)(1M)156Downloaded From JNTU World (http://JNTU-World.com)


17. If the impulse response h(n) = 2n u( -n) then determine the corresponding system iscausal or stable. (1M)18. Test the given discrete system for linearity , causality and time invariance

h(n) = n ex(n)( 2M)

ASSIGNMENT UNIT-51 (a) Draw the frequency response of N-point rectangular window.(b) Design a fifth order band pass linear phase filter for the following specifications.i. Lower cut-off frequency = 0.4 πrad/secii. Upper cut-off frequency = 0.6 πrad/seciii. Window type = HammingDraw the filter structure. [4+12]

2) Design a band pass filter to pass frequencies in the range 1-2 radians/second usingHanning window N=5. Draw the filter structure and plot its spectrum. [16]

3) (a) Compare the performances of rectangular window, hamming window and Keiserwindow

(b) The desired response of a low pass filter isHd(ej!) = _ e−j3!, −3π _ ω _ 3π/40 , 3π/4 _ |ω| _ π

Determine H(ej!) for M=7 using a Hamming window. [6+10]4) (a) Design a linear phase low pass filter with a cut-off frequency of π/2

radians/seconds. Take N=7(b) Derive the magnitude and phase functions of Finite Impulse Response filter wheni. impulse response is symmetric & N is odd

ii. impulse response is symmetric & N is even. [8+8]5) (a) Design a low pass filter by the Fourier series method for a seven stage with cut-off

frequency at 300 Hz if ts = 1msec. Use hanning window.(b) Explain in detail, the linear phase response and frequency response properties of

Finite Impulse Response filters. [8+8]6) (a) Outline the steps involved in the design of FIR filter using windows.(b) Determine the frequency response of FIR filter defined by y(n) = 0.25x(n)+ x(n-1)+

0.25x(n-2). Calculate the phase delay and group delay. [8+8]7) (a) Define Infinite Impulse Response & Finite Impulse Response filters and com-pare.157Downloaded From JNTU World (http://JNTU-World.com)


(b) Design a low pass Finite Impulse Response filter with a rectangular window for afive stage filter given: Sampling time 1 msec; fc = 200Hz.Draw the filter structure withminimum number of multipliers. [6+10]

ASSIGNMENT UNIT-71) a) What are the advantages of Multirate signal processing?

b) Differentiate between Decimator and Interpolator?2) Prove that spectrum of down sampler is sum of M uniformly shifted and stretched

version of X(ejw) scaled by a factor 1/M and also discuss the aliasing effect?

3) State and prove any one identity property in down sampler and any one identity

property in up sampler?4) Let x(n)=1,3,2,5,-1,-2,2,3,2,1,find

a) Up sample by 2 times and down sample by 4 times

b) Down sample by 4 times and up sample by 2 times c) Justify why these outputs are

not equal.158Downloaded From JNTU World (http://JNTU-World.com)

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