Digital Signal Octave Codes (0A) · Digital Signals Octave Codes (0A) 3 Young Won Lim 9/12/17 Based...

Young Won Lim9/12/17

Digital Signal Octave Codes (0A)

● Periodic Conditions

Young Won Lim9/12/17

Copyright (c) 2009 - 2017 Young W. Lim.

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Digital Signals Octave Codes (0A) 3 Young Won Lim

9/12/17

Based on M.J. Roberts, Fundamentals of Signals and SystemsS.K. Mitra, Digital Signal Processing : a computer-based approach 2nd edS.D. Stearns, Digital Signal Processing with Examples in MATLAB


9/12/17

Sampling and Normalized Frequency

ω0 t = 2π f 0 t

ω0 nT s = 2π f 0nT s

t = nT s

f 0 =1T 0

=2πT 0

nT s

= 2π nT s

T 0

T0 : signal period

Ts : sampling period

F 0 = f 0T s =f 0

f s

= 2π n F0

T s

T 0

=f 0

f s

normalization

normalization


9/12/17

Analog and Digital Frequencies

nω0T s = 2π n f 0T s

ω0 t = 2π f 0 t

Ω0n = 2π n F0

ω0T s = Ω0 f 0T s = F0

Analog Signal

Digital Signal

t = nT st = nT s


9/12/17

Multiplying by Ts – Normalization

ω0 t = 2π f 0 t

Ω0n = 2π n F0

⋅T sNormalization ⋅T s

Analog Signal

Digital Signal


9/12/17

Normalization

F0 = f 0⋅T s

= f 0 / f s

= T s / T 0

Multiplied by Ts

Ω0 = 2πF0

f 0 / f s

f 0⋅T s

Divided by fs


9/12/17

Normalized Cyclic and Radian Frequencies

Normalized Cyclic Frequency

Normalized Radian Frequency

F0 cycles/ sample =f 0 cycles / second

f s samples / second

Ω0 radians /sample =ω0 radians / secondf s samples / second


9/12/17

Periodic Relation

e j(2π(n+N 0)F0) = e j(2πnF0)

2πN 0 F0 = 2πm

N 0 =mF0

Digital Signal Period N0 : the smallest integer

= m⋅T 0

T s

Periodic Condition: integer m

e j2πN 0 F0 = 1

e j2πm = 1

Integer * Rational no: must be an Integer


9/12/17

Periodic Relation

N 0 =mF0

= m⋅T 0

T s

Integer * Rational no: must be an Integer

Integer

Rational numbers


9/12/17

Periodic Relation

e j(2 π(n+N0)F0) = e j (2π n F0)

2π N 0 F0 = 2π⋅m

N 0 =mF0

Digital Signal Period N0 : the smallest integer

= m⋅T 0

T s

Periodic Condition: some integers m

e j (2π f 0)(t+T 0) = e j(2π f 0)t

2π f 0T 0 = 2π⋅m

T 0 =1f 0

Analog Signal Period T0 : the smallest real number

Periodic Condition: for all integer m


9/12/17

Periodic Relation

2π N 0 F0 = 2π⋅m

N 0 =mF0

2π f 0T 0 = 2π⋅m

T 0 =1f 0

Integer N0 Real T

0

Rational F0 Real f

0

=mp /q

= mqp

m = p

N0 = q

m = 1

T 0 =1f 0

Minimum Integer N0 Minimum Real T

0


9/12/17

Periodic Relation

2π N 0 F0 = 2π⋅m

N 0 =mF0

2π f 0T 0 = 2π⋅m

T 0 =1f 0

f 0 =3619

T 0 = 1⋅1936

m = 1

2T 0 = 2⋅1936

m = 2

3T 0 = 3⋅1936

m = 3

F 0 =3619

N 0 = 36⋅1936

m = 36

2N 0 = 72⋅1936

m = 72

3N 0 = 108⋅1936

m = 108

given given

Integer N0 Real T

0

m: multiples of 36 m: all integers


9/12/17

Periodic Condition of a Sampled Signal

Fundamental Frequency

2πF 0n = 2πm

F 0n = m

F 0 =mn

Integers n , m

F 0 =mn

=f 0

f s Sampling Frequency

Rational Number

g(nT s) = A cos(2π f 0T sn+θ)

g [n] = A cos(2π F0 n + θ)

M.J. Roberts, Fundamentals of Signals and Systems

F 0 = f 0T s = f 0/ f s

F 0 =mn


9/12/17

A Cosine Waveform

n= [0:29]; n= [0:29];x= cos(2*pi*(n/10)); x= cos((2/10)*pi*n);

U of Rhode Island, ELE 436, FFT Tutorial

= 2π⋅1⋅n⋅110

f 0 = 1 (T 0 = 1)

T s = 0.1

= 2π⋅110

⋅n⋅1

f 0 = 0.1 (T 0=10)

T s = 1

nT s = n⋅110

nT s = n⋅1

F 0 = f 0T s = 0.1 F 0 = f 0T s = 0.1

2π f 0 nT s2π f 0 nT s

F 0 = f 0T s =f 0

f s

=T s

T 0


9/12/17

Many waveforms share the same sampled data

1.00000 0.80902 0.30902 -0.30902 -0.80902 -1.00000 -0.80902 -0.30902 0.30902 0.80902 1.00000 0.80902 0.30902 -0.30902 -0.80902 -1.00000 -0.80902 -0.30902 0.30902 0.80902 1.00000 0.80902 0.30902 -0.30902 -0.80902 -1.00000 -0.80902 -0.30902 0.30902 0.80902


x

`

3

30


9/12/17

Cosine Wave 1

n= [0:29];x= cos(2*pi*n/10);


ω0 = 2π f 0 =2π

T 0

t = [0:29]/10;y = cos(2*pi*t);stem(t, y)hold ont2 = [0:290]/100;y2 = cos(2*pi*t2);plot(t2, y2)

t = [0:29];y = cos(0.2*pi*t);stem(t, y)hold ont2 = [0:290]/10;y2 = cos(0.2*pi*t2);plot(t2, y2)

ω0 = 2π f 0 =2π

T 0


9/12/17

Normalized Frequency


subplot(2, 1, 1);t = [0:29]/10;y = cos(2*pi*t);stem(t, y)hold ont2 = [0:290]/100;y2 = cos(2*pi*t2);plot(t2, y2)axis([0, 10, -1, 1]);

subplot(2, 1, 2)t = [0:29];y = cos(0.2*pi*t);stem(t, y)hold ont2 = [0:290]/10;y2 = cos(0.2*pi*t2);plot(t2, y2)axis([0, 10, -1, 1]);

f 0 = 1 T 0 = 1

T s = 0.1f s = 10

f 0 = 0.1 T 0 = 10

T s = 1f s = 1


9/12/17

Cosine Wave 1


T 0 = 1

t = [0:29]/10;y = cos(2*pi*t);stem(t, y)hold ont2 = [0:290]/100;y2 = cos(2*pi*t2);plot(t2, y2)

f 0 = 1

T s = 0.1

F0 = f 0T s = 0.1

f 0 = 1 T 0 = 1

T s = 0.1f s = 10


9/12/17

Cosine Wave 2


t = [0:29];y = cos(0.2*pi*t);stem(t, y)hold ont2 = [0:290]/10;y2 = cos(0.2*pi*t2);plot(t2, y2)

f 0 = 0.1

T s = 1

F0 = f 0T s = 0.1

T 0 = 10f 0 = 0.1 T 0 = 10

T s = 1f s = 1


9/12/17

Sampled Sinusoids


g [n] = A eβn

g [n] = A zn z = eβ

g [n] = A cos(2π n /N 0 + θ)

g [n] = A cos(2π F0 n + θ)

m /N 0

g [n] = A cos(Ω0 n + θ)

= F0

= Ω0/2π

2πm /N 0

= 2π F0

= Ω0

N 0 ≠1F0


9/12/17

Sampling Period and Frequency

g [n] = A cos(2π F0 n + θ)

F 0 = f 0T s = f 0/ f s

g (t ) = A cos(2π f 0 t+θ)

g [n] = g(nT s)


g [n] = A cos(2π F0 n + θ)

T s =1f s

sampling period

sampling frequencysampling rate


1T s

= f s


9/12/17

Periodic Condition of a Sampled Signal

Fundamental Frequency

2πF 0n = 2πm

F 0n = m

F 0 =mn

Integers n , m

F 0 =mn

=f 0

f s Sampling Frequency

Rational Number


g [n] = A cos(2π F0 n + θ)


F 0 = f 0T s = f 0/ f s

F 0 =mn


9/12/17

Periodic Condition Examples

g [n] = A cos(2π F0 n + θ)

g (t ) = A cos(2π f 0 t+θ)

g [n] = 4 cos(72πn19 ) = 4cos(2π⋅

3619

⋅n)

g [n] = 4 cos(2π⋅3619

⋅(n + N 0)) N 0 = 19


Fundamental Period of

g (t ) = 4cos(2π⋅3619

⋅(t + T 0)) T 0 =1936


g [n]

g(t )

g [ t ] = 4cos(72π t19 ) = 4cos(2π⋅

3619

⋅t)T s = 1

N 0 ≠1F0

qN 0

= F 0

N 0

q=

1F0


9/12/17


g [n] = 4 cos(2π⋅3619

⋅(n + N 0)) N 0 = 19



g (t ) = 4cos(2π⋅3619

⋅(t + T 0)) T 0 =1936


g [n]

g(t )

N 0 ≠1F0

qN 0

= F 0

N 0

q=

1F0

F0 =3619

=qN0 the smallest integer : fundamental period

the number of cycles in N0 samples


9/12/17



F0 =3619

=qN0 the smallest integer : fundamental period

the number of cycles in N0 samples

“When F0 is not the reciprocal of an integer (q=1),

a discrete-time sinusoid may not be immediately recognizable from its graph as a sinusoid.”


9/12/17


g [n] = 4 cos(2π⋅3619

⋅(n + N 0))

N 0 = 19


Fundamental period of

g (t ) = 4cos(2π⋅3619

⋅(t + T 0))

T 0 =1936

Fundamental period of

g [n]

g(t )

3619

⋅(n + N 0)

3619

⋅(t + T 0)

119

⋅N0 = k

3619

⋅T 0 = k

integer integer

integer integer

N 0

T 0

integer

integer


9/12/17


g [n] = 4 cos(2π⋅3619

⋅n)


g [n] = 4 cos(2π⋅119

⋅n)

g [n] = 4 cos(2π⋅219

⋅n)

g [n] = 4 cos(2π⋅319

⋅n)

clf n = [0:36]; t = [0:3600]/100;y1 = 4*cos(2*pi*(1/19)*n);y2 = 4*cos(2*pi*(2/19)*n);y3 = 4*cos(2*pi*(3/19)*n);y4 = 4*cos(2*pi*(36/19)*n);yt1 = 4*cos(2*pi*(1/19)*t);yt2 = 4*cos(2*pi*(2/19)*t);yt3 = 4*cos(2*pi*(3/19)*t);yt4 = 4*cos(2*pi*(36/19)*t);

subplot(4,1,1);stem(n, y1); hold on;plot(t, yt1);subplot(4,1,2);stem(n, y2); hold on;plot(t, yt2);subplot(4,1,3);stem(n, y3); hold on;plot(t, yt3);subplot(4,1,4);stem(n, y4); hold on;plot(t, yt4);


9/12/17


N0 = 19

g [n] = 4 cos(2π⋅3619

⋅n)

g [n] = 4 cos(2π⋅119

⋅n)

g [n] = 4 cos(2π⋅219

⋅n)

g [n] = 4 cos(2π⋅319

⋅n)

1 cycle

2 cycles

3 cycles

36 cycles

T s = 1


9/12/17


g [n] = A cos(2π F0 n + θ)g (t ) = A cos(2π f 0 t+θ)


g1(t ) = 4cos (2π⋅1⋅t )

g2(t ) = 4cos (2π⋅2⋅t )

g3(t) = 4 cos (2π⋅3⋅t )

t ← nT1

t ← nT 2

t ← nT 3

g1[n] = 4 cos (2πnT s1 )g2[n ] = 4 cos (2πnT s 2 )g3 [n] = 4cos (2π nT s3 )

T 1 =110

T 2 = 120

T 3 = 130

n = 0, 1, 2, 3,⋯

n = 0, 1, 2, 3,⋯

n = 0, 1, 2, 3,⋯

1⋅t = 0, 0.1, 0.2, 0.3,⋯

2⋅t = 0, 0.1, 0.2, 0.3,⋯

3⋅t = 0, 0.1, 0.2, 0.3,⋯

t ← nT1

t ← nT 2

t ← nT 3

{ g1[n] }≡ { g2[n ] }≡ { g2 [n] }


9/12/17


g [n] = A cos(2π F0 n + θ)

g (t ) = A cos(2π f 0 t+θ) 2πF 0n = 2πm

g [n] = 4 cos(72 πn19 )

= 4cos(2π(3619 )n)

3619

n = m

3619

=mn

3619

=mn

=f 0

f s

smallest n = 19

1/N0

= F0

= Ω0/2π

= 4cos(2π(3619⋅(n + N 0)))N 0 = 19


smallestF 0 =

qN 0


9/12/17



4 cos (ω t1 )

4 cos (2ω t 2)

4 cos (ωn1 )

4 cos (ω2n2 )

t1 = 2 t 2

n1⋅1

n2⋅2

T s = 1

T s = 2

n1 = 2n2


9/12/17



clf n = [0:10]; t = [0:1000]/1000;y1 = 4*cos(2*pi*1*n/10);y2 = 4*cos(2*pi*2*n/20);y3 = 4*cos(2*pi*3*n/30);yt1 = 4*cos(2*pi*t);yt2 = 4*cos(2*pi*2*t);yt3 = 4*cos(2*pi*3*t);

subplot(3,1,1);stem(n, y1); hold on;plot(t, yt1);subplot(3,1,2);stem(n/20, y2); hold on;plot(t, yt2);subplot(3,1,3);stem(n/30, y3); hold on;plot(t, yt3);

T s =110

T s =120

T s =130

ω = 2π⋅1

ω = 2π⋅2

ω = 2π⋅3


9/12/17

References

[1] http://en.wikipedia.org/[2] J.H. McClellan, et al., Signal Processing First, Pearson Prentice Hall, 2003[3] M.J. Roberts, Fundamentals of Signals and Systems[4] S.J. Orfanidis, Introduction to Signal Processing[5] K. Shin, et al., Fundamentals of Signal Processing for Sound and Vibration Engineerings

[6] A “graphical interpretation” of the DFT and FFT, by Steve Mann

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