Procesamiento de Señales Digitales 1ra Edicion Thomas J. Cavicchi

7/21/2019 Procesamiento de Seales Digitales 1ra Edicion Thomas J. Cavicchi

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE

MUCHOS DE ESTOS LIBROSGRATIS EN

DESCARGA DIRECTA

SIGUENOS EN:

VISITANOS PARA DESARGALOS GRATIS.
https://plus.google.com/b/113394888343830071226/113394888343830071226https://www.facebook.com/pages/Solucionarios-de-Libros/345772498866324https://twitter.com/Libros_y_Soluhttp://librosysolucionarios.net/


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Digital Signal Processing

Solutions Manual

Thomas J. Cavicchi

Grove City College

John Wiley & Sons, Inc.

2000

http://librosysolucionarios.net


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Digital Signal ProcessingSolutions Manual

Thomas J. Cavicchi

Table of Contents

Begins on pageChapter 2......................................................1

Chapter 3....................................................50

Chapter 4..................................................120

Chapter 5..................................................220

Chapter 6..................................................283

Chapter 7..................................................408

Chapter 8..................................................494

Chapter 9..................................................582

Chapter 10................................................658

Note from author to instructors:I willplace corrections and/or improvements to these solutions on my Web page for this book.

To view these, you will need the followinglogin andpassword:

password: kh7yb2q5login: faculty

Ifyou have corrections, comments, or suggestions, please email me at the address also shown on the

main webpage for this book.



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{



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0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time t in sec.

xc(t)

Continuous-time input xc(t).



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0 5 10 15-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Time (secs)

Amplitude

Simulated output yc(t) for input xc(t).

0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time t in sec.

xc(t)

Continuous-time input xc(t-t0), where t0 = 5.



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0 5 10 15-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Time (secs)

Amplitude

Simulated output yc(t) for input xc(t-t0) where t0 = 5

0 5 10 15-3000

-2500

-2000

-1500

-1000

-500

0

500

1000

Time (secs)

Amplitude

Simulated output yc(t) for inputs xc(t) and xc(t-t0) where t0 = 5.



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0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

Time t (s)

yc(t)andestimates

* stems = lsim result; dt = 0.2s, Nintmax = 10Solid = exact sol., O stems = rectang. rule est.;



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-7 -6 -5 -4 -3 -2 -1 0-8

-7

-6

-5

-4

-3

-2

-1

0

Time t (s)

yc(t)

Result of continuous-time convolution.



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.



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0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

Time t (s)

xc(t)

Square-wave generated using "rem" function of Matlab.



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0 20 40 60 80 100-2

-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

y(n)

O stems = result using conv routine; * stems = closed-form expression; a = 0.4.

0 20 40 60 80 100-8

-6

-4

-2

0

2

4

6

8

Time index n

y(n)

O stems = result using conv routine; * stems = closed-form expression; a = 0.95.



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0 10 20 30 40 500

0.2

0.4

0.6

0.8

1

Time index n

h(n)

Stairs = dimpulse result, O stems = analytical solution result.Unit sample response of system y(n) - 0.2y(n-1) - 0.63y(n-2) = x(n).



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0 10 20 30 40 50-0.5

0

0.5

1

Time index n

y(n)

* stems = result of filter w/conv to neg z powers.O stems = res. of dlsim; + = res. of dimpulse;

0 10 20 30 40 50-0.5

0

0.5

1

Time index n

y(n)

O stems = result of filter; * stems = result of dimpulse = result of dlsim.



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0 5 10 15 20 25 30 35 400

0.01

0.02

0.03

0.04

0.05

0.06

No. of Samples

Amplitude

Unit sample response: stairs = dstep, stems = calc using text solution equation.

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

No. of Samples

Amplitude

Unit step response: stairs = dstep, stems = calc using text solution equation.



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0 5 10 15 20 25 30 35 400

0.01

0.02

0.03

0.04

0.05

0.06

No. of Samples

A

mplitude

* stems = eigenvalue methodh(n): stairs = dimpulse, O stems = calc using state-space soln. eq.;



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0 10 20 30 40 50-1

-0.5

0

0.5

1

1.5

2

Time index n

y(n)

O stems = result of dlsim; * stems = sum of modes.



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0 0.005 0.01 0.015 0.02 0.025 0.03 0.035-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

t, sec.

* = direct calculation from state-space eq. soln. in text.

solid = v output RL,cO = discretized

RL with dt = 0.002 sec

x = solution points using dlsim;



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-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Axis

ImagAxis

The sixth roots of z = -3 + j4.



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-2-1

01

2 -2

-1

0

1

2

-2

0

2

4

6

x = Re{z}

jy = jIm{z}

|ln(z)|


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-2 -10

12 -2

-1

0

1

2

-8

-6

-4

-2

0

2

4

x = Re{z}

jy = jIm{z}

Im{ln(z)}


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-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

x = Re{z}

y=

Im{z}

Solid = Re{w(z)}, --- = Im{w(z)} where w(z) = x^2+jy^2 where z=x+jy.

0.214

0.107

0.5360.429

0.2140.322



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-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

x = Re{z}

y=Im

{z}

Solid = Re{w(z)}, --- = Im{w(z)} where w(z) = z^2.

-0.536

-0.321

0.321

0.107

-1.07

-0.643

1.070.643



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-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

|z|^2 + 2ang(z) - jang(z)cos(4|z|) (analytic now

Solid = Re{w(z)}, --- = Im{w(z)} where



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-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

|z|^2 + 2ang(z) - jang(z)cos(4|z|) (analytic now

Solid = |w(z)|, --- = angle{w(z)} whereSolid = |w(z)|, --- = angle{w(z)} whereSolid = |w(z)|, --- = angle{w(z)} where



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0 5 10 15 20 25 30-80

-70

-60

-50

-40

-30

-20

-10

0

10* stems = result from "filter" w/conv to neg z powers.

Stairs = dlsim result; O stems = direct calc;



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-2-1

01

2 -2

0

2-2

0

2

4

6

x = Re{z}jy = jIm{z}

|X(z)| where X(z) = ZT{[(1-exp(-sig n)/n]u(n)}; sig = 0.8.


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0 0.5 1 1.5 2 2.5 3 3.50.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

Angle of z in radians

|X(z)|anditsestimate

Solid = exact |X(z)|; ---- = finite-sum estimate; |z| = 2.

0 0.5 1 1.5 2 2.5 3 3.50

1

2

3

4

5

6

7

8



Solid = exact |X(z)|; ---- = finite-sum estimate; |z| = 1.



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0 0.5 1 1.5 2 2.5 3 3.5

0.5

1

1.5

2

x 1044 ---- = finite-sum estimate; |z| = 0.7.





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0 5 10 15 20 25 30-3

-2

-1

0

1

2

3

4

5

6

Time index n

y(n)

O stems = analytical solution; * stems = result using dlsim.



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0 20 40 60 80 100-100

-50

0

50

100

150

Time index n

y(n)

O stems = analytical solution; * stems = result using dlsim.



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ZIR ZSR



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0 5 10 15 20 25 300

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Time index n

y(n)

O stems = analytical result; stairs = dlsim result.



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0 1 2 3 4 5 6 7-5

-4

-3

-2

-1

0

Time index n

h(n)

Lines = result from dimpulse; stems = results from analytical calculation.Unit sample response: num chosen for freqz.

0 1 2 3 4 5 6 7-5

-4

-3

-2

-1

0

Time index n

h(n)

Lines = result from dimpulse; stems = results from analytical calculation.Unit sample response: num chosen for dimpulse.



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0 5 10 15 20 25 30 35 400

1

2

3

4

5

6

Time index n

y(n)

O stems = analytical solution; * stems = dlsim with correct initial conditions.



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0 2 4 6 8 10-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

Time t (sec).

gc(t)

Solid = analytical result; ---- = result of impulse{num,den,t}.



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0 0.5 1 1.5 2 2.5 3 3.50

0.5

1

1.5

2

2.5

3

3.5

4

Frequency w (rad/s)

|Hc(jw)|

Magnitude of frequency response of given system



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0 1 2 3 4 5 6-2

0

2

4

6

8

10

12

14

Time t (sec)

xc(t)

xc(t) formed from Fourier series coefficients X(k) = alf^|k|, |k|


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0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

Time t (sec)

xc(t)

xc(t) formed from Fourier series coefficients X(k) = alf |k|, |k|


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-1 -0.5 0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (s)

xc(t)

Solid = triangle wave FS reconstr.; --- = exact triangle; Nk = 50



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-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Time t (s).

xc(t)

Reconstructed xc(t) using FS (cosine plus shifted triangle wave).



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0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (s)

xc(t)andFSestimate

One period of half-wave-rectified sine wave; solid = 50-term FS, --- = exact xc(t)



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4.12 65.6



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101

102

103

104

105

106

-30

-60

-90

0

Frequency (rad/sec)

Phasede

g

101

102

103

104

105

106

-20

0

20

40

Frequency (rad/sec)

GaindB

Hc(s) = 4.136e+005(s+8000)(s+3400)/[(s+5e+004)(s+1.5e+004)(s+300)



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101

102

103

104

105

106

-90

-180

0

90

Frequency (rad/sec

Phased

eg

101

102

103

104

105

106

-40

-20

0

20

Frequency (rad/sec

GaindB

Hc(s) = 6e+009s(s+1000)/[(s+500)(s+300)(s^2+1.6e+004s+6.464e+009)]



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0 50 100 150 200-4

-3

-2

-1

0

1

2

3

4

Time index n

Inputsignalx(n)

Input signal = 1.5cos(2 pi 0.02n) + 2cos(2 pi 0.2n+0.2pi) - sin(2 pi 0.4n).

-100 -50 0 50 100-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4Truncated digital high-pass filter unit sample response; M = 100

Time index n

h(n)



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0 50 100 150 200-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

Desiredoutputsignal

Desired output signal: -sin(2 pi 0.4n).

-100 0 100 200 300 400-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

Outputoffilter,y(n)

O stems = output of truncated filter; * stems = ideal output; M = 100.



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0 20 40 60 80 100-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

Outputoffilter,y(n)

O stems = output of truncated filter; * stems = ideal output; M = 3.

0 20 40 60 80 100-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

Outputoffilter,y(n)

O stems = output of truncated filter; * stems = ideal output; M = 100



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-2-1

01

2 -2

0

2-100

-50

0

50

100

150

200

x = Re{z}jy = jIm{z}

|X(z)| where X(z) = ZT{[ln(a)^n/n!]u(n)}; a = 0.05.


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0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12

14

16

18

20

Digital frequency f (cycles/sample)

|X(exp(j2pif)

|X(exp(j2pif)| where x(n) = [ln(a)^n]/n!, where a = 0.05.



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0 5 10 15 200

0.5

1

1.5

2

2.5

3

Time index n

x(n)

x(n) = [ln(a)^n]/n!, where a = 10.

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

7

8

9

10

Digital frequency f (cycles/sample)

|X(exp(j2pif)|

|X(exp(j2pif)| where x(n) = [ln(a)^n]/n!, where a = 10



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-60 -40 -20 0 20 40 600

0.05

0.1

0.15

0.2

0.25

0.3x(n) = DTFT-1{triangular spectrum}

Time index n

x(n)



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-0.5 0 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Digital frequency f

|X(exp(j2pif))|

Solid = exact X(exp(j2pif)); ----- = DTFT{truncated x(n)}



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-60 -40 -20 0 20 40 60-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

Time index n

y(n)

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

-0.5 0 0.50

0.05

0.1

0.15

0.2

0.25

Digital frequency f

|Y(exp(j2pif))|

Solid = exact Y(exp(j2pif)); ----- = DTFT{truncated y(n)}.



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3000 3050 3100 3150 3200

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

y(n)

O stems = exact y(n); * stems = convolution of causal sinusoid with h(n).



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400 420 440 460 480 500-30

-20

-10

0

10

20

30

Time index n

y(n)

O stems = analytical result; * stems = truncated "conv" result



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-3 -2 -1 0 1 2 3-1

0

1

Re{z}

jIm{z}

Pole-zero diagram of H(z) = -0.0625(z-2.646)(z+2.646)/[(z-0.5)(z-0.25)].

0 0.1 0.2 0.3 0.4 0.5-400

-300

-200

-100

0

Normalized frequency (Nyquist == 0.5)

Phase

(degrees)

0 0.1 0.2 0.3 0.4 0.5-15

-10

-5

0

Normalized frequency (Nyquist == 0.5)MagnitudeR

esponse(dB) H(z) = -0.0625(z-2.646)(z+2.646)/[(z-0.5)(z-0.25)].



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0 2 4 6 8 10-0.1

0

0.1

0.2

0.3

0.4

0.5

No. of Samples

Amplitude

Unit sample response. Lines = dimpulse; stems = analytical solution.



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0 0.1 0.2 0.3 0.4 0.5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Digital frequency f

angleofX(exp(j2pif))(radians)

Angle of DTFT of damped impulse train; a = 0.98, N = 20.

0 0.1 0.2 0.3 0.4 0.50.5

1

1.5

2

2.5

3

3.5

Digital frequency f

|X(exp(j2pif))|

Mag. of DTFT of damped impulse train; a = 0.98, N = 20.



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0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1L = 10 rep s of finite-length geom. seq. alf n for n on 0, N-1; alf = 0.9, N = 10.

Time index n

Lrepetitionsofx(n)



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0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1estimate using ZT-1 contour integration (* stems); |zc| = 1.5.

time index n

Comparison between x(n) (O stems) and

0 5 10 15 20 25 30-0.2

0

0.2

0.4

0.6

0.8

1

estimate using ZT-1 contour integration (* stems); |zc| = 0.9.

time index n

Comparison between x(n) (O stems) and



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188

0 5 10 15 20-1

-0.8-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

x(n)

x(n) = cos(2pi f0 n) where f0 = 0.5 cycle/sample.

0 5 10 15 200

2

4

6

8

10

12

14

16

18

20

DFT index k

|X(k)|

20-point DFT of cos(2pi f0 n) where f0 = 0.5 cycle/sample.



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0 5 10 15 20 25 30 35 40-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

x(n)


0 5 10 15 20 25 30 35 400

5

10

15

20

25

DFT index k

|X(k)|




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0 20 40 60 80 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

x(n)


0 20 40 60 80 1000

5

10

15

20

25

30

35

40

45

50

DFT index k

|X(k)|




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194



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195



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196



200/725

197

0 20 40 60 80 100 120 140-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Time index n

Re{hH(n)}

* stems = closed-form result.O stems = Re{ifft calc. of Hilbert trans. unit sample resp.};



201/725

198

0 20 40 60 80 100 120 140-16

-14

-12

-10

-8

-6

-4

-2

0

2x 10

-3

Time index n

Imag{hH(n)}

* stems = closed-form result.O stems = Im{ifft calc. of Hilbert trans. unit sample resp.};

0 20 40 60 80 100 120 140-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

cos(w1n)

x(n) = cos(w1n) where w1 = 2piM/N with M = 5 and N = 128.



202/725

199

0 20 40 60 80 100 120 140-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

Hilbert-filteredcos(w1n)andsin(w1n)

O stems = [y(n) = cos(w1n)*hH(n)]; w1 = 0.1pi, N = 128; O stems = sin(w1n).

0 20 40 60 80 100 120 140-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time index n

cos(w1n)

x(n) = cos(w1n) where w1 = 0.1pi with N = 128.



203/725

200

0 20 40 60 80 100 120 140-1

-0.5

0

0.5

1

Time index n

Hilbert-filteredcos(w1n)andsin(w1n)

M = 5, N = 128; O stems = sin(w1n).O stems = [y(n) = cos(w1n)*hH(n)]; w1 = 2piM/N,



204/725

201



205/725

202



206/725

203



207/725

204

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

vCc(t) for vBc(t)=uc(t)-uc(t-T) in RC circuit with tau=RC= 0.1s,T = 0.5s, and vCc(0) = 0V.

Time t (s)

vCc(t)(V)



208/725

205

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1


Time t (s)

vCc(t)(V)

0 0.5 1 1.5 20

0.5

1

1.5

2


Time t (s)

vCc(t)(V)



209/725

206

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (s)

vC

c(t)andvC(n)

Solid = vCc(t), stems = vC(n); T = 0.5 s, tau = 0.1 s, and dt = 0.005 s.

0 0.5 1 1.5 20

0.05

0.1

0.15

0.2

0.25T = 0.5s, and vCc(0) = 0V.

Time t (s)

vCc(t)(V)

vCc(t) for vBc(t)=uc(t)-uc(t-T) in RC circuit with tau=RC= 2s,



210/725

207

0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (s)

vCc(t)andvC(n)


0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time t (s)

vCc(t)andvC(n)




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208



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209



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210



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211



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212



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213



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214



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215



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216



220/725

217

+ +

+



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218



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219



223/725

220

0 5 10 15 200

0.5

1

Time index n

x(n-2)

x(n-n0) and phase{DTFT{x(n-n0)}} for n0 = 2

-3 -2 -1 0 1 2 3-500

0

500

Digital frequency (rad/sample)

-a],

dyc(t)/dt# = -ayc(n)t) + bxc(n)t) = -ay(n) + bx(n). Also, #t=n)t

dyc(t)/dt# = -ayc((n-1))t) + bxc((n-1))t) = -ay(n-1) + bx(n-1). Also, #t=(n-1))t

Thus,y(n) = ()t/2)[-ay(n)+bx(n)-ay(n-1)+bx(n-1)] + y(n-1), or

Y(z){1 + a)t/2(1+z-1) - z-1} = (b)t/2)(1+z-1)X(z), or

(b)t/2)(1+z-1)H(z) = 1-z-1+ (a)t/2)(1+z-1)

)t b = 2 1 - z-1

a)t/2 +

1 + z-1 334http://librosysolucionarios.net


338/725

335



339/725

336



340/725

337

0 2 4 6 8 10-0.5

0

0.5

1

1.5

2

Time t (s)

Simulation using predictor-corrector; dt = 0.1 s

---*--- = predictor-corrector estimate of yc(t)

solid = exact yc(t)



341/725

338

Time t (s)



solid = exact yc(t)

0 2 4 6 8 10-0.5

0

0.5

1

1.5

2



solid = exact yc(t)

0 2 4 6 8 10-1

-0.5

0

0.5

1

1.5

2

2.5

Time t (s)



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339



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340



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341



345/725

342



346/725

343



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344



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345



349/725

346



350/725

347



351/725

348



352/725

349



353/725

350



354/725

351



355/725

352



356/725

353



357/725

354



358/725

355



359/725

356



360/725

357



361/725

358

Clock

tsim

To Workspace1

Sine Wave

cos1

To Workspace2



362/725

359



363/725

360



364/725

361



365/725

362



366/725

363



367/725

364



368/725

365



369/725

366



370/725

367



371/725

368



372/725

369



373/725

370



374/725

371



375/725

372



376/725

373



377/725

374



378/725

375



379/725

376



380/725

370



381/725

378



382/725

379



383/725

380



384/725

381



385/725

382



386/725

383



387/725

384



388/725

385



389/725

386



390/725

387

Frequency (Hz)



391/725

388



392/725

389



393/725

390



394/725

391



395/725

392



396/725

393



397/725

394



398/725

395



399/725

396



400/725

397



401/725

398

+

+

+

2

4

6



402/725

399

+

+

+

2

4

6

2



403/725

400

2

+

+

+

2



404/725

401



405/725

402



406/725

403



407/725

404



408/725

405



409/725

406



410/725

407



411/725

408

0 10 20 30 40 50 60 700

5

10

15

20

25

30

35

k

|X(k)|

N = 64, L = 10, theta = 0.DFT mag of sampled sinusoid sin(w1n) where w1N=2piL+theta



412/725

409



413/725

410

0 10 20 30 40 50 60 700

5

10

15

20

25

30

35

k

|X(k)|

N = 64, L = 10, theta = 0.7854DFT mag of sampled sinusoid sin(w1n) where w1N=2piL+theta



414/725

411



415/725

412

0 0.1 0.2 0.3 0.4 0.50

0.5

1

1.5

2

2.5

3

3.5

4

Digital frequency f

|X(exp(j2pif))|

Mag of freq resp of h(n) = u(n)-u(n-4); curves=DTFT calc; * = DFT calc



416/725

413

0 5 10 15 20 25 30 350

100

200

300

400

500

DFT index k

|X(k)|

x(n) = n for 0


417/725

414

0 0.1 0.2 0.3 0.4 0.50

100

200

300

400

500

Digital frequency f

|X(exp(j2pif)

stems = |X(k)|. x(n) = n for 0


418/725

415



419/725

416

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

7

8

Digital frequency f

DFTF def, by analyt calc., by zero-padding, & by interp{DFT}; N = 16, M = 1024x(n) = triang. Stems = |X(k)|; other (identical) curves are: |X(exp(j2pif))| viax(n) = triang. Stems = |X(k)|; other (identical) curves are: |X(exp(j2pif))| viax(n) = triang. Stems = |X(k)|; other (identical) curves are: |X(exp(j2pif))| viax(n) = triang. Stems = |X(k)|; other (identical) curves are: |X(exp(j2pif))| viax(n) = triang. Stems = |X(k)|; other (identical) curves are: |X(exp(j2pif))| via



421/725

418

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

Digital frequency f

|XL

(k)|/Land|X(exp(j2pif))|

Stems = |XL(k)| via L-point FFT of L-repetitions-repeated h(n).Solid = |HL(exp(j2pif))| using property, --- = using direct sum calc of DTFT{hL(n)}Solid = |HL(exp(j2pif))| using property, --- = using direct sum calc of DTFT{hL(n)}Solid = |HL(exp(j2pif))| using property, --- = using direct sum calc of DTFT{hL(n)}



422/725

419



423/725

420



424/725

421



427/725

424

0 5 10 15 20 25 30-0.1

0

0.1

0.2

0.3

0.4

0.5

Time index n

h(n)

Digital FIR filter h(n).

0 10 20 30 40 50 60-1.5

-1

-0.5

0

0.5

1

1.5

2

2.53

3.5

Time index n

y(n)

O stems: y(n) = conv(x,h); * stems: y(n)=DFT-1{XH} with N = 30.



428/725

425

0 10 20 30 40 50 60-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

Time index n

y(n)

O stems: y(n) = conv(x,h); * stems: y(n)=DFT-1{XH} with N = 59



429/725

426

0 10 20 30 40 50 60 70-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

Time index n

x(n)

x(n) = cos(5pi n/(3N)) - 2sin(5pi n/(2N) + pi/4); N = 64

0 10 20 30 40 50 60 70-20

-15

-10

-5

0

5

10

Time index n

h(n)

h(n) = bandpass filter with cutoffs at f = 0.22 and 0.45.



430/725

427

0 0.1 0.2 0.3 0.4 0.50

10

20

30

40

50

60

Digital frequency f

|H(exp(j*2pif))|

|H(exp(j*2pif))|.

0 10 20 30 40 50 60 70-10

-8

-6

-4

-2

0

2

4

6

8

Time index n

ype

r(n)=N-periodicconvolution.

DFT-1{X(k)H(k)} = N-periodic convolution.



431/725

428

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

30

35

40

45

50

Digital frequency f (corresponding to DFT index)

|X(k)|

|X(k)| with zero-padding

0 0.1 0.2 0.3 0.4 0.50

10

20

30

40

50

60


|H(k)|

|H(k)| with zero-padding.



432/725

429

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

30

35

40

45

50


|Y(k)|

|Y(k)| where y(n) = x(n)*h(n); aperiodic convolution resul

0 20 40 60 80 100 120 140-15

-10

-5

0

5

10

Time index n

y(n)

* stems = zpadded DFT result.

O stems = y(n) = x(n)*h(n); aperiodic conv using conv(x,h)



433/725

430

0 10 20 30 40 50 60 70-10

-8

-6

-4

-2

0

2

4

6

8

Time index n

yN(n)

O stems = DFT-1{X(k)H(k)}; * stems = N-aliased y(n).



434/725

431



435/725

432

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time index n

x(n)andaliasedx(n)

O stems = original x(n); * stems = aliased x(n) (by DFT and formula); N = 10

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

Time index n

x(n)andaliasedx(n)

O stems = original x(n); * stems = aliased x(n) (by DFT and formula); N = 30.



436/725

433

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1

1.2

Time index n

x(n)andalias

edx(n)

O stems = original x(n); * stems = aliased x(n) (by DFT and formula); N = 10



437/725

434

0 10 20 30 40 50 60 700

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time index n

x(n)andDFT-1{MsamplesofDTFT}

N = 64, M = 32O stems = x(n); * stems = samples of DFT-1{M samples of XN(exp(jw))



438/725

435

0 100 200 300 400 500 600-40

-20

0

20

40

60

80

Time index n

x(n)

Sample of speech data.



439/725

436

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

Digital frequency f

|H(exp(j*2*pi*f))|

Bandpass filter magnitude frequency response

0 0.1 0.2 0.3 0.4 0.50

1000

2000

3000

4000

5000

6000

7000

Digi tal frequency f

|X(exp(j*2*pi*f))|(DFT

magnitude)

Magnitude of DFT{x(n)}.



440/725

437

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

30

35

40

45

50

Digital frequency f

|Y(exp(j*2*pi*f))|

Filtered data magnitude spectrum.

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

30

35

40

45

50

Digital frequency f

|Shifted-spectrumY|

Magnitude of frequency-shifted, filtered data spectrum.



441/725

438

0 0.1 0.2 0.3 0.4 0.50

5

10

15

20

25

30

35

40

45

50

Digital frequency f

|Y(exp(j*2*pi*f))|

Cleaned-up frequency-shifted, bandpass-filtered data magnitude spectrum.

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

Digital frequency f

|H(exp(j*2*pi*f))|

Lowpass filter magnitude frequency response



442/725

439



443/725

440

0 20 40 60 80 100-2

0

2

4

6

8

10

Time t in sec.

xc(ndt)

Sampled continuous-time function; dt = 1 s

0 20 40 60 80 1000

5

10

15

20

25

30

35

40

45

DFT index k

|X(k)|

DFT of N=100-point sample of continuous-time function



444/725

441

0 200 400 600 800 10000

50

100

150

200

250

300

350

400

450

DFT index k

|Xzp(k)|

Zero-padded DFT array for time-domain interpolation.



445/725

442



446/725

443

0 200 400 600 800 1000-2

0

2

4

6

8

10

Time index n

xe(n)

xe(n) = zero-interlaced version of x(n) used for time-domain interp.; L = 10.

0 200 400 600 800 10000

5

10

15

20

25

30

35

40

45

DFT index k

|Xe(k)|

M-point DFT of xe(n)



447/725

444

0 200 400 600 800 10000

50

100

150

200

250

300

350

400

450

DFT index k

|Xi(k)|

Xi(k) = lowpass version of M-point DFT of xe(n)



448/725

445

0 20 40 60 80 100-2

0

2

4

6

8

10

Time t in sec.

xc(ndt)

O stems = exact samples every dt = 1 s; * stems = downsample{upsample{x(ndt)}}.



449/725

446

0 5 10 15 20 25 30 35-3

-2

-1

0

1

2

3

Time index n

x(n)

O stems = ifft result; * stems = closed-form expression for x(n).



450/725

447

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6-4

-3

-2

-1

0

1

2

3

4

Time t (sec.)

xc(t)an

ditssamples

Solid curve = xc(t); stems = x(n) = xc(ndt).



451/725

448

0 20 40 60 80 100 120 140-3

-2

-1

0

1

2

3

Time index n

xe(n)

xe(n) = zero-interlaced version of x(n); L = 4.

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12

14

16

18

20

Digital frequency f (= w/(2pi))

|Xe(exp(jw)|

|Xe(exp(jw))| = |DTFT{xe(n)}|; L = 4.



452/725

449

0 20 40 60 80 100 120 1400

0.5

1

1.5

2

2.5

3

3.5

4

4.5

DFT index k

H(k)

H(k) = L times (0.5/L)-LPF for interpolation; L = 4

0 0.1 0.2 0.3 0.4 0.50

10

20

30

40

50

60

70

80

Digital frequency f (= w/(2pi)

|Xi(exp(jw)|

|Xi(exp(jw))| = L times (0.5/L)-LPF version of |Xe(exp(jw))|; L = 4.



453/725

450

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6-4

-3

-2

-1

0

1

2

3

4

Time t (sec.)

xc(t)anditssamples

Solid curve = xc(t); O stems = x(n) = xc(ndt); * stems = upsampled xi(n); L = 4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6-4

-3

-2

-1

0

1

2

3

4

Time t (sec.)

xc(t)anditssamples

M = 3, L = 4.Solid = xc(t); stems = M-downsampled after L-upsampled ~= xc(n(M/L)dt);



454/725

451



455/725

452



456/725

453

0 5 10 15 20 25 30 35 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time index n

h(n)andh`(n)

O stems = h(n); * stems = h`(n).



457/725

454



458/725

455



459/725

456

0 10 20 30 40 50-0.1

0

0.1

0.2

0.3

0.4

0.5

Time index n

h(

n)andh(n-n0)

* stems = h(n-n0) where n0 = 8; O stems = h(n).



460/725

457

0 20 40 60 80 10-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Lag index k

rhh1(k)andrhh(k)

O stems = rhh1(k) where n0 = 8; * stems = rhh(k).



461/725

458



462/725

459

0 20 40 60 80 100 120 140-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Time index n

x(n)andx(n-n0)

O stems = x(n); * stems = x1(n) = x(n-n0); n0 = 10



463/725

460

-150 -100 -50 0 50 100 150-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Lag index k

rxx1(k)

O stems = result using time-domain calc; * stems = DFT method.rxx1(k) where x1(n) is n0-delayed version of x(n); n0 = 10.



464/725


465/725

462

0 20 40 60 80 100 120 140-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Time index n

x1(n)andnoise-corrupte

dx1`(n)

O stems = x1(n); * stems = noise-corrupted x1`(n); SNR = 0 dB.

-150 -100 -50 0 50 100 150-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Lag index k

rxx1`(k)

of x(n); n0 = 10; SNR = 0 dB.rxx1 (k) where x1 (n) is n0-delayed and noise-corrupted version



466/725

463

0 20 40 60 80 100 120 140-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Time index n

x1(n)andnoise/conv-corruptedx1``(n)

Filter is highpass with cutoff = 0.02O stems = x1(n); * stems = noise/conv-corrupted x1``(n); SNR = 10 dB



467/725

464

-150 -100 -50 0 50 100 150-0.2

-0.1

0

0.1

0.2

0.3

0.4

Lag index k

rxx1``(k)

n0 = 10; SNR = 10 dB. Filter is highpass with cutoff = 0.02rxx1``(k) where x1``(n) is n0-delayed and noise/conv-corrupted version of x(n);

0 20 40 60 80 100 120 140-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Time index n

x1(n)andnoise/conv-corruptedx1``(n)

Filter is highpass with cutoff = 0.1O stems = x1(n); * stems = noise/conv-corrupted x1``(n); SNR = 10 dB



468/725465

-150 -100 -50 0 50 100 150-0.01

-0.005

0

0.005

0.01

0.015

Lag index k

rxx1``(k)

n0 = 10; SNR = 10 dB. Filter is highpass with cutoff = 0.1.rxx1``(k) where x1``(n) is n0-delayed and noise/conv-corrupted version of x(n);



469/725466



470/725467



471/725468



472/725469

0 50 100 150 200-4

-3

-2

-1

0

1

2

3

4

Time index n

y(n)

O stems = exact aperiodic conv y(n) = x(n)*h(n); * stems = overlap-add result.



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7/21/2019 Procesamiento de

Procesamiento de Señales Digitales 1ra Edicion Thomas J. Cavicchi

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