Digital Signal Processing, Department of Electrical and Information Technology LTH, Lund University

Digital Signal Processing in

multimedia

ETI265

2012

Problems and solutions

Problems from Digital Signal Processing: Principles,

Algorithms, and Applications. Problems and solutions

John G. Proakis, Dimitris G. Manolakis

Problems chapter 1

Problems chapter 2

Problems, chapter 3

Problems, chapter 4

Problems, chapter 5

Problems chapter 6

Extra problems A: A time continuous signal is given by iseconds)in(t)()( 10 tuetx t

a−=

a) Determine the Fourier transforms (F))aX and 2a |(F))| X

b) We want to sample the signal using the sampling frequency HzFs 100= . We first pre-filter the signal in an ideal low pass filter with cutoff frequency HzFF sc 502/ == . Determine how much of the signal energy that are cancelled by the pre-filter (anti-aliasing filter). c) The signal after the filter is sampled using the sampling frequency HzFs 100= . Determine the Fourier transform of the sampled signal. Determine also the Fourier transform of the sampled signal if no anti-aliasing filter is used. B: The system below is given. The input signal is )60002cos()10002cos()( tttx ππ += and the filter impulse response )8()()( −−= nununh . Sample rate is 8 kHz. Determine the output signal if the reconstruction is ideal.

C:

Problems, chapter 7

Problem chapter 9

Digital Signal Processing ETI265 2010, Solution to selected problems 1.2

a) 200,2001

201.0),

201.02cos()01.0cos( ==== pNfnn ππ

b) 7,71),

712cos()

10530cos( === pNfnn ππ

c) )alaising(,2,23),

232cos()3cos( === pNfnn ππ

d) cnonperiodifnn ,23),

232sin()3sin(

πππ ==

1.5

a, b) )612sin(3|)()( / ntxnx FsnTnta π== == 6,

612,

61

=== pNf πω

c) Np=6; d) Yes, Fs=200

2.1 a) ....},0,1,1,1,1,32,

31,0{...)(

↑=nx

b) folding ....},0,31,

32,1,1,1,1,0{...)(

↑=−nx

c) folding + delay ....},0,31,

32,1,1,1,1,0{...)4())4((

↑=+−=−− nxnx

d) delay + folding ....}0,0,31,

32,1,1,1,10{...)4())4(

↑=−−=−− nxnx

e) )4()()1(32)2(

31)( −−++++= nununnnx δδ

2.2 a) )2( −nx b) )4( nx − c) )2( nx + d) )2()( nunx − e) )3()1( −− nnx δ ...}05.05.01111{...)(

↑=nx

a) ...}05.05.011110{..)2(↑

=−nx

b) ...}011115.05.00{...)4(↑

=− nx

c) ...}05.05.011110{...)2(↑

=+nx

d) ...}01111{...)2()(↑

=− nunx

e) ...}010000{...)3()1(↑

=−− nnx δ

2.16 a) )()()()()()()( nxkhknxkhknxkhnynnnnknn∑∑∑∑∑∑∑ =−=−=

b) 1 }4,6,7,7,7,3,1{)(;}1,1,1,1,1{)(}4,2,1{)(↑↑↑

=== nynhnx

,7)(,5)(,35)( === ∑∑∑ nxnhnynnn

2 }1,4,2,4,1{)(;)()(}1,2,1{)( −==−=↑

nynxnhnx

,2)(,2)(,4)( === ∑∑∑ nxnhnynnn

3 }2,5.2,0,2,5.1,5.0,5.0,0{)(; −−−−=ny

11 )()25.05.02()( nuny nn −⋅=

,2)(,3/4)(3/8)( === ∑∑∑ nxnhnynnn

2.17 a) }1,3,6,10,14,18,15,11,6{)(;}1,1,1,1,1{)(}1,2,3,4,5,6{)(

↑↑↑=== nynxnh

b) }1,3,6,10,14,81,15,11,6{)(↑

=ny

c) }1,,2,2,2,1{)(↑

=ny d) }1,,2,2,2,1{)(↑

=ny

2.21 a)

⎪⎩

⎪⎨

⎧

=+

≠−−

=−

−==

==−=−=

++

−

+−

−

=

−

=

−

∑

∑∑∑

baifnb

baifabab

abab

abb

baknubkuaknxkhny

n

nnn

kn

k

n

knknn

k

knkn

kkk

)1(1

)(1)(

)()()()()(

11

1

11

1

0

0

b)

}1,2,3,3,0,0,1,1,1{)(

},1,1,0,0,1,1{)(}1,1,2,1{)(

↑

↑↑

−=

−==

ny

nhnx

c)

}1,2,2,1,5,8,9,8,6,3,1{)(

},1,2,3,2,1{)(}1,0,1,1,1,1,1{)(

−−−=

=−=↑↑

ny

nhnx

2.35 a) [ ])()()()()( 4321 nhnhnhnhnh ⋅−⋅=

b) [ ])3(5.2)2(2)1(25.1)(5,0)( −+−+−+= nunnnnh δδδ c) }4,0,3,0,0,1{)( −=

↑nx ...}0,2,5,5.7,25.6,2,25.1,5.0{)(

↑=ny

2.62 a) }1,3,5,7,5,3,1{)(

↑=nryy

b) }1,3,5,7,5,3,1{)(↑

=nryy , same as in a)

3.1 a) 215 463)()( −−− −++==∑ zzzznxzX n

n

b)

21||

21132

1211

1)21)

21()

21()

21()

21(

)21()

21()()(

1

5

1

511

0

515151

5

1

55

>−

=

=−

==⋅=

====

−

−

−

−−∞

=

−−−−∞

=

−∞

=

−∞

=

−

∑∑

∑∑∑

zROCz

z

zzzzzz

zzznxzX

n

n

n

n

n

n

nn

n

n

n

3.2 a) ?)1()( =+=∑ znzXn

From 2.21a )()1()()()( nuannuanuanx nnn +=∗= If a=1: )()1()()()( nunnununx +=∗= Then: use convolution corresponds to multiplication in the z-plane

2111 )1(1

11

11)()()( −−− −

=−

⋅−

=⋅=zzz

zUzUzX

And 2

2

21 )1()1(1)(

−=

−= − z

zz

zX

00:

10)1(:

2,12

2,12

==>=

==>=−

zZZeros

pzPoles

f) See slides

h) ))10()21()

21()(()

21())10()(()

21()( 1010 −−=−−= − nununununx nnn

)21(

)21(

211

)21(1

211

1)21(

211

1)(

9

1010

1

1010

1

1010

1

−

−=

=−

−=

−−

−=

−

−

−

−

−

zz

z

z

z

zz

zzX

poles: 21;0;0

21;0 109,...1

9 ===−= ppzz

zeros: 9,...,2,1,0;)21(;0)

21( 210101010 =⋅==− kezz kj π

3.8 Given: a) )()( kxnyn

k∑−∞=

= b) )()1()( nunnx +=

Task: Determine a) Y(z), b) X(z)

Slution: a) )()()()()()( nunxknukxkxnyk

n

k∗=−== ∑∑

∞

−∞=−∞=

)(1

1)()()( 1 zXz

zUzXzY −−==

b) )()1(1)()()()(0

nunknukununun

kk+==−=∗ ∑∑

=

21)1(1)()()()1()()( −−

=<==>+=∗z

zUzUnunnunu

21)1(1)()()1()( −−

=<==>+=z

zXnunnx

3.14 c) )7()6()(;11

)( 1

7

1

6

−+−=−

+−

= −

−

−

−

nununxz

zz

zzX

d)

21

11

2

11

2

2

2

2

)2/cos(21)2/sin(

1

11

1

11

1

11

21)( −−

−−

−

−−

−

−

−

−

+⋅−+=

++=

++=

++

=zz

zzz

zzz

zzzzX

ππ

)1()1(2/sin()()( −−+= nunnnx πδ 3.16a Task: Determine )()()( 11 nxnxny ∗=

a)

1111

111

1

5.012/1

13/1

25.013/1(

...)5.01

11

1(25.01

25.0)(

−−−−

−−−

−

−+

−+

−−

==−

+−−

=

zzzz

zzzzzY

)1(5.05.0)1(3/1)1(25.03/1)( 11 −⋅+−+−−= −− nunununy nn

c)

)()1(3/2)(5.03/1)(1

3/25.013/1

11

5.01211

5.011)( 1111

1

21

1

1

nununyzzzz

zzz

zz

zY

nn −+⋅=

++

−=

+−=

+++

−= −−−−

−

−−

−

−

3.35a Given: ,0,)3/cos()2/1()()3/1()( ≥== nnnxnh nn π Task: Determine y(n) Solution:

} } }

21

128/37/6

1

7/1

21

1

1

21

1

1

41

211)3/1(1

41

211

)4/11)3/1(1

1

41)3/cos(

2121

)3/cos(2/11)3/1(1

1)(

−−

−

−−−

−

−

−−

−

−

+−

++

−=

+−

−−

=+−

−−

=

zz

zCBz

A

zz

zz

zz

zz

zYπ

π

0)3/sin()21(

733)3/cos()

21(

76)

31(

71)(

41)3/cos(

2121

)3/sin(2/1)3/sin(2/1)8/1)3/cos(2/1()3/cos(2/11

)3/1(11

71

41

211

8/1176

)3/1(11

71)(

21

111

121

1

1

≥++=

+−

++−+

+−

=+−

++

−=

−−

−−−

−−−

−

−

nnnny

zz

zzz

zzz

zz

zY

nnn ππ

π

ππππ

3.40 Given: 11

1

1 3/111)(,

5.0125.0

5.011)( −−

−

− −=

−−

−=

zzY

zz

zzX

Task: Determine a) )()()(

zXzYzH = ,b) Diff. eq, c) Draw fig. d) Stable?

Solution: 11

11

1

3/112

25.013

3/111

25.011

5.011

)( −−

−−

−

−−

−=

−−

−=zz

zz

zzY

a) 0)3/1(2)4/1(3)( ≥−= nnh nn b) )1(5.0)()2(12/1)1(12/7)( −−=−+−− nxnxnynyny c) See Formula table. d) Stable because poles inside unit circle

3.49c: Given: 1)1(),()3/1()(),()1(5.0)( =−=+−= ynunxnxnyny n Task Determine y(n), Use One side z-transform Solution: )3/11/(1)(),()(5.0)( 11 −+−+ −=+= zzXzXzYzzY

)2/11(3

)3/11(2

5.015.0

)2/11()3/11(1

5.015.0)(

111

111

−−−

−−−+

−+

−−

−=

=−−

+−

=

zzz

zzzzY

)()3/1(2)()5.0(5.3)( nununy nn −=

4.8 Show that ⎩⎨⎧ ±±=

=∑−

= elsewhereNNkifN

e NknjN

n 0...2,,021

0

π

Solution: ⎩⎨⎧ ±±=

=−

−∑−

= elsewhereNNkifN

e

eeNknj

NNknj

NknjN

n 0...2,,0

1

12

221

0 π

ππ

4.9 a) }111111{)6()()(↑

=−−= nununx

2/565

0 )2/sin()3sin(

11)( ω

ω

ωω

ωωω j

j

jnj

ne

eeeX −

−

−−

=

=−−

==∑

b) ωω jn

eXnunx

5.011)(),(2)(

−=−=

c) ω

ωωj

jnn

eeXnununx

−

+−

−=+=+=

411

1256)();4()41()

41()4()

41()( 444

d) ωω

ω

ωωωω 20

00 )cos(21

)sin()(;1||),()sin()( jj

jn

eeaeaXanunanx −−

−

+−=<=

g) )2sin(4)sin(2)(},2,1,0,1,2{)( ωωω jjXnx −−=−−=

4.10 a) n

nnxπωδ )sin()()( 0−=

b) }25.0,5.0,25.0{)(↑

=nx

4.12 c) nn

nWnx cωπcos)2/sin(4)( ⋅=

4.14 a) 1)0( −=X b) negativeandrealXX )(,0)}0(arg{ ωπ+=

c) πωωπ

π

6)( −=∫−

dX

d) 9)( −=πX

e) formulaParsevalsnxndX ,38)(|)(| 22 πωωπ

π

== ∑∫−

5.2ab a) 2/)1(

10 )2/sin(

)2

1sin(

11)()( Mj

j

Mjnj

rect

M

nrect e

M

eeenwW ω

ω

ωω

ω

ωω −

−

+−−

==

+

=−

−==∑ 321

b) 2/2))2/sin(

)4

sin(()( Mj

trianel e

M

W ω

ω

ωω −=

5.17 a) )2()1(cos2)()( 0 −+−−= nxnxnxny ω

b) }1,cos2,1{)( 0ω−=nh

ωωω ωωωω jjj eeeH −−− −=+−= )cos2cos2(cos21)( 02

0 c) )3/6/3/cos(3)( πππ −+= nny

5.26 { ...}0002/1000...{)(↑

=ny

5.35 )22(4

125.01

)4

3cos(21)( 21

21

+=

+−

+−= −−

−−

Gainzz

zzGainzH

π

5.39 a

aaarzH dB 241cos();(

2

31+−−

=ω )1

2cos();( 232 +=

aaarzH dBω

5.52 )12.02cos(22

1)12.02cos(21)( 21

ππ

−=+−= −− GainzzGainzHFIR

)12.02cos(22

)12.02cos(21)12.02cos(21)12.02cos(21)(

2

221

21

ππ

ππ

−−+

=+−+−

= −−

−− rrGainzrzr

zzGainzHIIR

5.61 a) )1(1

)( 1 abza

bzH −=−

= −

b) )2

14cos();(2

3 aaaarzH dB

−−=ω

7.1

7.8

7.9

7.11

9.3

1

1

5.018)( −

−

−+

=z

zzH ))1(5.03)(8()1(5.0)(5.08)( 11 −⋅+=−−⋅= −− nunnununh nnn δ

9.4

1

1

1

1

2/1121

3/1135)( −

−

−

−

−+

++

+=z

zz

zzH

)1()21(2)()

21()1()

31(3)(5)( 11 −++−−+= −− nunununnh nnnδ

9.9 a)

44444 344444 214444 34444 21444 3444 21

parallellcascadeIIdirectform

zzzzz

zzzzH

)4/11(3/7

)2/11(3/10

)4/11()2/11(3/11

8/14/313/11)( 1111

1

21

1

−−−−

−

−−

−

−−

−=

−−⋅+

=+−⋅+

=

Figures, see the textbook