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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 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.
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