DSP Problems_dsp1_co thuc_full, dsp1, co thuc, xu li tin hieu, bkdn, dsp

7/31/2019 DSP Problems_dsp1_co thuc_full, dsp1, co thuc, xu li tin hieu, bkdn, dsp

1/51

BI TP CHNG 1


2/51

Bi 1 - Tnh tn s ly mu

Cho mt tn hiu lin tc c ph t 120-160 kHz. Vph 2 pha ca tn hiu ri rc c c bng cch ly

mu tn hiu trn vi 3 tn s ly mu khc nhau sau

y :

fs = 80 kHz

fs = 100 kHz

fs = 120 kHzTn s ly mu thch hp l bao nhiu trong 3 tn s

trn? Gii thch


3/51

Bi 2 - Quan h tn s

Cho tn hiu tng t:

a) Xc nh tn s ly mu nh nht trnh chng ph

b) Gi s tn hiu trn c ly mu vi tn s fS = 200

Hz, tn hiu ri rc sau ly mu l g ?c) Gi s tn hiu trn c ly mu vi tn s fS = 75 Hz,

tn hiu ri rc sau ly mu l g ?

d) Xc nh tn s (0 < f < fS) ca tn hiu sin c cc mutrn vi cc mu ca tn hi u c

ax (t) 3cos100 t


4/51


5/51

Bi 4 - Quan h tn s

Cho tn hiu tng t :

a) Xc nh tn s Nyquist

b) Gi s tn hiu c ly mu vi tc 5000 (mu/s),

tm tn hiu ri rc c c sau ly mu

c) Xc nh tn hiu tng t ya(t) khi phc t tn hiu

ri rc (gi s ni suy l tng)

a

x (t) 3cos2000 t+5sin6000 t+10cos12000 t


6/51

Cu hiCu 1:Nu s khc nhau gia tn hiu tng t v

tn hiu s.

Cu 2:Tn hiu tng t c chuyn thnh s,

sau chuyn li thnh tng t (khng qua DSP).

Hi tn hiu tng t ra c khc tn hiu tng tvo hay khng? Gii thch.

Cu 3:Phn tch cc u khuyt im ca x l s

so vi x l tng t


7/51

BI TP CHNG 2


8/51

Bi 1 - Cc php ton trn tn hiuri rc

a) Cho

V th ca:

b) V th ca tn hiu: x[n] = u[ 3 - n ]

c) Cho x[n] = 2u[n+2].Tm v v z[n] = x[3-2n]

d) Cho y[n] = anu[n], a>1.Tm v v z[n] = y[-2n+2]

[ ] [ ]n

x n a u n 1a

[ ] [ 3]y n x n


9/51

Bi 2 - Cc php ton trn tn hiuri rc

Cho

a)V th tn hiu x[n]

b) V th tn hiu x[-n+4], x[-n-4],

c) Biu din x[n]theo tn hiu dirac v tn hiu bc nhy

elsewhere,0

3n0,1

1n3,3

n1

]n[x


10/51

Bi 3 - Tn hiu ri rctun hon

Cc tn hiu sau c tun hon khng? Nu

c, tnh chu k c bn

a)

b)

4[ ] cos(1 2 )x n n

35[ ]

nj

x n e


11/51

Bi 4Tnh nhn qu ca h ri rc

Xt tnh nhn qu ca cc h thng ri rc sau:

a)

b)

n

k

]k[x]n[y

]4n[x3]n[x]n[y


12/51

Bi 5Tnh n nh ca h ri rc

Xt tnh n nh ca cc h thng ri rc sau:

a)

b)

[ ] cos( [ ])y n x n

[ ] [ ]n

k

y n x k


13/51

Bi 6Tnh tuyn tnh bt bin ca h

Xt tnh tuyn tnh bt bin ca cc h thng ri rcsau:

[ ] cos( [ ])y n x n

[ ] [ ]n

k

y n x k

0

[ ] [ ]n

k

y n x k

[ ] [ ] [ ]y n x n u n


14/51

Bi 7Tnh p ng xung ca h LTI

Cho h LTI c quan h vo-ra sau:

y[n] 0.9y[n-1] = x[n] + 2x[n-1]+3x[n-2]

a) Tnh p ng xung bng phng php th

b) V s thc hin h trn

c) Xt tnh n nh ca h


15/51

Bi 8Tnh tng chp tuyn tnh

a) Xc nh p ng ca h LTI c p ng xung sau:

h[n] = an u[n] i vi tn hiu vo l:

x[n] = u[n] u[n-10]

Gi :S dng tnh cht tuyn tnh bt bin

.

[ ] [ ]x n u n

[ ] [ 2] 1nh n a u n a

2

[2 ] [ 3]

1 1

na a

u n u n

a a

b) Chng minh rng khi cho tn hiu

i qua h thng LTI c p ng xung l:

th tn hiu ra l:


16/51

Bi 9Tnh tng chp tuyn tnh

Cho h LTI c s nh hnh sau:

Xc nh h[n], cho bit:

]2n[]n[h

]n[u)1n(]n[h]n[h

2/1,4/1,2/1]n[h

4

32

1

h1[n]h2[n]

h3[n] h4[n]


17/51

Bi 10Xc nh quan h vo-ra

Cho h LTI c s nh sau:

a) Xc nh phng trnh vo-ra

Z-1

Z-1

23

4

b) H trn c n nh khng?


18/51

Bi 11 - Gii phng trnh sai phn

0n],n[y

]1n[x2]n[x]2n[y4]1n[y3]n[y

Tm

ca h sau:

vi x[n] = 4n u[n] v cc iu kin u bng 0


19/51

BI TP CHNG 3


20/51

Bi 1Tnh bin i Z thun

Tm bin i Z v min hi t ca cc tn hiu sau y:

0n0

0n2]n[h)d

0n)(

0n)(

]n[v)c

4n0

5n)(]n[y)b

4,1,6,0,0,0,0,3]n[x)a

nn

31

n21

n

31

n

2

1


21/51

Bi 2Tnh bin i Z thun

Tm bin i Z v min hi t ca cc tn hiu sau y davo cc tnh cht v bng bin i Z:

])10n[u]n[u(]n[h)d

]1n[un]n[v)c

]n[u2)1(]n[y)b

]n[u)n1(]n[x)a

n

21

1n

3

1

2

1

nn


22/51

Bi 3Tnh bin i Z ngc

a) Tm bin i Z ngc sau bng phng php

khai trin thnh chui ly tha:

1|z|zz21

z21)z(X

21

1

b) Tm cc bin i Z ngc sau bng phng php khai

trin ring phn:

2|z|)z5.01)(z2z21(

zz61

4

1)z(X)c

2|z|z2z31

z31)z(X)b

2|z|

)z1)(z21(

1)z(X)a

121

21

21

1

211


23/51

Bi 4Tnh bin i Z ngc

Tm cc tn hiu x[n] nhn qu nu X(z) nh sau :

21

21

2

2

1

76

2

211

z4z41

zz21)z(X)d

z1

z21

)z(X)c

z1zz)z(X)b

zz1

1)z(X)a


24/51

Bi 5im cc v im khng

a) Tm X(z) bit X(0) = 1v cc im cc khng nh sau:

b) Tm x[n] nhn qu t X(z) trn

x

x

x

-1/2 -1/4 1/2

2/1r


25/51

Bi 6ng dng ZT tnh tng chp

Tnh tng chp ca cc cp tn hiu sau y:

]1n[u2]n[h,]n[nu]n[x)d

]n[u).ncos(]n[h,]n[u)(]n[x)c

]n[u)(]n[]n[h,]n[u]n[x)b

]n[u])(1[]n[h,]1n[u]n[x)a

n

n

21

n

21

n

21n

41


26/51

Bi 7Tnh cht o hm

a) Tm bin i Z ca tn hiu nx[n] v n2x[n] theo X(z)

b) ng dng kt qu cu (a) tnh cc bin i Z sau:

]1n[una]n[x)c]n[nusinna]n[x)b

]1n[u))(nn(]n[x)a

n

0

n

1n

312

21


27/51

Bi 8Phn tch h thng

a) H thng nhn qu sau c n nh khng?

)z2.01)(z5.01)(z1(zz2z21)z(H 111

321

b) Tnh p ng xung v p ng bc ca h trn


28/51

Bi 9Phn tch h thng

Cho h thng sau:

2

25

21

5

3

2

211

zz1

zz)z(H

a) Tnh p ng xung ca h trn

b) Tnh p ng bc trng thi 0 v p ng bc viiu kin u l y(-1) = 1 v y(-2) = 2

c) V s thc hin h trn


29/51

Bi 10Tng hp h thng

Ta mun thit k mt h thng tha mn iu kin sau:

Khi cho x[n] = (0.5)nu[n] -0.25(0.5)n-1u[n-1]

i vo h thng th u ra thu c y[n] = (1/3)nu[n]

a) Tm h[n] ca h trn

b) Tm H(z) ca h trn

c) Xc nh phng trnh sai phn m t quan h vo-ra

d) Xy dng s thc hin h trn

e) H trn c n nh khng? V sao?


30/51

Bi 11Tng hp h thng

p ng bc ca mt h LTI l:

s[n] = (1/3)n-2 .u[n+2]

a) Tm H(z) ca h trn

b) Tm h(n) ca h trn

c) Biu din cc im cc khng trn mt phng z

d) H trn c n nh khng? C nhn qu khng?


31/51

Bi 12Gii phng trnh sai phn

Tm y[n]vi n>=0trong cc trng hp sau:

1]2[y;0]1[y;]n[u]n[x

;]n[x]2n[y]n[y)d

;1]1[y;]n[u]n[x

;]n[x]1n[y]n[y)c

0]2[y;1]1[y;0]2n[y5.0]1n[y5.1]n[y)b

1]2[y]1[y

;0]2n[y]1n[y]n[y)a

41

n

31

21

41

21


32/51

BI TP CHNG 4


33/51

Bi 1Bin i Fourier thun

Tm bin i DTFT ca cc tn hiu sau y:

]6n[u]n[u]n[w)e

]n[u)nsin(a]n[h)d]n[u2]n[v)c

4|n|04|n|n)(2]n[y)b

2,1,0,1,2]n[x)a

0

n

n

2

1


34/51

Bi 2Bin i Fourier thun

Tm bin i DTFT ca cc tn hiu sau y:

1,0,0,1,0,0,1,0,0,1,0,0,1]n[x)c1,0,1,0,1,0,1,0,1]n[x)b

1,1,1,1,1]n[x)a

3

2

1

Tm mi quan h gia cc DTFT trn.


35/51

Bi 3Tnh bin i Fourier ngc

Tnh tn hiu x[n], cho bit bin i DTFT nh hnh v sau:

2

1


36/51

Bi 4Cng thc DTFT thun & ngc

Cho tn hiu sau y:

1,2,3,2,1]n[x

Tnh cc i lng sau m khng cn tnh DTFT:

)(X)d(d)(X)c()(X)b()0(X)a(


37/51

Bi 5Tnh cht ca DTFT

Cho tn hiu x[n] c DTFT sau y:

jae1

1)(X

Tnh DTFT ca cc tn hiu sau y:

]2n[xe)d()n3.0cos(]n[x)c(]1n[x]n[x)b(]n2[x)a(

2/nj


38/51

Bi 6Ph bin v ph phaTm v v ph bin v ph pha ca cc tn hiu sau y:

]3n[]2n[3]1n[3]n[81

]n[x)g

]2n[]n[2

1]n[x)f

]2n[]n[2

1]n[x)e

]1n[]1n[2

1]n[h)d

]1n[]1n[21]n[x)c

]1n[]n[2

1]n[x)b

]1n[]n[21]n[x)a


39/51

Bi 7p ng trng thi bn & png nht thi

Xc nh p ng trng thi bn v p ng nht thi

ca h sau:

]2n[x]n[x21

]n[y

i vi tn hiu vo l:

n60n

2cos35]n[x 0


40/51

Bi 8p ng i vi tn hiu vodng hm m phc v sin/cos

]10n[x]n[x]n[y

Cho b lc FIR:

n2

n5

2cos510]n[x)2(

n10

n3

sin3n10

cos]n[x)1(

a) Tnh v v p ng bin v p ng pha

a) Tnh p ng ca h thng i vi cc tn hiu vo sau:


41/51

Bi 9B lc

Xc nh cc h s ca b lc FIR sau:

]2n[xb]1n[xb]n[xb]n[y210

sao cho tha mn cc iu kin sau:

1. H(0) = 1

2. B lc ny ngn hon ton thnh phn tn s 3/20


42/51

Bi 10B lcp ng tn s ca mt b lc thng di l tng nh sau:

||8

30

8

3||

8

1

8||0

)(H

a) Tnh p ng xung

b) Chng minh rng c th biu din p ng xung ny

di dng tch ca v p ng xung ca

mt b lc thng thp

)4/ncos(


43/51

BI TP CHNG 5


44/51

Bi 1 - Tnh DTFT ca tn hiu tun hon

0[ ] [ ] 2 [ 2]x n n n

0 ( )X ( )X

Cho tn hiu x[n] tun hon vi chu k l N = 3

v mt chu k l:

Tm

Kim tra kt qu bng cch tnh DTFT ngc khi phc

li x[n]

v


45/51


46/51

Bi 3 - Tnh DFT ca tn hiu di hu hn

Cho 5 mu u tin ca DFT 8 mu ca mt tn hiu thc l:

Xc nh 3 mu cn li

0,0518.0j125.0,0,3018.0j125.0,25.0


47/51

Bi 4 - Tnh DFT ca tn hiu di hu hn

Tnh DFT N mu ca cc tn hiu sau:

1Nn0oddn0

evenn1]n[x)c(

]}Nn[u]n[u{a]n[x)b(

]n[]n[x)a(n


48/51

Bi 5 - Tnh cht tng chp vng

Cho cc tn hiu sau v cc DFT 5 mu ca chng:

]n[]n[s)c(

]1n[]n[x)b(

]4n[4]3n[3]2n[2]1n[]n[x)a(

2

1

a) Tm y[n] sao cho Y[k] = X1[k].X2[k]

b) Dy x3[n] c tn ti khng, nu S[k] = X1[k].X3[k]


49/51

Bi 6 - Tnh cht dch vng

7n4,0

3n0,1]n[x

Cho DFT 8 mu ca tn hiu x[n] sau, l X[k]:

Tnh DFT ca cc tn hiu sau theo X[k]:

7n6,0

5n2,1

1n0,0

]n[x)b(

7n5,1

4n1,0

0n,1

]n[x)a(

2

1

Bi Th FFT 2 h


50/51

Bi 7 - Thut ton FFT c s 2 phnthi gian

n,0

3n0,2/1]n[x

Tnh DFT 8 mu ca tn hiu x[n] sau:

s dng thut ton FFT c s 2 phn thi gian.

Yu cu v lu v ghi y cc gi tr ca tt c cc nt

trong lu

Bi 8 Th t t FFT 2 h


51/51

Bi 8 - Thut ton FFT c s 2 phnthi gian

V lu thut ton FFT phn thi gian tnh ti ch vi

N = 16trong hai trng hp:

1/ Cc gi tr vo sp xp theo th t t nhin

2/ Cc gi tr vo sp xp theo trt t o bit

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