Answer Test 1 UTHM

8/10/2019 Answer Test 1 UTHM

1/12

Answer Test 1 UTHM

1.

elsewhere

n

nn

nx n

;0

41;2

12;1

][ 1

i) Sketch and calculate Energy of )(nx

-3 -2 -1 0 1 2 3-5

-4

-3

-2

-1

0

1

2

3

4

5

n

x(

n)

J

nxnxExEnergy

n n

26

1644101

42210)1()()(,3

2

22222222


2/12

ii) Sketch and calculate Energy of )2()( nxny

422101)(nx

422101)2()(, nxnaLet

101224)()2()( nanxny

-2 -1 0 1 2 3 4 5-5

-4

-3

-2

-1

0

1

2

3

4

5

n

y(

n)=

x(2

-n

)


3/12

J

nynyEyEnergyn n

26

1014416

)1(01224)()(,4

1

22222222

iii) Sketch and calculate Energy of ]2[][][ nnxnz

000001)(

000001422101)2()(

nz

nnx

-3 -2 -1 0 1 2 3 4-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

n

z(

n)


4/12

J

nznzEzEnergyn n

1

000001

00000)1()()(,2

2

22222222

2. }6,2,3,5,2,4{)(

nx

a) }8,2,4,5{)(

nh

)()()( nhnxny

Using sum by column:

x(n) 4 2 -5 3 2 -6h(n) 5 4 -2 8

20 10 -25 15 10 -30

16 8 -20 12 8 -24

-8 -4 10 -6 -4 12

32 16 -40 24 16 -48

y(n) 20 26 -25 23 48 -68 -4 28 -48

b)

31

32

32

31 1)3/()( ntrinh

)()()( nhnxny

Using sum by column:


5/12

x(n) 4.00 2.00 -5.00 3.00 2.00 -6.00

h(n) 0.33 0.67 1.00 0.67 0.33

1.33 0.67 -1.67 1.00 0.67 -2.00

2.67 1.33 -3.33 2.00 1.33 -4.00

4.00 2.00 -5.00 3.00 2.00 -6.00

2.67 1.33 -3.33 2.00 1.33 -4.00

1.33 0.67 -1.67 1.00 0.67 -2

y(n) 1.33 3.33 3.67 2.33 0.33 -0.33 -1.67 -3.67 -3.33 -2.00

3. a)

)5.05.0()5.05.0()1(

)(

)5.0)(1(

)(

)5.0)(1(

)(

2

2

2

jz

B

jz

B

z

A

z

zX

zzz

z

z

zX

zzz

zzX

2

5.0)1()1(

1)5.0(

2

1

2

A

A

zz

zA

z

1

))(5.05.0(

5.05.0

)5.05.05.05.0)(15.05.0(5.05.0

)5.05.0)(1(5.05.0

B

jj

jB

jjjjB

jzz

zB

jz


6/12

)(4

cos5.02)()1(2)(

)5.05.0()5.05.0()1(

2)(

)5.05.0(

1

)5.05.0(

1

)1(

2)(

nun

nunx

jz

z

jz

z

z

zzX

jzjzzz

zX

nn

b) ]1[2

1][

1

nuny

n

)1(2

12

)1(2

1

2

1)(

1

nu

nuny

n

n

21

22)(

z

zzY

)()( nunx

1)(

z

zzX

21

21

2

21

21

2

22)1(2

1

2

)(

)()(

z

z

z

z

z

zz

z

z

zz

zX

zYzH

04)(2

12)1(

2

12)( nununh

nn


7/12

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

2(1/2)^n 4 2 1 0.5 0.25 0.125 0.0625 0.03125 0.015625 0.007813

u(n+1) 1 1 1 1 1 1 1 1 1 1

sig1 4 2 1 0.5 0.25 0.125 0.0625 0.03125 0.015625 0.007813

u(n) 0 1 1 1 1 1 1 1 1 1

sig2 0 2 1 0.5 0.25 0.125 0.0625 0.03125 0.015625 0.007813

sig 4 0 0 0 0 0 0 0 0 0

4. )30sin(5)20sin(3)2cos(2)( ttttx

S = 100;

01.0100/1 st

N = 10

n = [0 1 2 3 4 5 6 7 8 9]

)30sin(5)20sin(3)2cos(2)( sss ntntntnx

))01.0(30sin(5))01.0(20sin(3))01.0(2cos(2)( nnnnx

)3.0sin(5)2.0sin(3)02.0cos(2)( nnnnx

2)0sin(5)0sin(3)0cos(2)0( x

2857.0)3.0sin(5)2.0sin(3)02.0cos(2)1( x

1198.48558.55886.20351.39021.66394.62727.30821.02857.02)( nx


8/12

SKEMA JAWAPAN TEST 1

ADVANCED DIGITAL SIGNAL PROCESSING

MEE 10603

Q1.

1,5,9,1,3,11,5,2)(nx

i) 4,20,36,4,12,44,20,8)(4 nx (2)

ii)

)(

)(

)5.1()(

223

23

n

e

x

nx

nxny

(1)

2,5,11,3,1,9,5,1)2()(

2,2,5,5,11,11,3,3,1,1,9,9,5,5,1,1)3()(

1,1,5,5,9,9,1,1,3,311,11,5,5,2,2)3()(

1,1,5,5,9,9,1,1,3,3,11,11,5,5,2,2)()(,2

ncny

nbnc

nanb

xnaLet

e

n

(2)

1,2,3,6,1,6,3,2,1

5.0,5.2,5.4,5.0,5.1,5.5,5.2,11,5.2,5.5,5.1,5.0,5.4,5.2,5.0

1,5,9,1,3,11,5,25.02,5,11,3,1,9,5,15.0

)(5.0)(5.0)( nynyny eeee

(2)

iii) ))3()()2()(()( nununnxnz


9/12

n -3 -2 -1 0 1 2 3 4 5 6

d(n+2) 0 1 0 0 0 0 0 0 0 0

u(n) 0 0 0 1 1 1 1 1 1 1

-u(n-3) 0 0 0 0 0 0 -1 -1 -1 -1

+ 0 1 0 1 1 1 0 0 0 0x(n) 2 -5 11 -3 1 -9 -5 1 0 0

z(n) 0 -5 0 -3 1 -9 0 0 0 0

(4)

9,1,30,5)(nz (1)

Q2. i) )1(2)()1()2()3()( nunnunrnrnx

n -5 -4 -3 -2 -1 0 1 2 3 4 5 6

r(n+3) 0 0 0 1 2 3 4 5 6 7 8 9

-r(n+2) 0 0 0 0 -1 -2 -3 -4 -5 -6 -7 -8

u(n+1) 0 0 0 0 1 1 1 1 1 1 1 1

d(n) 0 0 0 0 0 1 0 0 0 0 0 0

-2u(n-1) 0 0 0 0 0 0 -2 -2 -2 -2 -2 -2

x(n) 0 0 0 1 2 3 0 0 0 0 0 0

(4)

3,2,1)(nx

ii) )2()1()()1()(1 nnunrnrnh

n -5 -4 -3 -2 -1 0 1 2 3 4 5 6

r(n+1) 0 0 0 0 0 1 2 3 4 5 6 7

-r(n) 0 0 0 0 0 0 -1 -2 -3 -4 -5 -6

-u(n-1) 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

d(n-2) 0 0 0 0 0 0 0 1 0 0 0 0

h1(n) 0 0 0 0 0 1 0 1 0 0 0 0

(4)

1,0,1)(1 nh

iii) )1(2)2()3()(2 nnununh


10/12

n -5 -4 -3 -2 -1 0 1 2 3 4 5 6

u(n+3) 0 0 1 1 1 1 1 1 1 1 1 1

-u(n+2) 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 -1

-2d(n-1) 0 0 0 0 0 0 -2 0 0 0 0 0

h2(n) 0 0 1 0 0 0 -2 0 0 0 0 0

(4)

2,0,0,0,1)(2 nh

)()()()()( 21 nhnxnhnxny

x(n) 1 2 3

h1(n) 1 0 1

1 2 3

0 0 01 2 3

x(n)*h1(n) 1 2 4 2 3

(2)

x(n) 1 2 3

h1(n) 1 0 0 0 -2

1 2 3

0 0 0

0 0 0

0 0 0

-2 -4 -6

x(n)*h2(n) 1 2 3 0 -2 -4 -6

(2)

3,8,8,4,1,3,2,1

6,4,2,0,3,2,13,2,4,2,1

)()()()()( 21 nhnxnhnxny

(2)

Q3 a)


11/12

)3()1(

)3)(1(

1

32

1)(

23

zC

zB

zA

zzz

zzzzX

(1)

3

1

)3)(1(

1

)3)(1(

1

0

z

zzA (2)

4

1

)4)(1(

1

)3(

1

1

zzz

B (2)

12

1

)4)(3(

1

)1(

1

3

z

zzC (2)

3

1

12

1

1

1

4

11

3

1)(

zzzzX

)1()3(12

1)1()1(

4

1)1(

3

1)( 11 nununnx nn (3)

b)

)()(

)2(3)()1(2)(

nunx

nxnxnxny

)(3

12

)(3

)()(2)(

2

2

zXz

z

zXz

zXzzXzY

(2)

2

312

)(

)()(

zz

zX

zYzH (2)


12/12

1)(

z

zzX (1)

1

3

11

2

1

32

312

1)()()(

23

23

2

zz

z

z

z

z

zz

zz

z

zzXzHzY

(2)

)1(3)1()2(2)( nunununy (2)

Q4a)

The Fourier transform of a rectangular pulse Fig. 3.9.a of duration T seconds is obtained as

)()sin(

.1)()(

2/2/

2/

2/

fTsincTfT

fTT

fj2

ee

dtedtetrfR

fTj2fTj2

T

T

ftj2ftj2

where sinc(x)= sin(x)/x.

b) Fig. shows the spectrum of the rectangular pulse. Note that in the frequency domain most

of the pulse energy is concentrated in the main lobe within a bandwidth of 2/Pulse.Duration

i.e. BW=2/T. However, there is pulse energy in the side lobes that may interfere with other

electronic devices operating at the side lobe frequencies.

The rectangular pulse is a particularly important signal in digital signal analysis and digital

communication. A rectangular pulse is inherent whenever a signal is segmented and processed

frame by frame, each frame is the result of multiplication of the signal and a rectangular

window. Furthermore, the spectrum of a rectangular pulse can be used to calculate the

bandwidth required by pulse radar systems or by digital communication systems that transmit

pulses.

b)

More than two samples is required per cycle of a sinewave and hence for a signal with a

bandwidth of B Hz the sampling rate need to be greater than 2B.

c)

Bit rate rb = 44100162=1411200 bps

Bandwidth =2 rb = 2822400 Hz

Documents

Answer Test 1 UTHM