GATE BY RK Kanodia signals and systems

8/22/2019 GATE BY RK Kanodia signals and systems

1/41

30. y t u t h t( ) ( ) * ( )= , where h te t

e t

t

t( )

,

,=

-

2

3

0

0

(A)1

21

5

6

1

3

2 3e u t e u tt t- -- - + - -( ) ( )

(B)1

21

5

6

1

3

2 3e u t e u tt t( ) ( )- - + - --

(C)1

2

1

65 3 22 2 3e e e u tt t t+ - - -[ ] ( )

(D)1

2

1

65 3 22 2 3e e e u tt t t+ - - --[ ] ( )

Statement for Q.31-34:

The impulse response of LTI system is given.

Determine the step response.

31. h t e t( ) | |= -

(A) 2 + - -e et t (B) e u t et t( )- + + - -1 2

(C) e u t e u tt t( ) [ ] ( )- + + - -1 2 (D) e e e u tt t t+ - --[ ] ( )2

32. h t t( ) ( )( )= d 2

(A) 1 (B) u t( )

(C) d( )( )3 t (D) d( )t

33. h t u t u t( ) ( ) ( )= - - 4

(A) tu t t u t( ) ( ) ( )+ - -1 4 (B) tu t t u t( ) ( ) ( )+ - -1 4

(C) 1 + t (D) ( ) ( )1 + t u t

34. h t y t( ) ( )=

(A) u t( ) (B) t

(C) 1 (D) tu t( )


The system described by the differential equations

has been specified with initial condition. Determine the

output of the system and choose correct option.

35.

dy t

dx y t x t( )

( ) ( )+ =10 2 , y x t u t( ) , ( ) ( )0 1- = =

(A) 15

101 4( ) ( )+ -e u tt (B) 15

101 4( )+ -e t

(C) - + -15

101 4( ) ( )e u tt (D) - + -15

101 4( )e t

36.d y t

dt

dy t

dty t

dx t

dt

2

25 4

( ) ( )( )

( )+ + = ,

y ( )0 0- = ,dy t

dt

( )

0

1-

= , x t t u t( ) sin ( )=

(A)5

34

3

34

1

6

13

61

4sin cost t e et t+ + -- - , t 0

(B)5

34

3

34

13

51

1

6

4sin cost t e et t+ - +- - , t 0

(C)3

34

5

34

13

51

1

6

4sin cost t e et t+ - +- - , t 0

(D)3

34

5

34

1

6

13

51

4 4sin cost t e et t+ + -- - , t 0

37.d y t

dt

dy t

dty t x t

2

26 8 2

( ) ( )( ) ( )+ + = ,

y ( ) ,0 1- = -dy t

dt

( )

0

1-

= , x t e u tt( ) ( )= -

(A)2

3

5

2

5

6

2 4e e et t t- - -- + , t 0

(B)2

3

5

2

5

6

2 4+ +- -e et t , t 0

(C) 4 5 3 2 4+ +- -( )e et t , t 0

(D) 4 5 3 2 4- +- -( )e et t , t 0

38.d y t

dty t

dx t

dt

2

2

3( )( )

( )+ = ,

y ( ) ,0 1- = -dy t

dt

( )

0

1-

= , x t te u tt( ) ( )= -2

(A) sin cost t te tt+ - +-4 3 3 , t 0

(B) 4 3sin cost t te t- - - , t 0

(C) sin cost t te tt

- + +-

4 33

, t 0(D) 4 3sin cost t te t+ - - , t 0

39. The raised cosine pulse x t( ) is defined as

x tt t

( )(cos ) ,

,

=+ -

1

21

0

wp

w

p

wotherwise

The total energy of x t( ) is

(A)3

4

p

w(B)

3

8

p

w

(C) 3pw

(D) 32

pw

40. The sinusoidal signal x t t( ) cos ( )= +4 200 6p is

passed through a square law device defined by the

input output relation y t x t( ) ( )= 2 . The DC component in

the signal is

(A) 3.46 (B) 4

(C) 2.83 (D) 8

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2/41

41. The impulse response of a system is h t t( ) ( . )= -d 0 5 .

If two such systems are cascaded, the impulse response

of the overall system will be

(A) 0.58( . )t - 0 25 (B) d( . )t - 0 25

(C) d( )t - 1 (D) 0 5 1. ( )d t -

42. Fig. P5.1.40 show the input x t( ) to a LTI system and

impulse response h t( ) of the system.

The output of the system is zero every where

except for the

(A) 0 5<


3/41

SOLUTIONS

1. (A)2

60p

pT

= T =p

30

2. (C) T12

5=

ps, T2

2

7=

ps, LCM

2

5

2

72

p pp,

=

3. (D) Not periodic because of t.

4. (D) Not periodic because least common multiple is

infinite.

5. (C) y t( ) is not periodic although sin t and 6 2cos pt are

independently periodic. The fundamental frequency

cant be determined.

6. (C) This is energy signal because

| |E x t dt-

= < ( ) = --

e u t dtt4 ( ) = =-

e dtt40

1

4

7. (A) | |x t( ) = 1, | |E x t dt-

= = ( )2

So this is a power signal not a energy.

| |P T x t dtTT

T

-

= =lim ( )12 12

8. (D) v t( ) is sum of 3 unit step signal starting from, 1, 2,

and 3, all signal ends at 4.

9. (A) The function 1 does not describe the given pulse.

It can be shown as follows :

10. (B)

11. (C)

12. (D) Multiplication by 5 will bring contraction on

time scale. It may be checked by x x( . ) ( )5 0 8 4 = .

13. (A) Division by 5 will bring expansion on time scale.

It may be checked by y t x x( ) ( )=

=

20

54 .

14. (C) y t

t

t( )

,

,

,

=

- < < -

- < 3 3 0or , y n nk

n[ ] = = +

=-

- 1 13

3,

y n n u n[ ] ( ) [ ]= + 1

31. (A) For n n- <


10/41

(Time variant)

At any discrete time, n no= the response depends only

on the excitation at that same time. (Causal)

If the excitation is a constant, the response is

unbounded as n approaches infinity. (Unstable)

34. (C) y n v m y n kv mm

n

m

n

1

1

2

1

[ ] [ ] , [ ] [ ]= ==-

+

=-

+

y n ky n2 1[ ] [ ]= (Homogeneous)

y n v mn

n

1

1

[ ] [ ],==-

+

y n w mn

n

2

1

[ ] [ ]==-

+

y n v n w mm

n

3

1

[ ] ( [ ] [ ])= +=-

+

= + = +=-

+

=-

+

v m w n y n y nm

n

m

n

[ ] [ ] [ ] [ ]1 1

1 2 (Additive)

Since the system is homogeneous and additive it is also

linear

y n v n y n v m nm

n

om

n

1

1

2

1

[ ] [ ] , [ ] [ ]= = -=-

+

=-

+

y n n v m v q n y nom

n n

oq

no

1

1 1

2[ ] [ ] [ ] [ ]- = = - ==-

- +

=-

+

(Time Invariant)

At any discrete time, n no= , the response depends on

the excitation at the next discrete time in future.

(Anti causal)

If the excitation is a constant, the response increases

without bound. (Unstable)

35. (A) y n v n y kv n k v n1 2[ ] [ ] , [ ] [ ]= = =

ky n k v n y n1 2[ ] [ ] [ ]= (Not Homogeneous Not linear)

y n v n y n v n no1 2[ ] [ ] , [ ] [ ]= = -

y n n v n n y no o1 2[ ] [ ] [ ]- = - = (Time Invariant)

At any discrete time n no= , the response depends only

on the excitation at that time (Causal)

If the excitation is bounded, the response is bounded.

(Stable).

36. (B) y n x n[ ] [ ]= 2 2

Let x n v n1[ ] [ ]= then y n v n122[ ] [ ]=

Let x n kv n2[ ] [ ]= then y n k v n22 22[ ] [ ]=

ky n y n[ ] [ ] 2 (Not homogeneous Not linear)

Let x n v n1[ ] [ ]= then y n v n122[ ] [ ]=

Let x n v n no2[ ] [ ]= - then y n v n no222[ ] [ ]= -

y n n v n n y no o1 22[ ] [ ] [ ]- = - = (Time invariant)

At any discrete time, n no= , the response depends only

on the excitation at that time. (Causal)


(Stable).

37. (B) y n v n1 10 5[ ] [ ]= - , y n kv n2 10 5[ ] [ ]= -

y n ky n2 1[ ] [ ] (Not Homogeneous so not linear)

y n v n y n v n no1 2

10 5 10 5[ ] [ ] , [ ] [ ]= - = - -

y n n v n n y no o1 210 5[ ] [ ] , [ ]- = - - = (Time Invariant)

At any discrete time, n no= the response depends only

on the excitation at that discrete time and not on any

future excitation. (Causal)


(Stable).

38. (B) y n x n y n[ ] [ ] [ ]= + - 1 , y n x n y n[ ] [ ] [ ]- = - + -1 1 2

y n x n x n y n[ ] [ ] [ ] [ ]= + - + -1 2 , Then by induction

y n x n x n x n k[ ] [ ] [ ] [ ]= - + - + - +1 2K K

= -=

x n kk [ ]0Let m n k= - then y n x m x m

m n m

n

[ ] [ ] [ ]= ==

-

=-

y n v m y n kv m ky nm

n

m

n

1 2 1[ ] [ ] , [ ] [ ] [ ]= = == - = -

(Homogeneous)

( )y n v m w m v m w mm

n

m m

n

3[ ] [ ] [ ] [ ] [ ]= + = += - = -

=-

= +y n y n1 2[ ] [ ] (Additive)

System is Linear.

y n v m y v n nm

om

n

1 2[ ] [ ] , [ ]= = -=-

=- y n1 [ ] can be written as

y n n v m v q n y nom

n n

oq

no

1 2[ ] [ ] [ ] [ ]- = = - ==-

-

=-

(Time Invariant)

At any discrete time n no= the response depends only

on the excitation at that discrete time and previous

discrete time. (Causal)

If the excitation is constant, the response increase

without bound. (Unstable)

39. (C) Only statement (b) is false. For example

S1 : y n x n b[ ] [ ]= + , and S2 : y n x n b[ ] [ ]= - , where b 0

S x n S S x n S x n b x n{ } { { }} { }[ ] [ ] [ ] [ ]= = + =2 1 2Hence S is linear.

40. (B) For example

S1 : y n nx n[ ] [ ]= and S2 : y n nx n[ ] [ ]= + 1

If x n n[ ] [ ]= d then S S n S2 1 2 0 0{ { }}d[ ] [ ]= = ,

Chap 5.2Discrete-Time Systems

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11/41


Determine the Laplace transform of given signal.

1. x t u t( ) ( )= - 2

(A)- -e

s

s2

(B)e

s

s-2

(C)e

s

s-

+

2

1(D) 0

2. x t u t( ) ( )= + 2

(A)1

s(B) -

1

s

(C)e

s

s-2

(D)- -e

s

s2

3. x t e u tt( ) ( )= +-2 1

(A)1

2s +(B)

e

s

s-

+ 2

(C)e

s

s- +

+

( )2

2(D)

-

+

-e

s

s

2

4. x t e u tt( ) ( )= - +2 2

(A)e

s

s2 2 1

2

( )- -

-(B)

e

s

s-

+

2

2

(C)1

2

2 2-

-

- -e

s

s( )

(D)e

s

s-

-

2

2

5. x t t( ) sin= 5

(A)5

52s +(B)

s

s2 5+

(C)5

252s +(D)

s

s2 25+

6. x t u t u t( ) ( ) ( )= - - 2

(A)e

s

s- -2 1(B)

1 2- -e

s

s

(C)2

s(D)

-2

s

7. x td

dtte u tt( ) { ( )}= -

(A)1

1 2s s( )+(B)

s

s( )+ 1 2

(C)e

s

s-

+ 1

(D)e

s

s-

+( )12

8. x t tu t t u t( ) ( ) * cos ( )= 2p

(A)1

42 2s s( )+ p(B)

2

42 2 2p

ps s( )+

(C)1

42 2 2s s( )+ p(D)

s

s

3

2 24+ p

9. x t t u t( ) ( )= 3

(A)

34s (B)

-34s

(C)6

4s(D) -

64s

10. x t u t e u tt( ) ( ) * ( )= - --1 12

(A)e

s

s- +

+

2 1

2 1

( )

(B)e

s

s- +

+

2 1

1

( )

(C)e

s

s- +

+

( )2

2(D)

e

s

s- +

+

2 1

2

( )

CHAPTER

5.3

THE LAPLACE TRANSFORM

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11. x t e dt

( ) cos= -3

0

2t t t

(A)- +

+ +

( )

(( ) )

s

s s

3

3 42(B)

( )

(( ) )

s

s s

+

+ +

3

3 42

(C)s s

s

( )

( )

+

+ +

3

3 42(D)

- +

+ +

s s

s

( )

( )

3

3 42

12. x t td

dte t u tt( ) { cos ( )}= -

(A)- + +

+ +

( )

( )

s s

s s

2

2 2

4 2

2 2(B)

( )

( )

s s

s s

2

2 2

4 2

2 2

+ +

+ +

(C)( )

( )

s s

s s

2

2 2

2 2

4 2

+ +

+ +(D)

- + +

+ +

( )

( )

s s

s s

2

2 2

2 2

4 2

Statement for Q.1324:

Determine the time signal x t( ) corresponding to

given X s( ) and choose correct option.

13. X ss

s s( ) =

+

+ +

3

3 22

(A) ( ) ( )2 2e e u tt t- -+ (B) ( ) ( )2 2e e u tt t- --

(C) ( ) ( )2 2e e u tt t- -- (D) ( ) ( )2 2e e u tt t- -+

14. X ss s

s s( ) =

+ +

+ +

2 10 11

5 6

2

2

(A) 2 3 2d( ) ( ) ( )t e e u tt t+ -- -

(B) 2 2 3d( ) ( ) ( )t e e u tt t+ -- -

(C) 2 2 3d( ) ( ) ( )t e e u tt t+ +- -

(D) 2 2 3d( ) ( ) ( )t e e u tt t- +- -

15. X ss

s s( ) =

-

+ +

2 1

2 12

(A) ( ) ( )3 2e te u tt t- --

(B) ( ) ( )3 2e te u tt t- -+

(C) ( ) ( )2 3e te u tt t- --

(D) ( ) ( )2 3e te u tt t- -+

16. X ss

s s s( ) =

+

+ +

5 4

3 23 2

(A) ( ) ( )2 3 2+ +- -e e u tt t

(B) ( ) ( )2 3 2+ -- -e e u tt t

(C) ( ) ( )3 3 2+ -- -e e u tt t

(D) ( ) ( )3 3 2+ +- -e e u tt t

17. X ss

s s s( )

( )( )=

-

+ + +

2

2

3

2 2 1

(A) ( ) ( )e te u tt t- --2 2 (B) ( ) ( )e te u tt t- -+2 2

(C) ( ) ( )e te u tt t- -- 2 2 (D) ( ) ( )e te u tt t- -+ 2 2

18. X s ss s( ) =

+

+ +

3 2

2 102

(A) 3 31

33e t e t u tt t- --

cos sin ( )

(B) 3 31

33e t e t u tt t- --

sin cos ( )

(C) ( cos sin ) ( )3 3 3e t e t u tt t- --

(D) ( sin cos ) ( )3 3 3 3e t e t u tt t- -+

19. X ss s

s s s( )

( )( )=

+ +

+ + +

4 8 10

2 2 5

2

2

(A) ( sin cos ) ( )2 2 2 2 22e e t e t u tt t t- - -+ -

(B) ( cos sin ) ( )2 2 2 2 22e e t e t u tt t t- - -+ -

(C) ( cos sin ) ( )2 2 2 22e e t e t u tt t t- - -+ -

(D) ( sin cos ) ( )2 2 2 22e e t e t u tt t t- - -+ -

20. X ss s

s s s( )

( )( )=

+ +

+ + +

3 10 10

2 6 10

2

2

(A) ( cos sin ) ( )e e t e t u tt t t- - -+ +2 3 32 2

(B) ( cos sin ) ( )e e t e t u tt t t- - -+ -2 3 32 6

(C) ( cos sin ) ( )e e t e t u t

t t t- - -+ -2 3 3

2 2(D) ( cos sin ) ( )9 6 32 3 3e e t e t u tt t t- - -- +

21. X ss s e

s s

s

( )( )

=+ + +

+ +

-2 11 16

5 6

2 2

2

(A) 2 3 2 22 3d( ) ( ) ( )t e e u tt t+ - -- -

(B) 2 2 2 3 2 2 3 2d( ) ( ) ( )( ) ( )t e e e e u tt t t t+ - + +- - - - - -

(C) 2 2 22 3 2 3d( ) ( ) ( ) ( ) ( )t e e u t e e u tt t t t+ - + - -- - - -

(D) 2 2 22 3 2 2 3 2d( ) ( ) ( ) ( ) ( )( ) ( )t e e u t e e u tt t t t+ - + - -- - - - - -

22. X s s

d

ds s s( ) = +

+ +

2

2 2

1

9

1

3

(A) et

tt

t u tt- + +

322

33

93sin cos ( )

(B) ( sin cos ) ( )e t t t t u tt- + +3 22 3 3

(C) et

t t t u tt- + +

3 22

33 3sin cos ( )

(D) ( sin cos ) ( )e t t t t u tt- + +3 2 3 2 3

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13/41

23. X ss

( )( )

=+ +

1

2 1 42

(A) e t u tt-0 5. sin ( ) (B)1

2e t u ttsin ( )

(C)1

4

0 5e t u tt- . sin ( ) (D) e t u ttsin ( )

24. X s ed

ds s

s( )( )

=+

-2

2

1

1

(A) - --te u tt ( )1 (B) - --te u tt ( )1

(C) - - -- -( ) ( )( )t e u tt2 22 2 (D) te u tt- -( )1


Given the transform pair below. Determine the

time signal y t( ) and choose correct option.

cos ( ) ( )2t u t X s

L .

25. Y s s X s( ) ( ) ( )= + 1

(A) [cos sin ] ( )2 2 2t t u t- (B) cossin

( )22

2t

tu t+

(C) [cos sin ] ( )2 2 2t t u t+ (D) cossin

( )22

2t

tu t-

26. Y s X s( ) ( )= 3

(A) cos ( )2

3

t u t

(B)1

3

2

3

cos ( )t u t

(C) cos ( )6t u t (D)1

36cos ( )t u t

27. Y s X s( ) ( )= + 2

(A) cos ( ) ( )2 2t u t- (B) e t u tt2 2cos ( )

(C) cos ( ) ( )2 2t u t+ (D) e t u tt-2 2cos ( )

28. Y sX s

s( )

( )=

2

(A) 4 2cos ( )t u t (B)

1 2

4

- cos

( )

t

u t

(C) t t u t2 2cos ( ) (D)cos

( )2

2

t

tu t

29. Y sd

dse X ss( ) [ ( )]= -3

(A) t t u tcos ( ) ( )2 3 3- - (B) t t u tcos ( ) ( )2 3-

(C) - - -t t u tcos ( ) ( )2 3 3 (D) - -t t u tcos ( ) ( )2 3


Given the transform pair

x t u ts

s

L( ) ( ) +

2

22.

Determine the Laplace transform Y s( ) of the given

time signal in question and choose correct option.

30. y t x t( ) ( )= - 2

(A)2

2

2

2

se

s

s-

+(B)

2

2

2

2

se

s

s

+

(C)2 2

2 12( )

( )

s

s

-

- +(D)

2 2

2 12( )

( )

s

s

+

+ +

31. y t x tdx t

dt( ) ( ) *

( )=

(A) 42

3

2 2s

s( )+(B) 4

22 2( )s +

(C)-

+

4

2

3

2 2

s

s( )(D)

4

22 2( )s +

32. y t e x tt( ) ( )= -

(A)2 1

1 22( )

( )

s

s

+

+ +(B)

2 1

2 22( )s

s s

+

+ +

(C)2 1

2 42( )s

s s

+

+ +(D)

2 1

22( )s

s s

+

+

33. y t tx t( ) ( )= 2

(A)8 4

2

2

2 2

-

+

s

s( )(B)

4 8

2

2

2 2

s

s

-

+( )

(C)4

1

2

2

s

s +(D)

s

s

2

2 1+


Determine the bilateral laplace transform and

choose correct option.

34. x t e u tt( ) ( )= +- 2

(A)e

s

s2 1

1

( )+

+, Re ( )s > -1

(B)1

1 + s, Re ( )s < - 1

(C)e

s

s2 1

1

( )+

+, Re ( )s < - 1

(D)1

1 + s, Re ( )s > -1

Chap 5.3The Laplace Transform

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35. x t u t( ) ( )= - + 3

(A)1 3- -e

s

s

, Re ( )s > 0

(B)- -e

s

s3

, Re ( )s < 0

(C) 1

3- -

es

s

, Re ( )s < 0

(D)- -e

s

s3

, Re ( )s > 0

36. y t t( ) ( )= +d 1

(A) es, Re( )s > 0 (B) es, Re ( )s < 0

(C) es, all s (D) None of above

37. x t t u t( ) sin ( )=

(A)1

1 2( )+ s , Re ( )s < 0

(B)1

1 2( )+ s, Re ( )s > 0

(C)-

+

1

1 2( )s, Re ( )s < 0

(D)-

+

1

1 2( )s, Re ( )s > 0

38. x t e u t e u t e u tt

t t( ) ( ) ( ) ( )= + + --

-2

(A) 6 2 22 1 1

2

2s s

s s+ -

+ -( )( ), Re ( ) .s < - 0 5

(B)6 2 2

2 1 1

2

2

s s

s s

+ -

+ -( )( ), - >1 Re ( )s > 1

(C)1

0 5

1

1

1

1s s s++

++

-., -1 < Re ( )s < 1

(D)1

0 5

1

1

1

1s s s++

+-

-., - 0

(B)-

+ +

s

s s( )( )1 92, - 1

(B)e

s

s2 1

21 4

( )

( )

-

- +, Re ( )s < 1

(C)e

s

s( )

( )

-

- +

2

21 4, Re ( )s > 1

(D)e

s

s( )

( )

-

- +

2

21 4, Re ( )s < 1

43. x t ed

dte u tt t( ) [ ( )]= --2

(A)1

1

-

+

s

s, Re ( )s < - 1

(B)1

1

-

+

s

s, Re ( )s > -1

(C)s

s

-

+

1

1, Re ( )s < - 1

(D)s

s

-

+

1

1, Re ( )s > -1


Determine the corresponding time signal for given

bilateral Laplace transform.

44. X se

s

s

( ) =+

5

2with ROC: Re ( )s < -2

(A) e u tt- + +2 5 5( ) ( )

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15/41

(A)1

2e t u ttsin ( ) (B) 2e t u tt- cos ( )

(C) 21

2e t u t e t u tt t- -+cos ( ) sin ( )

(D)1

21 2 1e t u t e t u tt t- -- + -cos ( ) sin ( )

58.d y t

dt

d y t

dt

dy t

dtx t

3

3

2

24 3

( ) ( ) ( )( )+ + =

All initial condition are zero, x t e t( ) = -10 2

(A)5

35 5

5

3

2 3+ - +

- - -e e e u tt t t ( )

(B)5

35 5

5

3

2 3- + +

- - -e e e u tt t t ( )

(C)5

35 1 5 2

5

33u t u t u t u t( ) ( ) ( ) ( )- - + - + -

(D) 53

5 1 5 2 53

3u t u t u t u t( ) ( ) ( ) ( )+ - - - + -

59. The transform function H s( ) of a causal system is

H ss s

s( ) =

+ -

-

2 2 2

1

2

2

The impulse response is

(A) 2d( ) ( ) ( )t e e u tt t- + --

(B) 2d( ) ( ) ( )t e e u tt t- +-

(C) 2d( ) ( ) ( )t e u t e u tt t+ - --

(D) 2d( ) ( ) ( )t e e u tt t+ +-

60. The transfer function H s( ) of a stable system is

H ss

s s( ) =

-

+ +

2 1

2 12


(A) 2 1 3 1u t tu t( ) ( )- + - - +

(B) ( ) ( )3 2te e u tt t- --

(C) 2 1 3 1u t tu t( ) ( )+ - +

(D) ( ) ( )2 3e te u tt t- --

61. The transfer function H s( ) of a stable system is

H ss s

s s s( )

( )( )=

+ -

+ - +

2

2

5 9

1 2 10


(A) - + +-e u t e t e t u tt t t( ) ( sin cos ) ( )3 2 3

(B) - - + --e u t e t e t u tt t t( ) ( sin cos ) ( )3 2 3

(C) - - +-e u t e t e t u tt t t( ) ( sin cos ) ( )3 2 3

(D) - + + --e u t e t e t u tt t t( ) ( sin cos ) ( )3 2 3

62. A stable system has input x t( ) and output

y t e t u tt( ) cos ( )= -2 . The impulse response of the system

is

(A) d( ) ( cos sin ) ( )t e t e t u tt t- +- -2 2

(B) d( ) ( cos sin ) ( )t e t e t u tt t- + -- -2 2 2

(C) d( ) ( cos sin ) ( )t e t e t u tt t- +2 2

(D) d( ) ( cos sin ) ( )t e t e t u tt t- + +2 2 2

63. The relation ship between the input x t( ) and output

y t( ) of a causal system is described by the differential

equation

dy t

dty t x t

( )( ) ( )+ =10 10

The impulse response of the system is

(A) - - +-10 1010e u tt ( ) (B) 10 10e u tt- ( )

(C) 10 1010e u tt- - +( ) (D) - -10 10e u tt ( )

64. The relationship between the input x t( ) and output

y t( ) of a causal system is defined as

d y t

dt

dy t

dty t x t

dx t

dt

2

22 4 5

( ) ( )( ) ( )

( )- - = - + .

The impulse response of system is

(A) 3 2 2e u t e u tt t- + -( ) ( )

(B) ( ) ( )3 2 2e e u tt t- +

(C) 3 2 2e u t e u tt t- - -( ) ( )

(D) ( ) ( )3 2 2e e u tt t- - -

*******

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16/41

SOLUTIONS

1. (B) X s x t e dtst( ) ( )= -

0

= =- -

e dte

s

sts

2

2

2. (A) X s x t e dtt( ) ( )= -

3

0

= + -

u t dtt( )2 3

0

= =-

e dt st3

0

1

3. (A) X s e e dts

t st( ) = =+

- -

2

0

1

2

4. (C) X s x t e dtst( ) ( )= -

0

= - + -

e u t e dtt st2

0

2( )

= =-

--

-

e dte

s

t ss

( )( )

2

0

2 2 2 1

2=

-

-

- -1

2

2 2e

s

s( )

5. (C) X se e

je dt

s

j t j tst( )

( )=

-=

+

--

5 5

0

22

5

25

6. (B) X s e dtst( ) = -0

2

=- -1 2e

s

s

7. (B) p t te u t P ss

t L( ) ( ) ( )( )

= =+

- 1

1 2

x td

dtp t X s

s

s

L( ) ( ) ( )( )

= =+ 1 2

8. (A) p t tu t P ss

L( ) ( ) ( )= =1

2

q t t u t Q ss

s

L( ) cos ( ) ( )= =+

242 2

pp

x t p t q t X s P s Q sL( ) ( ) * ( ) ( ) ( ) ( )= =

=+

X ss s

( )( )

1

42 2p

9. (C) p t tu t P ss

L( ) ( ) ( )= =1

2

q t tp t Q s dds

P ss

L( ) ( ) ( ) ( )= - = = -23

x t tq t X sd

dsQ s

s

L( ) ( ) ( ) ( )= - = =6

4

t u tn

s

n L

n( )

!

+ 1

10. (D) p t u t P ss

L( ) ( ) ( )= =1

q t u t Q se

s

Ls

( ) ( ) ( )= - =-

12

r t e u t R ss

t L( ) ( ) ( )= =+

-2 1

2

v t e u t V se

s

t Ls

( ) ( ) ( )( )

= - =-- +

22

21

x t q t v t X s Q s V sL( ) ( ) * ( ) ( ) ( ) ( )= =

= +

- +

X s

e

s

s

( )

( )2 1

2

11. (B) p t e t u t P ss

s

t L( ) cos ( ) ( )( )

= =+

+ +-3

22

3

3 4

p ds

p dP s

s

tL( ) ( )

( )t t t t

- - +

10

=+

+ +X s

s

s s( )

( )

[( ) ]

3

3 42

12. (A) p t e t u t P ss

s

t L( ) cos ( ) ( )( )

= =+

+ +- 1

1 12

q td

dtp t Q s

s s

s

L( ) ( ) ( )( )

( )= =

+

+ +

1

1 12

x t tq t X sd

dsQ sL( ) ( ) ( ) ( )= = -

=- + +

+ +X s

s s

s s( )

( )

( )

2

2 2

4 2

2 2

13. (B) X ss

s s

A

s

B

s( )

( )=

+

+ +=

++

+

3

3 2 1 22

As

s s=

+

+=

= -

3

22

1

, Bs

s s=

+

+= -

= -

3

11

2

x t e e u tt t( ) [ ] ( )= -- -2 2

14. (A) X ss s s s

( )( ) ( ) ( ) ( )

= -+ +

= -+

++

21

2 32

1

2

1

3

x t t e e u tt t( ) ( ) ( ) ( )= + -- -2 3 3d

15. (C) X ss

s s

A

s

B

s( )

( ) ( )=

-

+ +=

++

+

2 1

2 1 1 12 2

B ss

= - = -= -

( )2 2 31

, A = 2

x t( ) = x t e te u tt t( ) [ ] ( )= -- -2 3

16. (B) X ss

s s s

A

s

B

s

C

s( ) =

+

+ += +

++

+

5 4

3 2 1 23 2

A sX ss

= ==

( )0

2, B s X ss

= + ==-

( ) ( )1 11

,

C s X ss

= + = -=-

( ) ( )2 32

x t e e u tt t( ) [ ] ( )= + -- -2 3 2

17. (C) X ss

s s s( )

( )( )=

-

+ + +

2

2

3

2 2 1

=+

++

++

A

s

B

s

C

s( ) ( ) ( )2 1 1 2


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17/41

A s X ss

= + == -

( ) ( )2 12

, C s X ss

= + = -=-

( ) ( )1 221

A B+ = 1 B = 0

x t e te u tt t( ) [ ] ( )= -- -2

18. (A) X ss

s s( ) =

+

+ +

3 2

2 102=

+

+ +-

+ +

3 1

1 3

1

1 32 2 2 2( )

( ) ( )

s

s s

x t e t e t u tt t( ) cos sin ( )= -

- -3 3

1

33

19. (C) X ss s

s s s( )

( )( )=

+ +

+ + +

4 8 10

2 2 5

2

2

=+

++

+ ++

+ +

A

s

B s

s

C

s( )

( )

( ) ( )2

1

1 2 1 22 2 2 2

A s X ss

= + == -

( ) ( )2 22

A B B+ = =4 2

5 2 2 10A B C+ + = C = -2

x t e e t e t u tt t t

( ) [ cos sin ] ( )= + -- - -

2 2 2 22

20. (B) X ss s

s s s( )

( )( )=

+ +

+ + +

3 10 10

2 6 10

2

2

=+

++

+ ++

+ +

A

s

B s

s

C

s( )

( )

( ) ( )2

3

3 1 3 12 2

A s X ss

= + == -

( ) ( )2 12

, A B B+ = =3 2

10 6 2 10A B C+ + = C = -6

x t e e t e t u tt t t( ) [ cos sin ] ( )= + -- - -2 3 32 6

21. (D) X ss s e

s s

s

( )

( )

=+ + +

+ +

-2 11 16

5 6

2 2

2

= ++

++

++

-+

- -

22 3 2 3

2 2A

s

B

s

e

s

e

s

s s

( ) ( ) ( ) ( )

As s s

s ss

=+ + +

+ +=

= -

( )( )

( )

2 2 11 16

5 62

2

2

2

Bs s s

s ss

=+ + +

+ += -

= -

( )( )

( )

3 2 11 16

5 61

2

2

3

x t t e e u t e e u tt t t t( ) ( ) [ ] ( ) [ ] (( ) ( )= + - + -- - - - - -2 2 2 3 2 2 3 2d - 2)

22. (C) P s

s

p t t u tL( ) ( ) sin ( )=

+

=1

9

1

3

32

Q sd

dsP s q tL( ) ( ) ( )=

2

2= - =( ) ( ) sin ( )1

332 2

2

t p tt

t u t

R s sQ s r td

dtq t qL( ) ( ) ( ) ( ) ( )= = - -0

= +2

33 32

tt u t t t u tsin ( ) cos ( )

V ss

v t e u tL t( ) ( ) ( )=+

= -1

3

3

x t v t r t( ) ( ) ( )= + = + +

-2

33 32 3

tt u t t t u t e u ttsin ( ) cos ( ) ( )

23. (C) Ps

aap atL

( )

1

1 4

1

22

2( )sin ( )

se t u tL t

+ + -

x t e t u tL t( ) sin ( ). -1

4

0 5

24. (C) P ss

p t te u tL t( )( )

( ) ( )=+

= -1

1 2

Q sd

dsP s q t tp t t e u tL t( ) ( ) ( ) ( ) ( )= = - = - -2

X s e Q s x t q ts L( ) ( ) ( ) ( )= = --2 2

= - - --x t t e u tt( ) ( ) ( )( )2 22

25. (A) sX s X sdx t

dtx tL( ) ( )

( )( )+ +

= - +y t t t u t( ) ( sin cos ) ( )2 2 2

26. (B) Xs

aax atL

( )

X s t u tL( ) cos ( )31

3

2

3

27. (D) X s e x tL t( ) ( )+ -2 2

28. (B) P sX s

sx dL

t

( )( )

( )= - t t

L

t

u d

t =

- cos ( )sin

2

2

2t t t

P s

sd

tu tL

t( ) sin cos

( ) =-

2

2

1 2

40

tt ,

29. (C) P s e X s p t x ts L( ) ( ) ( ) ( )= = --3 3

= - -cos ( ) ( )2 3 3t u t

Q sd

dsP s q t p tL( ) ( ) ( ) ( )= = -

= - - -t t u tcos ( ) ( )2 3 3 .

30. (A) x t e X sL s( ) ( )- -

2 2 , Y sse

s

s

( ) = +

-2

2

2

2

31. (A) p td

dtx t P s sX sL( ) ( ) ( ) ( )= =

y t x t p t Y s P s X s s X sL( ) ( ) * ( ) ( ) ( ) ( ) ( ( ))= = = 2

32. (A) e x t X ss

s

t L- + =+

+ +( ) ( )

( )

( )1

2 1

1 22

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18/41

33. (B) 2 24 8

2

2

2 2tx t

d

dsX s

s

s

L( ) ( )( )

- =-

+

34. (A) X s x t e dtst( ) ( )= -

-

= = = +- -

-

- +

-

+

e e dt e dte

s

t st t ss

2

1

2

2 1

1

( )( )

, Re( )s > - 1

35. (B) X s u t e dt e dte

s

st sts

( ) ( )= - + = =--

-

-

-

-

33 3

Re ( )s < 0

36. (C) Y s t e dt est s( ) ( )= + =-

-

d 1 , All s

37. (B) X se e

je dt

jt jtst( )

( )=

- - -

20

= --

- +

121

20 0

je dt

je dtt j s t j s( ) ( ) =

+1

1 2s, Re ( )s > 0

38. (D) X s e e e e dt e e dtt

st t st t st( ) = + +-

-

- -

-

- 20 0

0

=+

++

--

1

0 5

1

1

1

1s s s.

Re ( ) .s > -0 5, Re ( )s > -1, Re ( )s < 1

- -1, Re ( ) .s > 0 5

Therefore 0.5 < Re ( )s < 1

X ss

s s s( )

( ) ( ) .= -

-

- ++

++

-

1

1 4

1

1

1

0 52

40. (C) x t e e u tt( ) ( )( )= +- - +3 3 3 3

p t e u t P ss

t L( ) ( ) ( )= =-

3 1

3

q t p t Q s e P s e

s

L ss

( ) ( ) ( ) ( )= + = =-

33

33

X se

s

s

( )( )

=-

-3 1

3, Re ( )s > 3

41. (B) p t q t P s Q sL( ) * ( ) ( ) ( )

X ss

s s( ) =

-

+ +

2 9

1

1

Re ( )s > -1, Re ( )s < 0

- 1

43. (A) p t e u t P s s

t L

( ) ( ) ( )= - =

-

+-2 1

2, Re ( )s < - 2

q td

dtp t Q s sP sL( ) ( ) ( ) ( )= =

x t e q t X s Q ss

s

t L( ) ( ) ( ) ( )= = - =-

+1

1

1

Re ( )s < - 1 thus left-sided .

44. (C) Left-sided

P ss

p t e u tL t( ) ( ) ( )=+

= - --1

2

2

X s e P s x t p ts L( ) ( ) ( ) ( )= = +5 5

= - - +- +x t e u tt( ) ( ( ))( )2 5 5

45. (A) Right-sided

P ss

p t e u tL t( )( )

( ) ( )=-

=1

3

3

X sd

dsP s x t t e u tL t( ) ( ) ( ) ( )= =

2

2

2 3

46. (D) Left-sided

x t u t u t t( ) ( ) ( ) ( )= - - + - + + +1 2d

47. (C) Right-sided, P ss

p t u tL( ) ( ) ( )= =1

Q s e P s q t p t u ts L( ) ( ) ( ) ( ) ( )= = - = --3 3 3

R sd

dsQ s r t tq t tu tL( ) ( ) ( ) ( ) ( )= = - = - - 3

V ss

R s v t r dLt

( ) ( ) ( ) ( )= =-

1t t

= - = - -v t tdt tt

( ) ( )3

21

29

X ss

v s x t tLt

( ) ( ) ( ) ( )= = - -

-

1 1

292

=-

- + -

-x t t t u t( ) ( ) ( ) ( )

1

627

9

23 33

48. (B) X ss

s s s s( )

( )=

- -

+ +=

-

++

+

4

3 2

3

1

2

22

Left-sided, x t e u t e u tt t( ) ( ) ( )= - - -- -3 2 2

49. (A) X ss s

( )( ) ( )

=+

-+

5

1

1

1 2

Left-sided, x t u t te u tt( ) ( ) ( )= - - + --5


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50. (D) x sX ss

s sx( ) lim ( )0

5 20

2

+

= =

+ -=

51. (A) x sX ss s

s ss( ) lim ( )0

2

2 31

2

2

+

= =

+

+ -=

52. (D) x sX s e s ss ss

s( ) lim ( ) ( )0 6

2 20

2 3 2

2

+

-= = ++ -

=

53. (A) x sX ss s

s ss( ) lim ( ) = =

+

+ +=

0

3

2

2 3

5 10

54. (C) x sX ss

s ss( ) lim ( ) = =

+

+ +=

0 2

2

3 12

55. (B) x sX se s

s ss

s

( ) lim ( )( )

= =+

+ +=

-

0

3 2

2

2 1

5 4

1

4

56. (C) sY s y Y s s( ) ( ) ( ) ( )- + =-0 10 10

y X ss

( ) , ( )0 11- = =

Y ss s s s

( )( ) ( )

=+

++

=10

1

1

1

1

y t u t( ) ( )=

57. (C) s Y s s sY s Y s2 2 2 2 5 1( ) ( ) ( )- + - + =

( ) ( )s s Y s s2 2 5 3 2+ + = +

Y ss

s s

s

s s

( )( )

( ) ( )

=+

+ +

=+

+ +

+

+ +

2 3

2 5

2 1

1 2

1

1 22 2 2 2 2

= +- -y t e t u t e t u tt t( ) cos ( ) sin ( )21

2

58. (B) s Y s s Y s sY ss

3 24 310

2( ) ( ) ( )

( )+ + =

+

Y ss s s s

A

s

B

s

C

s

D

s( )

( )( )( ) ( ) ( )=

+ + += +

++

++

+

10

1 2 3 1 2 3

A sY ss

= ==

( )0

5

3, B s Y s

s= + = -

= -( ) ( )1 5

1,

C s Y ss

= + == -

( ) ( )2 52

, D s Y ss

= + ==

( ) ( )35

30

= - + +

- - -y t e e e u tt t t( ) ( )

5

35 5

5

3

2 3

59. (D) For a causal system h t( ) = 0 for t < 0

H ss s

( ) = ++

+-

21

1

1

1

h t t e e u tt t( ) ( ) ( ) ( )= + +-2d

60. (D) H ss s

( )( )

=+

-+

2

1

3

1 2, System is stable

h t e te u tt t( ) ( ) ( )= -- -2 3 .

61. (A) H ss

s

s s( )

( )

( )

( ) ( )=

-

++

-

- ++

- +

1

1

2 1

1 3

3

1 32 2 2 2

System is stable

h t e u t e t e t u tt t t( ) ( ) ( cos sin ) ( )= - + + -- 2 3 3

62. (A) X ss

( ) =+

1

1, Y s

s

s( )

( )

( )=

+

+ +

2

2 12

H sY s

X s

s s

s( )

( )

( )

( )( )

( )= =

+ +

+ +

1 2

2 12

= -+

+ +-

+ +1

2

2 1

1

2 12 2( )

( ) ( )

s

s s

h t t e t e t u tt t( ) ( ) ( cos sin ) ( )= - +- -d 2 2

63. (B) sY s Y s X s( ) ( ) ( )+ =10 10

H sY s

X s s( )

( )

( )= =

+

10

10

h t e u tt( ) ( )= -10 10

64. (B) Y s s s X s s( )( ) ( )( )2 2 5 4- - = -

H sY s

X s

s

s s s s( )

( )

( )= =

-

- -=

++

-

5 4

2

3

1

2

22

h t e u t e u tt t( ) ( ) ( )= +-3 2 2 .

***********

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20/41

Statement forQ.1-12:

Determine the z-transform and choose correct

option.

1. x n n k k[ ] [ ] ,= - >d 0

(A) z zk , > 0 (B) z zk- >, 0

(C) z zk , 0 (D) z zk- , 0

2. x n n k k[ ] [ ] ,= + >d 0

(A) z zk- , 0 (B) z zk , 0

(C) z k- , all z (D) zk , all z

3. x n u n[ ] [ ]=

(A)1

11

1->

-zz, | | (B)

1

11

1-

.

( . ), . (B)

z

z zz

5 5

4

0 25

0 250

-

->

.

( . ),

(C)z

z zz

5 5

3

0 25

0 250 25

-

-, | | (B)

4

4 1

1

4

z

zz

-

zz, | | (D)

1

1 4

1

4-, | | (B)

z

zz

33

-

zz, | | (D)

3

33

- 112

58. Consider the following three systems

y n y n x n x n x n1 0 2 1 0 3 1 0 02 2[ ] . [ ] [ ] . [ ] . [ ]= - + - - + -

y n x n x n2 0 1 1[ ] [ ] . [ ]= - -

y n y n x n x n3 0 5 1 0 4 0 3 1[ ] . [ ] . [ ] . [ ]= - + - -

The equivalent system are

(A) y n1[ ] and y n2[ ] (B) y n2[ ] and y n3[ ]

(C) y n3[ ] and y n1[ ] (D) all

59. The z-transform of a causal system is given as

X zz

z z( )

.

. .=

-

- +

-

- -

2 15

1 15 0 5

1

1 2

The x[ ]0 is

(A) -15. (B) 2

(C) 1.5 (D) 0

60. The z-transform of a anti causal system is

X zz

z z( ) =

-

- +

12 21

3 7 12 2

The value of x[ ]0 is

(A) -7

4(B) 0

(C) 4 (D) Does not exist

61. Given the z-transforms

X zz z

z z( )

( )=

-

- +

8 7

4 7 32

The limit of x[ ] is

(A) 1 (B) 2

(C) (D) 0

62. The impulse response of the system shown in fig.

P5.4.62 is

(A) 2 1 11

2

22

n

n u n n-

+ - +( ( ) ) [ ] [ ]d

(B)2

21 1

1

2

nn u n n( ( ) ) [ ] [ ]+ - + d

(C) 2 1 11

2

22

n

n u n n-

+ - -( ( ) ) [ ] [ ]d

(D)2

21 1

1

2

nn u n n[ ( ) ] [ ] [ ]+ - - d

63. The system diagram for the transfer function

H zz

z z( ) =

+ +2 1

is shown in fig. P5.4.63. This system diagram is a

(A) Correct solution

(B) Not correct solution

(C) Correct and unique solution

(D) Correct but not unique solution

*****************

Chap 5.4The z-Transform

Pa285

X z( ) Y z( )

z-1

z-1

+ z-1

Fig. P5.4.62

X z( ) Y z( )

z-1

z-1

+

++

Fig. P5.4.63


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25/41

24. (A) X zz z

z z

z

z z

( ) =-

+

=-

+ --

-

- -

2

2 1

1

1 2

3

3

2

1 3

13

2

=+

-

--

-

2

1 2

1

11

2

11z z

, ROC :1

22< >, gives z1 2>

x n2[ ] is left-sided signal

z z2 221

2< and z3 2< gives

1

223< 1

2(Right-sided) = -

-

x n u n u nn

n

[ ] [ ] [ ]2

2

1

3

| |z 1

4.1

21

2

0

2

x t dt( ) =

The signal, that satisfy these condition, is

(A) 2 sin pt and unique

(B) 2 sin pt but not unique

(C) 2 sin pt and unique

(D) 2 sin pt but not unique

26. Consider a continuous-time LTI system whose

frequency response is

H j h t e dtj t( ) ( )sin

ww

ww= =-

-

4

The input to this system is a periodic signal

x tt

t( )

,

,=

-

2 0 4

2 4 8

with period T = 8. The output y t( ) will be

(A) 14

2+

sin

pt(B) 1

4

2+

cos

pt

(C) 14 4

+

+

sin cos

p pt t(D) 0

27. Consider a continuous-time ideal low pass filter

having the frequency response

H j( ), | |

, | |w

w

w=

>

1 80

0 80

When the input to this filter is a signal x t( ) with

fundamental frequency wo = 12 and Fourier series

coefficients X k[ ], it is found that x t y t x tS( ) ( ) ( ) = .

The largest value of| |k, for which X k[ ] is nonzero, is

(A) 6 (B) 80

(C) 7 (D) 12

28. A continuous-time periodic signal has a

fundamental period T = 8. The nonzero Fourier series

coefficients are as,

X X j X X[ ] [ ] , [ ] [ ]*1 1 5 5 2= - = = - = ,

The signal will be

(A) 44

24

cos sinp p

t t

-

(B) 24

44

cos sinp p

t t

+

(C) 24

24

cos sinp pt t

+

(D) None of the above


Consider the following three continuous-time

signals with a fundamental period of T = 1

x t t( ) cos= 2p , y t t( ) sin= 2p , z t x t y t( ) ( ) ( )=

29. The Fourier series coefficient X k[ ] of x t( ) are

(A) 12

1 1( [ ] [ ])d dk k+ + -

(B) 12

1 1( [ ] [ ])d dk k+ - -

(C) 12

1 1( [ ] [ ])d dk k- - +


30. The Fourier series coefficient of y t( ), Y k[ ] will be

(A)j

k k2

1 1( [ ] [ ])d d+ + +

(B)j

k k2

1 1( [ ] [ ])d d+ - -

(C)j

k k2

1 1( [ ] [ ])d d- - +

(D) 12 1 1j k k( [ ] [ ])d d+ + +

31. The Fourier series coefficient of z t Z k( ) , [ ] will be

(A) 14

2 2j

k k( [ ] [ ])d d- - +

(B) 12

2 2j

k k( [ ] [ ])d d- - +

(C) 12

2 2j

k kd d[ ] [ ])+ - -


32. Consider a periodic signal x t( ) whose Fourier series

coefficients are

X k

k

j

k[ ]

,

,

| |==

2 0

1

2otherwise

Consider the statements

1. x t( ) is real. 2. x t( ) is even 3.dx t

dt

( )is even

The true statements are

(A) 1 and 2 (B) only 2

(C) only 1 (D) 1 and 3

Chap 5.7The Continuous-Time Fourier Series

Pa311


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36/41


A waveform for one peroid is depicted in figure in

question. Determine the trigonometric Fourier series

and choose correct option.

33.

(A)2 1

22

1

33

1

44

p(cos cos cos cos ....)t t t t+ + + +

(B)2 1

22

1

33

1

44

p(sin sin sin sin ....)t t t t- + - +

(C)2 1

22

1

22

1

33

p(sin cos sin cos sin ....)t t t t t+ - - + +

(D)2 1

33

1

33

1

55

p(sin cos sin cos sin ....)t t t t t+ + + + +

34.

(A)A A

t t t2

4 1

22

1

33+ + + +

psin sin sin ....

(B)A A

t t t2

4 1

33

1

55+ + + +

pcos cos cos ....

(C)4 1

33

1

55

At t t

psin sin sin ....+ + +

(D)

4 1

2 2

1

3 3

A

t t tp cos cos cos ....+ + +

35.

(A)A A

t t t2

2 1

33

1

55+ - +

p(sin sin sin ....)

(B)A A

t t t2

2 1

22

1

33+ - +

p(cos cos cos ....)

(C)A A

t t t2

2 1

33

1

55+ - +

p(cos cos cos ....)

(D)A A

t t t t2

2 1

33

1

33+ + + +

p(sin cos sin cos ....)

36.

(A)1

2

12 1

93

1

255

2+ + + +

pp p p(cos cos cos ....)t t t

(B) 312 1

93

1

255

2+ + + +

pp p p(cos cos cos ....)t t t

(C)1

2

12 1

93

1

255

2+ - + -

pp p p(sin sin sin ....)t t t

(D) 312 1

93

1

255

2+ - + -

pp p p(sin sin sin ....)t t t

*****

Page312

UNIT 5 Signal & System

x t)

p t-p

A

-A

Fig. P5.7.34

2 2

-p p

x t( )

pt

-p

A

Fig. P5.7.35

x t)

2

-1

-1

1 t

Fig. P5.7.36

x t( )

1

-1

p t-p

Fig. P5.7.33


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37/41

SOLUTIONS

1. (D) X kT

A t e dtA

T

jk t

T

T

[ ] ( )= =-

-

1

2

2

d w o ,

A = 10 , T = 5, X k[ ] = 2

2. (C) X kT

x t e dtT

Ae dtjk t

T

T

jk t

T

T

[ ] ( )= =-

-

-

-

1 1

2

2

4

4

w wo o

=-

=

-

-

A

T

e

jk

A

k

kjk t

T

Tw

o

o

w p

p

4

4

2sin

3. (B) T = 2p , wp

po= =

2

21, x t

A t

( )

,

,

=<


38/41

13. X k k j k[ ] cos sin=

+

10

192

4

19

p p

(A)19

25 5 19 2 2 9( [ ] [ ]) [ ] [ ]),| |d d (d dn n n n n+ + - + + - -

(B)1

2

5 5 2 2 9( [ ] [ ]) [ ] [ ]),| |d d (d dn n n n n+ + - + + - -

(C)9

25 5 9 2 2 9( [ ] [ ]) [ ] [ ]),| |d d (d dn n n n n+ + - + + - -

(D)1

25 5 2 2 9( [ ] [ ]) [ ] [ ]),| |d d (d dn n n n n+ + - + + - -

14. X kk

[ ] cos=

p

21

(A)21

24 4 10(d d[ ] [ ]),| |n n n+ + -

(B)1

2

4 4 10(d d[ ] [ ]),| |n n n+ + -

(C)21

24 4 10(d d[ ] [ ]),| |n n n+ - -

(D)1

24 4 10(d d[ ] [ ]),| |n n n+ - -


Consider a periodic signal x n[ ] with period N and

FS coefficients X k[ ]. Determine the FS coefficients Y k[ ]

of the signal y n[ ] given in question.

15. y n x n n[ ] [ ]= - o

(A) e X kj

Nn k

2 po

[ ] (B) e X kj

Nn k-

2po

[ ]

(C) kX k[ ] (D) -kX k[ ]

16. y n x n x n[ ] [ ] [ ]= - - 2

(A) sin [ ]4p

Nk X k

(B) cos [ ]

4p

Nk X k

(C) 1

4

-

-

e X kj

Nk

p

[ ] (D) 1

4

-

e X kj

Nk

p

[ ]

17. y n x n x n N [ ] [ ] [ ]= + + 2 , (assume that N is even)

(A) 2 2 1X k[ ],- for 02

1 -

k

N

(B) 2 2 1X k[ ],- for 02

kN

(C) 2 2X k[ ], for 02

1 -

k

N

(D) 2 2X k[ ], for 02

kN

18. y n x n x n N [ ] [ ] [ ]= - - 2 , (assume that N is even)

(A) ( ( ) ) [ ]1 1 21- - +k X k (B) ( ( ) ) [ ]1 1- - k X k

(C) ( ( ) ) [ ]1 1 1- - +k X k (D) ( ( ) ) [ ]\1 1 2- - k X k

19. y n x n[ ] [ ]*= -

(A) - X k*[ ] (B) - -X k*[ ]

(C) X k*[ ] (D) X k*[ ]-

20. y n x nn[ ] ( ) [ ]= -1 , (assume that N is even)

(A) X kN

-

2(B) X k

N+

2

(C) X kN

- +

21 (D) X k

N+ -

21


Consider a discrete-time periodic signal

x nn

n[ ]

,

,=

1 0 7

0 8 9

with period N = 10. Also y n x n x n[ ] [ ] [ ]= - - 1

21. The fundamental period of y n[ ] is

(A) 9 (B) 10

(C) 11 (D) None of the above

22. The FS coefficients of y n[ ] are

(A)1

101

8

5-

ej k

p

(B)1

101

8

5-

-

ej k

p

(C)1

101

4

5-

ej k

p

(D)1

101

4

5-

-

ej k

p

23. The FS coefficients of x n[ ] are

(A) -

-

j

ek

Y k k

jk

2 10 010

pp

cosec [ ],

(B)j

ek

Y k kj

k

2 10010

-

pp

cosec [ ],

(C) -

-

1

2 10

10ek

Y kj

kpp

sec [ ]

(D)1

2 10

10ek

Y kj

k-

pp

sec [ ]

Page318


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Consider a discrete-time signal with Fourier

representation.

x n X kDTFS

[ ] [ ];

p

10

In question the FS coefficient Y k[ ] is given.

Determine the corresponding signal y n[ ] and choose

correct option.

24. Y k X k X k[ ] [ ] [ ]= - + +5 5

(A) 25

sin [ ]p

n x n

(B) 2

5cos [ ]

pn x n

(C) 22

sin [ ]p

n x n

(D) 2

2cos [ ]

pn x n

25. Y k

k

X k[ ] cos [ ]=

p

5

(A)1

25 5( [ ] [ ])x n x n+ + + (B)

1

22 2( [ ] [ ])x n x n+ + -

(C)1

210 10( [ ] [ ])x n x n+ + + (D) None of the above

26. Y k X k X k[ ] [ ] * [ ]=

(A)( [ ] )x n 2

2p(B) j x n2 2p( [ ] )

(C) ( [ ] )x n 2 (D) 2 2p( [ ] )x n

27. Y k X[ ] Re{ [= k]}

(A)x n x n[ ] [ ]+ -

2(B)

x n x n[ ] [ ]- -

2

(C)x n x n[ ] [ ]- -

2p(D)

x n x n[ ] [ ]+ -

2p

28. Consider a sequence x n[ ] with following facts :

1. x n[ ] is periodic with N = 6

2. x nn

[ ] ==

20

5

3. ( ) [ ]- =-

1 12

7n

n

x n

4. x n[ ] has the minimum power per period among the

set of signals satisfying the preceding three condition.

The sequence would be..

(A) ... , , , , , ...1

2

1

6

1

2

1

6

1

2

(B) ... , , , , , ...0 11

2

1

3

1

4

(C) ... , , , , , ...1

3

1

6

1

3

1

6

1

3

(D) { }... , , , , , ...0 1 2 3 4

29. A real and odd periodic signal x n[ ] has fundamental

period N = 7 and FS coefficients X k[ ]. Given that

X j[ ]15 = , X j[ ]16 2= , X j[ ]17 3= . The values of

X X X[ ], [ ], [ ],0 1 2- - and X [ ]-3 will be

(A) 0 2 3, , ,j j j (B) 1, 1, 2, 3

(C) 1, -1, -2, -3 (D) 0, -j , -2j, -3j

30. Consider a signal x n[ ] with following facts

1. x n[ ] is a real and even signal

2. The period of x n[ ] is N = 10

3. X[ ]11 5=

4.1

1050

2

0

9

X kn

[ ]=

=

The signal x n[ ] is

(A) 5

10

cosp

n

(B) 5

10

sinp

n

(C) 105

cosp

n

(D) 10

5sin

pn

31. Each of two sequence x n[ ] and y n[ ] has a period

N = 4. The FS coefficient are

X X X X[ ] [ ] [ ] [ ]0 31

21

1

22 1= = = = and

Y Y Y Y [ ], [ ], [ ], [ ]0 1 2 3 1=

The FS coefficient Z k[ ] for the signal

z n x n y n[ ] [ ] [ ]= will be

(A) 6 (B) 6| |k

(C) 6| |k (D) ej k

p

2

32. Consider a discrete-time periodic signal

x n

n

n

[ ]

sin

sin

=

11

20

20

p

p

with a fundamental period N = 20. The Fourierseries coefficients of this function are

(A)1

205 6( [ ] [ ])u k u k+ - - , | |k 10

(B)1

205 5( [ ] [ ])u k u k+ - - , | |k 10

(C) ( [ ] [ ])u k u k+ - +5 6 , | |k 10

(D) ( [ ] [ ])u k u k+ - -5 6 , | |k 10

************

Chap 5.8The Discrete-Time Fourier Series

Pa319


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40/41

11. (D) N = 7, W o =2p

7,

x n X k e ej kn

n

j n

[ ] [ ]( )

= =

=-

-

2

7

3

3 12

72

p p

- +

1 21

2

7ej n( )

p

=

-4

2

71cos

pn

12. (C) N = 12, W o =p

6, X k e

j k

[ ] =-

p

6

x n e e ej k j kn

k

j

[ ] = =-

=-

p p p

6 6

6

66

-

=-

k n

k

( )1

6

6

=

-

-

- - -

-

e e

e

j n j n

j n

( ) ( ) ( )

( )

46

19

61

61

1

1

p p

p

=-

-

sin ( )

sin ( )

3

41

121

p

p

n

n

13. (A) N = 19, Wo =2p19

X k k j k[ ] cos sin=

+

10

192

10

19

p p

= +

+ +

- - - - -1

2

52

195

2

192

2

19e e e ej k j k j k( ) ( ) ( )

p p p-

j k( )22

19

p

By inspection

x n n n n n[ ] ( [ ] [ ]) ( [ ] [ ])= + + - + + - -19

25 5 19 2 2d d d d ,

Where | |n 9

14. (A) N = 21, W o =2p

21

X k k[ ] cos=

8

21

p= +

- - -1

2

42

214

2

21e ej k j k( ) ( )

p p

Since X kN

x n e jk n

n N

[ ] [ ]= -=1 Wo , By inspection

x nn

n

[ ],

, { , , ...... . }

==

- -

21

24

0 10 9 9 10otherwise

15. (B) Y kN

x n n en

N jN

kn

[ ] [ ]= -=

- -

10

12

o

p

= =-

=

- -

-

12

0

12

Ne x n e e

jN

kn

n

N jN

knp p

o

[ ]j

Nkn

X k

2 po

[ ]

16. (C) Y k X k e X k ej

Nk j

Nk

[ ] [ ] [ ]= - = -

-

-

22

4

1

p p

X k[ ]

17. (C) Note that y n x n x n N [ ] [ ] [ ]= + + 2 has a period

of N 2 and N has been assumed to be even,

Y kN

x n x n N ej

Nkn

n

N

[ ] ( [ ] [ ])= + +-

=

-

2 24

0

2 1 p

= 2 2X k[ ] for 0 2 1 -k N( )

18. (B) y n x n x n N [ ] [ ] [ ]= - - 2

Y k e X k e X kj

N

Nk

j k[ ] [ ] ( ) [ ]= -

= --

-1 1

2

2

pp

=

0

2

, k

X k k

even

[ ], odd

19. (C) y n x n[ ] [ ]*= -

Y kN

x n e X kj

Nkn

n

N

[ ] [ ] [ ]* *= - =-

=

-

12

0

1p

20. (A) With N even

y n x n e x n e x nnj n

jN

N

[ ] ( ) [ ] [ ] [ ]= - = =

1

2

2pp

Y kN

e x n ej

N

Nj

Nkn

n

N

[ ] [ ]=

-

=

-

12

2

2

0

1p p

= = --

-

=

-

1 22

2

0

1

Nx n e X k N

jN

n kN

n

N

[ ] [ ]

p

21. (B) y n[ ] is shown is fig. S5.8.21. It has fundamental

period of 10.

22. (B) Y k y n en

j kn

[ ] [ ]==

-

110 0

92

10

p

= -

= -

-

-

110

1 110

1

2

108

8

5e ej k j

p p

k

23. (A) y n x n x n[ ] [ ] [ ]= - - 1

Y k X k e X k X kY k

e

j k

j

[ ] [ ] [ ] [ ][ ]

= - =

-

-

-

2

10

51

p

p

k

Chap 5.8The Discrete-Time Fourier Series

Pa321

2 31 4 5 87

9

10 11

y n

n

-1

1

Fig. S5.8.21


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41/41

=

-

-

X k

e Y k

e e

j k

j k j

[ ][ ]

p

p p

10

10 10

=

k

j k

e Y k

jk

p

p

10

210

[ ]

sin

=-

-

j

e k Y kj k

2 10

10

pp

cosec [ ]

24. (D) W op

=10

, Y k X k X k[ ] [ ] [ ]= - + +5 5

= +

=

-y n e e x n n

j n j n

[ ] [ ] cos( ) ( )5

105

10 22

p p p

x n[ ]

25. (B) Y k k X ke e

X k

j k j k

[ ] cos [ ] [ ]=

=

+

-p

p p

5 2

5 5

= +

-1

2

210

210e e X k

j k j k( ) ( )

[ ]p p

= - + +y n x n x n[ ] ( [ ] [ ])1

22 2

26. (C) Y k X k X k[ ] [ ] * [ ]= y n x n x n x n[ ] [ ] [ ] ( [ ])= = 2

27. (A) Y k[ ] =Re{ [ ] }X k =y n[ ] Ev{ [ ]}x n =+ -x n x n[ ] [ ]

2

28. (A) N = 6, W o =2p

6,

From fact 2, x nn

[ ] ==

20

5

= =

=1

6

1

30

1

3

2

60

0

5

e x n Xj k

n

p( )

[ ] [ ] ,

From fact 3, ( ) [ ]- ==

1 12

7n

n

x n

= =

=1

6

1

63

1

6

2

63

0

5

e x n Xj k

n

p( )

[ ] , [ ]

By Parsevals relation, the average power in x n[ ] is

P X kk

==

[ ] 20

5

,

The value of P is minimized by choosing

X X X X[ ] [ ] [ ] [ ]1 2 4 5 0= = = =

Therefore

x n X X en

n[ ] [ ] [ ] ( )= + = + -

0 31

31

1

6

2

63

p

= + -1

31

1

6( )n

x n[ ] =

1 1 1 1 1

29. (D) Since the FS coefficient repeat every N. Thus

X X X X X X[ ] [ ], [ ] [ ], [ ] [ ]1 15 2 16 3 17= = =

The signal real and odd, the FS coefficient X k[ ] will be

purely imaginary and odd. Therefore X[ ]0 0=

X X X X X X[ ] [ ], [ ] [ ], [ ] [ ]- = - - = - - = -1 1 2 2 3 3

Therefore (D) is correct option.

30. (C) Since N = 10, X X[ ] [ ]11 1 5= =

Since x n[ ] is real and even X k[ ] is also real and even.

Therefore X X[ ] [ ]1 1 5= - = .

Using Parsevals relation X k X kN k

[ ] [ ]2 2

1

8

50 = ==-

X X X X kk

[ ] [ ] [ ] [ ]- + + + ==

1 1 0 502 2 2 22

8

X X kk

[ ] [ ]0 02 2

2

8

+ ==

Therefore X k[ ] = 0 for k = 0 2 3 8, , , . .... .

x n X k e X k ej

Nkn

N

j kn

k

[ ] [ ] [ ]= =

=-

2 2

10

p p

1

8

= +

=-

5 105

10

2

10e ej n j n

2p pp

cos n

31. (A) z n x n y n X l Y k lDTFS

k N

[ ] [ ] [ ] [ ] [ ]= -=< >

= -=Z k X l Y k ll

[ ] [ ] [ ]0

3

= + - + - + -Z k X Y k X Y k X Y k X Y k[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]0 1 1 2 2 3 3

= + - + - + -Y k Y k Y k Y k[ ] [ ] [ ] [ ]2 1 2 2 3

Since Y k[ ] is 1 for all values of k.

Thus Z k[ ] = 6, for all k.

32. (A) N = 20 We know that

1 5

0 5 10

11

2010, | |

, | |

sin

si

;n

n

kDTFS

<

p

p

np

20k

Using duality

sin

sin

, | |;1120

20

1

20

110

p

p

pn

n

kDTFS

5

0 5 10, | |<

k

*********


GATE BY RK Kanodia signals and systems

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