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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( )
Statement for Q.35-38:
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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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<
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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- <
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(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)
If the excitation is bounded, the response is bounded.
(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)
If the excitation is bounded, the response is bounded.
(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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Statement for Q.1-12:
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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23. X ss
( )( )
=+ +
1
2 1 42
(A) e t u tt-0 5. sin ( ) (B)1
2e t u tt- sin ( )
(C)1
4
0 5e t u tt- . sin ( ) (D) e t u tt- sin ( )
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
Statement for Q.2529:
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
Statement for Q.3033:
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+
Statement for Q.3443:
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
Statement for Q.4449:
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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(A)1
2e t u tt- sin ( ) (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
The impulse response is
(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
The impulse response is
(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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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
Chap 5.3The Laplace Transform
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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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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 .
***********
Page278
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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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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
Statement for Q.29-31:
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- - +
(D) None of the above
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[ ] [ ])+ - -
(D) None of the above
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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Statement for Q.33-36:
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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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
( )
,
,
=<
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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+ - -
Statement for Q.15-20:
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
Statement for Q.21-23:
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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Statement for Q.24-27:
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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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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=
-
-
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
*********
UNIT 5 Signal & SystemGATE EC BY RK Kanodia