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PAPERElectrical EngineeringSIGNALS & SYSTEMSCopyright By NODIA
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A AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar
Nagar, J aipur 302039Ph :+91 - 141 - 2101150www.nodia.co.inemail :
[email protected] SOLVED PAPER - EESI GNALS &SYSTEMS
www.nodia.co.inYEAR 2013ONE MARKQ.
1Aband-limitedsignalwithamaximumfrequencyof5kHzistobesampled.
According to the sampling theorem, the sampling frequency which is
not valid is(A) 5kHz(B) 12kHz(C) 15kHz(D) 20kHzQ.
2Foraperiodicsignal/ sin cos sin v t t t t 30 100 10 300 6 500 4 p
= + + + ^ ^ h h,the fundamental frequency in/ rad s(A) 100(B)
300(C) 500(D) 1500Q. 3Two systems with impulse responses h t1^ h
and h t2^ h are connected in cascade. Then the overall impulse
response of the cascaded system is given by(A) product of h t1^ h
and h t2^ h(B) sum of h t1^ h and h t2^ h(C) convolution of h t1^ h
and h t2^ h(D) subtraction of h t2^ h from h t1^ hQ.
4WhichoneofthefollowingstatementsisNOTTRUEforacontinuoustime causal
and stable LTI system?(A) All the poles of the system must lie on
the left side of thejw axis(B) Zeros of the system can lie anywhere
in the s-plane(C) All the poles must lie withins 1 =(D) All the
roots of the characteristic equation must be located on the left
side of thejw axis.Q. 5Theimpulseresponseofasystemish t tu t = ^ ^
h h.Foraninputu t 1 - ^ h,the output is(A) t u t22^ h(B) t tu
t211--^^hh(C) tu t2112--^^hh(D) tu t2112-- ^ hYEAR 2013TWO MARKSQ.
6The impulse response of a continuous time system is given by h t t
t 1 3 d d = - + - ^ ^ ^ h h h. The value of the step response at t
2 =is(A) 0(B) 1(C) 2(D) 3 www.nodia.co.inGATE SOLVED PAPER - EESI
GNALS &SYSTEMSYEAR 2012ONE MARKQ. 7If[ ] (1/ 3) (1/ 2) [ ], x n
u nn n= - thentheregionofconvergence(ROC)ofitsz-transform in the
z-plane will be(A)z313 < < (B)z3121< be( ) F s1
andtheLaplacetransformofitsdelayedversion( ) f t t - be( ) F
s2.Let* ( ) F s1 bethecomplexconjugateof( ) F
s1withtheLaplacevariablesets j s w = + .If ( )( )( ) * ( )G sF sF s
F s122 1= , then the inverse Laplace transform of( ) G sis an
ideal(A) impulse( ) t d (B) delayed impulse( ) t d t -(C) step
function( ) u t (D) delayed step function( ) u t t -Q.
15Theresponse( ) h t ofalineartimeinvariantsystemtoanimpulse( ) t d
,under initially relaxed condition is( ) h t e et t 2= +- -. The
response of this system for a unit step input( ) u tis(A)( ) u t e
et t 2+ +- -(B) ( ) ( ) e e u tt t 2+- -(C) ( . . ) ( ) e e u t 15
05t t 2- -- -(D)( ) ( ) e t e u tt t 2d +- -YEAR 2010ONE MARKQ.
16For the system/ ( ) s 2 1 + , the approximate time taken for a
step response to reach 98% of the final value is(A) 1 s(B) 2 s(C) 4
s(D) 8 sQ. 17The period of the signal( ) 0.8 sin x t t 84pp= +` j
is(A) 0.4p s(B) 0.8p s(C) 1.25 s(D) 2.5 sQ. 18The system
represented by the input-output relationship( ) y t ( ) , x d t 0
>t 5t t =3 -#is(A) Linear and causal(B) Linear but not causal(C)
Causal but not linear(D) Neither liner nor causalQ. 19The second
harmonic component of the periodic waveform given in the figure has
an amplitude of(A) 0(B) 1(C)/ 2p(D)5 www.nodia.co.inGATE SOLVED
PAPER - EESI GNALS &SYSTEMSYEAR 2010TWO MARKSQ. 20( ) x t
isapositiverectangularpulsefrom1 1 t t to =- =+ withunitheightas
showninthefigure.Thevalueof( ) ( ) X d X where2w w w33-"
#istheFourier transform of( )} x tis.(A) 2(B) 2p(C) 4(D) 4pQ.
21Given the finite length input[ ] x n and the corresponding finite
length output[ ] y n of an LTI system as shown below, the impulse
response[ ] h n of the system is(A) -[ ] {1, 0, 0,1} h n= (B) -[ ]
{1, 0,1} h n=(C) -[ ] {1,1,1,1} h n=(D) -[ ] {1,1,1} h n=Common
Data Questions Q.22-23.Given( ) f tand( ) g tas show belowQ. 22( )
g tcan be expressed as(A)( ) ( ) g t ft 2 3 = - (B)( ) g t f t23 =
-` j(C)( ) g t f t 223= -` j(D)( ) g t f t2 23= -` jQ. 23The
Laplace transform of( ) g tis(A)( )s e e1 s s 3 5- (B)( )s e e1 s s
5 3-- -(C)(1 )seess32---(D)( )s e e1 s s 5 3-YEAR 2009ONE MARKQ.
24A Linear Time Invariant system with an impulse response( ) h
tproduces output ( ) y twhen input( ) x tis applied. When the
input( ) x t t -is applied to a system with impulse response( ) h t
t - , the output will be(A)( ) yt (B)( ( )) y t 2 t -(C)( ) y t t -
(D)( ) y t 2t -GATE SOLVED PAPER - EESI GNALS &SYSTEMS
www.nodia.co.inYEAR 2009TWO MARKSQ. 25A cascade of three Linear
Time Invariant systems is causal and unstable. From this, we
conclude that(A) each system in the cascade is individually causal
and unstable(B) at least on system is unstable and at least one
system is causal(C) at least one system is causal and all systems
are unstable(D) the majority are unstable and the majority are
causalQ. 26TheFourierSeriescoefficientsofaperiodicsignal( ) x t
expressedas ( ) x t a e/k j kt Tk2=33 p=-/ are given by2 1 a j2=
--,0.5 0.2 a j1= +-, a j20= , . . a j 05 021= - , a j 2 12= +and a
0k=fork 2 >Which of the following is true ?(A)( ) x thas finite
energy because only finitely many coefficients are non-zero(B)( ) x
thas zero average value because it is periodic(C) The imaginary
part of( ) x tis constant(D) The real part of( ) x tis evenQ. 27The
z-transform of a signal[ ] x n is given byz z z z 4 3 2 6 23 1 2 3+
+ - +- -It is applied to a system, with a transfer function( ) H z
z 3 21= --Let the output be[ ] y n. Which of the following is true
?(A)[ ] y n is non causal with finite support(B)[ ] y n is causal
with infinite support(C)[ ] ; y n n 0 3 > =(D)[ ( )] [ ( )][ (
)] [ ( )] ;Re ReIm ImY z Y zY z Y z = = = =-(B)[ ] ; , , [ ] , [ ]
[ ] h n n n h h h 0 1 1 1 1 0 1 2 < > = - - = = =(C)[ ] 0; 0,
3, [0] 1, [1] 2, [2] 1 h n n n h h h < > = =- = =(D)[ ] ; , ,
[ ] [ ] [ ] [ ] h n n n h h h h 0 2 1 2 1 1 0 3 < > = - - = =
- =- =Q. 49Thediscrete-timesignal[ ] ( ) x n X
znz23nnn02=+3=/,wheredenotesa transform-pair relationship, is
orthogonal to the signal(A)[ ] ( ) y n Y z z32 nnn1 10) =3=-`j/
(B)[ ] ( ) ( ) y n Y z n z 5( ) nnn2 202 1) = -3=- +/(C)[ ] ( ) y n
Y z z 2 nnn3 3) =33 -=--/(D)[ ] ( ) y n Y z z z 2 3 14 44 2) = + +-
-Q. 50A continuous-time system is described by( ) y t e( ) x t=-,
where( ) y tis the output and( ) x tis the input.( ) y tis
bounded(A) only when( ) x tis bounded(B) only when( ) x tis
non-negative(C) only for t 0 #if( ) x tis bounded for t 0 $(D) even
when( ) x tis not boundedQ. 51The running integration, given by( )
( ) y t x t dt' 't=3 -#(A) has no finite singularities in its
double sided Laplace Transform( ) Y s(B) produces a bounded output
for every causal bounded input(C) produces a bounded output for
every anticausal bounded input(D) has no finite zeroes in its
double sided Laplace Transform( ) Y sYEAR 2005TWO MARKSQ. 52For the
triangular wave from shown in the figure, the RMS value of the
voltage is equal toGATE SOLVED PAPER - EESI GNALS &SYSTEMS
www.nodia.co.in(A) 61(B) 31(C) 31(D) 32Q. 53The Laplace transform
of a function( ) f tis( )( 2 2)F ss s ss s 5 23 622=+ ++ + as, ( )
t f t " 3approaches(A) 3(B) 5(C) 217(D) 3Q. 54The Fourier series
for the function( ) sin f x x2=is(A) sin sin x x 2 + (B)cos x 1 2
-(C) sin cos x x 2 2 + (D). . cos x 05 05 2 -Q. 55If( ) u t
istheunitstepand( ) t d
istheunitimpulsefunction,theinversez-transform of( ) F z1 z1=+ for
k 0 >is(A) ( ) ( ) k 1 kd - (B)( ) ( ) k 1 kd --(C) ( ) ( ) u k
1 k- (D)( ) ( ) u k 1 k--YEAR 2004TWO MARKSQ. 56The rms value of
the resultant current in a wire which carries a dc current of 10 A
and a sinusoidal alternating current of peak value 20 is(A) 14.1
A(B) 17.3 A(C) 22.4 A(D) 30.0 AQ. 57The rms value of the periodic
waveform given in figure is(A) 2 6 A(B) 6 2 A(C)/ 4 3A(D)
1.5A*********** www.nodia.co.inGATE SOLVED PAPER - EESI GNALS
&SYSTEMSSOLUTI ONSol . 1Option (A) is correct.Given, the
maximum frequency of the band-limited signalfm5kHz =According to
the Nyquist sampling theorem, the sampling frequency must be
greater than the Nyquist frequency which is given asfN2 2 5 10kHz
fm #= = =So, the sampling frequencyfs must satisfyfsfN$fs10kHz
$only the option (A) does not satisfy the condition therefore,5kHz
is not a valid sampling frequency.Sol . 2Option (A) is
correct.Given, the signalv t ^ hsin cos sin t t t 30 100 10 300 6
5004= + + +p^ hSo we have 1w 100 / rad s = ; 2w 00 / rad s 3 =and
3w 00 / rad s 5 =Therefore, the respective time periods
areT1sec210021wp p= =T2sec230022wp p= =T3sec5002p=So, the
fundamental time period of the signal isL.C.M., T T T1 2 3 ^ h , ,2
,2 ,2HCFLCM100 300 500p p p=^ ^hhor, T0 1002p=Thus, the fundamental
frequency in rad/ sec is 0w 100 / rad s102p= =Sol . 3Option (C) is
correct.Ifthetwosystemswithimpulseresponseh t1^ handh t2^
hareconnectedin cascaded configuration as shown in figure, then the
overall response of the system is the convolution of the individual
impulse responses.GATE SOLVED PAPER - EESI GNALS &SYSTEMS
www.nodia.co.inSol . 4Option (C) is correct.For a system to be
casual, the R.O.C of system transfer function H s ^ h which is
rational should be in the right half plane and to the right of the
right most pole.For the stability of LTIsystem. All poles of the
system should lie in the
lefthalfofS-planeandnorepeatedpoleshouldbeonimaginaryaxis.Hence,
options (A), (B), (D) satisfies both stability and causality an
LTIsystem.But, Option (C) is not true for the stable system as,S 1
=have one pole in right hand plane also.Sol . 5Option (C) is
correct.Given, the inputx t ^ hu t 1 = - ^ hIts Laplace transform
isX s ^ h se s=-The impulse response of system is givenh t ^ ht u t
= ^ hIts Laplace transform isH s ^ h s12=Hence, the overall
response at the output isY s ^ hX s H s = ^ ^ h h se s3=-Its
inverse Laplace transform isy t ^ h tu t2112=--^^hhSol . 6Option
(B) is correct.Given, the impulse response of continuous time
systemh t ^ ht t 1 3 d d = - + - ^ ^ h hFrom the convolution
property, we knowx t t t0d -*^ ^ h hx t t0= - ^ hSo, for the inputx
t ^ hu t =^ h (Unit step funn)The output of the system is obtained
asy t ^ hu t h t =*^ ^ h hu t t t 1 3 d d = - + -*^ ^ ^ h h h6 @u t
u t 1 3 = - + - ^ ^ h hAt t 2 = y 2 ^ hu u 2 1 2 3 = - + - ^ ^ h h1
=Sol . 7Option (C) is correct.[ ] x n[ ] u n3121n n= -b b l l[ ] [
1] ( ) u n u n u n313121n n n= + - - --b b b l l lTaking
z-transformX z6 @[ ] [ ] z u n z u n31311nnnnn n= + - -3333---=-
=-b b l l/ / www.nodia.co.inGATE SOLVED PAPER - EESI GNALS
&SYSTEMS[ ] z u n21 nnn-33-=-bl/ z z z313121nnnnnnnnn 010= +
-333-=--=---=b b b l l l/ / / zzz 313121I II IIInnmmnn 0 1 0= + -3
3 3= = =b b b l l l1 2 3 44 44 1 2 3 44 44 1 2 3 44 44/ / /Taking m
n =-Series I converges if z 311 Series I I converges ifz311 t 5t t
=3 -#Causality :( ) y tdepends on( ), x t t 5 0 >system is
non-causal.For example t 2 =( ) y 2depends on( ) x 10(future value
of input)Linearity :Output is integration of input which is a
linear function, so system is linear.Sol . 19Option (A) is
correct.Fourier series of given function( ) x t cos sin A a n t b n
tnn n 010 0w w = + +3=/( ) x t a ( ) x t =-odd functionSo,A 00=a
0n=bn( )sinTx t n t dt2T00w =# www.nodia.co.inGATE SOLVED PAPER -
EESI GNALS &SYSTEMS( ) ( ) sin sinTn t dt n t dt21 1//TTT0 02
02w w = + -= G # # cos cosT nn tnn t 2//TTT0002002wwww=---c c m m =
G( ) ( ) cos cos cosn Tn n n21 20wp p p = - + -6 @( )n21 1 np= -- 6
@bn ,,nnn40oddevenp =*So only odd harmonic will be present in( ) x
tFor second harmonic component ( ) n 2 =amplitude is zero.Sol .
20Option (D) is correct.By parsvals theorem( ) X d21 2pw w33-#( ) x
t dt2=33-#( ) X d2w w33-#2 2 # p = 4p =Sol . 21Option (C) is
correct.Given sequences[ ] x n -{ , }, n 1 1 0 1 # # = -[ ] y n -{
, , , , }, n 1 0 0 0 1 0 4 # # = -If impulse response is[ ] h n
then[ ] y n[ ]* [ ] h n x n =Length of convolution ( [ ]) y nis 0
to 4,[ ] x n is of length 0 to 1 so length of[ ] h n will be 0 to
3.Let[ ] h n -{ , , , } a b c d =Convolution[ ] y n -{ , , , , } a
ab b c c d d = - + - + - + -By comparinga1 =ab - + b a 0 1 & =
= =b c - + c b 0 1 & = = =c d - + d c 0 1 & = = =So,[ ] h n
-{ , , , } 1 1 1 1 =GATE SOLVED PAPER - EESI GNALS &SYSTEMS
www.nodia.co.inSol . 22Option (D) is correct.We can observe thatif
we scale( ) f tby a factor of 21 and then shift, we will get ( ) g
t .First scale( ) f tby a factor of 21( ) g t1( / 2) f t =Shift( )
g t1 by 3,( ) g t ( ) g t f t3231= - =-` j( ) g t f t2 23= -` jSol
. 23Option (C) is correct.( ) g tcan be expressed as( ) g t ( 3) (
5) u t u t = - - -By shifting property we can write Laplace
transform of( ) g t( ) G ssese1 1 s s 3 5= -- -(1 )seess32= ---Sol
. 24Option (D) is correct.Let( ) ( ) x t X sL( ) ( ) y t Y sL( ) (
) h t H sLSo output of the system is given as( ) Y s ( ) ( ) X s H
s =Now for input( ) x t t - ( ) e X s (shiftingproperty)sLt -( ) h
t t - ( ) e H ssLt -So now output is'( ) Y s ( ) ( ) e X s e H ss
s$ =t t - -( ) ( ) e X s H ss 2=t -( ) e Y ss 2=t -'( ) y t ( 2 ) y
t t = -Sol . 25Option (B) is correct.Let three LTI systems having
response( ), ( ) H z H z1 2 and( ) H z3 areCascaded as showing
belowAssume( ) H z11 z z2 1= + +(non-causal)( ) H z21 z z3 2= +
+(non-causal) www.nodia.co.inGATE SOLVED PAPER - EESI GNALS
&SYSTEMSOverall response of the system( ) H z ( ) ( ) ( ) H z H
z H z1 2 3=( ) H z ( )( ) ( ) z z z z H z 1 12 1 3 23= + + + +To
make( ) H zcausal we have to take( ) H z3 also causal.Let( ) H z3z
z 16 4= + +- -( )( )( ) z z z z z z 1 1 12 1 3 2 6 4= + + + + + +-
-( ) H z causal "Similarly to make( ) H zunstable atleast one of
the system should be unstable.Sol . 26Option (C) is correct.Given
signal( ) x t a e/k j kt Tk2=33p=-/Let 0wis the fundamental
frequency of signal( ) x t( ) x t a ek jk tk0=33w=-/ T20apw =( ) x
t a e a e a a e a ej t jt jt j t221 0 1 22 0 0 0 0= + + + +w w w
w----(2 ) (0.5 0.2) 2 j e j e jjt jt 2 0 0= - + + + +w w - - (0.5
0.2) (2 ) e j ejt j t 2 0 0+ - + +w w2e e j e ej t j t j t j t 2 2
2 2 0 0 0 0= + + - +w w w w - -6 6 @ @ 0.5 0.2 2 e e j e e jjt jt
jt jt 0 0 0 0+ - - +w w w w - + -6 6 @ @2(2 2 ) (2 2 ) 0.5(2 ) cos
sin cos t jj t t0 0 0w w w = + + - 0.2 (2 ) 2 sin jj t j0w +4 2 2 2
. 2 cos sin cos sin t t t t j 040 0 0 0w w w w = - + + + 6 @[ ( )]
Imx t 2 (constant) =Sol . 27Option (A) is correct.Z-transform of[ ]
x n is( ) X z z z z z 4 3 2 6 23 1 2 3= + + - +- -Transfer function
of the system( ) H z z 3 21= --Output( ) Y z ( ) ( ) H z X z =( ) Y
z (3 2)(4 3 2 6 2 ) z z z z z1 3 1 2 3= - + + - +- - -12 9 6 18 6 8
6 4 12 4 z z z z z z z z z4 2 1 2 3 1 2 3= + + - + - - - + -- - - -
-z z z z z z 12 8 9 4 18 18 44 3 2 2 3= - + - - + -- - -Or
sequence[ ] y n is[ ] y n[ ] [ ] [ ] [ ] n n n n 12 4 8 3 9 2 4 d d
d d = - - - + - - - 18 [ 1] 18 [ 2] 4 [ 3] n n n d d d + + + - +[ ]
y n0 =Y ,0 n
