Signals and Systems-1
EE_GATE_2020_05
Signals and Systems-1
Subject No.of Sections Total Marks 2
Duration Start Time End Time
45 mins 08-05-2019 14:00:00 15-02-2020 23:59:00 25
One Mark Questions
Section
Total Questions : 5
Max Marks : 5
What are the frequencies (in Hertz) present in the signal x(t) = cos(2t)sin(4t)
(C)2,6
(D) Cant determined
Correct Answer: A
Not Attempted: 0
Hide Solution
Sol Challenge
Given x(t) = cos(2t)sin(4t) x(t) = 0.5 sin(6t) + 0.5 sin(2t)
6
So, the frequencies present in the x(t) are –
2 Hz
2721
Consider the following input output equation y(n) = 2x(n-1) + x(n-3). The impulse response for above system is
(A) Having finite support and stable (B) Having infinite support and stable (C) Having finite support and unstable (D) Having infinite support and unstable
Correct Answer: A
Not Attempted: 0
Hide Solution
Sol Challenge
y(n)=2x(n − 1) + x(n-3) Impulse response is given by
h(n) = 2 (n-1) + 8(n-3)
h(1)
+
1
0 1 2 3 Width of h(n) is finite Hence, h(n) is of finite support
Ž h(n) = finite
f1=-00
Hence h(n) is stable
3.
بر ارا|
Let the impulse response of one system is given as h, (n)={1 -1, 1), find another system h (n) which is connected in parallel to hi(n) make whole system memory less is
(A) h2(n) = {2, -1, 1}
(B) h (n) = {0, 2, -1}
(C) h (n) = {0,1,-1}
(D) h2(n)= {2, -2,2}
Correct Answer: C
Not Attempted: 0
Hide Solution
on
Video Solution
Sol Challenge
We know that for memory less system overall impulse response is (n) If two systems are connected in parallel, then overall impulse response
h(n)=h;(n) +h2(n) By eliminating option if we take option C
h(n)= {1,-1, 1} + {0, 1,-1} h(n)={1,0,0} = (n)
The Impulse response of LTI system is given as h(n) = eu(n)+ etu(-n). This system is stable if
(A) a is positive and B is positive (B) a is positive and B is negative (C)a is negative and B is positive (D) a is negative and B is negative
Correct Answer: C
Not Attempted: 0
Hide Solution
O Sol Challenge
fl=-00
A system is stable if į b[n)<o Since h(n)=e**u(n)+ePu(-n) ŠHON=%e* + %*
fl=
fl=0
n=-00
0 > 0
Hence for u(n) is existed between 0
and u(-n) is existed between -o Hence a <0 and B > 0 Hence (c) option is correct
5.
A continuous time periodic signal x(t) is real valued and has a fundamental period T = 8 and Exponential Fourier Series Coefficients are X1 = X 1 = 2; X3 = X* = 4j, then the signal x(t) is
(A) 4 cos * + esin
(B) 4 cos * – 8 sin 31
(C) - 4 cos TT + 8 sin 31
(D) – 4 cos *t - 8 sin 31
Correct Answer: B
Not Attempted:0
Hide Solution
Video Solution
Sol Challenge
Exponential Fourier Series Expansion of x(t) is x(t)=
ej2anfot
n
f=–00
j2rnfot
N=-1,-3
x(0)= x, eins ={sie * * * 2*)+(xsef x se*
CD
= 4com)+(-8)sin() ::$(t)= 4co(1) - ssin( 31)
Two Marks Questions
Section
Total Questions : 10
Max Marks : 20
6.
Let the signal x(t) is given by
X(t)
نيا
Let y(t) is Even part of x(t), then the energy of y(t) is_
Correct Answer: 4
Answer Range: (4,4)
Not Attempted: 0
Hide Solution
O Sol Challenge
y(t) =Even part of X(t) = -
_ x(t)+x(-t)
X(-1)
-37-it
Hence, y(t)= X(t)+x(-t)
y(1)
y(t)
2/2
-3
- 2/2
-2/2
Energy of y(t) is Eye
Exo = faydt + $(–1°dt + ${–19 dt = fata
-4
= 1+1+1+1
Ey(t) = 4J
Let the impulse response of one system is shown in figure
hi(t)
-1
0
The impulse response of another system is hz(t) = 8(t-1) + 8(t-2) + 8(t-3). These two are cascaded. Then the over all impulse response is 1. Causal 2. Stable
(A) Both 1, and 2 are not correct (B) Only 1 is correct (C) Only 2 is correct (D) Both 1 and 2 are correct
Correct Answer: D
Not Attempted: 0
Hide Solution
Sol Challenge
Since, h(t) = hi(t)*h2(t)
h(t) = [8(t-1) + 8(t - 2) + 6(t-3)]*hj(t) h(t) = hı(t-1) +h(t-2) +h(t-3)
hi(t-1)
hi(t-2)
hi(t-3)
After addition
h(t)
---
---
-
---
--
-
-
(a) Since h(t) is 0 for t<0, h(t) is causal
(b) Since I In ft Ydt <
So it is stable
ja
---
-
Consider two systems with following Input output equation System-1: y(n) = x(n+3) - x(n) System-2: y(n)=x(n+2)-2x(n - 1) The overall impulse response of cascaded system has existence from Ni SnS N2, then the value of N1 + N2 is
Correct Answer: -4
Answer Range: (-4,-4)
Not Attempted : 0
Hide Solution
Sol Challenge
For system-1
yı(n)= x(n+3) – x(n) Hence impulse response
ht(n)= S(n+ 3)-6(n) For system-2
ya(n) = x(n + 2)-2x(n-1)
hy(n)= 8(n + 2)-28(n-1) Let both system are cascaded
h(n)=h(1)*h2(n) where h(n) is over all impulse response
:: h(n) = [8(n+2) –28(n-1)]*[8(n+3) - 6(n)] We know that 8(n + a)* 8(n + B) = 8(n +a+B) .: h(n) = 8(n + 2)*8(n + 3) - 8(n + 2)*8(n) - 28(n − 1)*8(n + 3) + 28(n-1)*(n)
h(n) = (n +5) - 8(n + 2) - 28(n + 2) +26(n-1) = (n + 5) - 36(n +2)+28(n-1)
h(n)= {1,0,0,-3,0,0.2}
Hence Existence of above h(n) is -5 $nsi
N=-5, N2 = 1
N1 + N2 =-4
Consider an LTI system with input and output related through the equation
y(t)= 1 et-t)x(t – 4 )dt
Let the impulse response of above system is cascaded with another system having impulse response 8(t - 4). The overall impulse response system is
(A) h(t) = e(t - 4) u(t-4)
(B) h(t)= (-1) u(t-4)
(C) h(t) = e-t-8) u(t-8)
(D) h(t) =-et-3) (-t-3)
Correct Answer: C
Not Attempted: 0
Hide Solution
Sol Challenge
vlt)- Te-t-)s(-Aldo
19-j
A
u
If x(t) = 8(t), then y(t) =h(t)
h(t) je 4-1647 – 4)dt
:h(t) = e(t - 4) u(t-4) When this system is cascaded with another system hz(t)
h;(t) H
hg(t) ]
h(t) = hi(t)*hz(t) ..h(t) = e(t - 4) ust-4)*8 (t-4) We know that X(t + a) * 8(t +B)
= x(t+a+B) h(t) = e-t-4 - 4) u(t - 4 - 4) h(t) = e(t - 8) ust-8)
10.
If x(t) = =+3cos(2#t)+ 4 sin(167t). Then power content in 5th and 8th
harmonics are P1, P2 respectively. Then the value of Pi + P2 is
Watts
Correct Answer: 8
Answer Range: (8,8)
Not Attempted : 0
Hide Solution
Sol Challenge
x(t) = 1 + 3 cos(20)+4sin(1671) (1) 7: = 27. 1. -
L.C.M(1.1) 1
===1sec H.C.F(1,8) 1
=f, = = = 1sec
(3): 2info=21 2nfo=167
27.n.1=27 27.1=16A
n=1
n=8
..a1=3
be=4
Power content in gth harmonic = { b3 = * 4?
er cont
-X44
00
P2 = 8 watts Power content in 5th harmonic P1 = 0 watts P1 + P2 = 8 Watts
11.
il >>
Let x(t) be a signal whose Fourier transform is shown in figure
X(1)
21
Then the Magnitude spectrum of y(t) = txt)
|Y(f)
Y(f)
Y(f)
Y(1)
-
-1
Correct Answer: A
Not Attempted: 0
Hide Solution
Sol Challenge
y(t)=txt) According to frequency differentiation property
-j tx(t) > x(@) :: tx(0) *>jxm)
|Y()
12.
The inverse Fourier transform of X(0)=e 2 is
(C) HIFT)
Correct Answer: D
Not Attempted: 0
Hide Solution
Video Solution
Sol Challenge
Put a = 1/2
1
Apply duality property
To(1+4+)
13.
The magnitude and phase spectrum of a signal x(t) is as shown below. The value of x(t) at t = 2 sec is
f|X(0)
+
X00)
EN
NA
BN
Correct Answer: 0.955
Answer Range: (0.9, 1)
Not Attempted: 0
Hide Solution
Sol Challenge
X(0)= X(O). ej_X(6)
Inverse
lier Iri
X(m) = 3 2142; 5*50so
=3 43*2; 0505 X(m)= 3; ; ; soso
=-3j; 0 50 51 Inverse Fourier Transform is x()= x(o)"des x1 = - Lj Biomaot | x-je-drs] 20-802, 0) x) = 2 ( - - )-em-1)
/27
of
210 - A[2–2011
x() 2 = 3 (2+1) = * = 0.955
14.
il >>
A periodic signal x(t) of period To is given by
1; tkd 0; datKT,/2
X(t)=
The d.c. component of x(t) is_
(A) d/T.
(B) d/2T
(C) 20/T,
(D)T/
Correct Answer: C
Not Attempted: 0
Hide Solution
Sol Challenge
-
1
To
-d od
T.
1 tatlo
1
. 2d
20=To \x{t)dt =T, dt = To
15.
The convolution of sinc(t) with sinc-(t/2) is
il >>
(A) sinct
(B) sinc?(t/2)
(C) sinc (t)
(D) sinc (t/2)
Correct Answer: B
Not Attempted : 0
Hide Solution
Sol Challenge
Arect ^ <>AT sin c(FT)
rect (t) sinc(t)
sinc(1) rect(f)
Atri() -AT sine°(ET) Tri(20) *sinc () sine ( ) <> Thi(2)
i(2)
X(t)=sin c(t),y(t)
x(t) * y(t) + X(f). Y(f)
= rect(f). 2Tri(21) = 2 Tri(21)
x(0) * ¥€) = sinco ()
0 Comments