Signals & Systems Test - 3 - PDF Flipbook
Signals & Systems Test - 3
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GATE
EEE
Signals
&
Systems
Test-03Solutions
SIGNALS & SYSTEMS
1. The discrete-time equation y(n + 1) + 0.5n y(n) = 0.5x(n + 1) is
not attributable to a
a) memoryless system
b) time-varying system
c) linear system
d) causal system
Answer: (a)
Solution:
Since the output for each value of the independent variable at a
given time is not dependent on the input at the same time, hence
the system is having memory. Therefore, the system is not
attributable to a memoryless system.
NOTE: A system is said to be memoryless if its output for each
value of the independent variable at a given time is dependent
only on the input at the same time.
i.Memoryless System is also called static system.
ii.A System having memory is also called dynamic system.
2. A system defined by y[n] = ∑nk= −∞ x[K] is an example of
a) invertible system
b) memoryless system
c) non-invertible system
d) averaging system
Answer: (a)
1
Solution
Y[n] = ∑nK=−∞ X[K]
If a system is invertible, then an inverse system exists that, when
cascaded with the original system, yields an output w[n] equal
to the input x[n] to the first system
3. If the response of LTI continuous time system to unit step input
is �12 − 1 e−2t�, then impulse response of the system is
2
a) �12 − 1 e−2t�
2
b) e−2t
c) (1 − e−2t)
d) Constant
Answer: (b)
Solution:
Impulse response = d [Step response]
dt
= d �12 − 1 e−2t�
dt 2
= e−2t
4. The phase response of a pass band waveform at receiver is given
by φ(f) = −2πα (f − fC) − 2πβfC where fC is the center
frequency, and α and β are positive constants. The actual signal
propagation delay from the transmitter to receiver is
2
a) α−β
α+β
b) αβ
α+β
c) α
d) β
Answer: (c)
Solution:
tp(t) = −2θπ(ff)�f = = β
fc
tg(t) = −1 dθd(ff)�f = = α
2π
fc
5. Which one of the following is a Dirichlet condition?
a) ∫t∞1 |x(t)| < ∞
b) Signal x(t) must have a finite number of maxima and minima
in the expansion interval
c) X(t) can have a finite number of finite discontinuities in the
expansion interval
d) X2(t) must be absolutely summable
Answer: (b)
Solution:
Dirichlet’s condition states that:
i) Over any period signal x(t) must be absolutely integrable i.e.
∫T|x(t)|dt < ∞
ii) In any finite interval of time x(t) is of bounded variations i.e.
there are no more than a finite number of maxima and minima
during any single period of signal.
3
iii) In any finite interval of time there are only finite number of
discontinuities. Furthermore, each of these discontinuous is
finite.
6. The relation between input x(t) and output y(t) of a continuous-
time system is given by dy(t) + 3y(t) = x(t). What is the forced
dt
response of the system when x(t) = k (a constant)?
a) K
b) k/3
c) 3k
d) 0
Answer: (b)
7. If F(s) = L[f(t)] = (S + k + 4)is given by
1)(S2
a) K/4
b) Zero
c) Infinite
d) Undefined
Answer: (d)
Solution:
F(s) has poles at s = – 1, s = ± j2.
As all the poles are not in the open left half of the s plane the
final value theorem cannot be applied and tl⟶im∞ f(t) is undefined.
4
8. A solution for the differential equation Ẋ (t) + 2x(t) = δ(t) with
initial condition X (0–) = 0 is
a) e – 2t u(t)
b) e 2t u(t)
c) e– t u(t)
d) et u(t)
Answer: (a)
Solution:
X (t) + 2 X (t) = δ(t) ……(1)
X (0–) = 0 ……(2)
Taking Laplace transform on both sides of (1) with zero initial
condition (2),
(S + 2) X(s) = 1
X(s) = 1/(S + 2)
∴ X(t) = e−2t u(t)
Note that the impulse response of the 1st order system is asked in
the question.
9. Which one of the following transfer functions represents the
critically damped system?
a) H1(S) = S2 1
+ 4S + 4
b) H2(S) = S2 + 1 + 4
3S
c) H3(S) = S2 + 1 + 4
2S
d) H4(S) = S2 1 + 4
+S
Answer: (a)
5
Solution:
H1(s) = 1
S2+4S+4
= (S 1
+ 2)2
Since both roots are negative real and equal, it is a critically
damped system.
10. What is the Laplace transform of a delayed unit impulse
function δ(t – 1)?
a) 1
b) Zero
c) Exp (– s)
d) S
Answer: (c)
Solution:
x(t) = δ (t – 1)
X(s) = e– S .1
X(s) = exp (– S)
11. Consider a system with transfer function H(S) = 3S2 − 2 2. The
S2 + 3S +
step response of the system is given by
a) C(t) = 5e– t – e– t – 1
b) C(t) = 3δ(t) – 10e–2t + e–t
c) C(t) = 4e–t – e–2t – 1
d) C(t) = 2( 1 – e–2t)
Answer: (a)
6
Solution:
H(s) = 3S2−2 2 = C(s)
S2 + 3S + R(s)
Since the input signal is step so R(s) = 1
s
So C(s) = �3s2 − 2� 2)
s(s + 1)(s +
= −1 − (S 1 1) + (S 5 2)
S + +
Taking its inverse Laplace transform
C(t) = – 1 – e–t + 5 e–2t
12. A system is defined by its impulse response h(n) = 2n u (n – 2).
The system is
a) stable and causal
b) causal but not stable
c) stable but not causal
d) unstable and non-causal
Answer: (b)
Solution:
The impulse response of the system
h(n) = 2n u (n – 2) = 0, n ≤ 1
= 2n, n ≥ 2
The system is causal because h(n) = 0 for n < 0.
But the system is not stable because the output y(n) = h(n) is
unbounded due to the factor 2n for the bounded input,
X(n) = δ (n) = 1, n = 0
= 0, n ≠ 0
7
13. Consider the compound system shown in the below figure. Its
output is equal to input with a delay of two units.
If the transfer function of the first system is given by H1(Z) =
Z − 0.5 then the transfer function of the second system would be
Z − 0.8
a) H2(Z) = Z−2 − 0.2Z−3
1 − 0.4Z−1
b) H2(Z) = Z−2 − 0.8Z−3
1 − 0.5Z−1
c) H2(Z) = Z−2 + 0.2Z−3
1 + 0.4Z−1
d) H2(Z) = Z−2 + 0.8Z−3
1 + 0.5Z−1
Answer: (b)
Solution:
y(n) = x (n – 2)
y(z) = Z–2 X(z)
⇒ H(z) = H1(z)H2(z) = Y(Z) = Z−2
X(Z)
⇒ �ZZ − 00..85� H2(Z) = Z−2
−
⇒ H2(Z) = Z −1 + 0.8Z −2
z − 0.5
⇒ H2(Z) = Z −1 + 0.8Z −2
1 − 0.5Z −1
14. Decimation is the process of
a) Retaining sequence values of Xp [n] other than zeroes
b) Retaining all sequence values of Xp [n]
c) Dividing the sequence value by 10
8
d) Multiplying the sequence value by 10
Answer: (d)
Solution:
Decimation is the process of time scaling in discrete time
sequence when scaling factor is greater than ‘1’. i.e. α > 1
15. The pole - zero diagram of a causal and stable discrete-time
system is shown in the figure. The zero at the origin has
multiplicity 4. The impulse response of the system is h[n]. If h
[0] = 1, we can conclude.
a) h(n) is real for all n
b) h(n) is purely imaginary for all n
c) h(n) is real for only even n
d) h(n) is purely imaginary for odd n
Answer: (a)
Solution:
H(z) = Z4 1
Z4 + 4
H(z) = H (Z – 1)
h(n) = h (– n)
so h(n) is real for all ‘n’
9
16. A discrete-time signal x[n] = δ [n – 3] + 2δ [n – 5] has z-
transform X (Z). If Y(z) = X (–z) is the z-transform of another
signal y[n], then
a) Y[n] = x[n]
b) Y[n] = x [–n]
c) Y[n] = – x[n]
d) Y[n] = – x [–n]
Answer: (c)
Solution:
(a)nx(n) ⟷ X(Z/a)
a=–1
(−1)nx(n) ⟷X (–Z)
But x(n) = δ (n – 3) + 2δ (n – 5)
Y(n) = (−1)nx(n) = (−1)n [δ (n – 3) + 2δ (n – 5)]
Y(n) = – δ (n – 3) – 2δ (n – 5) = – x(n)
17. For a Z-transform X(Z)= z�2Z − 65� 31�.
�Z − 12��Z −
Match List-I (The sequences) with List-II (The region of
convergence) and select the correct answer using the codes
given below the lists:
List-I
A. [(1⁄2)n + (1⁄3)n]u(n)
B. (1⁄2)nu(n) − (1⁄3)nu(−n − 1)
C. −(1⁄2)nu(−n − 1) − (1⁄3)nu(n)
D. −[(1⁄2)n + (1⁄3)n]u(−n − 1)
10
List-II
1. (1⁄3) < |Z| < (1⁄2)
2. |Z| < (1⁄3)
3. |Z| < (1⁄3) &|Z| > (1⁄2)
4. |Z| > (1⁄2)
Codes:
A BC D
a) 4 2 1 3
b) 1 3 4 2
c) 4 3 1 2
d) 1 2 4 3
Answer: (c)
Solution:
X(Z) = z�2Z − 56� = z z 1 + z z 1
�z − 21��z − 31� − 2 − 3
We know that, X(z) = z z a
−
⇒ x(n) = anu(n), |z| > |a|
Or x(n) = −anu(−n − 1), |z| < |a|
Then If |z| > 21,
x(n) = ��21�n + �31�n� u(n)
If 1 < |z| > 12,
3
Then x(n) = − �21�n u(−n − 1) + �31�nu(n)
If |z| < 13,
11
Then x(n) = − �21�n u(−n − 1) + �13�nu (– n – 1)
Then
If |z| < 1 & |z| < 1
3 2
x(n) = �12�n u(n) − �13�nu (– n – 1)
18. Algebraic expression for z-transform of x[n] is x[z]. What is
the algebraic expression for z-transform of ejω0nx[n]?
a) X (Z – Z0)
b) X�e−jω0Z�
c) X �ejω0Z�
d) X(Z) ejω0Z
Answer: (b)
Solution:
Property z-transform, X(n) ↔ X(z)
�ejω0�nx(n) ↔ X �ejzω0�
19. The difference equation for a system is given by Y (n + 2) + y
(n + 1) + 0.16y(n) = x (n + 1) + 0 32x(n) The transfer function
of the system is
a) Z + 0.32
Z2 + Z + 0.16
b) Z2 + 1 0.16
Z+
c) Z + 0.32
Z2 + 0.16
d) (Z − Z+ 0.32 0.16)
1)(Z2 +Z+
Answer: (a)
12
Solution:
Y (n + 2) + y(n + 1) + 0.16y(n) = x(n + 1) + 0.32 x(n)
Taking z-transform of both sides
(Z2 + z + 0.16) y(Z) = (Z + 0.32) X(Z)
Transfer function, H(Z) = (Z + 0.32)
(Z2 + Z + 0.06)
20. Let x(n) = �21�n, u(n) = X2(n), and Y = (ejω) be the Fourier
transform of y(n). Then y (ej0) is
a) 1/4
b) 2
c) 4
d) 4/3
Answer: (d)
Solution:
For the given, X(n) = �21�nu(n)
Y(n) = x2(n)= �12�2nu(n) = �14�nu(n)
If y(n) ⟶ Y(ejω),
Then according to the area property of DTFT
Y�ejθ� = ∑n∞=−∞ y(n) = ∑n∞=0 �41�n
RH side above is the geometric series of infinite number of
terms.
Y�ejθ� = 1 FT , |CR| < 1
− CR
= 1 1 = 4
− 1⁄4 3
13
21. Consider the system shown in the Fig below. The transfer
function Y(z)/X(z) of the system is
a) 1 + aZ−1
1 + bZ−1
b) 1 + bZ−1
1 + aZ−1
c) 1 + aZ−1
1 − bZ−1
d) 1 − bZ−1
1 + aZ−1
Answer: (a)
Solution:
Let X(K) ⟶X(Z), y(k) ⟶ Y(z)
The system block diagram is redrawn in write the following
relations:
E(Z) = X(Z) – bZ – 1 E(Z),
Y(Z) = E(Z) + aZ – 1 E(Z),
or E(Z) = (1 + 1 X(Z)
bZ−1)
14
or Y(Z) = (1 + aZ−1)E(z)
Y(z) = (1 + aZ−1) (1 + 1 X(z),
bZ−1)
∴ Transfer function = Y(Z) = �1 + aZ−1�
X(Z) (1 + bZ−1)
The answer can also be obtained probably quickly by using
Manson’s gain formula Gains of the two existing forward paths:
M1 = 1 and M2 = a Z–1
Gain of the (only one existing) individual loop = –b z–1
Δ = 1− ∑ Individual loop gains
= 1 + b Z–1 and Δ1 = 1, Δ2 = 1
∴ Y(Z) = M = M1∆1 + M2∆2 = 1 + az−1
X(Z) ∆ 1 + bz−1
22. A half duty cycle rectangular clock output is sampled at 5
times its rate. Spectrum of the sampled clock will be having
mirror image in negative frequency domain of a
a) Sampled sinc pulse
b) Periodic sinc pulse
c) Periodic sampled sinc pulse
d) None of the above
Answer: (c)
Solution:
The sampled signal will be a discrete time periodic signal thus
the Fourier transform will be a discrete frequency periodic
signal and the rectangular function has sinc function as its
transform hence option (c) is correct.
15
23. The modules of 1 + cosα + I sinα is
a) 2 sin α
2
b) 2 cos α
2
c) sin2α2 − 1
d) cos2α2 − 1
Answer: (b)
Solution:
Z = 1 + cos α + i sin α
|z| = �(1 + cos α)2 + sin2α
= √1 + 2 cos α + cos2α + sin2α
= �2(1 + cos α)
= �2 × 2cos2 �α2�
= 2 cos �α2�
24. The amplitude spectrum of a Gaussian pulse is
a) uniform
b) a sine functions
c) Gaussian
d) an impulse function
Answer: (c)
Solution:
A normalized gaussian pulse is defined as
X(t) = e−πt2 for all ‘t’, with X(0) = 1
16
A = Area under X(t) = ∫−∞∞ X(t)dt = 1.
X(t) ⟶ X(f) = e−πf2
X(f) is also a Gaussian pulse in frequency domain having the
same expression as X(t) in time-domain. Even if the Gaussian
pulse is not normalized i.e., X(0) ≠ 1 and A ≠ 1, its its fourier
Transform is also a Gaussian of the form B e−Cπf2, where B and
C are constants.
25. An LTI system has the input signal x(n). Which of the
following sequence of operations is most approximate to get
output y(n) = x(n – M/L)
a) Interpolation by L, Delay by M, Decimation by L
b) Delay by M, Interpolation by L, Decimation by M
c) Decimation by L, Delay by M, Interpolation by L
d) Interpolation by L, Decimation by L, delay by M
Answer: (b)
26. Which of the following function have a jump discontinuity at x
= 0?
a) g(x) = ln |x|
b) g(x) = 1/x2
c) sinx(x), x ≠ 0
0, x ≠ 0
d) None of these
Answer: (c)
17
Solution:
27. The impulse response of a linear time invariant system is h(n)
= {1, 2, 1, –1}. The response for the input Signal x(n) = {1, 2, 3,
1}is
a) {1, 8, 4, 8, 3, – 1, 2}
b) {1, 4, 8, 3, 8, – 2, – 2}
c) {1, 4, 8, 8, 3, – 2, – 1}
d) {1, 8, 3, 8, 8, 4, –1}
Answer: (c)
Solution:
X(n) = {1, 2, 3, 1}
h(n) = {1, 2, 1, –1]
18
The sum of each diagonal element will give the response of the
input.
X(n) * h(n) = {1, 4, 8, 8, 3, – 2, – 1}.
28. Consider the sequence x[n] = [– 4 – j5, l + j2, 4) The conjugate
anti - symmetric part of the sequence is
a) [– 4 – j 2.5 j2 4 – j 2.5]
b) [ – j 2.5 1 j 2.5]
c) [–j 5 j2 0]
d) [– 4 1 4]
Answer: (a)
Solution:
A complex sequence x(n) is conjugate symmetric (even), if X(n)
= x* (– n).
A complex sequence x(n) is conjugate anti symmetric (odd), if
X(n) = – x* (– n).
If a complex sequence x(n) is neither conjugate even nor
conjugate odd, x(n) can be decomposed into even conjugate part
Xc(n) and odd conjugate part X0(n).
19
Xc(n) = x(n)+x∗(−n)
2
Xc(n) = x(n)−x∗(−n)
2
For the given sequence x(n) = [– 4 – j5, 1 + j2,4]
X*(n) = [– 4 – j5 1 + j2 4]
X*(– n) = [ 4 1 + j2 – 4 – j5]
∴ X0(n) = �−28 − j5 , j4 , 8 − j25�
2 2 2
∴X0(n) = [– 4 – j2.5 j2 4 – j2.5]
29. There is a function f(x), such that f(0) = 1 and f ′ (0) = – 1 is
positive for all values of x, then
a) f ′′ (x) < 0 for all x
b) – 1< f ′′ (x) < 0 for all x
c) – 2 < f ′′ (x) < –1 for all x
d) None of these
Answer: (d)
Solution:
f(x) = e−x ⇒ f (0) = e –0 = 1
f ′(x) = −e−x ⇒ f ′ (0) = –e –0 = –1
f(x) = e – x is + ve for ∀ x
f ′′(x) = e – x is + ve for ∀ x
30. Homogeneous solution of y(n) – 9 y (n – 2) = x (n – 1) is
16
a) C1 �34�n+C2 �− 43�n
b) C1 �− 34�n−1+C2 �43�n−1
20
c) C1 �43�n
d) C1 �− 34�n
Answer: (a)
Solution:
Taking z-transform of given difference equation
Y(z) – 9 Z−2Y(z) = Z−1Y(z)
16
⇒ y(z) = Z−1 = z
9 − 196�
1 − 16 Z−2 �Z2
So for homogeneous solution or complementary solution.
Z2 − 9 = 0
16
⇒ Z = ± 3
4
So homogenous solution is given as C1 �43�n + C2 �− 43�n
21
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