Electrical Engineering - Signals and Systems
Exam Duration: 45 Mins Total Questions : 30
The raised cosine pulse x(t) is defined as \(sgn(t)=\begin{cases} \frac { 1 }{ 2 } (cos\omega ),\quad -\frac { \pi }{ \omega } \le t\le \frac { \pi }{ \omega } \\ 0\quad ,otherwise \end{cases}\) The total energy of x(t) is
- (a)
\(\frac { 3\pi }{ 4\omega } \)
- (b)
\(\frac { 3\pi }{ 8\omega } \)
- (c)
\(\frac { 3\pi }{ \omega } \)
- (d)
\(\frac { 3\pi }{ 2\omega } \\ \)
The impulse response of a continuous time LTI system is h(t)=\(e^{ -t }\) u(t-2) The system is
- (a)
casual and stable
- (b)
casual but not stable
- (c)
Stable but not casual
- (d)
neither casual nor stable
Consider x(k)=\(7\left( \frac { 1 }{ 3 } \right) ^{ k }\)u(-k-1)-6\(\left( \frac { 1 }{ 2 } \right) ^{ k }\) The system is
- (a)
casual
- (b)
anti-casual
- (c)
non-casual
- (d)
None of these
The impulse response of a casual linear time invariant system is given as h(t) Now, consider the following two statements:
Statement (i) principle of superposition holds
Statement (II) h(t)=0 (for t<0)
Which one of the following staements is correct?
- (a)
Statement (I) is correct and statement (II) is wrong
- (b)
Statement (II) is correct and statement (I) is wrong
- (c)
Both statement (I) and statement (II) are wrong
- (d)
Both statement (I) and statement (II) are correct
Consider a signal:
x(t) =2sin t+4sin4t+6cos 4t+2cos 2t with period 1/2
Thje signl power is
- (a)
30w
- (b)
36 w
- (c)
60 w
- (d)
12 w
For a particular signal average power in its six harmonic components as 10 mW each and fundamental component also has 10 mV power Then average power in the periodic signal will be
- (a)
70
- (b)
60
- (c)
10
- (d)
5
The z-transform of u[n] i
- (a)
\(\frac { 1 }{ 1-z^{ -1 } } |z|>1\)
- (b)
\(\frac { 1 }{ 1-z^{ -1 } } |z|<1\)
- (c)
\(\frac { Z }{ 1-z^{ -1 } } |z|<1\)
- (d)
\(\frac { Z }{ 1-z^{ -1 } } |z|>1\)
The Z-transform of \(\left( \frac { 1 }{ 4 } \right) ^{ n }\omega |n|-u|n-5|\)
- (a)
\(\frac { z^{ 5 }-0.25 }{ z^{ 4 }(z-0.25) } ,z>0.25\)
- (b)
\(\frac { z^{ 5 }-0.25^{ 5 } }{ z^{ 4 }(z-0.25) } ,z>0\)
- (c)
\(\frac { z^{ 5 }-0.25^{ 5 } }{ z^{ 3 }(z-0.25) } ,z>0.25\)
- (d)
\(\frac { z^{ 5 }-0.25^{ 5 } }{ z^{ 3 }(z-0.25) } ,z>\) all z
Given the Z-transform pair
\(X[n]\overset { z }{ \leftrightarrow } \frac { 32 }{ z^{ 2 }-16 } |z|<4\)
The Z-transform of the signal X[n-2] is
- (a)
\(\frac { z^{ 4 } }{ z^{ 2 }-16 } \)
- (b)
\(\frac { (z+2)^{ 2 } }{ (z+2)^{ 2 }-16 } \)
- (c)
\(\frac { 1 }{ z^{ 2 }-16 } \)
- (d)
\(\frac { (z-2)^{ 2 } }{ (z-2)^{ 2 }-16 } \)
The z-transform of the signal x[n]*x[n-3] is
- (a)
\(\frac { z^{ -3 } }{ (z^{ 2 }-16)^{ 2 } } \)
- (b)
\(\frac { z^{ 7 } }{ (z^{ 2 }-16)^{ 2 } } \)
- (c)
\(\frac { z^{ 5 } }{ (z^{ 2 }-16)^{ 2 } } \)
- (d)
\(\frac { z }{ (z^{ 2 }-16)^{ 2 } } \)
The Fourier transform of signal shown below isZ
- (a)
\(\frac { 2sin\omega -2 }{ \omega } \)
- (b)
\(\frac { 2cos\omega -2 }{ \omega } \)
- (c)
\(2j\omega cos\omega \)
- (d)
\(2j\omega sin\omega \)
The Fourier transform of signal sgn(t) is
- (a)
\(\frac { -2 }{ j\omega } \)
- (b)
\(\frac { 4 }{ j\omega } \)
- (c)
\(\frac { 2 }{ j\omega } \)
- (d)
\(\frac { 1 }{ j\omega } +1\)
Consider a continuous-time periodic signal x(t) with fundamental period T and Fourier series coefficients X[k] Determine the Fourier series coefficient of the signal y(t) given in question and choose the correct option
The Fourier series coefficient of the signal y(t)-Re[x(t)] is
- (a)
\(\frac { X[k]+X[-k] }{ 2 } \)
- (b)
\(\frac { X[k]-X[-k] }{ 2 } \)
- (c)
\(\frac { X[k]+X*[-k] }{ 2 } \)
- (d)
\(\frac { X[k]-X*[-k] }{ 2 } \)
The trapezoidal pulse x(t) is time scaled producting \(y(t)=x\left( \frac { t }{ 5 } \right) \) The sketch for y(t) is
- (a)
- (b)
- (c)
- (d)
The inverse Fourier transform of \(e^{ -2|\omega | }\) is
- (a)
\(\frac { 2 }{ \pi (4+t^{ 2 }) } \)
- (b)
\(\frac { 1 }{ 2\pi (4+t^{ 2 }) } \)
- (c)
\(\frac { 1 }{ \pi (4+t^{ 2 }) } \)
- (d)
None of these
The DTFT of \(\left( \frac { 1 }{ 2 } \right) ^{ -n }u[n-1]\) is
- (a)
\(\frac { e^{ j\omega } }{ 2-e^{ -j\Omega } } \)
- (b)
\(\frac { 2e^{ j\omega } }{ 2-e^{ -j\Omega } } \)
- (c)
\(\frac { e^{ j\omega } }{ 2-e^{ -j\Omega } } \)
- (d)
\(\frac { 2e^{ j\omega } }{ 2-e^{ -j\Omega } } \)
Find the signal \(X\left( 1-\frac { 3 }{ 2 } t \right) \)with the help of following diagram
- (a)
- (b)
- (c)
- (d)
Consider a stable and usual system with impulse response h(t) and system function H(s) is rational,contains a pole at s=-2 and does not have a zero at origin and location of all other zeros is unknown ,poles are present at some unknown location other than origin ,Then
1.H(s)=H(-s)
2.\(\frac { d }{ dt } h(t)\) contains at least one pole in its Laplace transform
- (a)
1-true,2-true
- (b)
1-false,2-true
- (c)
1-true,2-false
- (d)
1-false,2-false
Find the signal power of x(t) with the help of following diagram
- (a)
0.25 W
- (b)
0.75 W
- (c)
0.5 W
- (d)
1 W
The value of z{[k -1] u[k]) is
- (a)
\(\frac { z(z+2) }{ (z-1)^{ 2 } } \)
- (b)
\(\frac { 2z-z^{ 2 } }{ (z-1)^{ 2 } } \)
- (c)
\(\frac { z^{ 2 } }{ (z-1)^{ 2 } } \)
- (d)
None of these
x[n]=O;n<-l,n>O
x[-1] = -1, x [0] = 2 is the input and
y [n] =0, n < - 1,n > 2
y[-1] = -1 = y[1], y[O] = 3, y[2] = - 2
is the output of a discrete time LTI system. The system impusle of response h[n] will be
- (a)
h[n] =0, n <0, n > 2 h[0] =1, h[1], h[2] =-1
- (b)
h[n]=0,n<-1,n>1 h[-1] =1, h[0] = h[1] = 2
- (c)
h[n] =0; n <0, n >3 h[0] =- ,h[1] =2, h[2] =1
- (d)
h[n]=0;n<-2,n>1 h[-2] = h[1] = h[-1] = - h[0] =3
z-transform of the signal is given as \(\frac { 4z^{ 4 }+3z^{ 3 }+2z^{ 2 }+z+1 }{ z^{ 4 }+2z^{ 3 }+3z^{ 2 }+4z+1 } \) after applying a particular property, the z-transform was changed to \(\frac { 4z+3z+2z^{ 2 }+z+1 }{ z^{ 4 }+2z^{ 3 }+3z^{ 2 }+4z^{ 3 }+z^{ 4 } } \) The property used is
- (a)
time scaling
- (b)
time shift
- (c)
time reversal
- (d)
time expansion
The raised cosine pulse x(t) is defined as \(x(t)=\left\{ \frac { 1 }{ 2 } \left( cos\omega t+1,\quad -\frac { \pi }{ \omega } \le t\le \frac { \pi }{ \omega } \right) \right\} \) otherwise The total energy of x(t) is
- (a)
\(\frac { 3\pi }{ 4\omega } \)
- (b)
\(\frac { 3\pi }{ 8\omega } \)
- (c)
\(\frac { 3\pi }{ 2\omega } \)
- (d)
\(\frac { 3\pi }{ \omega } \)
An LTI system with impulse response \(h_{ 1 }\left[ n \right] =-2\left( \frac { 1 }{ 4 } \right) ^{ n }u\left[ n \right] \) is connected in parallel with another causal LTI system, with impulse response h2 [n]. The resulting parallel interconnection has the frequency response \(H(e^{ j\Omega ) }=\frac { -12+5e^{ -j\Omega } }{ 12-7e^{ -j\Omega }+e^{ -j2\Omega } } \) The h2 [n] is
- (a)
\(\left( -\frac { 1 }{ 3 } \right) ^{ n }u\left[ -n-1 \right] \)
- (b)
\(\left( \frac { 1 }{ 3 } \right) ^{ n }u\left[ n \right] \)
- (c)
\(\left( -\frac { 1 }{ 3 } \right) ^{ n }u\left[ n \right] \)
- (d)
\(\left( \frac { 1 }{ 3 } \right) ^{ n }u\left[ -n-1 \right] \)
The running integrator, given by \(y(t)=\int _{ \infty }^{ \infty }{ X(t') } dt'\)
- (a)
has no finite singularities in its double sided Laplace transform Y(s)
- (b)
produces a bounded output for every casual bounded input
- (c)
produces a bounded output for every anti-casual bounded input
- (d)
has no finite zeros in its double sided Laplace transform Y(s)
A casual system has input x[n]=(-3)0 u[n] and output
\(y[n]=\left[4(2)^n-\left(1\over 2\right)^n\right]u[n].\)
the impulse response of this system is
- (a)
\(\left[7(2)^n-10\left(1\over 2\right)^n\right]u[n].\)
- (b)
\(\left[7\left(1\over 2\right)^n-10\left(1\over 2\right)^n\right]u[n].\)
- (c)
\(\left[10\left(1\over 2\right)^n-7(2)^n\right]u[n].\)
- (d)
\(\left[10(2)^n-7\left(1\over 2\right)^n\right]u[n].\)
In the following network, the switch is closed at t = 0 and the sampling starts from t = 0. The sampling frequency is 10 Hz.
The samples x(n), where n =0,1,2, ... , is given by
- (a)
5(1-e-0.05n)
- (b)
5e-0.05n
- (c)
5(1- e-5n)
- (d)
5e-5n
A sequence x(n) has non-zero values as shown in figure.
The sequence
\(y(n)=\begin{cases} \times \left( \frac { n }{ 2 } -1 \right) \qquad for\quad n\quad even \\ 0\qquad for\quad n\quad odd \\ \quad \end{cases}\) will be
- (a)
- (b)
- (c)
- (d)
The system under consideration in an RC low-pass filter RC low-pass filter with \(R=1.0\Omega \) and C=1.0 \(\mu F\).
Let tf (f) be the group delay function of the given RC, low-pass filter and f2 =100 Hz. Then, t g(f2) in ms is
- (a)
0.717
- (b)
7.17
- (c)
71.7
- (d)
4.505
Let P be linearity, Q be time invariance, R be causality and S be stability.
The system y [n] = rect (x [n]) has the properties
- (a)
P, Q and R
- (b)
Q, R and S
- (c)
R, S and P
- (d)
S, P and Q