Electrical Engineering - Signals and Systems Question Paper 2
Exam Duration: 45 Mins Total Questions : 30
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
The following circuit is
- (a)
Memory less system
- (b)
casual system
- (c)
dynamic system
- (d)
static system
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
The z-transform of [n+k],k>0 is
- (a)
\(z^{- k },z\neq 0\)
- (b)
\(z^{ k },z\neq 0\)
- (c)
\(z^{ -k },all\quad z\)
- (d)
\(z^{ k },all\quad z\)
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)
\(\frac { Z }{ 1-z^{ -1 } } |z|>1\)
Given that \(f(\tau )\) and \(g(\tau )\) are the one-sided z-transforms of discrete time functions \(f(n\tau )\) and \(f(g\tau )\) the z-transform of
\(\sum { f(k\tau )g(n\tau -k\tau ) } \) is given by
- (a)
\(\sum { f(k\tau )g(n\tau -k\tau ) } z^{ -n }\)
- (b)
\(\sum { (k\tau )Z^{ n }\delta g(n\tau -k\tau ) } z^{ -n }\)
- (c)
\(\sum { (k\tau )g(n\tau -k\tau ) } z^{ -n }\)
- (d)
\(\sum { (n\tau -k\tau )g(n\tau ) } z^{ -n }\)
The Laplace transform of signal \(t\frac { d }{ dt } (e^{ -1 }costu(t)]\) is
- (a)
\(\frac { -(s^{ 2 }+4s+2) }{ (s^{ 2 }+2s+2)^{ 2 } } \)
- (b)
\(\frac { (s^{ 2 }+4s+2) }{ (s^{ 2 }+2s+2)^{ 2 } } \)
- (c)
\(\frac { (s^{ 2 }+2s+2) }{ (s^{ 2 }+4s+2)^{ 2 } } \)
- (d)
\(\frac { -(s^{ 2 }+2s+2) }{ (s^{ 2 }+4s+2)^{ 2 } } \)
The time signal corresponding to \(\frac { s+3 }{ s^{ 2 }+3s+2 } \) is
- (a)
\((2e^{ -2t }+e^{ -t })u(t)\)
- (b)
\((2e^{ -2t }+e^{ -2t })u(t)\)
- (c)
\((2e^{ -2t }+e^{ -t })u(t)\)
- (d)
\((2e^{ -t }+e^{ -2t })u(t)\)
The time signal corresponding to \(\frac { s^{ 2 }-3 }{ (s+2)(s^{ 2 }+23+1) } \) is
- (a)
\((e^{ -2t }-2te^{ -1 })u(t)\)
- (b)
\((e^{ -2t }+2te^{ -1 })u(t)\)
- (c)
\((e^{ -2t }-2te^{ -21 })u(t)\)
- (d)
\((e^{ -2t }+2te^{ -21 })u(t)\)
The z-transform of cos \(\left( \frac { \pi }{ 3 } n \right) \) u[n]is
- (a)
\(\frac { z }{ 2 } \frac { (2z-1) }{ (z^{ 2 }-z+1) } ,0<|z|<1\)
- (b)
\(\frac { z }{ 2 } \frac { (2z-1) }{ (z^{ 2 }-z+1) } ,|z|<1\)
- (c)
\(\frac { z }{ 2 } \frac { (1-2z) }{ (z^{ 2 }-z+1) } ,0<|z|<1\)
- (d)
\(\frac { z }{ 2 } \frac { (1-2z) }{ (z^{ 2 }-z+1) } ,|z|<1\)
The Fourier series coefficient of time domain signal x(t) is shown below.
The fundamental frequency of signal is \(\omega _{ 0 }=\pi \) The signal is
- (a)
3cos \(3\pi t+2cos\quad 2\pi t+cos\quad \pi t\)
- (b)
3sin \(3\pi t+2sin\quad 2\pi t+sin\quad \pi t\)
- (c)
6 sin \(3\pi t+4sin2\pi t+sin\pi t\)
- (d)
6 cos \(3\pi t+4cos2\pi t+cos\pi t\)
The Fourier series coefficient of time domain signal x(t) is shown below
The fundamental frequency of signal is \(\omega _{ 0 }=\pi \)The signal x(t) is
- (a)
\(\frac { sin9\pi t }{ sin\pi t } \)
- (b)
\(\frac { sin9\pi t }{ \pi sin\pi t } \)
- (c)
\(\frac { sin18\pi t }{ 2sin\pi t } \)
- (d)
None of these
Unit impulse response of a system is given as
c(t)=-4-t+6e-2t
The step response of the same system for \(t\ge 0\) is equal to
- (a)
3e-2t+4e-t-1
- (b)
-3e-2t+4e-t-1
- (c)
-3e-2t-4e-t-1
- (d)
-3e-2t+4e-t+1
Verify whether x(t)=Ae-at u(t), where a>0, is an energy signal or not.
- (a)
True
- (b)
False
- (c)
Insufficient data
- (d)
Cannot say
The relationship between the input x(t) and output y(t) of a casual system is defined as
\(\frac { d^{ 2 }y(t) }{ dt^{ 2 } } -\frac { dy(t) }{ dt } -2y(t)=4\times (t)+5\frac { dx(t) }{ dt } \)
The impulse response of system is
- (a)
(3e-t-2e2t)u(-t)
- (b)
(3e-t+2e2t)u(t)
- (c)
3e-tu(t)-2e2tu(-t)
- (d)
3e-tu(t)+2e2tu(-t)
The bilateral Laplace transform of x(t)=e-tu(t+2) will be
- (a)
\(\frac { e^{ 2(s+1) } }{ s+1 } ,Re(s)>-1\)
- (b)
\(\frac { 1 }{ 1+s } ,Re(s)>-1\)
- (c)
\(\frac { e^{ 2(s+1) } }{ s+1 } ,Re(s)<-1\)
- (d)
\(\frac { 1 }{ 1+s } ,Re(s)<-1\)
The Laplace transform of x(t) can be interpreted as the transform of x(t) after multiplication by real ......... signal
- (a)
Fourier, exponential
- (b)
z-transform, impulse
- (c)
z-transform, exponential
- (d)
Fourier, impulse
System function H(s) =\(\frac { 1 }{ s+3 } \) For a signal sin 2t, the 5+3 steady state response is
- (a)
1/8
- (b)
infinite
- (c)
zero
- (d)
8
If H(f)=\(\frac { y(t) }{ x(t) } \), then for this to be true x(t) is
- (a)
\(exp\left( \frac { j2nf }{ t } \right) \)
- (b)
\(exp\left( -\frac { j2nf }{ t } \right) \)
- (c)
exp(j2nft)
- (d)
exp(-j2nft)
If a system has all its poles and zeros inside the unit circle, then system is
- (a)
minimum phase system
- (b)
mixed phase system
- (c)
maximum phase system
- (d)
None of the above
Consider the discrete time system shown in the figure, where the impulse response of G(z) is g(0) =0, g(1) =g(2) =1, g(3) =1,g(3) =g(4) = ... =0
This system is stable for range of values of K
- (a)
[-1,1/2]
- (b)
[-1,1]
- (c)
[-1/2, 1]
- (d)
[-1/2, 2]
A system with x(t) and output y(t) is defined by the input output relation \(Y(t)=\int _{ -\infty }^{ -2t }{ X\left( \tau \right) d\tau } \) The system wi II be
- (a)
causal, time invariant and unstable
- (b)
causal, time invariant and stable
- (c)
non-causal, time invariant and unstable
- (d)
non-causal, time variant and unstable
A periodic triangular wave is shown in the figure. Its Fourier components will consist only if
- (a)
neither odd nor even function
- (b)
an odd function
- (c)
an even function
- (d)
both odd and even functions
The Fourier series coefficients for a discrete time signal \(x\left[ n \right] =\left\{ \underset { \uparrow }{ 1 } ,1,0,0 \right\} \) periodic with N =4 is given by
- (a)
\(C_{ k }=\frac { 1 }{ 2 } \left[ 1+e^{ -j\pi k/2 } \right] \)
- (b)
\(C_{ k }=\frac { 1 }{ 4 } \left[ 1+e^{ -j\pi k/2 } \right] \)
- (c)
\(C_{ k }=\frac { 1 }{ 2 } \left[ 1+e^{ -j\pi k } \right] \)
- (d)
\(C_{ k }=\frac { 1 }{ 4 } \left[ e^{ 2\pi k }+e^{ -j\pi k/4 } \right] \)
Determine the CTFT of a continuous time signal x(t) = e-A|t|, A >0
- (a)
\(X(j\omega )=\frac { A }{ A^{ 2 }+\omega ^{ 2 } } \)
- (b)
\(X(j\omega )=\frac { 2A }{ 4A^{ 2 }+\omega ^{ 2 } } \)
- (c)
\(X(j\omega )=\frac { 2A }{ A^{ 2 }+\omega ^{ 2 } } \)
- (d)
None of these
If u(t), r(t) denote the unit step and \(u(t)\times r(t)\) their conv, then the function \(u(t+1)\times r(t-2)\) given by
- (a)
\(\left( \frac { 1 }{ 2 } \right) (t-1)(t-1)(t-2)\)
- (b)
\(\left( \frac { 1 }{ 2 } \right) (t-1)(t-2\)
- (c)
\(\left( \frac { 1 }{ 2 } \right) (t-1)^{ 2 }u(t-1)\)
- (d)
None of these
Compute the convolution
y(n)=x(n)*h(n) of
x(n)=anu(n),h(n)=bnu(n)
- (a)
\({b^{n+1}-a^{n+1}\over b-a}u(n)\)
(for all values of a and b) - (b)
bn(n+1)u(n)
- (c)
\({b^{n+1}-a^{n+1}\over b-a}\)
- (d)
\({b^{n+1}-a^{n+1}\over b-a}u(n), a\neq b\)
bn(n+1)u(n), a=b
Consider the transform pair given below:
\(x(t)u(t)\overset { LT }{ \longleftrightarrow } \frac { 2(s) }{ s^{ 2 }+2 } \)
Determine the Laplace transform Y(s) of the given time signal in questions and choose correct option.
The Laplace transform of 2tx(t) is
- (a)
\(\frac { 8-4s^{ 2 } }{ \left( s^{ 2 }+2 \right) ^{ 2 } } \)
- (b)
\(\frac { 4s^{ 2 }-8 }{ \left( s^{ 2 }+2 \right) ^{ 2 } } \)
- (c)
\(\frac { 4s^{ 2 } }{ s^{ 2 }+1 } \)
- (d)
\(\frac { s^{ 2 } }{ s^{ 2 }+1 } \)
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)
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- (b)
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- (c)
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- (d)
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A sequence x(n) has non-zero values as shown in figure.
The Fourier transform of y(2n) will be
- (a)
\(e^{ -j2\omega }\left[ cos4\omega +2cos2\omega +2 \right] \)
- (b)
\(\left[ cos2\omega +2cos\omega +2 \right] \)
- (c)
\(e^{ -j2\omega }\left[ cos2\omega +2cos\omega +2 \right] \)
- (d)
\(e^{ j2\omega }\left[ cos2\omega +2cos\omega +2 \right] \)
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