Signals and Systems
Exam Duration: 45 Mins Total Questions : 30
The total energy of x(t) is
- (a)
0
- (b)
13
- (c)
13/3
- (d)
26/3
The system \(y(t)=x\left( \frac { t }{ 2 } \right) \) has the properties
- (a)
K, L, M, N
- (b)
K, L
- (c)
K, M
- (d)
K, N
The continuous time convolution integral \(y(t)=e^{ -3t }u(t+3)\) is
- (a)
\(\frac { 1 }{ 3 } [1-e^{ -3(t+3) }]u(t+3)\)
- (b)
\(\frac { 1 }{ 3 } [1-e^{ -3(t+3) }]u(t)\)
- (c)
\(\frac { 1 }{ 3 } [1-e^{ 3t) }]u(t)\)
- (d)
\(\frac { 1 }{ 3 } [1-e^{ -3t) }]u(t+3)\)
The laplace transform of signal u(t)-u(t-2) is
- (a)
\(\frac { e^{ -2s }-1 }{ s } \)
- (b)
\(\frac { 1-e^{ 2s } }{ s } \)
- (c)
\(\frac { 2 }{ s } \)
- (d)
\(\frac { -2 }{ s } \)
Which one of the systems described by the following is time variant ?
- (a)
y(n)=nx(n)
- (b)
y(n)=x(n)-x(n-1)
- (c)
y(n)=x(-n)
- (d)
y(n)=x(n)cos 2\(nf_{ 0 }\)
Consider Laplace transform of two signls as \(\frac { 1 }{ s+2 } \) and \(\frac { 1 }{ s+1 } \) The resultant signal obtained after frequency convolation o these two signals along with constant multiplier \(\frac { 1 }{ 2\pi j } \) is
- (a)
\(e^{ -2t }+e^{ -t }\)
- (b)
\(e^{ -2t }\)
- (c)
\(e^{ -3t }\)
- (d)
\(e^{ -t }\)
The Z-transform of \(x|n|\) =[2,4,5,7,0,] is
- (a)
\(2z^{ 2 }+4z+5+7z+z^{ 2 },z\neq \infty \)
- (b)
\(2z^{ 2 }+4z^{ -1 }+5+7z+z^{ 3 },z\neq \infty \)
- (c)
\(2z^{ 2 }+4z^{ -1 }+5+7z+z^{ 3 },0<|z|\neq \infty \)
- (d)
\(2z^{ 2 }+4z+5+7z^{ -1 }+z^{ 3 },0<|z|\neq \infty \)
The time signal corresponding to \(\frac { z^{ 2 }-z^{ -2 } }{ 2 } \)X(z) is
- (a)
\(\frac { 1 }{ 2 } (x[n+2]-x[n-2]\)
- (b)
X[n+2]-x[n-2]
- (c)
\(\frac { 1 }{ 2 } (x[n-2]-x[n-2]\)
- (d)
X[n-2]-x[n+2]
The time signal corresponding to \(\frac { 2s^{ 2 }+10s+11 }{ s^{ 2 }+5s+6 } \) is
- (a)
\(2\eth (t)+(e^{ -3t }-e^{ -2t })u(t)\)
- (b)
\(2\eth (t)+(e^{ -2t }-e^{ -3t })u(t)\)
- (c)
\(2\eth (t)+(e^{ -2t }+e^{ -3t })u(t)\)
- (d)
\(2\eth (t)-(e^{ -2t }+e^{ -3t })u(t)\)
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)
\((e^{ -2t }+2te^{ -21 })u(t)\)
Given the Z-transform pair
\(X[n]\overset { z }{ \leftrightarrow } \frac { 32 }{ z^{ 2 }-16 } |z|<4\)
The Z-transform of the signal \(y[n]=\frac { 1 }{ 2^{ n } } x[n]\)
- (a)
\(\frac { (z+2)^{ 2 } }{ (z+2)^{ 2 }-16 } \)
- (b)
\(\frac { z^{ 2 } }{ z^{ 2 }-4 } \)
- (c)
\(\frac { (z-2)^{ 2 } }{ (z-2)^{ 2 }-16 } \)
- (d)
\(\frac { z^{ 2 } }{ z^{ 2 }-64 } \)
Given the Z-transform pair
\(3^{ n }n^{ 2 }u[n]\overset { z }{ \leftrightarrow } X(z)\)
The time of signal corresponding to \(X(z^{ -1 })\) is
- (a)
\(n^{ 2 }3^{ -n }u[-n]\)
- (b)
\(n^{ 2 }3^{ -n }u[-n]\)
- (c)
\(\frac { 1 }{ n^{ 2 } } 3^{ \frac { 1 }{ n } }u[n]\)
- (d)
\(\frac { 1 }{ n^{ 2 } } 3^{ \frac { 1 }{ n } }u[-n]\)
The z-transform of \(\left( \frac { 2 }{ 3 } \right) ^{ |n| }\) is
- (a)
\(\frac { -5z }{ (2z-3)(3z-2) } ,\frac { 3 }{ 2 } <|z|<-\frac { 2 }{ 3 } \)
- (b)
\(\frac { -5z }{ (2z-3)(3z-2) } ,\frac { 3 }{ 2 } <|z|<-\frac { 3 }{ 2 } \)
- (c)
\(\frac { -5z }{ (2z-3)(3z-2) } ,\frac { 2 }{ 3 } <|z|<-\frac { 3 }{ 2 } \)
- (d)
\(\frac { -5z }{ (2z-3)(3z-2) } ,\frac { 3 }{ 2 } <z|<-\frac { 2 }{ 3 } \)
In the question, the Fourier series coefficient of time domain signal have been again Determine the corresponding time domain signal and choose correct option
The Fourier series coefficient o time domain signal x(t) is
\(X[k]=\left( \frac { -1 }{ 3 } \right) ^{ k }\)
The fundamental frequency of signal is \(\omega _{ 0 }=1\) The signal is
- (a)
\(\frac { 4 }{ 5+3sint } \)
- (b)
\(\frac { 5 }{ 4+3sint } \)
- (c)
\(\frac { 5 }{ 4+3cost } \)
- (d)
\(\frac { 5 }{ 5+3cost } \)
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)
6 cos \(\left( 2\pi t+\frac { \pi }{ 4 } \right) -3cos\left( 3\pi t-\frac { \pi }{ 4 } \right) \)
- (b)
4 cos \(\left( 4\pi t+\frac { \pi }{ 4 } \right) -2cos\left( 3\pi t-\frac { \pi }{ 4 } \right) \)
- (c)
2cos \(\left( 2\pi t+\frac { \pi }{ 4 } \right) -2cos\left( 3\pi t-\frac { \pi }{ 4 } \right) \)
- (d)
4 cos \(\left( 4\pi t+\frac { \pi }{ 4 } \right) +2cos\left( 3\pi t-\frac { \pi }{ 4 } \right) \)
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 cofficient of the signal y(t)=x(4(t-1) is
- (a)
\(\frac { 8\pi }{ T } X[k]\)
- (b)
\(\frac { 4\pi }{ T } X[k]\)
- (c)
\(e^{ -jk\frac { 8\pi }{ t } }X[k]\)
- (d)
\(e^{ jk\frac { 8\pi }{ t } }X[k]\)
The Fourier transform of signal (sin \(2\pi t)e^{ -t }u(t)\) is
- (a)
\(\frac { 1 }{ 2 } \left[ \frac { 1 }{ 1+j\omega -2\pi } -\frac { 1 }{ 1+j\omega +2\pi } \right] \)
- (b)
\(\frac { 1 }{ 2j } \left[ \frac { 1 }{ 1+j\omega -2\pi } -\frac { 1 }{ 1+j\omega +2\pi } \right] \)
- (c)
\(\frac { 1 }{ 2j } \left[ \frac { 1 }{ 1+j\omega -2\pi } -\frac { 1 }{ 1+j\omega -2\pi } \right] \)
- (d)
\(\frac { 1 }{ 2 } \left[ \frac { 1 }{ 1+j\omega +2\pi } -\frac { 1 }{ 1+j\omega -2\pi } \right] \)
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 u(t) is
- (a)
\(n\delta \omega \)
- (b)
\(\frac { 1 }{ j\omega } \)
- (c)
\(n\delta (\omega )+\frac { 1 }{ j\omega } \)
- (d)
None of these
Compute the convolution y[n]=x[n]*h[n]. When x[n]=[1,2,4] and h[n]=[1,1,1,1,1]
- (a)
[1,3,7,7,7,7,6,4]
- (b)
[1,3,3,7,7,6,4]
- (c)
[1,2,4]
- (d)
[1,3,7]
The discrete time Fourier coefficient of x[n]=[....0,1,2,3,0,1,2,3,0,1,.....) is
- (a)
\(X[0]=\frac { 3 }{ 2 } ,X[1],=X*[3]=\frac { 1 }{ 2 } (1-j),X[2]=\frac { -1 }{ 2 } \)
\(x[k]=\frac { 1 }{ 4 } \sum _{ n=4 }^{ 4 }{ x[n](-j)^{ kn }=\frac { 1 }{ 4 } \sum _{ n=4 }^{ 4 }{ n(-j)^{ kn } } } \)
\(\frac { 1 }{ 4 } (-j)^{ 6 }+2(-j)+3(-j)]\)
\(X[0]=\frac { 3 }{ 2 } ,X[1]=\frac { 1 }{ 2 } (-1+j),X[2]=\frac { -1 }{ 2 } \)
\(x[3]=\frac { 1 }{ 2 } (-1-j)\) - (b)
\(X[0]=\frac { 3 }{ 2 } ,X[1],=X*[3]=\frac { -1 }{ 2 } (1-j),X[2]=\frac { -1 }{ 2 } \)
- (c)
\(X[0]=\frac { 3 }{ 2 } ,X[1],=X*[3]=\frac { -1 }{ 2 } (1-j),X[2]=\frac { -3 }{ 2 } \)
- (d)
None of the above
Consider the Voltage waveform shown below:
The Equation for V(t) is
- (a)
ut(-1)+u(t-2)+u(t-3)
- (b)
u(t-1)+2u(t-2)+3u(t-3)
- (c)
u(t-1)+u(t-3)+u(t-4)
- (d)
u(t-1)+u(t-2)+u(t-2)-3u(t-4)
The DTFT of \(a^{ |n| },|a|<1\) is
- (a)
\(\frac { 1-a^{ 2 } }{ 1-2asin\Omega +a^{ 2 } } \)
- (b)
\(\frac { 1-a^{ 2 } }{ 1-2acos\Omega +a^{ 2 } } \)
- (c)
\(\frac { 1-a^{ 2 } }{ 1-2/asin\Omega +a^{ 2 } } \)
- (d)
None of the 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)
\(\frac { 2e^{ j\omega } }{ 2-e^{ -j\Omega } } \)
The final value of \(L^{ -1 }\frac { 2s+1 }{ s^{ 4 }+8s^{ 3 }+16s^{ 2 }+s } \) is
- (a)
infinity
- (b)
2
- (c)
1
- (d)
zero
A continuous time signal x(t) is applied to the input of a continuous time LTI system with unit impulse response h(t). Find the output y(t). Given that x(t)=e2tu(-t) and h(t) = u(t - 3)
- (a)
1/2
- (b)
-1/2
- (c)
0
- (d)
2
Given a sequence x[n], to generate the sequence y [n] = x[3 - 4nl, which one of the following procedures would be correct?
- (a)
First delay x[n] by 3 samples to generate z, [n] then pick every 4th sample of z, [n] to generate Z2 [n] and then finally time reverse Z2 [n] to obtain y[n]
- (b)
First advance x[n] by 3 samples to generate z, [n], then pick every 4th sample of z, [n] to generate z2 [n] and then finally time reverse Z2 [n] to obtain y[n]
- (c)
First pick every 4th sample of x[n] to generate v, [n], time reverse v, [n] to obtain v2 [n] and finally advance v2 [n] by 3 samples to obtain y[n]
- (d)
First pick every 4th sample of x[n] to generate v, [n], time reverse v, [n to obtain v2 [n] and finally delay v2 [n] by 3 samples to obtain y[n]
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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Let P be linearity, Q be time invariance, R be causality and S be stability.
The system \(y\left[ n \right] =\sqrt { X\left[ n \right] } \) has the properties
- (a)
Q, R and S
- (b)
R, S and P
- (c)
S, P and Q
- (d)
P, Q and R
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 H(f) denotes the frequency response of the Re, low-pass filter. Let f, be the highest frequency, such that \(0\le \left| f \right| \le f_{ 1 },\frac { \left| H\left( f_{ 1 } \right) \right| }{ H(0) } \ge 0.95\) Then, f1 (in Hz) is
- (a)
327.8
- (b)
163.9
- (c)
52.2
- (d)
104.4
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