GATE Electronics and Communication Engineering - Signals and Systems
Exam Duration: 45 Mins Total Questions : 30
The system t\(\frac{d}{dt}\) y(t) - 8y(t) = x(t) has the prioperties
- (a)
K, L, M, N
- (b)
K, L, M
- (c)
K, M
- (d)
K, N
The continuous time convolution integral y(t) = u(t)*h(t), where h(t) = \(\begin{cases} { e }^{ 2t } \ \ , \\ { e }^{ -3t }, \end{cases}\begin{matrix} t<0 \\ t>0 \end{matrix}is\)
- (a)
\(\frac { 1 }{ 2 } { e }^{ -2t }u\left( -t-1 \right) +\frac { 5 }{ 6 } -\frac { 1 }{ 3 } { e }^{ -3t }u\left( -t \right) \)
- (b)
\(\frac { 1 }{ 2 } { e }^{ 2t }u\left( -t-1 \right) +\frac { 5 }{ 6 } -\frac { 1 }{ 3 } { e }^{ -3t }u\left( -t \right) \)
- (c)
\(\frac { 1 }{ 2 } { e }^{ -2t }+\frac { 1 }{ 6 } \left[ 5-3{ e }^{ 2t }-2{ e }^{ 3t } \right] u\left( t \right) \)
- (d)
\(\frac { 1 }{ 2 } { e }^{ -2t }+\frac { 1 }{ 6 } \left[ 5-3{ e }^{ 2t }-2{ e }^{ 3t } \right] u\left( -t \right) \)
The Laplace transform of signal tu(t)*cos2\(\pi\)t u(t) is
- (a)
\(\frac { 1 }{ s\left( { s }^{ 2 }+4{ \pi }^{ 2 } \right) } \)
- (b)
\(\frac { 2\pi }{ s\left( { s }^{ 2 }+4{ \pi }^{ 2 } \right) } \)
- (c)
\(\frac { 1 }{ s^{ 2 }\left( { s }^{ 2 }+4{ \pi }^{ 2 } \right) } \)
- (d)
\(\frac { { s }^{ 3 } }{ { s }^{ 2 }+4{ \pi }^{ 2 } } \)
The Laplace transform of signal u(t-1)*e-2tu(t-1) is
- (a)
\(\frac { { e }^{ -2\left( s+1 \right) } }{ 2s+1 } \)
- (b)
\(\frac { { e }^{ -2\left( s+1 \right) } }{ s+1 } \)
- (c)
\(\frac { { e }^{ -\left( s+2 \right) } }{ s+2 } \)
- (d)
\(\frac { { e }^{ -2\left( s+1 \right) } }{ s+2 } \)
If \(y(t)={ e }^{ x(t) }\), then the relation is
- (a)
dynamic
- (b)
static
- (c)
memory
- (d)
None of these
\(H(z)=\frac { 1 }{ 1-\frac { 1 }{ 2 } Z^{ -1 } } +\frac { 1 }{ 1-2Z^{ -1 } } \left| Z \right| >2\) is
- (a)
causal
- (b)
non-causal
- (c)
anti-causal
- (d)
Cannot be determined
Impulse train is a kind of signal which has ..... discontinuity.
- (a)
infinite
- (b)
zero
- (c)
one
- (d)
finite
The DTFT of 2\(\delta \)[4-2n] is
- (a)
\({ 2e }^{ -j2\Omega }\)
- (b)
\({ 2e }^{ j2\Omega }\)
- (c)
1
- (d)
None of these
The impulse response h(t) of a linear time-invariant continuous time system is described by h(t) - exp (\(\alpha\)t)u(t) + exp(\(\beta\)t)u(-t), where u(-t), where u(t) denotes the unit step function and \(\alpha\) and \(\beta\) real constants. This system is stable if
- (a)
\(\alpha\) is positive and \(\beta\) is positive
- (b)
\(\alpha\) is negative and \(\beta\) is positive
- (c)
\(\alpha\) is positive and \(\beta\) is negative
- (d)
\(\alpha\) is negative and \(\beta\) is positive
Choose the function f(t);\(-\infty \) <t <\(+\infty \), for which a Fourier series cannot be defined.
- (a)
3 sin (25 t)
- (b)
4 cos (20t + 3) +2 sin(10t)
- (c)
exp(-|t|) sin (25t)
- (d)
1
The region of convergence of the z-transform of a unit step function is
- (a)
|z|>1
- (b)
|z|<1
- (c)
(real part of z)>0
- (d)
(real part of x)<0
Which one is most appropirate dynamic system out of the following?
- (a)
y(n) = y(n-1) + y(n+1)
- (b)
y(n) = y(n-1)
- (c)
y(n) = x(n)
- (d)
y(n) + y(n-1) + y(n+3) = 0
Which one is causal system?
- (a)
y(n) = 3x(n) - 2x(n-1)
- (b)
y(n) = 3x(n) + 2x(n+1)
- (c)
y(n) = 3x(n+1) + 2x(n-1)
- (d)
y(n) = 3x(n+1) + 2x(n-1) + x(n)
Compute the convolution y[n] = x[n]*h[n]. When x[n] = {1, 2, 4}, h[n]={1, 1, 1, 1, 1}
- (a)
{1, 3, 7, 7, 7, 6, 4}
- (b)
{1, 3, 3, 7, 7, 6, 4}
- (c)
{1, 2, 4}
- (d)
{1, 3, 7}
A system has N different poles. Then the system can have
- (a)
N ROC's
- (b)
(N-1) ROC's
- (c)
(N+1) ROC's
- (d)
None of these
For the above system \(G_{ 1 }\left( s \right) =\frac { s^{ -1 } }{ 1+2{ s }^{ -1 }+{ s }^{ -2 } } \). Find G2(s) for the given system to be invertible.
- (a)
\(\frac { s }{ 1+2s+{ s }^{ 2 } } \)
- (b)
\(\frac { 1+2s+{ s }^{ 2 } }{ s } \)
- (c)
\(\frac { 1 }{ 1+2s+{ s }^{ 2 } } \)
- (d)
\(1+2s+{ s }^{ 2 }\)
The output y(t) of a linear time-invariant system is related to its input x(t) by the following equation y(t) = 0.5x(t-td+T)+x(t-td)+0.5x(t-td-T). The filter transfer function h(\(\omega\)) of such a system is given by
- (a)
(1+cos\(\omega\) T) e-j\(\omega\)td
- (b)
(1+ 0.5 cos \(\omega\) T) e-j\(\omega\)td
- (c)
(1+cos\(\omega\) T) ej\(\omega\)td
- (d)
(1-0.5 cos\(\omega\) T) e-j\(\omega\)td
The impulse response h|n| of a linear time-invariant system is given as \(h\left[ n \right] =\begin{cases} -2\sqrt { 2 } ,\quad n=1,\quad -1 \\ 4\sqrt { 2 } ,\quad \ \ \ n=2,\quad -2 \\ 0,\quad \ \ \ \ \ \ \ otherwise \end{cases}\) .If the input to the above system is the sequence \({ e }^{ \frac { j\pi n }{ 4 } }\), then the output is
- (a)
\(4\sqrt { 2 } { e }^{ \frac { j\pi n }{ 4 } }\)
- (b)
\(4\sqrt { 2 } { e }^-{ \frac { j\pi n }{ 4 } }\)
- (c)
\(4\ { e }^{ \frac { j\pi n }{ 4 } }\)
- (d)
\(-4\ { e }^{ \frac { j\pi n }{ 4 } }\)
Let x(t) and y(t) (with Fourier transform X(f) and Y(f), respectively be related as shown in figure. Then Y(f) is
- (a)
\(-\frac { 1 }{ 2 } X\left( \frac { f }{ 2 } \right) { e }^{ -j2\pi f }\)
- (b)
\(-\frac { 1 }{ 2 } X\left( \frac { f }{ 2 } \right) { e }^{ j2\pi f }\)
- (c)
\(- X\left( \frac { f }{ 2 } \right) { e }^{ j2\pi f }\)
- (d)
\(-X\left( \frac { f }{ 2 } \right) { e }^{ -j2\pi f }\)
A system described by the following differential equation \(\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } +3\frac { dy }{ dt } +2y=x\left( t \right) \)is initially at rest. For input x(t) = 2u(t), the output y(t) is
- (a)
(1-2e-t+e-2t) u(t)
- (b)
(1+2e-t-e-2t) u(t)
- (c)
(0.5 + e-t + 1.5e-2t) u(t)
- (d)
(0.5 +2e-t + 2e-2t) u(t)
Number of complex addition and complex multiplication required for 128 point DFT using FFT algorithm.
- (a)
16256, 16384
- (b)
448, 896
- (c)
16384, 16256
- (d)
896, 448
The z-transform of a signal is given by \(\frac { { z }^{ -1 }{ \left( 1-z^{ 4 } \right) } }{ 4\left( { 1-z }^{ -1 } \right) ^{ 2 } } \), Its final value is
- (a)
\(\frac { 1 }{ 4 } \)
- (b)
0
- (c)
1
- (d)
\(\infty \)
The bilateral Laplace transform of e(3t+6) u(t+3) is
- (a)
\(\frac { { e }^{ 3s } }{ s-3 } \), Re(s)>3
- (b)
\(\frac { { e }^{ 3s } }{ s-3 } \), Re(s)<3
- (c)
\(\frac { { e }^{ 3(s-1) } }{ s-3 } \), Re(s)>3
- (d)
\(\frac { { e }^{ 3(s-1) } }{ s-3 } \), Re(s)<3
The Laplace transform of x(t) is \(X\left( s \right) =\frac { { e }^{ -2s }\left( { 2s }^{ 2 }+1 \right) }{ { s }\left( s+1 \right) \left( s+4 \right) } \). The final value of x(t) is
- (a)
2
- (b)
\(\frac{1}{4}\)
- (c)
-3
- (d)
Does not exist
A stable system has input x(t) and output y(t) = e-2t cos t u(t). The impulse response of the system is
- (a)
\(\delta \) (t) - (e-2t cos t + e-2t sin t) u(t)
- (b)
\(\delta \) (t) - (e-2t cos t + e-2t sin t) u(t-2)
- (c)
\(\delta \) (t) - (e-2t cos t + e2t sin t) u(t)
- (d)
\(\delta \) (t) - (e2t cos t + e2t sin t) u(t+2)
The impulse response h(t) of a linear time-invariant continuous time system is given by h(t) = exp(-2t) u(t), where u(t) denote the unit step function.The output of this system to the sinusoidal input x(t) = 2cos(2t) for all time t, is
- (a)
0
- (b)
2-0.25 cos(2t-0.125\(\pi\))
- (c)
2-0.5 cos(2t-0.125\(\pi\))
- (d)
2-0.5 cos(2t-0.25\(\pi\))
Consider the transform pair given below \(cos2tu(t)\underleftrightarrow { L } X(s)\)
The time signal corresponding to \(\frac { X(s) }{ s^{ 2 } } \) is
- (a)
4cos 2tu(t)
- (b)
\(\frac { 1-cos2t }{ 4 } u(t)\)
- (c)
\(t^{ 2 }cos2tu(t)\)
- (d)
\(\frac { cos\quad 2t }{ t^{ 2 } } u(t)\)
Consider the signals x[n] and y[n] given below figure:
The signal y[1- n] is
- (a)
- (b)
- (c)
- (d)
Two discrete time systems \(S_{ 1 },S_{ 2 }\) are connected in cascade to form a new system as shown in figure below
Consider the following statements
1.If \(S_{ 1 }\) and \(S_{ 2 }\) are linear and time invariant,then interchanging their order does not change the system
2.If \(S_{ 1 }\) and \(S_{ 2 }\) are linear and time varying interchanging their order does not change the system
True statements are
- (a)
Both 1 and 2
- (b)
Only 1
- (c)
Only 2
- (d)
None of these
Consider a periodic signal x[n] with period N and FS coefficient X[k]. The FS co0rfficient Y[k] of the signal y[n] = x[n] - x[n-N/2] is (assume that N is even)
- (a)
(1-(-1)k+1)X[2k]
- (b)
(1-(-1)k)X[k]
- (c)
(1-(-1)k+1)X[k]
- (d)
(1-(-1)k)X[2k]