Signals and Systems
1
GATE ECE 2015 Set 3
Numerical
+2
-0
Consider a continuous-time signal defined as $$x(t) = \left( {{{\sin \,(\pi t/2)} \over {(\pi t/2)}}} \right)*\sum\limits_{n = - \infty }^\infty {\delta (t - 10n)} $$ Where ' * ' denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in samples/sec) for x(t) is __________________.
Your input ____
2
GATE ECE 2010
MCQ (Single Correct Answer)
+2
-0.6
The Nyquist sampling rate for the signal $$s(t) = {{\sin \,(500\pi t)} \over {\pi \,t}} \times {{\sin \,(700\pi t)} \over {\pi \,t}}$$ is given by
3
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
The minimum sampling frequency (in samples /sec) required to reconstruct the following signal from its samples without distortion $$x(t) = 5{\left( {{{\sin \,\,2\,\pi \,1000\,t)} \over {\pi \,t}}} \right)^3} + 7{\left( {{{\sin \,\,2\,\pi \,1000\,t} \over {\pi \,t}}} \right)^2}$$
would be
4
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where $$g(t)\, = \,\,\sum\limits_{k = - \infty }^\infty {{{( - 10)}^k}\,\delta (t - 0.5x{{10}^{ - 4}}k)} $$
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be
GATE ECE Subjects
Signals and Systems
Network Theory
Control Systems
Digital Circuits
General Aptitude
Electronic Devices and VLSI
Analog Circuits
Engineering Mathematics
Microprocessors
Communications
Electromagnetics