1
GATE ECE 2015 Set 3
MCQ (Single Correct Answer)
+2
-0.6
The complex envelope of the bandpass signal $$x(t)\, = \, - \sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right)\sin \left( {\pi t - {\pi \over 4}} \right),$$ centered about f = $${1 \over {2\,}}\,Hz,$$ is
A
$$\left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{j{\pi \over 4}}}$$
B
$$\left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{ - j{\pi \over 4}}}$$
C
$$\sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{j{\pi \over 4}}}$$
D
$$\sqrt 2 \left( {{{\sin (\pi t/5)} \over {\pi t/5}}} \right){e^{ - j{\pi \over 4}}}$$
2
GATE ECE 2014 Set 2
Numerical
+2
-0
The value of the integral $$\int\limits_{ - \infty }^\infty {\sin \,{c^2}} $$ (5t) dt is
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3
GATE ECE 2014 Set 1
MCQ (Single Correct Answer)
+2
-0.6
For a function g(t), it is given that $$\int_{ - \infty }^\infty {g(t){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}} $$ for any real value $$\omega $$. If y(t)=$$\int_{ - \infty }^t {g(\tau )d\tau ,\,then\,\int_{ - \infty }^\infty {y(t)\,dt} \,} $$ is
A
0
B
- j
C
$$ - {j \over 2}$$
D
$${j \over 2}$$
4
GATE ECE 2012
MCQ (Single Correct Answer)
+2
-0.6
The Fourier transform of a signal h(t) is $$H(j\omega )$$ =(2 cos $$\omega $$) (sin 2$$\omega $$) / $$\omega $$. The value of h(0) is
A
1/4
B
1/2
C
1
D
2
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