1
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
The signal x(t) is described by $$x\left( t \right) = \left\{ {\matrix{ {1\,\,\,for\,\, - 1 \le t \le + 1} \cr {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise} \cr } } \right.$$

Two of the angular frequencies at which its Fourier transform becomes zero are

A
$$\pi ,\,2\pi $$
B
$$0.5\,\pi ,\,1.5\,\pi $$
C
$$0,\,\pi $$
D
$$2\,\pi ,\,2.5\,\pi $$
2
GATE ECE 2005
MCQ (Single Correct Answer)
+2
-0.6
For a signal x(t) the Fourier transform is X(f). Then the inverse Fourier transform of X(3f+2) is given by
A
$${1 \over 2}\,x\left( {{t \over 2}} \right){e^{j3\pi t}}$$
B
$${1 \over 3}\,x\left( {{t \over 3}} \right){e^{ - j4\pi t/3}}$$
C
$$3\,x(3t){e^{ - j4\pi t}}$$
D
$$x(3t + 2)$$
3
GATE ECE 2004
MCQ (Single Correct Answer)
+2
-0.6
Let x(t) and y(t) (with Fourier transforms X(f) and Y(f) respectively) be related as shown in Fig.(1) & (2). GATE ECE 2004 Signals and Systems - Fourier Transform Question 14 English

Then Y(f) is

A
$$ - {1 \over 2}X(f/2){e^{ - j2\pi f}}$$
B
$$ - {1 \over 2}X(f/2){e^{j2\pi f}}$$
C
$$ - X(f/2){e^{j2\pi f}}$$
D
$$ - X(f/2){e^{ - j2\pi f}}$$
4
GATE ECE 2000
MCQ (Single Correct Answer)
+2
-0.6
The Hilbert transform of $$\left[ {\cos \,{\omega _1}t + \,\sin {\omega _2}t\,} \right]$$ is
A
$$\sin \,{\omega _1}t + \,\cos {\omega _2}t$$
B
$$\sin \,{\omega _1}t + \,\cos {\omega _2}t$$
C
$$\cos \,{\omega _1}t + \,\sin {\omega _2}t$$
D
$$\sin {\omega _1}t + \,\sin {\omega _2}t$$
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