1
GATE ECE 2012
+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
2
GATE ECE 2008
+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
$$2\,\pi ,\,2.5\,\pi$$
3
GATE ECE 2005
+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
$$x(3t + 2)$$
4
GATE ECE 2004
+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).

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
$$- X(f/2){e^{ - j2\pi f}}$$
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