Signal and systems miscellaneous
- Let x(t) and y(t) (with Fourier transforms X(f) and Y(f) respectively) be related as shown in figure.
Then Y(f) is—
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From fig. we see that y(t) is a compressed version of x(t). But in frequency domain it will be represented by expansion i.e., shifting of two unit left side.
First shifting then
Scaling and then
Amplitude reflection i.e.,
x(t) = – x(t)we get – X f e–j2πf 2 Correct Option: D
From fig. we see that y(t) is a compressed version of x(t). But in frequency domain it will be represented by expansion i.e., shifting of two unit left side.
First shifting then
Scaling and then
Amplitude reflection i.e.,
x(t) = – x(t)we get – X f e–j2πf 2
- The Fourier transform of V(t) = cos ω0t is given by—
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NA
Correct Option: D
NA
- Match List-I {x(n)} with List-II {X(z)} and select the correct answer using the codes given below the Lists—
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A. αn u(n) ←z→ 1 , |z| > |α| (1 – αz–1) B. - αn u(- n - 1) ←z→ 1 , |z| < |α| (1 – αz–1) c. – nαn u(- n - 1) ←z→ αz–1 , |z| < |α| (1 – αz–1) D. nαn u(n) ←z→ αz–1 , |z| > |α| (1 – αz–1)2
Hence, alternative (D) is the correct choice.Correct Option: D
A. αn u(n) ←z→ 1 , |z| > |α| (1 – αz–1) B. - αn u(- n - 1) ←z→ 1 , |z| < |α| (1 – αz–1) c. – nαn u(- n - 1) ←z→ αz–1 , |z| < |α| (1 – αz–1) D. nαn u(n) ←z→ αz–1 , |z| > |α| (1 – αz–1)2
Hence, alternative (D) is the correct choice.
- A periodic triangular wave is shown in the figure. Its Fourier components will consist only of—
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NA
Correct Option: D
NA
- Convolution of x(t + 4) with impulse function δ(t – 6) is equal to—
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Let the convolution of x(t + 4) and δ(t – 6) is y(t)
y(t) = x(t + 4)* δ(t – 6)x(t)*δ(t) = x(t) x(t)*δ(t + t 0) = x(t + t 0)
or
y(t) = x(t – 6 + 4)
or
y(t) = x(t – 2)
Hence, alternative (C) in the correct choice.Correct Option: C
Let the convolution of x(t + 4) and δ(t – 6) is y(t)
y(t) = x(t + 4)* δ(t – 6)x(t)*δ(t) = x(t) x(t)*δ(t + t 0) = x(t + t 0)
or
y(t) = x(t – 6 + 4)
or
y(t) = x(t – 2)
Hence, alternative (C) in the correct choice.