Signal and systems miscellaneous
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For the signal X(n) = 1 n u(n– 3) the X(z) will be— 5
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Given that x(n) = 1 n u(n– 3) can be written as 5 x(n) = 1 n – 3 · 1 3 u(n – 3) 5 5
By using shifting property
If x(n) ← z→ X(z)
x(n – n0) ←z→ z– n0.x(z)
Therefore,1 3 · 1 n – 3 5 5 1 3 z– 3 |z| > 1 5 1 – 1 z– 1 5 5
Hence, alternative (B) is correct choice.Correct Option: B
Given that x(n) = 1 n u(n– 3) can be written as 5 x(n) = 1 n – 3 · 1 3 u(n – 3) 5 5
By using shifting property
If x(n) ← z→ X(z)
x(n – n0) ←z→ z– n0.x(z)
Therefore,1 3 · 1 n – 3 5 5 1 3 z– 3 |z| > 1 5 1 – 1 z– 1 5 5
Hence, alternative (B) is correct choice.
- Energy of a signal depends upon—
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According to the Parsevel’s theorem for energy signals.
i.e., E = ∞ |F(ω)|2df - ∞
From this relation it is clear that energy of a signal depends only upon the magnitude spectrum of Fourier transform.Correct Option: A
According to the Parsevel’s theorem for energy signals.
i.e., E = ∞ |F(ω)|2df - ∞
From this relation it is clear that energy of a signal depends only upon the magnitude spectrum of Fourier transform.
- Mathematically discrete time unit impulse can be obtained as—
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NA
Correct Option: B
NA
- The relation between the u|n| and δ[m] can be given by—
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NA
Correct Option: B
NA
- y(t) = 2 x(t) is an invertible system whose inverse system H–1 is—
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From given figure
y(t) = H x(t)
H– 1 2x(t) = y′(t)
y(t)=2x(t)
Again from figure
y′(t) = y(t). H– 1or y′(t) = 2. x(t) 1 2
or y′(t) = x(t)
Therefore, it is clear that the system is invertible.
Hence, alternative (D) is the correct choice.Correct Option: D
From given figure
y(t) = H x(t)
H– 1 2x(t) = y′(t)
y(t)=2x(t)
Again from figure
y′(t) = y(t). H– 1or y′(t) = 2. x(t) 1 2
or y′(t) = x(t)
Therefore, it is clear that the system is invertible.
Hence, alternative (D) is the correct choice.