Signals and systems electrical engineering miscellaneous Difficult Questions and Answers | Page - 18

Signals and systems electrical engineering miscellaneous

x [n] = 0; n < – 1, n > 0, x [– 1] = – 1, x[0] = 2 is the input and y [n] = 0; n < – 1, n > 2, y[– 1] = – 1 = y[1], y [0] = 3, y [2] = – 2 is the output of a discret e-time LTI system. The system impulse response h[n] will be

1. h[n] = 0; n < 0, n > 2, h[0] = 1, h[1] = h[2] = – 1
1. h[n] = 0; n < – 1, n > 1, h[– 1] = 1, h[0] = h[1] = 2
1. h[n] = 0; n < 0, n > 3, h[0] = – 1, h[1] = 2, h[2] = 1
1. h[n] = 0; n < – 2, n > 1, h[– 2] = h[1] = h[– 1] = – h[0] = 3

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NA

Correct Option: A

NA

x (t) is a real valued function of a real variable with period T. Its trigonometric Fourier Series expansion contains no terms of frequency
ω = 2π (2k)/T; k = 1, 2,.... Also, no sine terms are present. Then x (t) satisfies the equation

1. x (t) =– x (t – T)
1. x (t) = x (T – t) = – x (– t)
1. x (t) = x (T – t) = – x (t – T/2)
1. x (t) = x (t – T) = x (t – T/2)

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Since, the fourier expansion of x(t) contains no sine terms, therefore,
x(t) = x(–t)
or, x(t) = x(T – t)
as x(t) is periodic with T
Now, as signal x(t) contains odd harmonics.
Then x(t) = -x t - T
2

Thus x(t) = x(T - t) = -x t - T
2

Correct Option: D

Since, the fourier expansion of x(t) contains no sine terms, therefore,
x(t) = x(–t)
or, x(t) = x(T – t)
as x(t) is periodic with T
Now, as signal x(t) contains odd harmonics.
Then x(t) = -x t - T
2

Thus x(t) = x(T - t) = -x t - T
2

Which of the following is true?

1. A finite signal is always bounded
1. A bounded signal always possesses finite energy
1. A bounded signal is always zero outside the interval [– t₀, t₀] for some t₀
1. A bounded signal is always finite

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A signal having amplitude within finite boundaries called bounded signal and it can have finite energy or infinite energy. If this signal is power signal, then it possess infinite energy. But in any case, this bounded signal will be finite for all the values of time.

Correct Option: B

A signal having amplitude within finite boundaries called bounded signal and it can have finite energy or infinite energy. If this signal is power signal, then it possess infinite energy. But in any case, this bounded signal will be finite for all the values of time.

X(z) = 1 – 3z^{– 1}, Y (z) = 1 + 2z^–2 are Z- transforms of two signals x[n], y[n] respectively. A linear time invariant system has the impulse response h[n] defined by these two signals as h[n] = x[n– 1]*y[n] where * denotes discrete time convolution. Then the output of the system for the input δ [n – 1]

1. has Z – transform z^{– 1} X (z) Y(z)
1. equals δ[n - 2] – 3δ[n – 3] + 2δ[n – 4] – 6δ[n – 5]
1. has Z – transform 1 – 3z^{– 1} + 2z^–2 – 6z^–3
1. does not satisfy any of the above three.

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NA

Correct Option: B

NA

The frequency spectrum of a signal is shown in the figure. If this signal is ideally sampled at intervals of 1 ms, then frequency spectrum of the sampled signal will be

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Here, frequency of signal, ω_c = 1 kHz
Sampling frequency, ω_s = 1 = 1kHz
1ms

For an ideal sampler, ω_s > 2ω_c
But here ω_s = ω_c
so, at ω_s/2, next sampling will start resulting sampled signal will be as follows:

Correct Option: D

Here, frequency of signal, ω_c = 1 kHz
Sampling frequency, ω_s = 1 = 1kHz
1ms

For an ideal sampler, ω_s > 2ω_c
But here ω_s = ω_c
so, at ω_s/2, next sampling will start resulting sampled signal will be as follows: