Test: Discrete-Time Fourier Transform - Electrical Engineering (EE) MCQ

y[k] = x [3k-2]
Now, y [0] = x [-2] = 1
Or, y [1] = x [1] = 1
Or, y [2] = x [4] = 0
∴ y[k] = 1, for k = 0, 1 and 0 otherwise.

Given that, N₁ = 18, N₂ = 14
We know that period of X [n] (say N) = LCM (N₁, N₂)
∴ Period of X [n] = LCM (18, 14) = 126.

Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z^-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z^-n.

We know that any signal be it discrete or continuous is said to be even or symmetric when that signal f(x) = f (-x). Here given signal is X (n). It is a discrete time signal. So, the signal will be even symmetric if X (n) = X (-n).

Given that, N₁ = 4, N₂ = 16, N₃ = 8
We know that period of X [n] (say N) = LCM (N₁, N₂, N₃)
∴ Period of X [n] = LCM (4, 16, 8) = 16.

DTF System operates with a discrete signal, on the other hand time invariant system is a system whose output does not depend explicitly on time. For continuous time system, the inputs as well as output both are CT signals.

The steady state value of the DT signal F(s) exists since all poles of the given Laplace transform have negative real part.
∴F (∞) = lim_s→0 s F(s)

= 0.

We know that for X [n] = X₁ [n] × X₂ [n] to be periodic, both X₁ [n] and X₂ [n] should be periodic with finite periods.
Here X₂ [n] = cos (πn/8), is periodic with fundamental period as 8/n
But X1 [n] = cos (n/8) is non periodic.
∴ X [n] is a non-periodic signal.

We know that Nyquist frequency is twice the maximum frequency, i.e. f_s = 2 f_m.
The maximum frequency present in the signal is ω_m = 300 π or f_m = 150 Hz. Therefore the Nyquist frequency f_s = 2 f_m = 300 Hz.