Questions: Discrete-Time Fourier Transform (DTFT)

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A discrete-time system has poles at z = 0.5 and z = 1.5. Can its DTFT be computed?

AYes — the DTFT is always defined for any discrete-time system by evaluating X(z) at z = e^(jω)
BNo — the pole at z = 1.5 places it outside the unit circle, so the ROC of the Z-transform excludes the unit circle and the DTFT does not exist
CYes — the DTFT averages the contributions of stable and unstable poles, converging to a finite result
DNo — the DTFT cannot be computed for systems with more than one pole, regardless of their location
Question 2 Multiple Choice

What is the fundamental reason why the DTFT X(e^(jω)) is exactly periodic with period 2π in ω?

ADigital frequency must be normalized to fit within a bounded range, so periodicity is a mathematical convention
BMultiplying a discrete sequence x[n] by e^(j(ω+2π)n) produces exactly the same result as multiplying by e^(jωn), because e^(j2πn) = 1 for all integers n
CThe Z-transform is defined on a circle, making all evaluations naturally periodic in the angular variable
DThe DFT (which approximates the DTFT) is computed over N discrete points, introducing periodicity as a numerical artifact
Question 3 True / False

The DTFT of any stable discrete-time system (all poles strictly inside the unit circle) always exists and is a continuous, periodic function of digital frequency ω.

TTrue
FFalse
Question 4 True / False

The 2π-periodicity of the DTFT is a limitation of discrete sampling that can be overcome by using a higher sampling rate, which extends the usable frequency range beyond 2π radians.

TTrue
FFalse
Question 5 Short Answer

Explain how the 2π-periodicity of the DTFT is the frequency-domain expression of aliasing in sampled signals.

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