Discrete-Time Fourier Transform (DTFT) Quiz Essentials Quiz

Explore core principles of Discrete-Time Fourier Transform (DTFT) with this quiz focused on properties, fundamental definitions, and practical analysis techniques in digital signal processing. Strengthen your understanding of DTFT's mathematical formulation, periodicity, and its relationship to signal representation.

1. DTFT Mathematical Definition

   Which of the following correctly defines the Discrete-Time Fourier Transform (DTFT) X(e^{jω}) for a discrete signal x[n]?

   1. X(e^{jω}) = ∑_{n=-∞}^{∞} x[n] e^{-jωn}
   2. X(e^{jω}) = ∑_{n=0}^{N-1} x[n] e^{-j2πkn/N}
   3. X(e^{jω}) = ∫_{-∞}^{∞} x(t) e^{-jωt} dt
   4. X(e^{jω}) = ∫_{-π}^{π} x(t) dt

   Explanation: The DTFT is defined by the infinite sum X(e^{jω}) = ∑_{n=-∞}^{∞} x[n] e^{-jωn}, which transforms a discrete-time signal into a continuous frequency domain representation. The option involving integration with respect to t is the formula for the continuous-time Fourier Transform and not applicable to discrete signals. The finite sum with 2πkn/N belongs to the Discrete Fourier Transform (DFT), not the DTFT. Lastly, integrating x(t) from -π to π is unrelated to the DTFT and fails to account for the frequency component.

2. DTFT Periodicity Property

   What is the periodicity property of the DTFT in the frequency domain?

   1. The DTFT is periodic with period 2π
   2. The DTFT is periodic with period π/2
   3. The DTFT is periodic with period 1
   4. The DTFT is non-periodic in the frequency domain

   Explanation: The frequency response of the DTFT is inherently periodic with a period of 2π, which means X(e^{jω}) = X(e^{j(ω+2π)}). The period 1 is incorrect and confused with normalization in some contexts. Non-periodicity does not apply to the DTFT—unlike the continuous Fourier transform, periodicity is a defining feature. A period of π/2 is too short and does not reflect the properties of the DTFT.

3. Inverse DTFT Representation

   How can the original sequence x[n] be recovered from its DTFT X(e^{jω})?

   1. By the inverse formula: x[n] = (1/2π) ∫_{-π}^{π} X(e^{jω}) e^{jωn} dω
   2. By the inverse formula: x[n] = ∑_{k=0}^{N-1} X[k] e^{j2πkn/N}
   3. By the inverse formula: x[n] = ∫_{0}^{1} X(e^{jω}) dω
   4. By the formula: x[n] = (1/π) ∫_{-π}^{π} X(e^{jω}) e^{-jωn} dω

   Explanation: The inverse DTFT is correctly given by x[n] = (1/2π) ∫_{-π}^{π} X(e^{jω}) e^{jωn} dω, which reconstructs the discrete sequence from its frequency representation. The finite sum is for the inverse DFT, not the inverse DTFT. Integrating only from 0 to 1 ignores the correct limits and normalization. Using (1/π) as a normalization constant is incorrect; the standard normalization is 1/2π.

4. Frequency Response Example

   For the unit impulse sequence δ[n], what is its DTFT X(e^{jω})?

   1. X(e^{jω}) = 1
   2. X(e^{jω}) = e^{-jωn}
   3. X(e^{jω}) = δ(ω)
   4. X(e^{jω}) = 0

   Explanation: The DTFT of the unit impulse δ[n] is always 1, since the sum reduces to a single nonzero sample at n=0. The function e^{-jωn} is the kernel of the DTFT, not the result. Zero is incorrect because the impulse is not a zero signal. δ(ω) denotes a Dirac delta function in the frequency domain, unrelated to the discrete case.

5. DTFT Linearity Property

   If x1[n] has the DTFT X1(e^{jω}) and x2[n] has the DTFT X2(e^{jω}), what is the DTFT of the signal 3x1[n] - 2x2[n]?

   1. 2X1(e^{jω}) + 3X2(e^{jω})
   2. 3X1(e^{jω}) - 2X2(e^{jω})
   3. X1(e^{jω}) * X2(e^{jω})
   4. 3x1[n] - 2x2[n]

   Explanation: The DTFT is a linear transform, so the DTFT of a linear combination is the same linear combination of the individual DTFTs: 3X1(e^{jω}) - 2X2(e^{jω}). The product X1(e^{jω}) * X2(e^{jω}) applies in convolution but not here. Writing the combination in terms of time-domain signals simply repeats the original without answering in the frequency domain. Adding 2X1 to 3X2 reverses the coefficients and is not correct.