Which of the following is NOT TRUE, given that “→” represents fourier transform "*" represents convolution,  x(n) → X(ej^ω) and y(n) → Y(ej^ω)

• x(n) → X(e^jω): Denotes that the Fourier transform of x(n) is X(e^jω).
• y(n) → Y(e^jω): Denotes that the Fourier transform of y(n) is Y(e^jω).

The question asks us to identify which of the provided statements is NOT TRUE.

Correct Option: Option 2

Detailed Explanation:

Let us analyze the given options one by one to determine why Option 2 is the correct answer.

Option 2: $x\left(n-n_{d}\right) → e^{j \omega n_{d}} \left(e^{j \omega}\right)$

This statement is NOT TRUE. Let us break it down:

• When a signal is delayed in the time domain by n_d (i.e., x(n - n_d)), the Fourier transform of the delayed signal introduces a phase shift in the frequency domain.
• The correct Fourier transform for x(n - n_d) is given by:
  x(n - n_d) → X(e^jω) × e^-jωn_d
• In the given statement, the phase factor e^jωn_d is incorrectly written as e^jωn_d (positive exponent), which is incorrect. The correct phase factor must have a negative exponent.

Therefore, Option 2 is incorrect because it does not accurately represent the Fourier transform of a delayed signal. The correct representation should involve the term e^-jωn_d, not e^jωn_d.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: a × x(n) + b × y(n) → a × X(e^jω) + b × Y(e^jω)

This statement is TRUE. The Fourier transform is a linear operation, which means that the Fourier transform of a linear combination of signals is equal to the same linear combination of their respective Fourier transforms. Hence, this property is valid.

Option 3: $e^{jω_{0}} × x(n) → X\left(e^{j\left(ω - ω_{0}\right)}\right)$

This statement is TRUE. When a signal is modulated in the time domain by e^jω₀ (complex exponential multiplication), the effect in the frequency domain is a shift in the frequency spectrum by ω₀. The Fourier transform accurately reflects this frequency shift, making this statement valid.

Option 4: x(n) × y(n) → X(e^jω) × Y(e^jω)

This statement is TRUE. Convolution in the time domain corresponds to multiplication in the frequency domain. Hence, the Fourier transform of the convolution of x(n) and y(n) is the product of their respective Fourier transforms. This property is fundamental to the Fourier transform and is valid.