Question

Consider two continuous-time signals \(x(t)\) and \(y(t)\) as shown below. If \(X(f)\) denotes the Fourier transform of \(x(t)\), then the Fourier transform of \(y(t)\) is \(\_\_\_\_\). \begin{center} \includegraphics[width=8cm]{44.png} \end{center}

• A. \(-4X(4f)e^{-j\pi f}\)
• B. \(-4X(4f)e^{-j4\pi f}\)
• C. \(-\frac{1}{4} X\left(\frac{f}{4}\right)e^{-j\pi f}\)
• D. \(-\frac{1}{4} X\left(\frac{f}{4}\right)e^{-j4\pi f}\)

Hint:

For transformed signals, use the scaling property to modify the frequency and the time-shifting property to adjust the phase of the Fourier transform.

Answer 1

The Correct Option is B

Step 1: Recall Fourier transform properties.
For a scaled signal \(x(at)\), the Fourier transform is scaled in frequency: \[ \mathcal{F}[x(at)] = \frac{1}{|a|} X\left(\frac{f}{a}\right). \] For a time-shifted signal \(x(t - t_0)\), the Fourier transform is: \[ \mathcal{F}[x(t - t_0)] = X(f) e^{-j2\pi f t_0}. \] Step 2: Analyze the given transformation.
The signal \(y(t)\) is related to \(x(t)\) as: \[ y(t) = -4x(4t - 4). \] Here: The scaling factor is \(a = 4\), introducing a frequency scaling by \(4f\). The time shift is \(t_0 = 1\), introducing a phase shift \(e^{-j2\pi f \cdot 1}\). The amplitude factor of \(-4\) multiplies the entire Fourier transform. Step 3: Apply the Fourier transform properties.
Using the above properties, the Fourier transform of \(y(t)\) is: \[ \mathcal{F}[y(t)] = -4 X(4f) e^{-j4\pi f}. \] Final Answer: \[\boxed{{(2) } -4X(4f)e^{-j4\pi f}}\]