Given the function x(t)=t its Fourier transform, X(f) is: A. B. C. D. X(f)= sin(f) 2nf X(f) = j X(f) = j -∞ ≤ x < -1 −1≤x≤+1 cos(f) 2(π f)² cos(2л f) π f 1

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Given the function \( x(t) \):

\[
x(t) = 
\begin{cases} 
0 & , -\infty \leq x < -1 \\
t & , -1 \leq x \leq +1 \\
0 & , 1 < x < +\infty 
\end{cases}
\]

its Fourier transform, \( X(f) \), is:

- **Option A:**
  \[
  X(f) = \frac{\sin(\pi f)}{2 \pi f} - \frac{\cos(\pi f)}{2(\pi f)^2}
  \]

- **Option B (circled):**
  \[
  X(f) = j \frac{\cos(2 \pi f)}{\pi f} - j \frac{\sin(2 \pi f)}{2(\pi f)^2}
  \]

- **Option C:**
  \[
  X(f) = j \frac{\sin(2 \pi f)}{2(2 \pi f)^2} - j \frac{\cos(2 \pi f)}{2 \pi f}
  \]

- **Option D:**
  \[
  X(f) = \frac{1}{(\pi f)^2} \sin(\pi f) - \frac{1}{\pi f} \cos(\pi f)
  \]

In the image, option B is highlighted with an oval, indicating it might be the correct Fourier transform for the given function \( x(t) \).
Transcribed Image Text:Given the function \( x(t) \): \[ x(t) = \begin{cases} 0 & , -\infty \leq x < -1 \\ t & , -1 \leq x \leq +1 \\ 0 & , 1 < x < +\infty \end{cases} \] its Fourier transform, \( X(f) \), is: - **Option A:** \[ X(f) = \frac{\sin(\pi f)}{2 \pi f} - \frac{\cos(\pi f)}{2(\pi f)^2} \] - **Option B (circled):** \[ X(f) = j \frac{\cos(2 \pi f)}{\pi f} - j \frac{\sin(2 \pi f)}{2(\pi f)^2} \] - **Option C:** \[ X(f) = j \frac{\sin(2 \pi f)}{2(2 \pi f)^2} - j \frac{\cos(2 \pi f)}{2 \pi f} \] - **Option D:** \[ X(f) = \frac{1}{(\pi f)^2} \sin(\pi f) - \frac{1}{\pi f} \cos(\pi f) \] In the image, option B is highlighted with an oval, indicating it might be the correct Fourier transform for the given function \( x(t) \).
Expert Solution
Step 1: What is fourier transform

The Fourier transform is a mathematical tool used in signal processing, mathematics, physics, and engineering to analyze and represent a function in terms of its frequency components. It allows you to decompose a complex waveform into a sum of simple sinusoidal functions, each with a specific frequency, amplitude, and phase.

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