Fourier Transform of the Exponential Decay Signal

This solution explains how to find the Fourier Transform of the signal defined as $x(t) = e^{at}u(-t)$, where $a > 0$. The Fourier Transform is a fundamental tool in signal processing used to analyze signals in the frequency domain.

Understanding the Signal

The signal $x(t)=e^{at}u(-t)$ consists of two parts:

• $e^{at}$: An exponential function. Since $a > 0$, this term grows as $t$ increases.
• $u(-t)$: The unit step function. This function is equal to 1 when its argument is non-negative (i.e., $-t \ge 0$ or $t \le 0$) and 0 when its argument is negative (i.e., $-t < 0$ or $t > 0$).

Combining these, the signal $x(t)$ is $e^{at}$ for $t \le 0$ and $0$ for $t > 0$. This represents an exponentially decaying signal as $t$ goes from $0$ towards negative infinity.

Fourier Transform Definition

The Fourier Transform of a continuous-time signal $x(t)$ is defined as: $ X(f) = \mathcal{F}\{x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt $ where $f$ is the frequency and $j$ is the imaginary unit.

Derivation Steps

1. Substitute the signal $x(t)$ into the Fourier Transform integral: $ X(f) = \int_{-\infty}^{\infty} \left( e^{at}u(-t) \right) e^{-j2\pi ft} dt $
2. Apply the property of the unit step function $u(-t)$: Since $u(-t)$ is 1 only for $t \le 0$ and 0 otherwise, the integral limits change from $(-\infty, \infty)$ to $(-\infty, 0]$. $ X(f) = \int_{-\infty}^{0} e^{at} e^{-j2\pi ft} dt $
3. Combine the exponential terms: Using the property $e^a e^b = e^{a+b}$, we can combine the exponents. $ X(f) = \int_{-\infty}^{0} e^{(a - j2\pi f)t} dt $
4. Evaluate the integral: The integral of $e^{kt}$ is $\frac{1}{k}e^{kt}$. Here, $k = a - j2\pi f$. $ X(f) = \left[ \frac{e^{(a - j2\pi f)t}}{a - j2\pi f} \right]_{t=-\infty}^{t=0} $
5. Apply the limits of integration: Evaluate the expression at the upper limit ($t=0$) and subtract the value at the lower limit ($t \to -\infty$).
   • At $t=0$: $\frac{e^{(a - j2\pi f)(0)}}{a - j2\pi f} = \frac{e^0}{a - j2\pi f} = \frac{1}{a - j2\pi f}$
   • As $t \to -\infty$: Since $a > 0$, the term $at$ in the exponent $(a - j2\pi f)t$ approaches $-\infty$. Therefore, $e^{(a - j2\pi f)t} \to 0$.
6. Final Result: Subtracting the value at the lower limit from the value at the upper limit gives: $ X(f) = \frac{1}{a - j2\pi f} - 0 $ $ X(f) = \frac{1}{a - j2\pi f} $

Conclusion

The Fourier Transform of the signal $x(t)=e^{at}u(-t)$ for $a>0$ is $\frac{1}{a - j2\pi f}$. This matches option 3.