Calculating the Fourier Transform of $x(t) = e^{4t} u(-t)$
The problem requires finding the Fourier Transform (FT) of the function $x(t) = e^{4t} u(-t)$. The FT is a mathematical tool used to decompose a function of time into its constituent frequencies.
Fourier Transform Definition
The Fourier Transform $X(j\omega)$ of a time-domain signal $x(t)$ is defined by the integral:
$X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt$Applying the Definition
Substitute the given function $x(t) = e^{4t} u(-t)$ into the FT definition:
$X(j\omega) = \int_{-\infty}^{\infty} (e^{4t} u(-t)) e^{-j\omega t} dt$The unit step function $u(-t)$ is equal to 1 for $t < 0$ and 0 for $t \geq 0$. Therefore, the integral limits change:
$X(j\omega) = \int_{-\infty}^{0} e^{4t} e^{-j\omega t} dt$Evaluating the Integral
Combine the exponential terms:
$X(j\omega) = \int_{-\infty}^{0} e^{(4 - j\omega)t} dt$Now, perform the integration:
$X(j\omega) = \left[ \frac{e^{(4 - j\omega)t}}{4 - j\omega} \right]_{-\infty}^{0}$Evaluate the expression at the limits:
$X(j\omega) = \frac{e^{(4 - j\omega)(0)}}{4 - j\omega} - \lim_{t \to -\infty} \frac{e^{(4 - j\omega)t}}{4 - j\omega}$Simplify the first term ($e^0 = 1$):
$X(j\omega) = \frac{1}{4 - j\omega} - \lim_{t \to -\infty} \frac{e^{4t} e^{-j\omega t}}{4 - j\omega}$As $t \to -\infty$, $e^{4t} \to 0$. Since $e^{-j\omega t}$ is a complex exponential with a constant magnitude of 1, the limit term becomes 0:
$X(j\omega) = \frac{1}{4 - j\omega} - 0$Final Result
The Fourier Transform is:
$X(j\omega) = \frac{1}{4 - j\omega}$This result matches Option 4.
Important Questions from Fourier Transform
Fourier transform of the unit impulse δ(t) is
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The function f(t) has a Fourier transform F(ω). The Fourier transform of F(t) is
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