Question 1165583: Determine the Fourier Transform of f(t) = e^−𝑐|𝑡|, with c a positive number.
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! The Fourier Transform $\mathcal{F}\{f(t)\}$ of a function $f(t)$ is defined as:
$$\mathcal{F}\{f(t)\} = F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$$
Given the function $f(t) = e^{-c|t|}$, where $c$ is a positive constant, we need to split the integral due to the absolute value $|t|$.
### 1. Split the Integral
$$F(\omega) = \int_{-\infty}^{\infty} e^{-c|t|} e^{-j\omega t} dt$$
$$F(\omega) = \int_{-\infty}^{0} e^{-c(-t)} e^{-j\omega t} dt + \int_{0}^{\infty} e^{-c(t)} e^{-j\omega t} dt$$
$$F(\omega) = \int_{-\infty}^{0} e^{(c - j\omega)t} dt + \int_{0}^{\infty} e^{-(c + j\omega)t} dt$$
### 2. Evaluate the First Integral ($I_1$)
$$I_1 = \int_{-\infty}^{0} e^{(c - j\omega)t} dt$$
Let $A = c - j\omega$. Since $c > 0$, the real part of $A$ is $c$, so the integral converges as $t \to -\infty$.
$$I_1 = \left[ \frac{1}{c - j\omega} e^{(c - j\omega)t} \right]_{-\infty}^{0}$$
Evaluating the limits:
$$I_1 = \frac{1}{c - j\omega} e^0 - \lim_{t \to -\infty} \frac{1}{c - j\omega} e^{(c - j\omega)t}$$
$$I_1 = \frac{1}{c - j\omega} - 0$$
$$I_1 = \frac{1}{c - j\omega}$$
### 3. Evaluate the Second Integral ($I_2$)
$$I_2 = \int_{0}^{\infty} e^{-(c + j\omega)t} dt$$
Let $B = c + j\omega$. Since $c > 0$, the real part of $B$ is $c$, so the integral converges as $t \to \infty$.
$$I_2 = \left[ \frac{-1}{c + j\omega} e^{-(c + j\omega)t} \right]_{0}^{\infty}$$
Evaluating the limits:
$$I_2 = \lim_{t \to \infty} \left( \frac{-1}{c + j\omega} e^{-(c + j\omega)t} \right) - \left( \frac{-1}{c + j\omega} e^{0} \right)$$
$$I_2 = 0 - \left( \frac{-1}{c + j\omega} \right)$$
$$I_2 = \frac{1}{c + j\omega}$$
### 4. Combine the Results
The Fourier Transform $F(\omega)$ is $I_1 + I_2$:
$$F(\omega) = \frac{1}{c - j\omega} + \frac{1}{c + j\omega}$$
Find a common denominator:
$$F(\omega) = \frac{(c + j\omega) + (c - j\omega)}{(c - j\omega)(c + j\omega)}$$
Simplify the numerator ($-j\omega$ and $+j\omega$ cancel) and the denominator (difference of squares: $(a-b)(a+b) = a^2 - b^2$):
$$F(\omega) = \frac{2c}{c^2 - (j\omega)^2}$$
Recall that $j^2 = -1$:
$$F(\omega) = \frac{2c}{c^2 - (-1)\omega^2}$$
### Final Result
$$\mathcal{F}\{e^{-c|t|}\} = F(\omega) = \frac{2c}{c^2 + \omega^2}$$
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