System output using Fourier transforms

For an LTI system, convolution in time becomes multiplication in frequency:

$$y(t) = x(t) * h(t) \;\Longleftrightarrow\; Y(\omega) = X(\omega)\,H(\omega)$$

Step 1: Fourier transform of the input

$$\begin{aligned} X(\omega) &= \int_{0}^{\infty} e^{-t}e^{-j\omega t}dt = \left[\frac{e^{-(1+j\omega)t}}{-(1+j\omega)}\right]_0^{\infty} \\ &= \frac{1}{1 + j\omega} \end{aligned}$$

Step 2: Fourier transform of the impulse response

$h(t) = e^{t}u(-t)$ is non-zero only for $t < 0$:

$$\begin{aligned} H(\omega) &= \int_{-\infty}^{0} e^{t}e^{-j\omega t}dt = \left[\frac{e^{(1-j\omega)t}}{1-j\omega}\right]_{-\infty}^{0} \\ &= \frac{1}{1 - j\omega} \end{aligned}$$

Step 3: Output spectrum

$$Y(\omega) = \frac{1}{(1+j\omega)(1-j\omega)} = \frac{1}{1 + \omega^2}$$

Step 4: Inverse transform (partial fractions)

$$\frac{1}{(1+j\omega)(1-j\omega)} = \frac{A}{1+j\omega} + \frac{B}{1-j\omega}$$

Multiply out: $1 = A(1-j\omega) + B(1+j\omega)$. Comparing terms: $A + B = 1$ and $-A + B = 0$, so $A = B = \tfrac{1}{2}$.

$$Y(\omega) = \frac{1}{2}\cdot\frac{1}{1+j\omega} + \frac{1}{2}\cdot\frac{1}{1-j\omega}$$

Using the pairs from Steps 1 and 2:

• $\frac{1}{1+j\omega} \leftrightarrow e^{-t}u(t)$
• $\frac{1}{1-j\omega} \leftrightarrow e^{t}u(-t)$

$$y(t) = \frac{1}{2}e^{-t}u(t) + \frac{1}{2}e^{t}u(-t)$$

Answer:

$$y(t) = \frac{1}{2}e^{-|t|}, \quad -\infty < t < \infty$$

Check: the standard pair $e^{-a|t|} \leftrightarrow \frac{2a}{a^2+\omega^2}$ with $a = 1$ gives $\frac{1}{2}e^{-|t|} \leftrightarrow \frac{1}{1+\omega^2}$. Also $y(0) = \frac{1}{2\pi}\int\frac{d\omega}{1+\omega^2} = \frac{1}{2\pi}\cdot\pi = 0.5$.

          y(t)
          0.5
          /\
       .-'  '-.
 ___.-'        '-.___
 ---------+----------> t
          0

The output is two-sided because $h(t)$ is anti-causal (non-zero for $t < 0$).