Support of the convolution of two rectangular signals

Question

I'm trying to convolve two rectangular signals in the frequency domain $$H_1(\omega) = u[\omega +.2\pi] - u[\omega -.2\pi]$$ and $$H_2(\omega) = u[\omega +.1\pi] - u[\omega -.1\pi]$$

My result is a trapezoid that ranges from $[-.2\pi, .4\pi]$, but I am expecting a trapezoid centered at 0 with range $[-.3\pi, .3\pi]$.

I am pretty sure I'm getting something wrong with the bounds of integration, but I'm not sure what I'm doing wrong. My strategy is to flip $H_{2}$ and slide it from negative infinity through $H_{1}$. What am I missing in determining the bounds correctly?

Answer 1

We have rectangular pulses

$$H_1 (\omega) = u \left(\omega + \frac{\pi}{5}\right) - u \left(\omega - \frac{\pi}{5}\right)$$

and

$$H_2 (\omega) = u \left(\omega + \frac{\pi}{10}\right) - u \left(\omega - \frac{\pi}{10}\right)$$

and would like to convolve them. When convolving piecewise constant functions, a useful "trick" is to differentiate, convolve and then integrate. Differentiating $H_1$,

$$H_1 ' (\omega) = \delta \left(\omega + \frac{\pi}{5}\right) - \delta \left(\omega - \frac{\pi}{5}\right)$$

and, convolving with $H_2$,

$$(H_1' * H_2) (\omega) = H_2 \left(\omega + \frac{\pi}{5}\right) - H_2 \left(\omega - \frac{\pi}{5}\right) = \begin{cases} \,~~ 1, & \text{if } \omega \in \left[ -\frac{3 \pi}{10}, -\frac{\pi}{10} \right]\\ -1, & \text{if } \omega \in \left[ \frac{\pi}{10}, \frac{3\pi}{10} \right]\\ \,~~0, & \text{otherwise}\end{cases}$$

Finally, integrating,

$$(H_1 * H_2) (\omega) = \begin{cases} \, \frac{3 \pi}{10} + \omega, & \text{if } \omega \in \left[ -\frac{3 \pi}{10}, -\frac{\pi}{10} \right]\\ \qquad\frac{\pi}{5}, & \text{if } \omega \in \left[ -\frac{\pi}{10}, \frac{\pi}{10} \right]\\ \, \frac{3 \pi}{10} - \omega, & \text{if } \omega \in \left[ \frac{\pi}{10}, \frac{3\pi}{10} \right]\\ \,\, \qquad 0, & \text{otherwise}\end{cases}$$

Answer 2

Let $G(\omega) = H_1(\omega) \star H_2(\omega)$, then

$$ G(\omega) = \int_{\phi = - \pi}^{\pi} H_1(\phi) H_2(\omega - \phi) d\phi \tag{1} $$

The nonzero ranges for $H_1$ and $H_2$ are found as:

$$ H_1(\theta) = \begin{cases} { 1 ~~~ , ~~~~ -0.2\pi < \theta < 0.2 \pi \\ 0 ~~~ , ~~~~ \text{ otherwise } } \end{cases} $$

$$ H_2(\theta) = \begin{cases} { 1 ~~~ , ~~~~ -0.1\pi < \theta < 0.1 \pi \\ 0 ~~~ , ~~~~ \text{ otherwise } } \end{cases} $$

Also the nonzero range of the convolution is (convolution bounds) is also known to be: $ -0.3 \pi < \omega < 0.3 \pi $.

Then the nonzero range of integration in Eq(1) for the integral is found as:

$$ \{ -\pi < \phi < \pi \} \cap \{ -0.2 \pi < \phi < 0.2 \pi \} \cap \{ -0.1 \pi < \omega - \phi < 0.1 \pi \} $$

$$ \{ -\pi < \phi < \pi \} \cap \{ -0.2 \pi < \phi < 0.2 \pi \} \cap \{ \omega -0.1 \pi < \phi < \omega + 0.1 \pi \} $$

which yields:

$$ \max(-\pi, -0.2\pi, \omega - 0.1\pi) < \phi < \min(\pi, 0.2\pi, \omega + 0.1\pi) $$

$$ \max(-0.2\pi, \omega - 0.1\pi) = \beta < \phi < \alpha =\min(0.2\pi, \omega + 0.1\pi) $$

Then the convolution integral becomes $$ G(\omega) = \int_{\phi = \beta}^{\alpha} d\phi = \alpha - \beta \tag{2} $$

$$ = \min(0.2\pi, \omega + 0.1\pi) - \max(-0.2\pi, \omega - 0.1\pi) $$

Now using the range of $\omega$ as $-0.3 \pi < \omega < 0.3 \pi $ , evaluate the piece-wise result:

$$ G(\omega) = \begin{cases} { 0 ~~~~~~~~~~~~~~~~~~,~~~ -\pi < \omega < -0.3\pi \\ \omega + 0.3\pi ~~~~~,~~~ -0.3\pi < \omega < -0.1\pi \\ 0.2\pi ~~~~~~~~~~~~ ,~~~ -0.1\pi < \omega < 0.1\pi \\ 0.3\pi - \omega ~~~~~ ,~~~ 0.1\pi < \omega < 0.3\pi \\ 0 ~~~~~~~~~~~~~~~~~~,~~~ 0.3\pi < \omega < \pi }\end{cases} \tag{3} $$