# Support of the convolution of two rectangular signals

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?

• Hi rafafan2010. Now you have two answers that yield the same result in two different approaches. Please select the one that feels best and upvote the other, so that the question is closed... Feb 10, 2019 at 9:17
• @RodrigodeAzevedo Yes, I'll update that. Feb 11, 2019 at 14:27
• @Fat32 I'll select the one I think is best. Thanks for the suggestion. Feb 11, 2019 at 14:28

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,

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

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:

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