I started studying signal convolution recently and the first sample problem I got is to find convolution of sine and unit step function (Heaviside function). Here is what I have right now.
$y(t)=H(t)*sin(t)=\int_{-\infty}^{\infty} H(\tau) sin(t – \tau)d\tau = \int_{0}^{\infty}sin(t – \tau)d\tau$
Now when I apply substitution, I have $\lambda = t – \tau, d\lambda = -d\tau$. Then I have
$y(t) = -\int_{t}^{t-\infty} sin(\lambda)d\lambda = \int_{t – \infty}^{t} sin(\lambda)d\lambda = -cos(t) + cos(t – \infty)$.
This, of course, makes no sense. Wolfram Alpha says the answer is simply $-cos(t)$. What am I doing wrong? My textbook does not have the solution for this sample problem. Am I calculating the integral in a wrong way?
Best Answer
The two functions you are trying to convolve are both not integrable. The natural domain of the convolution is $L^1(\mathbf R)$, to this class neither of your two functions belong. The convolution can still be done but you need to view both, $H(t)$ and $g(t) = \sin(t)$ as distributions. I do not want to go into detail about the existence of the convolution in this case (this requires a careful analysis of the domain of definition of $H$ and $g$). If we accept that the two distributions can be convolved we need three ingredients (note that all the operations are now on distributions and they need to be well defined and explained in that context):
Now we can perform the following computation:
$$ H * \sin = H * \partial (-\cos) = \partial H * (-\cos) = \delta * (-\cos) = -\cos $$