# Convolution integral with step function

for the following convolution integral

$$\int_{-\infty }^{\infty}\sigma (\tau)\tau A\sigma(t-\tau)\sin(t-\tau)d\tau \text,$$ where $$\\sigma(t)\$$ denotes the step function

We'll only get results for $$\t>0\$$, since thats when $$\\sigma(t)\$$ will have the value 1, which means that the integral will be evaluated for $$\\tau\$$ from $$\0\$$ to $$\t\$$.

I get the rest of the math in the given solution (partial integration, integration by substitution etc...). What I don't get is the following:

\begin{align} \int_{-\infty }^{\infty}\sigma (\tau)\tau A\sigma(t-\tau)\sin(t-\tau)d\tau &= \int_{0 }^{t}\sigma (\tau)\tau A\sigma(t-\tau)\sin(t-\tau)d\tau\\ &= A\sigma (t)\int_{0 }^{t}\tau \sin(t-\tau)d\tau \end{align}

Why can $$\\sigma(t)\$$ be taken out of the integral just like that? Isn't the integral of the step function $$\\sigma(t)\$$ the ramp function?

I have another, somewhat related question: Why is the stability in a bode plot evaluated at the cutoff frequency? The solution to a given problem in my script says that if the gain at the cutoff is $$\< 0\$$, then it's stable, if it's $$\> 0\$$ it is not. Intuitively that makes sense, but why at the cutoff frequency?

• wait, is $*$ the convolution operator or is it multiplication? Feb 16, 2020 at 11:13
• sorry for the confusion, it's supposed to denote the multiplication. Feb 16, 2020 at 11:17
• I'll remove it then, because all of the other multiplications don't have it. Feb 16, 2020 at 11:18
• if $\sigma(\tau)$ is a function of $\tau$ (which it is!) you can't simply "pull it out" of the integral. Feb 16, 2020 at 11:21
• Those are the steps in the solution. If you integrate $\sigma(\tau)$ from 0 to t you'd get the ramp function though, right? This is exactly what I don't understand. Feb 16, 2020 at 11:23

Consider the following integral:

$$\int_{-\infty}^{\infty}\sigma(\tau)\sigma(t-\tau)f(\tau)d\tau\tag{1}$$

where $$\f(t)\$$ is an arbitrary function.

The integrand is zero for $$\\tau<0\$$ (due to the factor $$\\sigma(\tau)\$$), and it is zero for $$\\tau>t\$$ (due to the factor $$\\sigma(t-\tau)\$$).

Now it's important to note that replacing the integration limits by $$\0\$$ (lower limit) and $$\t\$$ (upper limit) is only justified for $$\t>0\$$, because otherwise the factor $$\\sigma(\tau)\sigma(t-\tau)\$$ equals zero for all values of $$\\tau\$$, because $$\\sigma(\tau)\$$ and $$\\sigma(t-\tau)\$$ don't overlap in that case.

Consequently, the correct way to rewrite $$\(1)\$$ is

$$\sigma(t)\int_{0}^{t}f(\tau)d\tau\tag{2}$$

because without the factor $$\\sigma(t)\$$, the integral does not evaluate to zero for $$\t<0\$$ as it should.

I don't see the relation of your second question to this one, so I suggest you formulate it as a separate question.