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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?

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  • \$\begingroup\$ wait, is \$*\$ the convolution operator or is it multiplication? \$\endgroup\$ Feb 16, 2020 at 11:13
  • \$\begingroup\$ sorry for the confusion, it's supposed to denote the multiplication. \$\endgroup\$
    – Pilotf4
    Feb 16, 2020 at 11:17
  • \$\begingroup\$ I'll remove it then, because all of the other multiplications don't have it. \$\endgroup\$ Feb 16, 2020 at 11:18
  • \$\begingroup\$ if \$\sigma(\tau)\$ is a function of \$\tau\$ (which it is!) you can't simply "pull it out" of the integral. \$\endgroup\$ Feb 16, 2020 at 11:21
  • \$\begingroup\$ 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. \$\endgroup\$
    – Pilotf4
    Feb 16, 2020 at 11:23

1 Answer 1

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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.

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