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I am pretending that laplace does not exist because I am being tested on these concepts separately.

Essentially, I have solved for the step response of a first order circuit and found it to be:

$$v_{c}(t)=\left(\frac{-5}{99}e^{-5t}+\frac{106}{99}e^{\frac{-t}{20}}\right)u(t)$$

I'm only dealing with LTI systems so I know that the impulse is the derivative of the step, but I will be left with some terms attached to the delta function, and some terms attached to the unit step function.

$$h(t)= \left( \frac{-5}{99}e^{-5t}+\frac{106}{99}e^{\frac{-t}{20}} \right)\delta(t) + \left( \frac{25}{99}e^{-5t}+\frac{53}{990}e^{\frac{-t}{20}} \right)u(t)$$

If I want to use the impulse response in the convolution integral, how do I handle these delta terms to make it less... convoluted? Do they reduce to a constant?

Marcus Müller
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amateurhour
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  • A property of the delta function is $$\small \int_0^\infty f(t)\delta(t)=f(0) $$ Hence the convolution integral will include the constant value $$\small \frac{-5}{99}+\frac{106}{99}=\frac{101}{99}$$ – Chu Oct 24 '20 at 10:40

2 Answers2

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Since \$\delta(t) = 0\$ when \$t \ne 0\$, you should be able to replace:

$$f(t)\cdot\delta(t)$$ with $$f(0)\cdot\delta(t)$$

(I don't think this is always true in every case, e.g., inside of an integral, but I'm reasonably sure you can do it in your case)

Justin
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You don't have to worry about \$\delta(t)\$ since the integral of it results in \$u(t)\$. Even integrating it alone gives \$\int_0^x{\delta(\pm t)\text{d}t}=2u(x)-1\$. So whatever convolutions you'll have with \$h(t)\$ will include the step function in the result. BTW, the derivative is with \$-\frac{53}{990}\$ in the 2nd term.

a concerned citizen
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  • To the people downvoting: by not stating why you downvoted, you're not letting OP know *why* the selected answer is wrong. If I am wrong, so be it, but ultimately it's about selecting the correct answer for a question. A downvote means this should not be the one. Therefore, by downvoting a selected answer without an explanation *for the OP*, you're making it ad hominem, as opposed to correcting something wrong. And by doing that, you're showing you think only of yourselves, and casts a shadow on the reason for your presence on this site. – a concerned citizen Oct 21 '20 at 09:27