2
\$\begingroup\$

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?

\$\endgroup\$
1
  • \$\begingroup\$ 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 $$\small \frac{-5}{99}+\frac{106}{99}=\frac{101}{99}$$ \$\endgroup\$
    – Chu
    Oct 24, 2020 at 10:40

2 Answers 2

0
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ 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. \$\endgroup\$ Oct 21, 2020 at 9:27
1
\$\begingroup\$

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)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.