0
$\begingroup$

I was given the above homework problem and I know how the step response is achieved, but I'm not sure how to set up the integrals in order to evaluate the problem. I've attached my work, but it is wrong because I keep getting the wrong answer. The solution is given as

$$S(t) = u(-t)e^t + u(t)(2-e^{-t})$$

*Note: We have not yet covered laplace transform. In my class, we are expressing the step response as running sums/integrals of the impulse response.enter image description here

$\endgroup$
  • $\begingroup$ Given what is mentioned in your question, can you please add a representative plot of the step function and a plot of the given $h$? $\endgroup$ – A_A Feb 13 '18 at 5:34
  • $\begingroup$ Hi A_A, the step function is 1 for all t equal to or less than 0, and the plot is rudimentary. The plot of the step function or impulse response h is not given, as this is a completely analytical problem. $\endgroup$ – Lee Jordan Feb 13 '18 at 13:20
  • $\begingroup$ edit: step function is 1 when t is equal to or greater than 0. $\endgroup$ – Lee Jordan Feb 13 '18 at 23:22
  • $\begingroup$ No worries. The way the question was phrased initially, I thought it might helped if you simply sketched what it looks like. $\endgroup$ – A_A Feb 13 '18 at 23:25
0
$\begingroup$

HINT: In your work you implicitly assume that $t\ge 0$, so your solution is (probably) correct for $t\ge 0$, but you must also consider the case $t<0$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ hmmm but I thought I accounted for negative t when I'm integrating from -inf to 0? $\endgroup$ – Lee Jordan Feb 13 '18 at 13:21
  • $\begingroup$ @LeeJordan: No, you have to integrate from $-\infty$ to $t$, and if $t <0$ that interval does not include the value zero. $\endgroup$ – Matt L. Feb 13 '18 at 14:14
  • $\begingroup$ OOOOH thank you, Matt! I will do this as soon as I get home. I can just multiply the result of the integral by u(-t) afterwards and that will take care of the case that t>0. thank you so much! $\endgroup$ – Lee Jordan Feb 13 '18 at 14:19

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.