0
$\begingroup$

I have this linear system that is defined by the differential equation $$ y''(t) + 4y'(t) + 5y(t) = 2x(t) + 3x'(t) $$ with x(t) the input and y(t) the output. I'm asked to find two things: The output when x(t) = cos(t), and the impulse response

This is what I did: First, I found the frequency response,H(f) $$ H(f) = \frac{2 + 6\pi fj}{8\pi fj -4\pi^2 f^2+5} $$

Second, I found the Fourier Transform of cos(t) which equals to: $$ X(f) = \frac{1}{2}\delta (f-\frac{1}{2\pi }) + \frac{1}{2}\delta (f+\frac{1}{2\pi }) $$

Now, the output basically would be $$ Y(f) = H(f)\ X(f) $$ Is it possible to simplify the answer of Y(f)?

For the second part of the question, I found the impulse response using Laplace Transform $$ h(t) = [\frac{-17}{6} e^{5t} - \frac{1}{6} e^{-t}]u(t) $$

What do you think about my answer? Is it possible to use Z-transform to find h(t)?

$\endgroup$
  • $\begingroup$ The output is not $Y(f) = H(f)*X(f)$ if by $*$ you mean convolution. Do not mix up standard notation in system theory with the conventions of programming languages. $\endgroup$ – Dilip Sarwate Dec 28 '13 at 9:50
  • $\begingroup$ Hint: $\displaystyle \int_{-\infty}^\infty G(f)\delta(f-f_0)e^{-j2\pi f t}\, \mathrm df = G(f_0)e^{-j 2\pi f_0 t}$ provided that $G(f)$ is continuous at $f = f_0$. $\endgroup$ – Dilip Sarwate Dec 28 '13 at 15:29
  • $\begingroup$ @DilipSarwate No. Sorry, I didn't mean convolution. Just a multiplication. :) $\endgroup$ – Self Dec 28 '13 at 22:59
  • $\begingroup$ Then, please edit your question to express $Y(f)$ properly, and then use the hint I provided in my second comment to obtain the output. Note that the output is $y(t)$, not $Y(f)$ as you claim it is. $\endgroup$ – Dilip Sarwate Dec 29 '13 at 7:11
  • $\begingroup$ @Sultan: The $z$-transform is used for discrete-time systems. It would not be appropriate to use here. $\endgroup$ – Jason R Dec 30 '13 at 15:13

Your Answer

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

Browse other questions tagged or ask your own question.