# Solving linear system and find impulse response

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)?

• 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. – Dilip Sarwate Dec 28 '13 at 9:50
• 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 $f = f_0$. – Dilip Sarwate Dec 28 '13 at 15:29
• @DilipSarwate No. Sorry, I didn't mean convolution. Just a multiplication. :) – Self Dec 28 '13 at 22:59
• 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. – Dilip Sarwate Dec 29 '13 at 7:11
• @Sultan: The $z$-transform is used for discrete-time systems. It would not be appropriate to use here. – Jason R Dec 30 '13 at 15:13