# Which Method of Convolution (If Any) Is Most Appropriate Here?

I need to convolve (or otherwise get the impulse response h(t) of) the input signal $x(t) = 2u(t)$ and $y(t) = cos(4t) + 2e^{(t-1)}$. I have tried the Fourier Transform and the Laplace Transform, but they are both very complex to evaluate.

Is there a simpler way, or is taking the convolution the best approach? I'm occasionally confused on how to recognize which method is best when finding the impulse response, so any general advice on that would be appreciated.

Thanks for any and all help provided!

• This isn't as hard as it looks. You just have to be cautious about your variables and use integration by parts. – Ataraxia Aug 3 '13 at 17:59
• Write out the convolution formula: $\int_{-\infty}^{+\infty} 2u(\tau) y(t-\tau)d\tau = \int_{0}^{+\infty} 2y(t-\tau)d\tau$. – AnonSubmitter85 Aug 3 '13 at 20:05

I have tried the Fourier Transform and the Laplace Transform, but they are both very complex to evaluate.

I don't know what you have tried for Laplace transform, but the Laplace transform of $x*y$ is: $$\mathcal{L}\{(x*y)(t)\} = \mathcal{L}\{(x(t)\}\cdot \mathcal{L}\{y(t)\} = \frac{2}{s}\left( \frac{s}{s^2+4^2} + \frac{2}{e(s-1)}\right),$$ which if using partial fraction, can be represented as: $$\mathcal{L}\{(x*y)(t)\} = \frac12 \frac{4}{s^2+4^2} + \frac{4}{e} \frac{1}{s-1} - \frac{4}{e}\frac{1}{s}.$$ Now every term on the right side can be found in any Laplace transform table, so you can take the inverse Laplace transform to get back to $(x*y)(t)$.

• I got the former step, but was unable to complete the partial fractions. – Rome_Leader Aug 3 '13 at 20:45
• @Rome_Leader The partial fraction can be found using undetermined coefficient: $$\frac{1}{s(s-1)} = \frac{A}{s} + \frac{B}{s-1}.$$ – Shuhao Cao Aug 3 '13 at 21:09

$$x(t)*y(t)=\int_{-\infty}^\infty{x}(\tau)y(t-\tau)\,d\tau$$

For all $t<0$, $x(t)=0$ and for all $t>0$, $x(t)=2$.

$$x(t)*y(t)=\int_0^\infty(2\cos(4t-4\tau)+4e^{(t-\tau-1)})\,d\tau$$

From here treat $t$ as constant, and use u-substitution and the linearity of the integral to solve the integral.

• Are you sure you wrote that? correctly – Amzoti Aug 3 '13 at 20:02
• @Amzoti No, I'm not. Thanks for the catch! – Ataraxia Aug 3 '13 at 20:05
• @Ataraxiz, you are welcome - that looks better! Regards – Amzoti Aug 3 '13 at 20:08