# Impulse Response of a LTI system

The input to a L.T.I. circuit is $x(t) = 6\cos(t)\cos(3t)$, and the impulse response of the circuit is $$h(t) = \frac{\sin(3t)}{3t}$$ Obtain an explicit expression for the output y(t) as a function of time. The fourier transform of $x(t) =$

$$\sum_{n=-\infty}^\infty C_n e^{in2t}$$

I converted $h(t)$ to $$H(iw) = \frac{\pi}{3}\times\text{rect}\left(\frac{w}{6}\right)$$

However, I am confused on how I would use Fourier series coefficients to solve this problem.

• Have you tried Laplace? – Chu Mar 16 '16 at 7:48
• I want to try it using fourier only. – Jonathan Mar 16 '16 at 8:20
• Your equation for function in time domain x(t) (written as a sum with series coeff.) is not correct. You are missing 'pi' in exponent. – Haris778 Mar 16 '16 at 9:10

$$x(t)=3(\cos(2t)+\cos(4t))$$
Since you (should) know that your filter is an ideal low pass filter with cut-off frequency $\omega_c=3$ you know immediately that the term $\cos(4t)$ will be completely suppressed, whereas the term $\cos(2t)$ will appear at the output just with a scaling. I'm sure you can determine that scaling yourself. So your output signal is simply
$$y(t)=A\cos(2t)$$
where $A$ is determined by the original scaling of the input signal and by the scaling of the low pass filter.