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.

  • \$\begingroup\$ Have you tried Laplace? \$\endgroup\$ – Chu Mar 16 '16 at 7:48
  • \$\begingroup\$ I want to try it using fourier only. \$\endgroup\$ – Jonathan Mar 16 '16 at 8:20
  • \$\begingroup\$ 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. \$\endgroup\$ – Haris778 Mar 16 '16 at 9:10

There's no need for the explicit use of the Fourier or Laplace transform. I think the most straightforward way is to rewrite your input signal as a sum of cosine functions:


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


where \$A\$ is determined by the original scaling of the input signal and by the scaling of the low pass filter.


If you need just explicit expression for y(t), output in time domain, you can calculate Fourier transform of input and impulse response, multiple it in frequency domain and than calculate inverse Fourier to get time domain.


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