Problem description:

problem description

The following is what I've worked out:

work out

Now, to find the steady state response vo(t), which I assume is just vout(t), I would just have to multiply the transfer function by the input. My question is how exactly do I express the input in Fig. 3 as an equation ?

  • \$\begingroup\$ Hint: Note that the input is periodic in 6ms. \$\endgroup\$
    – Shamtam
    Commented Apr 16, 2014 at 18:09
  • 1
    \$\begingroup\$ I think you may need to use Fourier analysis to calculate the steady state response. \$\endgroup\$
    – Andy aka
    Commented Apr 16, 2014 at 18:41

1 Answer 1


If you replace \$s\$ by \$j\omega\$ you get the system's frequency response \$H(j\omega)\$, which you'll need later on. First you have to compute the Fourier series of the periodic input signal:

$$v_b(t)=\sum_{n=-\infty}^{\infty}c_ne^{jn\omega_0t},\quad \omega_0=\frac{2\pi}{T}$$

where \$T=6\$ms is the period of \$v_b(t)\$. The Fourier coefficients \$c_n\$ are given by


I haven't evaluated the integral, but it should be pretty straightforward because of all the straight lines in \$v_b(t)\$. Once you have the coefficients \$c_n\$ you need to realize that the response of the system to an exponential input \$e^{j\omega_0t}\$ is simply \$H(j\omega_0)e^{j\omega_0t}\$ (because the system is linear and time-invariant). So you finally get for the output signal


EDIT: In order to compute the Fourier coefficients you need to write down the piecewise definition of the input signal \$v_b(t)\$:

$$v_b(t)=\left\{\begin{array}{rc}3t-4,& 1\le t<2\\ -3t+8,&2\le t<3\\ -1,&3\le t<7\end{array}\right.$$

Then you split the integral (1) into three intervals:

$$c_n=\frac{1}{6}\left\{\int_1^2(3t-4)e^{-jn\omega_0t}dt+ \int_2^3(-3t+8)e^{-jn\omega_0t}dt- \int_3^7e^{-jn\omega_0t}dt \right\}$$

I guess you can take it from here.

  • \$\begingroup\$ I've tried splitting up the integral in 3, meaning: 1/2 * integral 1,2 (wte^(-jnwt)) + 1/3 * integral 2,3 (-wte^(~)) + 1/7 integral 3,7 (e^(~)). Mathcad seems to be giving me an error when trying to compute them. Could you please show how to set up the integral properly ? \$\endgroup\$
    – user2236
    Commented Apr 17, 2014 at 16:39
  • \$\begingroup\$ @Virtuoso I've added some information about the integral to my answer. \$\endgroup\$
    – Matt L.
    Commented Apr 18, 2014 at 11:55

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