# LTI system response to periodic input

I'm trying to find the zero-state response (ZSR) of an LTI system to a one sided periodic input, like a square wave that is equals to zero for $t < 0$.

I know that I can use the Fourier series of said input function to find the steady-state (SS) response, however I'm having trouble understanding how to use the Laplace transform to obtain the ZSR, which contains the SS component plus a transient one.

My guess is that I need to calculate the Laplace transform of the periodic input, and then solve the system for $s$ to obtain the transform of the output.

• Usually you can just find the response by adding the transient responses times some constants to the periodic "steady state" response. Those constants can be found by the constraints that $y(0)=0$, the higher the order of the system the more initial conditions are required, such as $y'(0)=0$. This is the case because the $y(t)=0\; \forall t\leq0$ – fibonatic May 10 '16 at 19:29

You can use the Laplace transform, but can also simply use convolution in the time domain. In any case, you will need the system's impulse response $h(t)$. Let the input signal $x(t)$ satisfy $x(t)=0$ for $t<0$, and $x(t+T)=x(t)$ for $t>0$ and $T>0$, as required. Furthermore, let $f(t)$ be the first period of $x(t)$:

$$f(t)=\begin{cases}x(t),&0<t<T\\0,&\text{otherwise}\end{cases}\tag{1}$$

Then

$$x(t)=\sum_{n=0}^{\infty}f(t-nT)=f(t)\star\sum_{n=0}^{\infty}\delta(t-nT)\tag{2}$$

where $\star$ denotes convolution, and $\delta(t)$ is the Dirac delta impulse. If $g(t)$ denotes the convolution $h(t)\star f(t)$, i.e., the system's response to $f(t)$, then the output signal can be written as

\begin{align}y(t)&=h(t)\star x(t)\\&=h(t)\star f(t)\star\sum_{n=0}^{\infty}\delta(t-nT)\\&=g(t)\star \sum_{n=0}^{\infty}\delta(t-nT)\\&=\sum_{n=0}^{\infty}g(t-nT)\tag{3}\end{align}

According to $(3)$, the output signal can be written as a sum of shifted responses to the finite length signal $f(t)$, which corresponds to the first period of the input signal.

The Laplace transform of the output signal can also be written in terms of the transforms of the input signal and of the function $f(t)$. Note that

$$f(t)=x(t)-x(t-T)\tag{4}$$

The Laplace transform of $(4)$ is

$$F(s)=X(s)(1-e^{-st})\tag{5}$$

If $H(s)$ is the system's transfer function, i.e. the Laplace transform of the impulse response $h(t)$, then the output is given by

$$Y(s)=H(s)X(s)=H(s)\frac{F(s)}{1-e^{-st}}\tag{6}$$