# Linear Systems Question (Several Parts)

1. Nov 17, 2007

### EugP

1. The problem statement, all variables and given/known data

Find the complete response of the system represented by

$$y''(t)+2y'(t)+y(t)=u(t),$$

$$y(0^-)=1,$$

$$y'(0^-)=2$$

and identify the zro-state and zero-input response components. Find the system transfer function and the system impulse response.

2. Relevant equations

3. The attempt at a solution

I thought I solved for the complete response, but apparently what I got was just the zero-state response. I thought that by taking the laplace transform of the whole function, I will get the complete response. I then remembered that I need to have both zero-state and zero-input responses to get complete response, but I don't know how to find zero-input response.

Here's what I did:

$$y''(t)+2y'(t)+y(t)=u(t)$$

$$s^2Y(s)-f(0^-)-f^{(1)}(0^-)+2sY(s)-f(0^-)+Y(s)=\frac{1}{s}$$

$$Y(s)[s^2+2s+1]-f(0^-)-f^{(1)}(0^-)-f(0^-)=\frac{1}{s}$$

$$Y(s)[s^2+2s+1]-1+2-1=\frac{1}{s}$$

$$Y(s)[s^2+2s+1]=\frac{1}{s}$$

$$Y(s)=\frac{1}{s(s^2+2s+1)}$$

At this point I thought this was already the complete response only in it's transformed for, so now I took the inverse laplace and got:

$$y(t)=(1-te^{-t}-e^{-t})u(t)$$

which is what the answer is for the zero-state response, not complete response.
So I tried figure out the zero-input response, but I just don't even know where to start. I looked everywhere and I can't seem to find the way to find it. Once I find the zero-input response, I will probably be able to solve the rest by myself.

Can anyone please tell me how to find the zero-input response?

2. Nov 18, 2007

### kataya

Why can't you solve for the initial conditions analytically? Like plug in zero for t in your current solution and then solve for the initial conditions? For example:

The first initial condition is y(0) = 1;

$$y(0) = 1 -( (0) + 1 )e^{(0)} = 1 - 1$$

So add 1 to the equation to get satisfy the first condition.

For the second one if you plug in zero into the derivative of your equation you will get -2, so add 4t to the equation so that when you take the derivative you get +4 and it won't mess with the first condition.

This could be horribly wrong, but you can test it easily enough in matlab by taking the new function of t:

$$y(t) = 2 + 4t -(t + 1)e^{-t}$$

and graphing it and comparing vs. step() of the original (zero initial condition) function.

Last edited: Nov 19, 2007
3. Nov 19, 2007

### EugP

Thanks for the reply, I figured out the solution.