Unit Step Response

1. May 3, 2007

The step response of an LTI system is given by $s(t) = e^{-t}u(t)$. Find its response to the input $u(t) - u(t-2)$.

Note: $u(t)$ is the unit step function.

So I have the solution, but I do not understand it.

$$y(t) = s(t) - s(t-2) = e^{-t}u(t) - e^{-(t-2)}u(t-2)$$

I really don't understand where this is coming from. If somone could guide me to this solution, that would be good.

Thanks!

2. May 3, 2007

mjsd

the only thing I can see here is that LTI system is a linear system and so a superposition of step function inputs give a superposition of response.
(linearity)

3. May 3, 2007

I need to see the steps filled in here to understand it. The professor just skipped too much for my understanding.

I don't understand this well enough I guess.

Here is my attempt.

We can categorize an LTI system as:

x(t)------>h(t)------>y(t)

If we let x(t) be the unit step function, then $y(t)=\int_{-\infty}^t dt$

Since the step response is given:
$$y(t)=e^{-t}u(t)$$
Therefore, $$h(t)=\frac{dy(t)}{dt}=-e^{-t}u(t)$$

So when the response of $x(t) = u(t)-u(t-2)$

Performming the Laplace transforms yield:
$$X(s)=\frac{1}{s}-\frac{e^{-s2}}{s}$$
$$H(s)=\frac{-1}{s}$$

$$Y(s)=\frac{e^{-s2}}{s(s+1)}-\frac{1}{s(s+1)}$$

The inverse transform yields:
$$y(t)=u(t-2)+e^{-(t-2)}u(t-2)-u(t)-e^{-t}u(t)$$
$$y(t)=u(t-2)(e^{t-2}+1)-u(t)(e^{-t}+1)$$

Why am I getting these extra factors of u(t-2) and u(t)?

Thanks!

4. May 4, 2007

mjsd

short ans: you've forgotten one very important thing, namely, the derivative of a unit step function, is the delta function.

longer ans: (by the way, symbols, x, y, h, s etc,.... were kind of mixed up in your post so I will define them again)

x(t) input signal, y(t) output signal, h(t) impulse response, s(t) unit step response, u(t) unit step function.
X(s), Y(s), H(s), S(s)... etc.. are Laplace Transformed quantities.

you are given $$x(t) = u(t) -u(t-2)$$ and that $$s(t) =e^{-t} u(t)$$ which is the response of a $$u(t)$$ input. ie. the results of
$$s(t)=\int_{-\infty}^{\infty} u(\tau) h(t-\tau) \,d\tau$$ where h(t) is the yet to be determined impulse response that characterise the system. Ok, now LTI system has a nice property called linearity that means
$$f(aX+bY) = a f(X) + b f(Y)$$
And this works nicely (in this case) because your x(t) is made up of just step fucntions and you are given the response for a step function s(t) already!
so if u(t) gives your s(t) then u(t) - u(t-2) will give you s(t) - s(t-2).
to see that u(t-a) does lead to s(t-a), just do a change of variable as follow

recall: $$s(t)=\int_{-\infty}^{\infty} u(\tau) h(t-\tau) \,d\tau$$
$$\int_{-\infty}^{\infty} u(\tau-2) h(t-\tau) \,d\tau$$
let $$\kappa = \tau-2, \; d\kappa = d\tau$$ then
$$\int_{-\infty}^{\infty} u(\kappa) h(t-(\kappa+2)) \,d\kappa =\int_{-\infty}^{\infty} u(\kappa) h((t-2)-\kappa) \,d\kappa =s(t-2)$$

Usually this is all you need to get your result....which may explain why your prof did not show more workings..... But to prove to you that this is indeed correct, we can try this again in the long way. ie. first acquire h(t) from s(t) then do the convolution integral (or go to frequency domain by Laplace transform). I shall cut the narrative for the following (to save me some time)... just showing you the maths, symbols are as described above

$$h(t) = \frac{d s(t)}{dt} = -e^{-t}u(t) + e^{-t}u'(t)= -e^{-t}u(t) + e^{-t}\delta(t)$$

$$H(s) = -\frac{1}{s+1}+1$$

$$X(s) = \frac{1}{s} - \frac{e^{-2s}}{s}$$

$$Y(s) = H(s)X(s) = \frac{1-e^{-2s}}{s+1} \; \Rightarrow\; y(t) = e^{-t}u(t) - e^{2-t}u(t-2)$$
try it!

Last edited: May 4, 2007
5. May 4, 2007

First off. You are so amazing, I can't even begin to tell you how much that helped. Thanks a million!

Second, yeah... I did not differentiate right :(
$$e^{-t}u(t) = \left\{ \begin{array}{c} e^{-t}, \,\,\,\, t \geq 0 \\ 0, \,\,\,\, t< 0 \end{array}$$
I then just blindly differentiated $e^{-t}$ and $0$. However I didn't take into account the fact that it is discontinuous at 0! So thankyou for pointing that out.