# Homework Help: Laplace Transform of Noncausal systems

1. Jul 27, 2010

### hatfarm

This is some homework from my Analog Signal Processing class, it's a cool class and I've not had much trouble, but this is giving me a bit of a problem and I was hoping someone could help me see what I'm doing... Here's the problem:

Sketch, Get the Laplace Transform of, and Sketch the complex plane w/ region of convergence of the following functions:

f(t) = u(t) - u(t+8)

g(t) = rect((t+8)/3)

Of course there is the Laplace transform integral of integral(e^(-st)f(t) dt,t,0,infinity), which is what I've used. Also, g(t) also is equal to u(t+19/2)-u(t+13/2), which makes it a bit easier to connect to known transforms.

Doing the integration, I get 0 for both signals, which is because they are not causal (they have a value before t=0).
Looking through the text, I see that a non-causal signal has the same transform as a(t)*u(t) <-multiplication, not convolution. Applying this, I would still get zero. However, someone was asking about this after class and the professor implied that this is not correct. Now, I'm not 100% sure if he was talking about the answer of zero, or if he was arguing about the Region of Convergence of zero. Am I right that the transform comes to zero, or am I missing something? If I am correct, I was thinking that with a transform of zero, the ROC would be all values, but I'm not sure if that is true, or if it is no value. I was thinking all values, because it wouldn't matter what you're putting in, nothing is getting through, so all values are in play. Am I lost, or am I just not seeing something? Thanks for the help.

2. Jul 27, 2010

### xcvxcvvc

you're doing a unilaterial laplace on anticauasal signals which is sort of useless. are you supposed to do this (maybe to "show what happens"?) Or are you supposed to do a bilateral laplace on these signals? Yes, the answers will be zero with the unilateral laplace, because the signals are not even in the limits of integration. You're integrating 0*e^whatever from 0 to infinity, which is zero.

3. Jul 27, 2010

### hatfarm

We're just kind of getting used to doing transforms. There are others that we are doing that aren't difficult because they are easily integrable or they are in our transform list.