# continous time system

Discussion in 'Homework Help' started by TAKYMOUNIR, Oct 10, 2012.

1. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
what the form equation of h(t) which make output y(t) if we have input x(t) ,h(t) is a linear time invariant system
thanks

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2. ### mlog Member

Feb 11, 2012
276
36
Have you studied convolution in class?

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Jun 23, 2008
351
1
no i did not

4. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
but it will be taught in class

5. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
can i solve it without using convolution

6. ### mlog Member

Feb 11, 2012
276
36
Maybe someone else has a better idea, but you could write x(t) and y(t) in the time domain (in terms of t), take the Laplace transforms to get X(s) and Y(s), divide Y(s) by X(s) to get H(s), and then take the inverse of the Laplace transform to find h(t). After you go through that process, you'll find out why convolution is so nice to use in this case. Often times Laplace transforms are the best approach. Here I'm not so sure. I am assuming you've studied Laplace transforms, but maybe not.

One thing you might notice is that between t=0 and t=1, you effectively have an integration of a step input within h(t). Think about what is happening between t=1 and t=2 and for t>2.

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7. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Without knowing the context of the problem (i.e., why this problem was given and how it fits into the material before and after it), it is difficult to give a good answer. I think mlog has a good suggestion.

My first impression is that the response seems hinky. You apply a setp at t=0 and hold it constant until t=2s. In response, the output ramps up and ramps down in that same t=2s. If it is a linear time-invariant system, then if you shift the input pulse right (or left) by 2s, you shift the output the same amount. This means that if you applied a constant input of 1 for all time, that the output would oscillate in a triangular fashion between 0 and 2. That doesn't sound like a linear time-invariant system to me. But maybe I'm forgetting something simple.

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