Response of System to sin(3t) and tri(t-1) Inputs

[mathrocks]

The question is as follows:

The step response of a linear time-invariant system is given as
s(t)=t*e^(-2t) u(t)
Determine the response of the system to the following inputs:
a. x(t)=sin(3t) u(t)
b. x(t)=tri(t-1)

As of now I see this problem and I have no idea what it means...I tried looking in my textbook but the book is poorly written so that confuses me more and oh yeah, my teacher is horrible. But I'm trying to put in an effort to learn on my own, hopefully it's possible...

Any help with this is truly appreciated.
Also, if there are any websites that you guys know of that can help me with Signals and Systems that would be great!

 

[quantumdude]

You know that the output is $s(t)=te^{-2t}U(t)$ when the input is $x(t)=U(t)$. From those you can find the transfer function $H(s)=S(s)/X(s)$. Then you can use the transfer function to get the other outputs, because $S(s)=H(s)X(s)$ holds generally.

 

[mathrocks]

You know that the output is $s(t)=te^{-2t}U(t)$ when the input is $x(t)=U(t)$. From those you can find the transfer function $H(s)=S(s)/X(s)$. Then you can use the transfer function to get the other outputs, because $S(s)=H(s)X(s)$ holds generally.

Is there another way to do this problem? I don't think we've done transfer functions yet. All we've done so far is basic signals stuff like graphing impulses, find signal energy and power, proving linearity, time-invariance, etc and convolution and basic laplace.

 

[quantumdude]

Perhaps you have seen the transfer function, but just haven't been told so.

Basically, S(s) is the Laplace transform of the output and X(s) is the Laplace transform of the input. H(s) is just their ratio. If you followed the procedure that I laid out you would be doing nothing more than taking Laplace transforms and inverse Laplace transforms.

 

[mathrocks]

Perhaps you have seen the transfer function, but just haven't been told so.

Basically, S(s) is the Laplace transform of the output and X(s) is the Laplace transform of the input. H(s) is just their ratio. If you followed the procedure that I laid out you would be doing nothing more than taking Laplace transforms and inverse Laplace transforms.

Ok, so X(s) is the Laplace transform of the function x(t) and S(s) is the Laplace transform of s(t)=t*e^(-2t) u(t)? So is H(s) the response of the input? I don't see when you would do inverse Laplace...

Sorry, but this stuff is 100% new to me, we definitely didn't talk about this yet. But it doesn't seem to hard to understand, I just need some guidance...

 

[quantumdude]

Ok, so X(s) is the Laplace transform of the function x(t) and S(s) is the Laplace transform of s(t)=t*e^(-2t) u(t)?

Yes.

So is H(s) the response of the input?

No, H(s) is the ratio of S(s) and X(s).

I don't see when you would do inverse Laplace...

You would do it when you try to determine the responses to the other inputs.

S(s)=H(s)X(s)