How to calulate response to input for LTI frequency response?

Mar 2013
96
1
Texas
How can I solve this problem. I cant find how to do it anywhere in my book. Thanks.

HOW.PNG
 
Apr 2012
263
107
Erewhon
If:

\(\displaystyle H(\omega)=\frac{1}{1+j\omega}\)

then:

\(\displaystyle H(s)=\frac{1}{1+s}\)

and:

\(\displaystyle Y(s)=\frac{1}{1+s}X(s)\)

or:

\(\displaystyle (1+s)Y(s)=X(s)\)

which is equivalent to the differential equation in the time domain:

\(\displaystyle \left[ \frac{d}{dt}+1 \right]y(t)=x(t)\)

or:

\(\displaystyle y'+y=x\)

So putting \(\displaystyle x(t)=\cos(t)\) we have:

\(\displaystyle y'(t)+y(t)=\cos(t)\ \ \ \dots[1]\)

Now how you solve this depends on what you know.

One approach is, as a LTI system driven by a sinusoid has an sinusoidal output at the same frequency as the input, use a trial solution of the form \(\displaystyle y(t)=A\sin(t)+B\cos(t)\).

Substitute this into \(\displaystyle [1] \)above and solve for \(\displaystyle A\) and \(\displaystyle B\).
 
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romsek

MHF Helper
Nov 2013
6,665
3,002
California
If:

\(\displaystyle H(\omega)=\frac{1}{1+j\omega}\)

then:

\(\displaystyle H(s)=\frac{1}{1+s}\)

and:

\(\displaystyle Y(s)=\frac{1}{1+s}X(s)\)

or:

\(\displaystyle (1+s)Y(s)=X(s)\)

which is equivalent to the differential equation in the time domain:

\(\displaystyle \left[ \frac{d}{dt}+1 \right]y(t)=x(t)\)

or:

\(\displaystyle y'+y=x\)

So putting \(\displaystyle x(t)=\cos(t)\) we have:

\(\displaystyle y'(t)+y(t)=\cos(t)\ \ \ \dots[1]\)

Now how you solve this depends on what you know.

One approach is, as a LTI system driven by a sinusoid has an sinusoidal output at the same frequency as the input, use a trial solution of the form \(\displaystyle y(t)=A\sin(t)+B\cos(t)\).

Substitute this into \(\displaystyle [1] \)above and solve for \(\displaystyle A\) and \(\displaystyle B\).
This isn't quite correct. Fourier Transforms imply steady state solutions whereas Laplace Transforms reveal transient behavior as well.

Simply what you want to do is set

$X(\omega) = \mathscr{F}\{\cos(t)\}$

$H(\omega) = \dfrac{1}{1+j \omega}$

$Y(\omega) = H(\omega)X(\omega)$

$y(t) = \mathscr{F}^{-1}\{Y(\omega)\}$

The steady state answer is that $y(t) = \dfrac{1}{2} (\sin (t)+\cos (t))$

Including transients, i.e. repeating above with Laplace transforms we get $y(t) = \dfrac{1}{2} (\sin (t)+\cos (t))-\frac{e^{-t}}{2}$

This second answer, including transients, is what is found by using the differential equation approach done by zzephod
 
Apr 2012
263
107
Erewhon
The first solution is what you get from my method, as it assumes the output is sinusoidal which forces out the transient response.

We can do this because the input is \(\displaystyle x(t)=\cos(t)\) not \(\displaystyle x(t)=u(t)\cos(t)\). That is we have an infinite expanse of time before \(\displaystyle t=0\) with a sinusoidal input, which is why the output is going to be sinusoidal. As I said it all depends on how much and what you know.
 
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Mar 2013
96
1
Texas
We have not learned Fourier transforms yet and have not applied Laplace to anything. Is there perhaps a different approach. What is the guy doing here?
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