Re: linear/non-linear system
- From: Tim Wescott <tim@xxxxxxxxxxxxxxxx>
- Date: Sat, 26 Aug 2006 16:13:09 -0700
VijaKhara wrote:
hi all,You can't make a Laplace transfer function of the system if it is non linear or time varying. You have to apply the definition of a linear system in the time domain.
1. For a system, if the input function is u1(t), the output is y1(t)
and if the input u2(t) the output y2(t). They ask if this system is
linear or not. I simply look for the transfer function:
H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is
linear, if not it nonlinear. But I am confused. What if the system is
linear but not time invariant. Does the proof above still work?
2. For my specific case:
u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1).
u2(t)=3(t-1)step(t-1), y2(t)=9step(t).
Step(t) is the step function/
As pointed out elsewhere, if you have two _definite_ signals you can't assure global linearity. You can only find out if a system is 100% linear from it's mathematical description, by manipulating symbols.
If it's a real system you can just assume that it's nonlinear, because all real systems are nonlinear when you push them hard enough. The only question you can ask is are they linear _enough_ that you can _pretend_ that they're linear for the purposes of your analysis.
No.
Is there any another simpler way to verify the linearity?
Because the Laplace transform is more general, and more useful when you're interested in the time-domain results. The Fourier transform is better for analyzing systems when you just care about their frequency response, and has some handy properties to analyze systems that are periodically time varying, so it's good for communications systems.
3. In control books, they often use Laplace transform to find the
transfer function. Why don't they use Fourier Transform? For some
problems, I feel FOurier Transform is more convenient because we can
compute the transform in both sides (negative and positive).
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
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