# LTI system and sinusoids/complex exponentials

1. Mar 6, 2014

### Bipolarity

Not sure if this is the right forum. This is a mathematical question, but it is from signals.

An LTI system is a mapping from one function space to another that is both linear and time invariant.
Given this, how might I show that if the input is a sinusoid, the output, is also a sinusoid? Similarly for complex exponentials? If one can show that a spectral decomposition can be performed on an LTI system, then this is trivial, but that requires analysis of the eigenvalues and eigenfunctions of the system, which I am not very familiar with.

Furthermore, it turns out that complex exponentials are eigenfunctions of any LTI system. Are they the only eigenfunctions? If so, is the frequency response of an LTI system equivalent to its spectrum (set of eigenvalues)?

I'd appreciate it if someone could provide insight on these matters, and perhaps recommendation of a textbook that would explain these concepts very well.

BiP

2. Mar 6, 2014

### marcusl

Think about how to express the output of an LTI system in terms of an arbitrary input x(t) (hint: it involves a convolution integral).

I'm not an EE, but I have one book (out of print) that covers this: Gupta, something like State Variables and LInear Systems.

3. Mar 6, 2014

### Bipolarity

The output is the input convolved with the impulse response. In the frequency domain, this would correspond to multiplication by the frequency response. But how does this prove that the output is a sinusoid? Is the frequency response a sinusoid, so that the product of two sinusoids is also a sinusoid?

BiP

4. Mar 6, 2014

### AlephZero

A sinusoid in the time domain becomes a delta function in the frequency domain.

When you multiply a delta function by any frequency response function, you get a delta function.

5. Mar 7, 2014

### marcusl

The system frequency response H(ω) is definitely not a sinusoid, it is a complex function of frequency. Do you see how AlephZero's comment then gives you the answer?

6. Mar 7, 2014

### Bipolarity

I'm sorry I still don't see it. I would greatly appreciate if you could elaborate, preferably showing the steps.

BiP

7. Mar 7, 2014

### marcusl

The output of a linear system is expressed by a convolution integral $$y(t)=\int_{-\infty}^\infty{x(t')h(t,t')dt'}$$where x(t) is the input signal and h is a time-varying impulse response (in physics problems, it would be called a Green function and it could be spatially varying instead of or in addition to being time varying). For a time invariant system, the impulse response doesn't depend on the actual time (by definition) so has the form $$y(t)=\int_{-\infty}^\infty{x(t')h(t-t')dt'},$$or $$Y(\omega)=X(\omega)H(\omega)$$in the frequency domain.

To proceed, write down the integral form of the expression that defines the eigenfunction/eigenvalue problem. You can use a trial solution to demonstrate that it is an eigenvalue. Alternately, transform it, too, to the frequency domain whereupon AlephZero's hint gives the answer. I encourage you to try it (you'll learn it better than just reading the answer from me).

BTW, the book I had in mind is S. Gupta, Transform and State Variable Methods in Linear Systems, 1966.

Last edited: Mar 7, 2014
8. Mar 13, 2014

### Vaedoris

One way to find the solution is to express the LTI system in State Space model (SS).

Then Laplace-transform the two equations of SS, make an algebraic substitution, and take the inverse Laplace transform of the substitution. Voila...

This is a great note (one that I now treasure) on how to do it:
http://web.mit.edu/2.14/www/Handouts/StateSpace.pdf