# Eigenfunctions of LTI systems

Discussion in 'Math' started by jrv9090, Nov 29, 2014.

1. ### jrv9090 Thread Starter New Member

Nov 26, 2014
16
0
Hello all,

I have been studying about LTI system and Eigen functions. Here is a doubt I like to clarify.

Say I have a LTI system defined by the system equation [y''(t) + 5y'(t) + 6y(t) = x(t) ]. It is evident that for the above system the transient portion will be of the form Aexp(-2t) + Bexp(-3t) [Same as Eigen functions of this system].

My understanding is as follows: that the transients are independent of x(t). Hence no matter what signal I send into the signal via x(t), the transient will persist. Also, by definition of Eigen functions, if I send an Eigen function into the system, I will get the same as output with different gain.

Say, I decide to send the signal exp(-2t) as input via x(t). Then the output has to be, by definition of Eigen functions, C *exp(-2t). I would like to know what has happened to the transient exp(-3t). Has it disappeared? If so how?

My guess is that it has something to do with initial conditions. If so, I don't see how. Kindly let me know.

Nov 9, 2007
5,005
515

Nov 26, 2014
16
0
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Can you show how (mathematically) the other term might have disappeared.

5. ### jrv9090 Thread Starter New Member

Nov 26, 2014
16
0
I don't know if the other term disappears or not.

1.By definition of Eigen functions, if I send exp(-2t) [which by the way is the Eigen function for the above system] into the system, the output should only contain exp(-2t) and not exp(-3t) or any other signal for that matter.
2.But since transient are independent of input signals, I believe transient response to contain both exp(-2t) and exp(-3t).

Statements 1 and 2 above contradict each other based on my understanding and not by any mathematical derivation. Hence I require clarification.

Mar 6, 2009
5,448
783

7. ### jrv9090 Thread Starter New Member

Nov 26, 2014
16
0
Can you please let me know which part is ambiguous???

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I would probably firstly answer the question as to what the solution would be if the input is of the form of one of the eigenfunctions.

I believe for

$x(t)=exp{-2t}$

The response is

$y(t)=exp{-3t}-\[1-t\] exp{-2t}$

So as you proposed in point 2 of your last post (#5) the response does indeed include both forms of the eigenvalues.

I don't know what your point 1 is based upon. Perhaps you could cite a reference.

If the system is not driven by a steady state input, one would expect there to be only the natural transient terms in the response at t=infinity. Since the natural transient terms are decaying exponentials, then at t=infinity these transients would have vanished and the system is "at rest". This would be the case if (say) there was simply some initial energy storage in the system such as a voltage on a capacitor or perhaps an impulse Dirac type input. Or indeed a decaying exponential.
One could also have a driven system where x(t) is a known function - such as a steady state sinusoid or a DC step & so on - which persists indefinitely. One would anticipate the resultant response would therefore include the superposition of terms resulting from both the steady state input and the natural [decaying transient] responses.

Last edited: Nov 29, 2014
9. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
But these are not eigenfunctions.

if y=exp(-2t) is a solution of the homogeneous equation this means that

L[y] = 0

Whereas the definition of an eigen function requires L[y] = hy where h is a number.

Another way of saying the output has to be C *exp(-2t) is to substitute this into the LTI and seeing if this is possible.

Now the A and B in your expression are purely arbitrary until determined by the boundary conditions and Aexp(-2t) + Bexp(-3t) = 0 is true for all values of A and B so taking A = B = 1 means no loss of generality

So by my reckoning substituting y=C *exp(-2t) into [y''(t) + 5y'(t) + 6y(t) = x(t) ]. yields

+4y -10y +6y = Cy

Thus there is no C for which this is possible

Note this will always happen if you try to substitute the solutions of the homogenous expression as the x(t) term in the inhomogenous equation so you cannot use these as trial functions to obtain a particular integral.

There are standard proceedures for dealing with this, tnk has showed that for real exponential solutions you use the product of t and the exponenetial.

A useful theorem connecting the solutions of the homogeneous and nonhomogenous equations is

If L[y] = 0 is the homogeneous equation with solutions y1 and y2

and L[y] = f(t) with solutions t1 and t2

Then
1)The difference of solutions of the homogeneous equation is a solution of the nonhomogeneous equation

that is y1-y2 = t1 or y1-y2 = t2

and

2)

The sum of one solution of the homogeneous equations and one solution of the non homogeneous equations is a solution of the homogeneous equation.

Last edited: Nov 30, 2014