# Complex Eigen values

1. Jul 18, 2012

### svishal03

Please can anyone provide an insight into what do complex eigen values physically indicate?

Vishal

2. Jul 18, 2012

### ehild

Hi svishal,

What part of Physics do you mean?

ehild

3. Jul 18, 2012

### ozone

complex eigenvalues usually represent some linear combination of sin and cos

4. Jul 18, 2012

Typically, real eigenvalues signify exponential solutions, imaginary eigenvalues signify oscillatory solutions and complex eigenvalues signify a combination of the two.

5. Jul 18, 2012

### svishal03

@ehild:I meant the part for finding the solution of SDOF (or a modal analysis of a multi degree of freedom) vibration problem.

@boneh3ad: if $ω = √k/ m$, where ω is the natural frequency and m the mass, suppose we get a real value for ω (say 1.1 or 3.0), what you mean by exponential here in terms of vibration?

Also, if we get an imaginary value i.e $ω^2= -2$ , then, what you mena by osciallting?Anyways the body is vibratiing (i.eosciallating), right?

6. Jul 19, 2012

### ehild

In case of vibrating systems, you transform the system of differential equations into a homogeneous system of linear equations by assuming the solution in form of y=yoeλt. The eigenvalues are imaginary in case of vibration without damping, and they are complex when damping is present.

ehild

7. Jul 19, 2012

### svishal03

How can you say so?Take a simple cantilever beam example with a mass at the free end.Then

$ω=√k/m$

$k = 3EI / L^3$

That does not give an imaginary Eigen value?

8. Jul 20, 2012

Sure it does. If you have no damping, for an unforced oscillation you have the governing equation
$$m\ddot{x} + kx = 0$$
This admits solutions of the form
$$x = Ce^{\lambda t}$$
where $\lambda$ are the eigenvalues. Here, your eigenvalues are
$$\lambda^2 = -\frac{k}{m}$$
which means that $\lambda$ is always imaginary since both $k$ and $m$ are real and positive. For an undamped, free oscillation, the eigenvalues are always purely imaginary. That means without the damping force, the oscillation will continue into infinity.

Now, for damping the governing equation is
$$m\ddot{x} +c\dot{x}+ kx = 0$$
which has the same general form of the solution, only the eigenvalues are now
$$\lambda = \frac{-c \pm \sqrt{c^2 - 4km}}{2m}$$
In this case, the eigenvalues are purely imaginary if $c = 0$ (the undamped case), are purely real if $c > 2\sqrt{km}$ and are complex if $c < 2\sqrt{km}$.

A purely imaginary eigenvalue means the system oscillates for all time. A purely real eigenvalue means that the solutions are exponential and decay directly to zero (since it is impossible to have a positive eigenvalue). A complex eigenvalue means you have an oscillation of decreasing amplitude until eventually you reach zero.

9. Jul 20, 2012

### AlephZero

There are two different ways to define "the eigenvalues" here.

If you are doing vibration analysis of a system without damping, you "know" the response is going to be an oscillation, so instead of Boned3ad's $x = Ce^{\lambda t}$ you usually start from $x = Ce^{i\omega t}$.

The first one gives you an imaginary value of $\lambda$, the second gives a real value of $\omega$.

If you have a general damping matrix, you would normally folllow Boned3ad's math (which leads to a quadratic eigenproblem). On the other hand if you assume a special forms of damping matrix (e.g. Rayleigh damping) it can be shown that the damped mode shapes are the same as the undamped, so you usually formulate the math in terms of the real values of $\omega$.

10. Jul 21, 2012

### svishal03

Thank you very much boneh3ad, it was very useful for me.

Thanks to Alephzero for pointing out what I was actually trying to ask. However the texts on Structural Dynamics (Anil K Chopra, UCB) do not start the way boneh3ad explained. I fell that is how it should be explianed which brings out the physical interpretation of real,imaginary,complex Eigen values.

Thanks to all.

11. Jul 22, 2012

### svishal03

@Boneh3d:

You said:

Did you mean a pure real eigen value would indicate that the system is stationary?

12. Jul 22, 2012

No, the oscillator (be it a beam, spring, circuit or whatever else) would move exponentially to the equilibrium position. Assuming the initial conditions are $\dot{x}=0$ and $x=0$ then it would be stationary, but in that case it would be stationary regardless of the eigenvalues.

13. Jul 22, 2012

### svishal03

When you say:
do you mean (perhaps I'm looking for a physical meaning of the word 'exponential' here) that it moves very rapidly to the equilibrium position or the oscillation decays to zero (stops moving) very rapidly?

14. Jul 22, 2012

Meaning it moves to zero at an exponential rate. Remember, the solutions were exponential and with negative, real eigenvalues, the solutions therefore approach zero exponentially.

15. Jul 23, 2012

### svishal03

Can we put it like this:

1)c= 0- undamped case
2)$c>2√km$means the system does NOT oscillate thus returns to equilibrium position without oscillating
3)$c<2√km$ system oscillates with a decreasing amplitude and eventually stops oscillating.

Last edited: Jul 23, 2012