# Complex resonant frequency not resonant without imaginary part. So can I still just take real part as solution?

I am working with a matrix on a harmonic oscillator problem and the lowest (absolute) frequency $\omega_0$ where the matrix becomes singular is the resonant frequency.

Now I obtained this frequency by calculating it numerically at let's say it is $\omega_0 = 0.3302 - 0.0121i$, ie. a complex frequency.

My question is what is the actual resonant frequency considering there is no such thing as a complex resonant frequency in real life? I know that in various situations you work with complex numbers for convenience and then take the real part at the end to obtain a solution.

However the imaginary part of $\omega_0$ here is necessary to make the matrix singular, if I just take $\omega_0 = 0.3302$ the matrix is not singular and $w_0$ will not be a resonant frequency.

Or have I got it wrong and is it actually fine to just take the real part (and the imaginary part can be sort of considered to be implied)?

• @Asked: without more context it's hard to say. If the complex numbers were introduced in order to simplify calculations -- exponentials in place of trigonometric functions -- and if the equations are linear, then the real and the imaginary parts are both solutions. But if the imaginary represents losses, the – Peter Diehr Mar 5 '16 at 12:18
• ... then it is part of the answer; this happens when the imaginary part represents the absorption in optics. One cannot tell by simply looking at an answer; the problem setup is required. – Peter Diehr Mar 5 '16 at 12:20
• The complex numbers were not introduced to simplify calculations. The problem is the resonant frequency of bubbles in water. – sonicboom Mar 5 '16 at 17:04
• then you already have your answer, and it is complex. Here is a paper that uses the complex angular resonant frequency - I'm more familiar with optical applications than with bubbles, but the complex value simply needs to be decoded into two independent parameters; the imaginary part is usually related to damping, as in the reference paper. – Peter Diehr Mar 5 '16 at 20:13
• ncbi.nlm.nih.gov/pmc/articles/PMC3855064; I need new fingertips! – Peter Diehr Mar 5 '16 at 21:07

## 1 Answer

If you were to plot the eigenvalues of your system matrix in the complex plane (real x-axis, y imaginary axis), your harmonics would appear as complex pairs in the left half plane, reflected from one another by the real axis. These are called poles of the system.

Draw a vector to either of these poles originating at the origin. The magnitude of this vector is the natural frequency of the harmonic. The angle the vector makes with the imaginary axis is a measure of loss due to damping. Poles with zero damping exist at an angle of ninety degrees - on the imaginary axis.

For complex poles with any degree of damping, the projection of the vector onto the imaginary axis provides the damped natural frequency. This is the frequency you would measure in the real world with whatever damping exists. With damping of zero the projection is equal to the magnitude of the vector. With increasing damping the projection gets smaller and smaller so the period between the settling oscillations gets further apart.

• If you had a pole on the imaginary axis which implies no damping then when you project it onto the real axis you get $0$ which seems incorrect. Did you mean to say poles with zero damping are those on the real axis? Because those ones will not change under projection onto the real axis? – sonicboom Mar 15 '16 at 10:55
• @sonicboom First of all complex poles for any physical system appear as pairs. Secondly, what you say is correct - the projection of complex poles on the imaginary axis onto the real axis is zero. That means that the damping is indeed zero. When I said "on the imaginary axis" I do mean the poles are on the imaginary axis and their projection onto the real axis is zero. – docscience Mar 15 '16 at 11:03
• Yes but then for undamped systems we can only have a damped natural frequency of zero? I would have thought that we could have a frequency at any $\omega \in \mathbb{R}$, the natural resonant frequency with no decrease in oscillations, if there was no damping but you seem to be saying $\omega$ has to be zero in this case? – sonicboom Mar 15 '16 at 11:21
• @sonicboom The (real) frequency is the magnitude of the vector. When you take the magnitude of the vector pointing to the poles on the imaginary axis you do get a real number. And damping is zero. In this case the natural frequency = damped natural frequency. – docscience Mar 15 '16 at 11:26
• The projection should be onto the imaginary axis in the case of damped natural frequency. See System Dynamics and Control by Kypuros, page 257. – sonicboom Mar 16 '16 at 9:23