# What is complex frequency? [closed]

I am learning EE, and about complex frequencies, but what is its physical meaning? What is it used for? Why is it? And only happen in the laplace transform?

• You might want to wait before you finish electromagnetism and then ask a question is something is still unclear. – gented Mar 18 '19 at 18:12
• @gented, Generally we don't use complex frequency in electromagnetics problems. They do appear when using the Laplace transform, which is more of a circuit theory topic, as far as EE's are concerned. – The Photon Mar 18 '19 at 18:20
• OP, the key is understanding what a particular s value tells you about the input or output signal. Once you get your head around that, you'll understand what complex frequencies represent. – The Photon Mar 18 '19 at 18:22
• Ah, physical meaning of complex numbers is a fascinating subject! I'll try to find time to write an answer today, if no one ventures to do that first. – kkm Mar 18 '19 at 20:43
• Try changing you question to "I am learning EE, and about complex frequencies, but what is its physical meaning?" (i.e. remove the other questions), and then flag it to try to get the HOLD removed. – user45664 Mar 19 '19 at 17:43

It is hard to know what you mean without any examples, but typically complex frequency means either loss or gain, depending on the sign of the imaginary part and the convention you are using.

For example, I could try to find solutions to:

$$\ddot{x}+\gamma \dot{x}+ \omega_0^2 x = 0$$, where $$\gamma,\omega_0 \in \mathbb{R}$$

by trying $$x=\exp\left(i\omega t\right)$$, I would find that such are valid provided $$-\omega^2+i\gamma\omega+\omega_0^2=0$$, which will necessitate for $$\omega$$ to have an imaginary part. So I would end up with

$$x=\exp\left(i\Re\left(\omega\right)t\right)\exp\left(-\Im\left(\omega\right)t\right)$$, so now you can see that depending on the sign of the imaginary part, $$x$$ will either grow (gain) or decay (loss).

Note that you can perfectly easily put complex frequency into $$\sin/\cos$$, e.g.

$$\sin\left(\left(\omega+i\omega'\right)t\right)=\sin\left(\omega t\right)\cosh\left(\omega ' t\right)+i\cos\left(\omega t\right)\sinh\left(\omega't\right)$$