# Complex Frequency

1. Aug 3, 2004

### techvineet

Complex.... Frequency

Hi,
Can anybody explain me what is complex frequency and what is real frequency. In control system my teacher taught me about frequency-domain. This is exactly where he used these words. :grumpy:

2. Aug 3, 2004

### Goalie_Ca

s=jw

j = sqrt(-1)

I assume you're talking about electric circuits here so here's how i'll explain it.
All resistors, caps, and inductors are dealt with as impedances. That is, once you transform the circuit into this s-domain it's all the impedances are treated like resistors. When you start out you will write out the differential equation of the regular node and then you will laplace transform the equation.

3. Aug 4, 2004

### turin

I will elaborate Goalie's post:

the general complex frequency, s, can be thought of in terms of its real and imaginary parts:

s = σ + jω

or in terms of its period and periodic decay rate:

s = T-1e.

The former is always the most appropriate way to use the complex frequency in a Laplace or Fourier transform, however the latter offers a bit more insight.

The imaginary part of s:

Im{s} = ω

is what you would normally think of as "frequency." It tells you how often the signal goes from positive to negative. The real part of s:

Re{s} = σ

is what you would normally think of as "decay rate." It tells you how quickly the amplitude of the oscillations change.

What does this have to do with control systems? In control systems, one of the key concerns is stability. The &sigma; (of the poles of the system response) is usually the most important consideration in that respect.

4. Aug 18, 2004

### flexten

If you have a system that can be calculated, the poles and zeros may be found in the complex frequency plane. However, as pointed out by O. Heavyside, this is largely a useless exersize, the spectral responce is of much more use even though the poles and zeros do completely describe the system.

Best

5. Aug 18, 2004

### evert

To analyse an analog filter circuit for intance you would replace C by 1/sC and and L by sL to construct the Laplace transform of the circuit. Then you can use normal algebra to simplify the equations.

For example a lowpass filter with only a resistor and a capacitor:
$$H(s) = \frac{ \frac{1}{sC} }{\frac{1}{sC}+R} = \frac{1}{1+sRC}$$

Next you replace s by $$jw = j 2 \pi f$$ in which f is the frequency you are interested in. The result is a complex frequency in which the relation between the input amplitude and the output magnitude is the gain for that frequency.

Below a screenshot and script that calculates the frequency response and runs on http://www.adacs.com/menu/PDAcalc_matrix.php [Broken] for the palm, PocketPC and windows.

001 R=1000;
002 C=10E-07;
003 Fc=1/(2*pi*R*C)
004
005 f=logspace(0.1,3);
006 w=1j*2*pi*f;
007 H=1./(1+w.*R*C);
008 semilogx(f,abs(H))
009 title('Frequency response')
010 xlabel('Frequency')
011 ylabel('Gain')
012 pause
013 clf
014 semilogx(f,angle(H))
015 title('Phase response')
016 xlabel('Frequency')
017 ylabel('Angle')