The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Using Complex Conjugate Numbers

Date: 09/14/1999 at 12:29:29
From: Larry Carroll
Subject: Conjugate Numbers

We are using electronic hardware to measure impedance, structural 
return loss, and other properties of CATV cables that we manufacture. 
The process has been automated by using a combination of HPBasic and 
dBaseIII programs (to capture, manipulate, and perform math operations 
on the data and download it to a database file where it is further 
manipulated and operated on.

The memory variables are generally complex numbers and in the program 
code I frequently see where a value is multiplied by a quantity that 
takes this form (example):

     Impedance = 75*(1+Dat(I))/(1-Dat(I))

where I varies from 1 to x (usually 21 or 201 depending on the 
frequency band we are interested in - 21 equates to 210 MHz and 201 
equates to 2.2 GHz.

This appears to be multiplying a real number by a complex number 
divided by the complex number's conjugate.

Is there a mathematical significance to this operation? What are we 
doing when we multiply a real number by (a+bi)/(a-bi)?

Date: 09/14/1999 at 13:17:13
From: Doctor Tom
Subject: Re: Conjugate Numbers

Hi Larry,

Of course it's tough to guess exactly what's going on, but I do 
remember from my college physics that the formula for the impedance of 
a capacitance/inductance ladder is given by an expression similar to 
what you wrote. Of course it's not obvious - you have to derive it.

I also remember that you can often model cables as an infinite series 
of components - each little length has some capacitance and some 
inductance, and to model the impedance of the entire cable, you "sum 
up the contributions."

I also vaguely remember that this has to do with the calculation of 
some frequency filter calculations, but my brain is even fuzzier 

- Doctor Tom, The Math Forum   

Date: 09/16/1999 at 21:27:37
From: Doctor Douglas
Subject: Re: Conjugate Numbers

Hi Larry,

Here's one way of looking at it. Let z = a+bi = r exp(i q), where z is 
a complex number. Clearly a and b are the real and imaginary parts of 
z, and r and q are the magnitude and phase of z. The numbers a, b, r, 
and q are all real. Denote the complex conjugate of z by z*. 
Obviously, z* = a-bi = r exp(-i q).

Now, consider the number

     (z/z*) = (a+bi)/(a-bi)
            = (r exp (i q))/(r exp (-i q))
            = exp(2 i q)

Note that you could have calculated this without going to the polar 
coordinates and just worked straight from the a and b, but this 
construction makes the algebra a bit simpler, and makes the result 
somewhat easier to understand.

So what is that result? Multiplying by (z/z*) is the same as 
multiplying by exp(2 i q), which has magnitude = unity, and 
phase = +2q. So the multiplication by (z/z*) doesn't change the 
magnitude, but advances the phase of the input by +2q (where q is 
expressed in radians).

Does this make sense in the microwave context?

If you need more help please write back. Good luck!

- Doctor Douglas, The Math Forum   
Associated Topics:
College Imaginary/Complex Numbers
College Physics

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.