Using Complex Conjugate NumbersDate: 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 there. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/