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# Finding the magnitude of complex fractions

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# Finding the magnitude of complex fractions

November 4, 2007 11:18 AM Subscribe

Finding the magnitude of a complex fraction. Is it simply |num|/|denom|? Or is there more to it?

Say I have a complex fraction (a+bi)/(c+di). If I need to find the magnitude of the entire fraction, can I simply take the magnitude of the numerator over the magnitude of the denominator? ie,

sqrt(a^2 + b^2) / sqrt(c^2 + d^2)

I've tried solving this in MathCAD, but the answer it gives is much more complicated than this, and I can't figure out how to work its logic through. The book I'm using doesn't go into how to solve this type of problem with the numerator != 1.

If it matters, the actual problem I'm solving (for a control systems class) is

|(iw+5)/(iw+2)(iw+4)|

but I'd prefer a method to just the solution.

I have access to MATLAB 2007a and MathCAD 13

Yes, this is homework help, but it's not homework answers. I need help in finding the analytical method

Say I have a complex fraction (a+bi)/(c+di). If I need to find the magnitude of the entire fraction, can I simply take the magnitude of the numerator over the magnitude of the denominator? ie,

sqrt(a^2 + b^2) / sqrt(c^2 + d^2)

I've tried solving this in MathCAD, but the answer it gives is much more complicated than this, and I can't figure out how to work its logic through. The book I'm using doesn't go into how to solve this type of problem with the numerator != 1.

If it matters, the actual problem I'm solving (for a control systems class) is

|(iw+5)/(iw+2)(iw+4)|

but I'd prefer a method to just the solution.

I have access to MATLAB 2007a and MathCAD 13

Yes, this is homework help, but it's not homework answers. I need help in finding the analytical method

ah-hah. ok, thanks.

MathCAD gave me an unusually long polynomial, that although it had recognizable parts, didn't look simple enough for the answer.

posted by toomanyplugs at 11:37 AM on November 4, 2007

MathCAD gave me an unusually long polynomial, that although it had recognizable parts, didn't look simple enough for the answer.

posted by toomanyplugs at 11:37 AM on November 4, 2007

Another way of thinking about it is to make the denominator real. You can do this by multiplying the top and bottom of the fraction by the complex conjugate of the denominator.

i.e.

(a + bi)/(c+di) = (a+bi)(c-di) / (c+di)(c-di)

= (a+bi)(c-di)/(c^2 + d^2)

Then find the magnitude in the usual way, now that the denominator is a real number.

posted by number9dream at 11:46 AM on November 4, 2007

i.e.

(a + bi)/(c+di) = (a+bi)(c-di) / (c+di)(c-di)

= (a+bi)(c-di)/(c^2 + d^2)

Then find the magnitude in the usual way, now that the denominator is a real number.

posted by number9dream at 11:46 AM on November 4, 2007

number9dream's method is probably better, given your stated problem. From your description, you appear to be doing the Bode plot of (s+5)/(s+2)(s+4); multiplying by the conjugate gives you the real and imaginary parts of the transfer function. You'll need them both for the second part of the Bode plot, the phase plot.

If G(s) = G(iω) = R(ω)e

Re[G] = R(ω) cos θ(ω),

Im[G] = R(ω) sin θ(ω),

R(ω) = sqrt(Re[G]

and θ(ω) = tan

Extending number9dream's example:

(a +bi)(c - di)/(c

real part = (ac + bd)/(c

imaginary part = (bc - ad)/(c

posted by Upton O'Good at 8:11 PM on November 5, 2007

If G(s) = G(iω) = R(ω)e

^{i θ(ω)},Re[G] = R(ω) cos θ(ω),

Im[G] = R(ω) sin θ(ω),

R(ω) = sqrt(Re[G]

^{2}+ Im[G]^{2}),and θ(ω) = tan

^{-1}(Im[G]/Re[G]).Extending number9dream's example:

(a +bi)(c - di)/(c

^{2}+ d^{2}) = ([ac + bd] +i[bc - ad])/(c^{2}+ d^{2})real part = (ac + bd)/(c

^{2}+ d^{2})imaginary part = (bc - ad)/(c

^{2}+ d^{2}).posted by Upton O'Good at 8:11 PM on November 5, 2007

This thread is closed to new comments.

1. |wz| = |w||z|, and

2. |1/z| = 1/|z|

Either one of those by itself is easy to verify, and together they justify the "norm of a quotient is the quotient of the norms" rule that you want.

See also: MathWorld (in particular equation 5). It's easier to think about these rules in polar form (re

^{iθ}) than in x+yI form.posted by Wolfdog at 11:35 AM on November 4, 2007