# Chapter 5 - Equation 5.17, p 108 - Evaluating the Transfer Function | FX Book Questions | Forum Register | Lost password?

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Minimum search word length is 3 characters - maximum search word length is 84 characters Home Forum FX Book Questions Chapter 5 - Equation 5.17, p 108 - Evaluating the Transfer Function Topic RSS Chapter 5 - Equation 5.17, p 108 - Evaluating the Transfer Function Member
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July 21, 2017 - 9:34 am
Member Since: July 21, 2017
Forum Posts: 3 Offline

Hi Will,

I'm struggling to understand why H(w) = sqrt (a + jb)(a - jb) on page 108.

From what I understand you can work out the magnitude of a complex number, R, from sqrt of A squared + B squared. The complex number that we have is 1.0 - j0.

Why in this instance does H(w) != sqrt (1 squared + 0 squared)?

Could you please explain how you reach H(w) = sqrt (a + jb)(a - jb)?

Many ThanksÂ

Jake July 21, 2017 - 2:15 pm
Member Since: January 29, 2017
Forum Posts: 698 Offline

I'm struggling to understand why H(w) = sqrt (a + jb)(a - jb) on page 108.Â Could you please explain how you reach H(w) = sqrt (a + jb)(a - jb)?

It's actually the magnitude of H(w) which is written |H(w)| and it is defined as the square root of the product of the complex number and its complex conjugate - there is nothing to "reach" here - it's the definition of the magnitude of the complex number. By multiplying the polynomial out, and knowing that j^2 = -1, you will arrive at the shortcut version, |H| = sqrt(a^2 + b^2).Â

For the number 1.0 - j0.0, the complex conjugate is 1.0 + j0.0

|H| = sqrt((1.0*1.0) - j0.0*1.0 +j0.0*1.0 -j^2(0.0) = sqrt(1.0 + 0.0) = sqrt(a^2 + b^2)

- Will Member
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July 25, 2017 - 6:25 am
Member Since: July 21, 2017
Forum Posts: 3 Offline

Ah okay it's the same. Thanks, I couldn't bring those two together. I can move on now!

Cheers

Jake

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