I am working on a problem for a given transfer function given below and am having trouble calculating the magnitude and phase of it. The problem is specifically that I do not know what method to use to seperate the imaginary value out of from the real to allow me to use their seperate values in calculating the magnitude and phase.

$$\frac{V_{out}}{V_{in}} = \frac{1}{1-(4\pi^2\cdot 0.04)+(j\cdot 2\pi\cdot 0.2)}$$

Ideally I'd like to end up with something like a+bi to allow me to do the \$\sqrt{a^2 + b^2}\$ and the tan equation for the phase but I do not know how to do this.

Could anyone suggest a method to use or some mathematical identity to use?

Implementing what was in the first comment from ocrdu simplifies to

$$\frac{V_{out}}{V_{in}} = \frac{1}{(-0.579+j\cdot 1.25)}$$

Then multiplying top and bottom by inverse of bottom gives:

$$\frac{V_{out}}{V_{in}} = \frac{(0.579-j\cdot1.25)}{(-0.579+j\cdot1.25)\cdot (0.579-j\cdot1.25)}$$

But how does this help me?

  • \$\begingroup\$ Work out the denominator to arrive at something looking like 1/1-aj, then multiply both numerator and denominator by 1+aj. \$\endgroup\$
    – ocrdu
    Sep 23, 2020 at 18:59
  • \$\begingroup\$ If you now work out the denominator, the j should magically disappear and leave you with a real number in the denominator. \$\endgroup\$
    – ocrdu
    Sep 23, 2020 at 19:11
  • \$\begingroup\$ BTW I think you should have multiplied by -0.579 - 1.25j (the complex conjugate), but I am suffering from a severe coffee underflow, so YMMV. \$\endgroup\$
    – ocrdu
    Sep 23, 2020 at 19:21
  • 2
    \$\begingroup\$ Why bothering to rearrange the transfer function? The magnitude of a quotient is the magnitude of the numerator over that of the denominator and the phase is the argument of the numerator minus that of the denominator. Nothing special to do in your case! \$\endgroup\$ Sep 23, 2020 at 20:15

3 Answers 3


\$(1 + ja)\cdot(1 - ja) = 1 + a^2\$


So the solution was simply to take the square as follows:


j squared is j * j = -1



  • \$\begingroup\$ j^2 = -1, not 1. \$\endgroup\$
    – ocrdu
    Oct 8, 2020 at 8:23
  • \$\begingroup\$ editted - thanks! \$\endgroup\$
    – smake5730
    Oct 8, 2020 at 9:03
  • \$\begingroup\$ Please note this answer is wrong. While doing a complex number modulus you need the coefficient of the imaginary part, j²=-1 must not be included \$\endgroup\$
    – carloc
    Jul 13, 2022 at 5:12

Multiplying top and bottom by the denominator's complex conjugate gives a real number in the denominator, which is what you want.

Like so:

$$\small{\frac{V_\mathrm{out}}{V_\mathrm{in}}=\frac{1}{-0.579+1.25j}=}$$ $$\small{\frac{-0.579-1.25j}{(-0.579+1.25j)\cdot(-0.579-1.25j)}=}$$ $$\small{\frac{-0.579-1.25j}{0.335+0.724j-0.724j+1.563} =}$$ $$\small{\frac{-0.579-1.25j}{1.898} =}$$ $$\small{-0.305-0.659j}$$

Please check the above for errors (yours and mine), but you get the idea.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.