# Transfer Function Magnitude, how to deal with 'j'?

I have the simple transfer function of an RC filter:

$H(s) = \frac{sRC}{1 + sRC}$

In order to find the magnitude, square the previous equation and take the square root:

$(H(s))^2 = \frac{(sRC)^2}{(sRC + 1)^2} ; s = j\omega$

$(H(s))^2 = \frac{(j\omega R C)^2}{(j\omega R C + 1)^2} \qquad(1)$

This is where I get stuck. I know that the correct magnitude should be of the form:

$|H(s)| = \sqrt{\frac{(\omega R C)^2}{(\omega R C)^2 + 1}} \qquad (2)$

However, when I actually try to calculate this out on paper I don't know how to deal with the $j$. I don't understand how to get from equation I get from equation (1) to equation (2). The $-1$ from $j^2$ causes the form to look like the following:

$(H(s))^2 = \frac{-top}{1 - bottom}$

If I were to take the square root of this I do not get equation (2).

What am I doing wrong?

• You didn't take the square of the denominator properly: it should be (1+sRC)^2. – Barry Nov 19 '17 at 23:30
• You're correct, I did not do that correctly. However, doing it again I get that: H(s)^2 = [(j^2)(w^2)(R^2)(C^2)] / [1 + 2jwRC + (j^2)(w^2)(R^2)(C^2)] ; This leaves me.. more confused than before? – Zearia Nov 19 '17 at 23:43

\begin{align*} H\left(s\right)\quad &=\quad \frac{s R C}{1 + s R C}\\\\ \mid\: H\left(s\right)\: \mid^{\:2}\quad &=\quad H\left(s\right)\cdot H\left(s^*\right)\\\\ &=\quad\frac{s R C}{1 + s R C}\cdot\frac{s^* R C}{1 + s^* R C}\\\\ &=\quad\frac{\left(\sigma^2+\omega^2\right) R^2 C^2}{1 + 2\sigma R C+\sigma^2 R^2 C^2+\omega^2 R^2 C^2} \end{align*}

Laplace uses $s=\sigma + j \omega$. But this collapses to Fourier along the imaginary axis portion, or the $j \omega$ part of that. Since you want the periodic and non-decaying part of the system response -- the frequency response -- you set $\sigma=0$ in $s$.

So the above devolves to:

\begin{align*} \mid\:H\left(j\omega\right)\:\mid\quad &=\sqrt{ \frac{\omega^2 R^2 C^2}{1 + \omega^2 R^2 C^2}}\\\\ &=\quad\frac{\omega R C}{\sqrt{1+\omega^2 R^2 C^2}} \end{align*}

I hope that helps.

• What I seem to be confusing about this whole process is that when you square the s, would you not get s^2 = (sigma^2) + [(j^2)(w^2)]? If sigma was set to 0 would that not still yield that s^2 = -(w^2)? – Zearia Nov 20 '17 at 1:27
• @Zearia By definition, the magnitude squared of a complex number is $\mathscr{C}\cdot\mathscr{C^*}$. See: Absolute value and argument. And no, we are talking about $s=\sigma+j \omega$. The magnitude squared is $\mid s\mid ^2=\left( \sigma+j \omega\right )\cdot \left( \sigma-j \omega\right)$. – jonk Nov 20 '17 at 2:03
• @Zearia If you don't see why, just plot a point on the complex plane and work out how you'd compute the hypotenuse. It should be very clear, then, I think. – jonk Nov 20 '17 at 2:16