# Second Order LTI Low Pass Filter solving for Voltage Out

I'm having a problem with understanding my textbook for solving the Voltage Out for the second order LTI Low pass RC Filter.

Why do you need to find the square root in the given equation with the red circle on it? Why are there also the $$\| |\$$ symbols in $$\V_o/V_i\$$?

My transfer function I have acquired is $$\frac{V_o(s)}{V_i(s)} = \frac{1}{s^2RR_1CC_1+s(RC+RC_1+R_1C_1)+1}$$

The circuit schematic looks like this;

Here is the clearer version where there is no red circle

Why do you need to find the square root in the given equation with the red circle on it?

If you have a simple transfer function like this: $$\\dfrac{1}{1 + j\omega RC}\$$

To find the magnitude of the output voltage divided by the input voltage....

You have to do some math manipulations. But firstly: -

Why are there also the || symbols in $$\Vo/Vi\$$?

$$\|V_O|\$$ is the magnitude of $$\V_O\$$.

• $$\|V_O|\$$ disregards phase angle information to make life simpler in amplitude bode plots.

So, going back to solving the TF to find the magnitude, we have to get rid of the complex number in the denominator and, the way is to square each term, add them together and, take the square root of the sum. Hence: -

$$|\dfrac{1}{1 + j\omega RC}| = \dfrac{1}{\sqrt{1^2 + \omega^2 R^2 C^2}}$$

It's basically a conversion from rectangular co-ordinates to polar co-ordinates: -

Image from here. It's Pythagoras and basic geometry.

$$\dfrac{1}{s^2RR_1CC_1+s(RC+RC_1+R_1C_1)+1}$$

Then for the bode plot information you replace s with $$\j\omega\$$: -

$$\dfrac{1}{j^2\omega^2RR_1CC_1+j\omega (RC+RC_1+R_1C_1)+1}$$

And, because $$\j^2 = 1\$$ we get this: -

$$\dfrac{1}{-\omega^2RR_1CC_1+j\omega (RC+RC_1+R_1C_1)+1}$$

$$= \dfrac{1}{1-\omega^2RR_1CC_1+j\omega (RC+RC_1+R_1C_1)}$$

Then you square the real terms and you square the imaginary terms and take the square root: -

$$= \dfrac{1}{\sqrt{(1-\omega^2RR_1CC_1)^2+(\omega (RC+RC_1+R_1C_1))^2}}$$

You square the individual terms, add the squares then square root the sum. It's the Pythagoras method for finding the hypotenuse of a right angled triangle.

• Could you elaborate it more with the equation above? I'm still getting quite confused. Thank you Commented Dec 10, 2021 at 14:57
• I can't properly see the equation above because it's obscured by your red circle. I used a simple formula because the method is easier to see. If you re-read my answer to your previous question I mentioned the same technique. Commented Dec 10, 2021 at 15:13
• Hello sir, I have reuploaded a clearer version where there is no red circle. I'm starting to understand it slowly but your response will be much appreciated. Commented Dec 10, 2021 at 16:01

A transfer function is really a phasor- it has a magnitude and phase component. To find the magnitude of a complex number, say, z = a + bi, where i is the imaginary unit, you need to find

$$|z| = \sqrt{a^2 + b^2}$$

Now your transfer function is expressed as a fraction, something like,

$$H = \dfrac{a + bi}{c + di}$$

Then the magnitude is

$$|H| = \bigg|\dfrac{a + bi}{c + di}\bigg|$$

Which is also equivalent to,

$$|H| = \dfrac{|a + bi|}{|c + di|} = \dfrac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}$$

In your problem, there's a 1 in the numerator so it obviously makes sense why you still see the 1. The denominator is where you have real and imaginary components and you need to group them accordingly. So 'c' represents all the real terms in the denominator and 'd' all the imaginary terms (those with a multiplicative 'j').

An intuitive way to think of this is, as a signal propagates through your systems, from the input to the output, what happens to the magnitude at a particular frequency? The same question could be posed for the phase, but this problem is only asking for the magnitude.