Calculate the amplitude of the fifth harmonic (square wave + HP filter)

[Twinflower]

Homework Statement

A high pass filter tuned for a corner freq. of 40 000 rad/sek and with a damping coeffisient of 1/sqrt(2) is being subjected to a square wave voltage with amplitide of 1 volt and 8000 rad/sek.
What is the amplitude of the fifth harmonic on the output side of the filter?

Homework Equations

This is a part of a large case regarding filters of different kinds, so I have alreade defined the filter components and dimensions. I have also found the equation needed to find the output amplitude factor when a specific frequency is being subjected to the filter:

$$|H| = \frac{\omega^2}{\sqrt((\omega_0^2 - \omega^2)^2 + (2 * \zeta * \omega_0 * \omega)^2)}$$

Where $\omega$ is the frequency subjected to the filter and $\omega_0$ is the corner frequency of the filter

So with values it goes like this:
$$|H| = \frac{8000^2}{\sqrt((40000^2 - 8000^2)^2 + (2 * \frac{1}{\sqrt(2)} * 40000 * 8000)^2)} = 0,0399690$$


Next I have found that the amplitude of the fifth harmonic when subjecting the filter to a square wave is:
$$\frac{4A}{5\pi}$$
Where A is the applied voltage.

The Attempt at a Solution

As the applied voltage is 1 volt, and the output amplitude factor is 0.0399, I jumped right on it like this:
$$\frac{4*0,0399}{5\pi} = 0,01018$$


But that's wrong.
The correct answer is $\frac{2\sqrt(2)}{5\pi}=0,180$

Where did I go wrong?

 

[Twinflower]

No one.. ?

 

[rude man]

Homework Statement

A high pass filter tuned for a corner freq. of 40 000 rad/sek and with a damping coeffisient of 1/sqrt(2) is being subjected to a square wave voltage with amplitide of 1 volt and 8000 rad/sek.
What is the amplitude of the fifth harmonic on the output side of the filter?

Homework Equations

This is a part of a large case regarding filters of different kinds, so I have alreade defined the filter components and dimensions. I have also found the equation needed to find the output amplitude factor when a specific frequency is being subjected to the filter:

$$|H| = \frac{\omega^2}{\sqrt((\omega_0^2 - \omega^2)^2 + (2 * \zeta * \omega_0 * \omega)^2)}$$

Where $\omega$ is the frequency subjected to the filter and $\omega_0$ is the corner frequency of the filter

So with values it goes like this:
$$|H| = \frac{8000^2}{\sqrt((40000^2 - 8000^2)^2 + (2 * \frac{1}{\sqrt(2)} * 40000 * 8000)^2)} = 0,0399690$$


Next I have found that the amplitude of the fifth harmonic when subjecting the filter to a square wave is:
$$\frac{4A}{5\pi}$$
Where A is the applied voltage.

The Attempt at a Solution

As the applied voltage is 1 volt, and the output amplitude factor is 0.0399, I jumped right on it like this:
$$\frac{4*0,0399}{5\pi} = 0,01018$$


But that's wrong.
The correct answer is $\frac{2\sqrt(2)}{5\pi}=0,180$

Where did I go wrong?

Looks like you used the wrong value for A. It's 1V, not 0.0399V.
BTW I don't know how you defined A. So when you recalculate don't be surprised if you're off by 2:1 either way.

The number 0.0399 is totally irrelevant.

 

[Twinflower]

The value A is 1 V on the input side of the filter, but according to |H|, it is 0.0399 V on the output side.

The correct answer indicates that the output voltage should be 0.707 V, which is miles away from my calculated one. (By curiosity, the only way you can achieve 0.707 V on the output of the HP filter, is when you are applying the same freq. as the border frequency and get -3dB damping).

The output voltage is anyway multiplied by four and divided by 5pi. That is the correct procedure to find the fifth harmonic, right?

 

[rude man]

What's the Fourier series for a square wave with amplitude A ?

 

[Twinflower]

4A/5pi

 

[I like Serena]

Hi again Twinflower!

The fifth harmonic of the square wave is indeed at ω=5·8000 rad/s and has amplitude A=1 V.

To find the response at this frequency, you need to calculate:
|V_out(ω=5·8000)| = |H(ω=5·8000)| |V_in(ω=5·8000)|

I believe you calculated H at the wrong frequency... ;)

 

[Twinflower]

OH!

THATS why I got that the "real answer" coicided with the border frequency! 40000 = 5*8000

thank you plenty, Serena! I have 10 boys to tell some news now :)

 

[I like Serena]

Yep. H is much easier to calculate with ω=40000.
I think they chose the 5th harmonic on purpose!

I have 10 boys to tell some news now :)

Uhh... which boys?
Don't tell me you have 10 dwarves...

 

[Twinflower]

hehe, we where like 10 guys a the study a few days back, and *noone* thought about the sin(5*omega*t)-part, only of the amplitude part.

So, i got some good news for them now, this has puzzled us for days. And boy do I feel stupid :)

edit: And btw, I only have one dwarf, which is plenty for now :) He can count to 16 or something, but doesn't understand magnituds of more than 3 (one candy, two candies, three candies --> mandy candies)

 

[I like Serena]

Yeah, it's like finding x.
Avoid difficult calculations - just point out where it is! :D

 

[Twinflower]

Like your avatar :)

 

[rude man]

4A/5pi

Right! So propagate that voltage thru your transfer function:
(4A/5π)*|H(jω)| where ω = 40,000.

And remember, 4A/5π is amplitude, not rms.

 

[Twinflower]

I actually forgot to multiply the input freq. with 5
Thank you by the way :)