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Low Pass Filter

by hemant03
Tags: filter, pass
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hemant03
#1
Jun2-12, 09:13 AM
P: 15
Hi,
I am trying to understand the impact of cut off frequency on the gain and trying to relate it in time domain but unable to do so.
For e.g., I have built a single pole RC filter wherein R and C values are chosen to match the cut off frequency of 1K (R = 10KΩ/C = 16nF). The input is a square wave 10V @ 1Khz. There should be loss of 3db at the output but I am unable to validate it.
Please help.

Thanks,
- Hemant
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tamtam402
#2
Jun2-12, 10:36 AM
P: 202
Do you observe an output voltage higher or lower than what you predicted?
hemant03
#3
Jun2-12, 10:56 AM
P: 15
The output is lower than the input as expected.
Input = 0-10V square wave @ 1KHz
Output = Hi-9.5V; Lo-0.4 @ 1Khz not a square wave though.

My expectation was to see the output amplitude being around 7 V but that's not the case.
Am I looking at it in the right way?

Thanks,

the_emi_guy
#4
Jun2-12, 11:04 AM
P: 589
Low Pass Filter

The harmonics of your 10V square wave have amplitudes of 10*4/(nπ).
Your 1KHz fundamental (n=1) has an amplitude of 12.7V. The filter will attenuate it to 9V.
tamtam402
#5
Jun2-12, 11:36 AM
P: 202
Are you familiar with Fourier series?

http://en.wikipedia.org/wiki/Fourier_series

Basically, your 1kHz square wave is formed by a combination of sine waves of frequency n*1000. 1kHz is your fundamental frequency, and the other wine waves that make the square wave are integer multiples of the fundamental frequency.

What does that mean? In a square wave, the fundamental frequency has a higher amplitude than the resulting square wave (see the first picture in the wikipedia link I gave you). Since you have a lowpass filter with a 1k cutoff frequency, only the fundamental sine wave can pass through your filter (with a -3dB gain).

Your output signal should look like a sine wave, since the next sine wave that forms your square wave has a frequency of 2kHz.

Sorry for the crappy expanation, english isn't my primary language.
tamtam402
#6
Jun2-12, 11:42 AM
P: 202
I'd like to add that the only signals that can pass through a hardware* filter without being deformed are sine waves. Why? Every other signal is actually formed by a combination (sum) of sine waves of different frequencies. Since these individual sine waves all have a different gain when going through a filter, the output wave (given by the sum of the individual sine waves) will have a different enveloppe.

*Obviously if you use a software to filter signals, it can be programmed to "fix" this issue.
hemant03
#7
Jun2-12, 12:36 PM
P: 15
Thanks for the help!
psparky
#8
Jun2-12, 02:57 PM
PF Gold
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P: 766
I can answer your question mathematically......

What's your transfer function?

Let me guess.......1/(JωRC+1)

That is found by taking your voltage divider across your cap

1/jwc/(1/jwc + R)

Multiply top and bottom by jwc...and you get...

1/(JωRC+1)

Your break frequency happens when ω= 1/RC.

So when ω= 1/RC.....you are left with

1/(j+1)

What is the magnitude and value of this lovely function?

Well....I'll simplify so you can see. 1/(1.41 at -45 degrees)

Simply further and you get .7 at -45 degrees.

Ok....now to your actual question.....where does the -3 dB come from?

You know that you find dB from this equation ~ 20 log |gain|

20 log .7 = ~ -3dB.
I also just proved that your phase angle will be -45 degrees at the break as well.

However....all that may go away if you are talking a non ideal situation...........
tamtam402
#9
Jun2-12, 05:27 PM
P: 202
No offense but have you read his question? He said he chose his R and C to get -3dB at 1kHz. Obviously he was able to find and solve the transfer function.



Quote Quote by psparky View Post
I can answer your question mathematically......

What's your transfer function?

Let me guess.......1/(JωRC+1)

That is found by taking your voltage divider across your cap

1/jwc/(1/jwc + R)

Multiply top and bottom by jwc...and you get...

1/(JωRC+1)

Your break frequency happens when ω= 1/RC.

So when ω= 1/RC.....you are left with

1/(j+1)

What is the magnitude and value of this lovely function?

Well....I'll simplify so you can see. 1/(1.41 at -45 degrees)

Simply further and you get .7 at -45 degrees.

Ok....now to your actual question.....where does the -3 dB come from?

You know that you find dB from this equation ~ 20 log |gain|

20 log .7 = ~ -3dB.
I also just proved that your phase angle will be -45 degrees at the break as well.

However....all that may go away if you are talking a non ideal situation...........
vk6kro
#10
Jun2-12, 06:40 PM
Sci Advisor
P: 4,044
That filter seems to be for 10 KHz.

1/ F = 2 * π * R * C = 2 * π * 1000 * 16 * 10^-9

F = 9947 Hz
gnurf
#11
Jun2-12, 07:29 PM
P: 330
Quote Quote by vk6kro View Post
That filter seems to be for 10 KHz.

1/ F = 2 * π * R * C = 2 * π * 1000 * 16 * 10^-9

F = 9947 Hz
Did you drop a zero somewhere, good sir?
vk6kro
#12
Jun2-12, 08:03 PM
Sci Advisor
P: 4,044
Yes, I see the 10 K now. Thanks.

I read the 1 K frequency as the resistance.
psparky
#13
Jun3-12, 09:01 AM
PF Gold
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P: 766
No offense but have you read his question? He said he chose his R and C to get -3dB at 1kHz. Obviously he was able to find and solve the transfer function.
I write that out mathematically because most students are told it drops 3 dB at the break without knowing WHY. They see all the graphs of filters with the 3 db drop and just take it as that. For all of you who already know this, I apologize for insulting your intelligence with such a trivial post. However, if just one person learned something....then my post was a success.

To this day I have not met one student or EE in person that knows why it drops 3 dB mathematically. Therefore, I assume it to be a weak point for most.


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