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Low Pass Filter 
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#1
Jun212, 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 


#2
Jun212, 10:36 AM

P: 202

Do you observe an output voltage higher or lower than what you predicted?



#3
Jun212, 10:56 AM

P: 15

The output is lower than the input as expected.
Input = 010V square wave @ 1KHz Output = Hi9.5V; Lo0.4 @ 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, 


#4
Jun212, 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. 


#5
Jun212, 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. 


#6
Jun212, 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. 


#7
Jun212, 12:36 PM

P: 15

Thanks for the help!



#8
Jun212, 02:57 PM

PF Gold
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........... 


#9
Jun212, 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.



#10
Jun212, 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 


#11
Jun212, 07:29 PM

P: 330




#12
Jun212, 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. 


#13
Jun312, 09:01 AM

PF Gold
P: 766

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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