Equiv. DC value of Square wave after Highpass filter

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jonnybgood

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If I had to consider a positive square wave of a certain amplitude it's DC equivalent would be the amplitude divided by 2.

Now if I were to pass it through a 2nd order high pass filter where I know the two poles (ex 50Hz, 200Hz), what is the best way to calculate the DC equivalent of the filtered output.

my try;
-The square wave is made up of odd harmonics;
- Can I take harmonics from a pole onwards and attenuate them accordingly by 20/40dB slope ?

What is the best way to test my answers in simulation?
 

What do you mean by "DC equivalent"? RMS voltage?
 

FvM said:
What do you mean by "DC equivalent"? RMS voltage?
Click to expand...

I want the value of the high pass output after I filter it with a lowpass filter with a very low cut off. I think it average or RMS value. So if I have a 10kHz square wave and I pass it through a high pass filter with (50Hz, 200Hz poles). I will get something also close to a squarewave at the output of the filter but with sloped crests and throughs. Now i decide to get the Avg or RMS value of that output by using a low pass filter of 0.1Hz cutoff for example. How can I derive the value analytically? Then i am going to confirm it with simulation using ideal components.
 

I think it average or RMS value.
Click to expand...
A clear problem specification is the prerequisite for an exact answer. The average of an AC voltage is zero. You can simply refer to the definition of RMS and calculate the square sum of fourier series after multiplying each component with the filter factor.
 

jonnybgood said:
What is the best way to test my answers in simulation?
Click to expand...

In case this is what your question is about...

Second-order butterworth filters, filtering a 200V AC square wave.

In the left-hand circuit, L and C values are correctly matched to the load resistance. It yields a 240 VAC sine (or close to it). The coil must be the proper value to admit sufficient current to the load. The capacitor must be the proper value to divert high frequencies.

The other circuits show what waveforms result from using 'off' values for L and C. (However these could conceivably yield sine waves, if the load and frequency were changed.)

 

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