Help! Power in square wave?

[franklan118]

Hi!
I got a question that confused me.



you may consider the T0 as 0.4

I have google it and some says P=amplitute^2 / 2

or P=a0^2 + sum of (an^2/2) , where n is from 1 to infinity.

Can anyone help plzzzzzzzzzzzz?


Always thanks

 

[godfreyl]

I have google it and some says P=amplitute^2 / 2
or P=a0^2 + sum of (an^2/2) , where n is from 1 to infinity.

They're both right (sort of).

Let's say the peak voltage = 1 Volt, and the load = 1 Ohm.
So peak current = 1 Amp and peak power = 1 Volt * 1 Amp = 1 Watt.
But it's only 1W for half the time, so the average power = 0.5W.
That agrees with "P=amplitude^2 / 2".

But the question is about harmonics, so we have to worry about the Fourier analysis.

Let's say T0 = 1mS, so the frequency is 1KHz.
Then that square wave is equivalent to:

• 0.5V DC
• plus 1khz sin wave with peak amplitude = .637V
• plus 3khz sin wave with peak amplitude = .637V / 3
• plus 5khz sin wave with peak amplitude = .637V / 5
• plus 7khz sin wave with peak amplitude = .637V / 7
• plus etc etc etc to infinity.

But they only want to know the percentage up to the third harmonic so you can work it out like this:

• DC: 0.5V so Power = 0.5V^2 / 1 Ohm = 0.25W = 50% of total.
• 1khz: .637V peak = .450V rms so Power = 0.45V^2 / 1 Ohm = 0.203W = 40.6% of total.
• 3khz: .637V / 3 peak = .150V rms so Power = 0.15V^2 / 1 Ohm = 0.022W = 4.4% of total.

Answer = 50% + 40.6% + 4.4% = 95%

 

[franklan118]

thanks for your excellent relay ‘godfreyl’. Now I understand how and where I should apply the formulas. Just another question, may sound silly.

Why there is only the odd numbered harmonics existing?

 

[godfreyl]

Why there is only the odd numbered harmonics existing?

Because it's a symmetrical waveform. Even harmonics make the waveform unsymmetrical.

e.g. See below. red is 1KHz + 2KHz, green is 1KHz + 3KHz.

 

[klak47]

They're both right (sort of).

Let's say the peak voltage = 1 Volt, and the load = 1 Ohm.
So peak current = 1 Amp and peak power = 1 Volt * 1 Amp = 1 Watt.
But it's only 1W for half the time, so the average power = 0.5W.
That agrees with "P=amplitude^2 / 2".

But the question is about harmonics, so we have to worry about the Fourier analysis.

Let's say T0 = 1mS, so the frequency is 1KHz.
Then that square wave is equivalent to:

• 0.5V DC
• plus 1khz sin wave with peak amplitude = .637V
• plus 3khz sin wave with peak amplitude = .637V / 3
• plus 5khz sin wave with peak amplitude = .637V / 5
• plus 7khz sin wave with peak amplitude = .637V / 7
• plus etc etc etc to infinity.

But they only want to know the percentage up to the third harmonic so you can work it out like this:

• DC: 0.5V so Power = 0.5V^2 / 1 Ohm = 0.25W = 50% of total.
• 1khz: .637V peak = .450V rms so Power = 0.45V^2 / 1 Ohm = 0.203W = 40.6% of total.
• 3khz: .637V / 3 peak = .150V rms so Power = 0.15V^2 / 1 Ohm = 0.022W = 4.4% of total.

Answer = 50% + 40.6% + 4.4% = 95%

How to get the value .637v for the peak amplitude? Does it come out from the amplitude of 1st Harmonic (which is 2/pi)?

 

[godfreyl]

Yes. 2 / pi = about 0.637.

 

[Ratch]

franklan118,

I don't think any of the answers are correct. Let me show you why. The square wave shown is a -0.5 to +0.5 p-p with a +0.5 DC offset. Power is proportional to the square of the voltage. The DC offset gives a power of 0.5² = 0.25. The RMS value of the first harmonic is 0.5/√2 = 0.353, and the square of that is 0.353² = 0.125. The RMS value of the third harmonic is (1/6)/√2 = 0.118 and the square of that is 0.118² = 0.0139. Adding the three powers together we get 0.25+0.125+0.0139 = 0.389. Now, dividing by 0.5, which is the total power of the square wave, we get 78%.

Ratch

 

[godfreyl]

The RMS value of the first harmonic is 0.5/√2 = 0.353,.....

No. You're assuming that the pk-pk amplitude of the first harmonic is the same as the pk-pk amplitude of the square wave, but it's not.

 

[Ratch]

godfreyl,

Yes, you are correct, and your answer is correct also.

Ratch