Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Help! Power in square wave?

Status
Not open for further replies.

franklan118

Newbie level 2
Newbie level 2
Joined
Mar 31, 2013
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,298
Hi!
I got a question that confused me.

Q20130331150643.jpg

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:p
 

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%
 
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?
 

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.

 

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)?
 
Last edited:

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
 

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.
 

godfreyl,

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

Ratch
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top