I wanna make a square rooter circuit so that take the square root of a sine wave, The idea is to add some DC-OFFSET to the sine wave so that I have a Completely positive sine wave.
This sounds highly bizarre? - if you want to DC shift a sinewave then there are many much simpler ways (and probably none more complicated and less effective).
Just use a single opamp to shift the sinewave to positive only.
I wanna make a square rooter circuit so that take the square root of a sine wave, The idea is to add some DC-OFFSET to the sine wave so that I have a Completely positive sine wave.
Doctor Power,
You don't have to add any DC offset. The √(sin(t)) is a completely positive function.
Ratch
Doc Power,
Interesting. What is the simulator?
Ratch
Doc Power,
OK, I $aw what they wanted for the full ver$ion. Did you do your simulation with the stunted "free" version?
Ratch
I just use the full version. I never used the free version And have no idea about it.
Where did you go fro the price list?
Any idea regarding my question please? How to add a dc offset to the AC signal so that the AC signal does not be affected by the DC offset? I do not want the domin of the Ac signal to be reduced due to DC offset volatge.
By the way no one knows if the circuit takes: √dc-offset + √Vac or it takes: √(dc-offset + Vac)?
I noticed if the circuit takes the later then the output will not be a correct square rooted signal, Are you agreed with me regarding to that?
Doc Power,
**broken link removed**
Did the company you work for buy the software?
I would run the sin through the √ process, then pipe both the signal and the offset into the two ports of an adder. The output of the adder should be the √(signal) plus the pure offset. **broken link removed**
Correct, you do not want the sum of the DC and the signal getting sqrt'ed. The signal only should be sqrt'ed. The method I suggested above should do that.
Hello drpower,
What is it you are trying to do, ie what are you trying to generate?
I ask because you state that you want to add some DC offset to a sine wave yet you dont have to take the square root of it to do that. There are much easier techniques. So i ask that question because of that.
If you want a completely positive wave, you wouldnt take the square root, you would take the square, or simpler yet take the absolute value. There are simple ways to take the absolution value and squaring is harder to do.
Or is it that you need to take the square root of a sine wave plus a DC offset?
So you know what an absolute value circuit looks like?
Hello again drpower,
Yes i think that makes sense, and explains what you are trying to do a little bit better.
It seems that you want to take the square root of sin(wt) and you found that since that wave goes both plus and minus that you run into trouble when it goes negative because a simple square rooter circuit can not deal with the negative input. So first try is to add a DC offset to the original signal sin(wt) but then you end up with sqrt(sin(wt)+k) instead of sqrt(sin(wt))+sqrt(k), so you need another approach.
The next idea is to take the absolute value of the sin(wt) and then take the square root of it. That way you have a completely positive output and it really is the square root of |sin(wt)|. So first you run sin(wt) through an absolute value circuit, then run it to the square root circuit.
If you need the output to go negative when sin(wt) goes negative then we'd need a little bit more to detect when the input goes negative, or build the circuit a little differently to handle the two cases plus and minus independently and then add them back together. I think you want all positive though if i understand you correctly.
So you know what an absolute value circuit looks like?
BTW, im not sure if you mentioned this yet, but what is the input frequency range going to have to be?
Hi again,
An absolute value circuit is not exactly just a full wave rectifier bridge made of only diodes, it's a op amp circuit that is a little more accurate for low level signals. A bridge rectifier has diode drops which impairs its ability to function accurately with signals that are comparable to the diode drops. The op amp equivalent circuit is better in that it can rectify lower level signals much more accurately.
Are you saying that you need to have the output go negative when the input goes negative? In that case you also need a circuit to detect the negative portion of the input and switch the output so that it goes negative too (after taking the absolute value and square root).
I'll take a look at your circuit and see what i can find out. What are the two adjustments doing for you?
Also, what range input are you looking for?
That circuit looks a little strange. Did you check it for numerical accuracy?
Basic tests:
When you input 0.25v, did you get 0.5v out?
When you input 1v, did you get 1v out?
When you input 4v, did you get 2v out?
When you input 9v, did you get 3v out?
I ask because certain parts of the circuit look very unorthodox (R2 for example).
They make pretty good multiplier chips these days i think so you may be better off with a design based on a multiplier chip.
You can also consider using a micro controller chip with built in AD converter and DA output.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?