[SOLVED] frequency doubling square wave

[preethi19]

Hi i built a multiplier in cadence and when i give the inputs x,y to be sin wave and having certain amplitude and freq the result say z is doubled in frequency and it works fine for sine wave. Now replacing isin with ipulse or isource in cadence and setting rise and fall time very small and pulse width 10u and period 20u i am able to generate a input square wave for x and y. But the output z is a square wave varying only in amplitude but is in the same frequency as the input and i cant see it doubling in freq when i run transient analysis. I am not changing anything except the input sources. Can anyone pls tell me how to fix this??? Thank you!!!

 

[Warpspeed]

Multiplying a sine wave with itself does produce twice the output frequency, but it only works with sine waves.

Multiplying a square wave by itself just produces a larger square wave.
A different frequency doubling technique is called for.

What are you trying to do ?
What is this for ?

 

[preethi19]

Thank you for the reply!!! I had designed it for my project work and i was reading a paper and i noticed they used different waveforms like triangle and ramp wave. So i was just trying with different waveforms to just see the result... so i also tried it with a square waveform.. But i thought from the name frequency doubler it was supposed to double the freq of any waveform.

 

[dick_freebird]

If you need to retain true "squareness" that's tough. But
to only double pulse frequency you could take several
approaches. One is phase shift and XOR. Another is a
one-shot that triggers on both edges.

 

[preethi19]

Oh i am not looking to freq double the square wave pulse. I was just trying different waveforms using a 4-Q multiplier. Like can anyone pls let me know why is a square wave unable to double in freq using a multiplier?? Because just like any other wave it has a particular amplitude and frequency. Like whats that actual difference between a square and a sine wave (apart from the shape ) that i am able to double a sine wave and not square. Pls correct me if i am wrong. But a square wave is combination of many sinewaves together. So say at a particular point of both inputs of multiplier x and y has a particular sine wave (building a square wave) why cant it double?? Doubling all the sine waves so that all those doubled sine waves combined will result in a freq doubled square wave???

 

[crutschow]

Mathematically multiply two sinewaves together. What is the result?
Mathematically multiply two squarewaves together. What is the result?
There's your answer.

 

[c_mitra]

Multiplying a sine wave with itself does produce twice the output frequency, but it only works with sine waves.

sin(wt)*sin(wt)=sin2(wt)=0.5*(1-cos(2wt))

So there is a level shift (all the values are +ve) and phase shift (sin becomes cos) and frequency doubling (wt becomes 2wt)

A square wave can be considered a sine wave in the first approximation (see, e.g., https://mathworld.wolfram.com/FourierSeriesSquareWave.html)

The same considerations will apply but the product will not remain a square wave (I give up; too messy to calculate) but the frequency will double. This is not going to work for a pulse, though.

For a pulse the Fourier transform is complicated (see, e.g., https://demonstrations.wolfram.com/RectangularPulseAndItsFourierTransform/) and what the OP is observing is consistent with theory.

 

[crutschow]

Here's a simulation of the a sinewave and a square-wave with equal peak values into both inputs of a 4-quadrant analog multiplier with a transfer function of (X*Y) / 10.
As you can see, the sinewave input gives a double frequency output and the square-wave input gives a DC output (with short spikes during the transitions.
Both outputs are as expected.

 

[FvM]

As you can see, the sinewave input gives a double frequency output and the square-wave input gives a DC output (with short spikes during the transitions.
Both outputs are as expected.

Yes, I believe, you get the answer by using pencil and paper in half a minute. It's simple if you analyze the waveforms in time domain.

Referring to fourier series confuses the problem in a first order. Why is it so? A multiplier is a non-linear circuit, so the frequency components don't superimpose independently, you need to consider the instantaneous time domain signal.

 

[c_mitra]

Yes, I believe, you get the answer by using pencil and paper in half a minute.

Well, I am not so fast but the sine wave is also not a sine wave, it is a DC with tons of ripple. Multiplicative of f(x) with itself will always be positive. Engineers call it something like synchronous rectification (am I correct??) ...

I do not know what to do with a pencil other than to chew it...

 

[c_mitra]

Here's a simulation of the a sinewave and a square-wave with equal peak values into both inputs of a 4-quadrant analog multiplier with a transfer function of (X*Y) / 10.
As you can see, the sinewave input gives a double frequency output and the square-wave input gives a DC output (with short spikes during the transitions.
Both outputs are as expected.

Yes, both outputs are as expected.

Real square waves have somewhat sloping sides and if you use square waves with not-so-vertical-walls-on-both-sides you will notice the difference. The short spikes you are seeing now will open up (curved like a parabola) but the flat tops stay like that (as before). If you extrapolate the curved side-walls over the flat-tops, they will now clearly indicate how the frequency doubling has taken place,

In less-than-accurate mathematical terms, the frequency doubling takes place because the negative half is folded over (just like the ripple has double the mains frequency).