[SOLVED] difference between four and two quadrant multiplier

Hi i am able to understand that two quadrant just has one input (bipolar) and another input to be unipolar. While 4 quadrant has both inputs bipolar. And also 2 quad multiplier is an unbalanced modulator while 4-q is balanced meaning 2-Q has carrier and sidebands while 4-Q has carrier supressed.
But can anyone pls explain the difference between both 2-Q and 4-Q multipliers (in terms of AM). In 2-Q one input is ac signal and other is DC. So the wave appears for postive cycle. so does that mean for the positive cycle the carrier is modulated resulting in the sidebands and during the negative cycle no modulation occurs and thus the unmodulated carrier is present at the negative cycle of the O/p. Can anyone pls tell if this is right...
Also in 4-Q what exactly happens that the carrier is suppressed? This is the main thing between those two which i'm not quite sure about. Can anyone pls help!!! Thank you!!!

It requires a 2f notch filter to suppress the carrier to improve the transient response.

Standard AM is implemented with a 2Q modulator.

I can see that Complete modulation is not achieved using a 2-Q multiplier. I can see for the negative cycle of the message signal the carrier is being clipped off... For the multiplier eqn = x *y/z... I can understand that during the negative cycle the 'y' the pmos gets turned off resulting in x * 0/ causing output to be 0. So is this kind of Amplitude modulation good and is it preferred?? A 2-Q modulator is also called an unbalanced modulator. Apart from carrier being suppressed can anyone tell any other difference between a 2-Q and 4-Q amplitude modulation???

Fourier transform of multiplier

Hi i have built a 2-q multiplier with function Ix*i1/Ib.... and a
4-q multiplier with function 2*ix*iy/Ib.....The input to the multiplier are two sine waves one for message and one for carrier wave. So i am able to understand time domain multiplication of the two sine waves. Its straightforward. But in the freq domain in 2-q i get the freq of the carrier and sidebands. While in 4-q the spectrum is carrier suppressed and has only the sidebands. I understand multiplication in time domain is convolution in freq domain. But i found convolution very confusing. Can anyone kindly explain me with equation how in 2-q carrier occurs and in 4-q carrier is suppressed??? Since i am unable to understand convolution i am quite confused on how to go about with the equation. Any hints any help will be really great!!! Thank you!!!

Re: Fourier transform of multiplier

Hi,

I'm not an RF specialist...but maybe I can help with FFT.

***
Imagine in 2q.
Multiplication is AM.
So the carrier gets amplificated with the signal input voltage.
Input signal voltage is always positive. Therefore the carrier is always existent...
Therefore you find it in the FFT result.

Now consider 4q.
If input signal is 0, then the result is 0 also. No carrier.
But now the input signal has positive and negative halfwaves.
This means at positive input signal there is a carrier....and at negative input signals also.
But now the output is inverted with respect to the positive input...therefore the carrier of the positive halfwave is cancelled by the carrier of the negative halfwave.

Klaus

Re: Fourier transform of multiplier

Convolution is very simple- it is useful to limit bandwidth. Basically you take one signal (say one sine wave), multiply with another (say one pulse) and add them together (integrate from 0 to t) and the integral (a function of t; the limit of the integral in the previous step) is the convolution of the signal with the modulator (the pulse). To limit the bandwidth, integrate from t-delta to t).

I hate to write equations because it takes too much time and effort. There is an animation in https://mathworld.wolfram.com/Convolution.html. Hope this helps!

Re: Fourier transform of multiplier


For 2-q output is not furnished during the negative cycle of the message signal. But its in time domain. But still if we observe the freq of the message signal of 2-q it will be the same freq as a complete sine wave freq would be in 4-q. So when both 2-q and 4-q have the same ways to convolve whats the difference? Or since 2-q does ouput during
-ve cycle has an impact?

Re: Fourier transform of multiplier

Everytime you limit the bandwidth, you get an artifact: sidebands. They are a new beast altogether. If you are interested in radio transmission, you also need to understand sidebands.

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Fourier transformation is a tool that takes a time domain function into a frequency domain. The peaks in YOUR graphs are one single frequency, corresponding to one sine wave.

Fourier transformation of a convolution is F[f*g]=F[f]F[g] where f(t) and g(t) are two time domain functions and F is the Fourier transformation.

Shamelessly copied from https://mathworld.wolfram.com/ConvolutionTheorem.html but I strongly urge you to take a look. Are you clear about the quadrature outputs?

Re: Fourier transform of multiplier

Thank you for all the help so far!!! I am able to understand 2-q and 4-q theoretically. But still not clear with eqns. I'll present my understanding. Pls correct me.

For 2-q the eqn is 2*ix*Iy/Ib... (ix-> bipolar i/p and Iy->unipolar input)

So when i run simulation i give isin to both. So both being sin waves. But actually during negative cycle of Iy, (since Iy being unipolar the pmos goes into cutoff and so Iy is 0 and so no carrier. So it seems like i can't represent Iy to be a sine wave since Iy is just positive half of a cycle. So in eqn 2*Iy*sin(wt)/Ib
(k=2/Ib for simplifying) so eqn becomes k*Iy*sin(wt).... So meaning the carrier freq is got at the output and not suppressed. no matter what changes occur to the sin wave in time domain say its amplitude is varied or anything but in freq domain that freq component will be retained of the output. So then what about side bands in the eqn. How to represent them????


Now 4-q eqn is 2*ix*iy/Ib... So both inputs are bipolar and giving sine waves to both and (k=2/Ib for simplification of the eqn) it becomes
k*sin(wt)*sin(wt)

From trig eqn sin(A)sin(B)=1/2[cos(A-B)-cos(A+B)]
two input sine waves with freq 10K and 5K

So sin(2*pi*10K*t)*sin(2*pi*5K*t)=
1/2[cos(2*pi*10K*t-2*pi*5K*t)- cos(2*pi*10K*t+2*pi*5K*t)]

result is 1/2[cos(2*pi*5K*t)-cos(2*pi*15K*t)]

So now der is freq component of 5K and 15K which is fc-fm and fc+fm and no carrier freq component so carrier suppresed. Forgive me if its wrong but i am trying to learn all this so any help would be really great. Kindly can somone explain with eqn if possible!!! Thanks a lot!!!

Re: Fourier transform of multiplier

For 2-q the eqn is 2*ix*Iy/Ib... (ix-> bipolar i/p and Iy->unipolar input)

So when i run simulation i give isin to both. So both being sin waves. But actually during negative cycle of Iy, (since Iy being unipolar the pmos goes into cutoff and so Iy is 0 and so no carrier.

Click to expand...

You claim to use unipolar Iy but you don't. To make Iy unipolar, you need to add an offset to the sine so that Iy(t) >= 0. That's how regular AM (with carrier) works. All the "negative cycle clipped off" stuff is erroneous. Clipping happens only in overmodulation.

Respectively, you need to correct the spectra in post #4. A sine signal has no DC component unless you add an offset. This is only the case for the message in a 2Q modulator.



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Review the AM example waveforms and notice that nothing is being clipped. https://en.wikipedia.org/wiki/Amplitude_modulation