Math for balanced mixer...

[Lewish]

Can anyone provide me with the math for a balanced mixer. I looked thru all of my old textbooks, and can only find the math for a simple mixer. I.e., 2 * Cos(A) * Cos(B).

Thanks.

 

[nbucska]

Look at:

http://www.onsemi.com/pub/Collateral/AN531-D.PDF

<nbucska@pcperipherals.com>

 

[Lewish]

Hi nbucska, thanks for the reply. However, the MC1496 is actually a double-balanced mixer. Otherwise known as a Gilbert Cell mixer, named after Barrie Gilbert. I have a 1975 datasheet for this part, having used many of them in the past, and still have a few dozen kicking around in the junk box. The 1975 datasheet clearly describes this as a double-balanced mixer.

Also, the equation I am looking for would have the result expressed as the product of 2 trig functions plus some other stuff.

Thanks for trying.

 

[nbucska]

The literature I quoted has the equations for both
high and low level signals. What do you need or what is
your application ? You could use SPICE or you could
write a simulation using an ideal xistor easily enough.

<nbucska@pcperipherals.com>

 

[Lewish]

Hi nbucska,
My application is a carrier suppressor after frequency shifting for a medical monitor.
Yes, I could do it in Spice, but can't get the data required for this application. I need to do a mathematical analysis to prove my theory. The application will be done with a DSP, so transistor simulation doesn't help. I need a &quot;raw&quot; math equation.

Thanks for any help you can give.

Lewis

 

[IRstuff]

I'm confused, isn't what you're asking for in Eq. 13 or 15 or 16 on the datasheet? Eq. 15 shows a product of 2 cosines, while 16 is re-written as a quadrature form?

TTFN

 

[Lewish]

I guess I am doing a poor job of explaining the problem. I want to replicate a balanced mixer inside a DSP. No transistor parameters involved.

The equation for a simple mixer is 2 * Cos(A) * Cos(B) which yields Cos(A+B)+Cos(A-B). This is what you see in Eq. 13 with the emitter resistances and the long-tailed pair resistance thrown in. I.E. the &quot;RE&quot; and the &quot;Re&quot;.

Doing away with any reference to the behavior of a bipolar transistor, what does the equation for a balanced mixer look like?

 

[nbucska]

The mixer works better if it is overdriven -- i.e. the signal levels are high.

It has two diff. inputs: A to xtors 1-2-3-4 and B to xtors 5-6.I suggest to assume at least signal A to be very high
relative to Vbe (~.65 V).

For very large positive A:
For small signals the output voltage is B*Rc/Re where Rc
is R from Coll to coll and Re is from Em to Em.

For large signal the output is square wave with the
amplitude of Iem*Re (iem is one emitter current).
For intermediate: trapesoidal with a limited rise/fall time.

The effect of overdriven A is to multiply B*Rc/Re with
F(A)= sign(A) where F(A)=1 if A positive, -1 when negative
and never 0 ( assume A to be so high that it crosses 0
&quot;instantly&quot;.)

I hope your DSP is fast enough. If not : Do U have to do it
in real time or can U do it on stored data ?


<nbucska@pcperipherals.com>

 

[Lewish]

To nbucska et all, I know how the 1496/1596 works. I have used lots of the 1596 in military projects. I want to know how to duplicate a balanced mixer in DSP code.
Yes, I have to do it in a DSP. The signal is a 55 KHz signal that has been sampled at 40 Megasamples/second. After a 100:1 decimation, I have plenty of time to process the data.
So...., the question still remains, what is the algebraic equation for a balanced mixer, independent of how it is implemented. The equation 2*Cos(A)*Cos(B) does not depend on electronic components to work. This is a pure Communication Theory question. No electronics necessary. Just like a Fourier Transform is not dependant on electronics to work.

 

[nbucska]

To simplify: Pick a gain for signal B; with this the output
is Vout=G.B.

With signal A included: Vout= sign(A)* G * B
where sign() is define as +/- 1 with the sign of A.

If this is not enough, perhaps we should know more about
your problem.



<nbucska@pcperipherals.com>

 

[melone]

Doesn't your forumala depend on what nonlinear method you choose for modulation? I am looking through my Analog and Digital Communication Theory book, and they show a circuit for a square-law (balanced) modulator. The point is that the balanced modulator is designed using the effects of a nonlinear device / circuit.

Basically, the block diagram is:

s(t) -> [summer with cos 2*PI*fc*t] -> [Nonlinear]

-s(t) -> [summer with cos 2*PI*fc*t] -> [Nonlinear]

where both outputs are then summed together and sent through a bandpass filter.

Hopefully, this will help.

 

[Lewish]

nbucska, the equation you give is for a synchronous detector mixer.

Please see the previous post. I have an amplitude modulated 55KHz signal and I want to mix it with a 55KHz carrier and extract the sum and difference signals without the 55KHz carrier. A double or triple balanced mixer would be better, but I don't think I have the processing time to implement those.

 

[nbucska]

Melone: The switch/(abs.value)/sign() is equivalent with
a polynomial of even degree (2,4,6,8,etc )i.e. nonlinear!!

Lewish: use B=55KHz AM, A=55KHz CW. You will get abs(B)
if the 55 KHz-s are in sync. You could get the same with rectification.

In circuit you would use a PLL to generate &quot;A&quot;.
<nbucska@pcperipherals.com>

 

[Lewish]

Hi melone, thanks for the input, but a balanced mixer doesn't have to be square-law or non-linear. That is just the most common way to do it. Jones of Honeywell filed in 1963 and received in 1966 a patent on a linear mixer based on feedback. Unfortunately that is too far back to be online to read. The concept was actually originated back around 1918-1925, but at that time there was no way to implement it as tubes are very non-linear. I am old enough to have played with tubes, and still have a few in stock.

I didn't realize when I started this thread that it would be this hard to find an answer.

 

[brennaj]

Hi Lewish,

I had a look in some text books and found the circuits for the balanced mixer and double balanced mixer. There were no equations given with either other than the classic product of cosines you gave above. In terms of the desired outputs from the mixer (sum and difference frequencies) they are the same. Do you want the DSP to mimic an ideal mixer or replicate the actual (including non-ideal) behaviour of a balanced mixer?

 

[Lewish]

Hi brennaj, I want the DSP to mimic an ideal linear balanced mixer. The result would be the sum and difference between the signal and the carrier, with the carrier canceled. The equation I gave above lets the carrier come thru also.

I have at least 5 textbooks on communication theory, and not one gives anything but the above equation. I wonder why. Someone told me that in the 1930s and 40s engineers knew how to derive all this stuff and it wasn't printed in the texts. I guess my college education was a little lacking, OR maybe I didn't pay close enough attention.

 

[brennaj]

I did the math and I don't see any carrier term. Check this:

cos(A+B)=cosAcosB-sinAsinB (Eq 1)
cos(A-B)=cosAcosB+sinAsinB (Eq 2)

Add Eq 1 to Eg 2

cos(A+B)+ cos(A-B)=2cosAcosB

Divide both sides by 2

0.5(cos(A+B)) + 0.5(cos(A-B))= cosAcosB

 

[jcm]

Hi Lewish,
My approach to model the balanced mixer would be to go all the way to the diode equation. A single diode mixer would look like I = esp(qV/nkT) - 1, where V is your A + B. Let's let qV/nkT = x.
Then the equation for the balanced mixer would be:

I = [exp(x) - 1] + [exp(-x) -1]

For the balanced mixer the odd powered terms cancel and the even powered terms remain.

Hope this helps.

 

[nbucska]

Brennaj:
Your eq. shows that in the case of overmodulation the carrier disappears.

For general case Y= cos(A) * ( 1 + cos(b)) where B is
the modulation and A the carrier.

Lewish:
If you want synchronous demodulation,
set up an oscillator and at every 0-xing of the input
measure the phase error and use this to keep the osc. in
sync (PLL) Use PID control in the PLL !

Rectify it syncronously ( invert whenever the OSC <0 ):
-----------------
The zero-xing:
Whenever a sample has different polarity from the previous one. If sample rate is small, you may use linear interpolation to calculate the time of 0-xing.

<nbucska@pcperipherals.com>

 

[brennaj]

Nbrucksa:

I don't think we're comparing apples to apples here.

The equation in the original post 2cosAcosB is the mixer equation and yields double side band suppressed carrier modulation. There is no carrier term as the math shows.

cosAcosB = 0.5cos(A+B) + 0.5cos(A-B)

This latest equation Y=cosA(1+cosB)is for standard amplitude modulation or double sideband with carrier modulation.

cosA(1 + cosB) = cosA + 0.5cos(A-B) + 0.5cos(A+B)

These are two different modulation types.