Synchronous Demodulator, why the carrier still appear?

Thread Starter

Felipe Kalinski Ferreira

Joined Jul 28, 2016
5
Hello my friends, I'm new here and I hope someone may can help with this mixer.
I'm trying to do a synchonous demodulator, and multplying the AM signal with the local carrier I'm hope that the carrier frequency does not appear in the result of this multiplication. The expected would be that the information would appear in the baseband and the other harmonic component would be around twice the frequency of the carrier. I'm using a single JFET mixer (in the atached file) and the frequency component of the carrier ALWAYS appear in the Vout. Please can someone help me, I'm just trying to understand why this happened. Thank you!

jfetMIXER.jpg
 

Thread Starter

Felipe Kalinski Ferreira

Joined Jul 28, 2016
5
The signals are:

AMsignal = (1+0.5cos(wm*t))*cos(wc*t)
Carrier = cos(wc*t)

I'm expecting to find something like this in the output, in relation to the harmonic content of the output signal:
Vout = a*cos(wm*t)+b*cos((2*wc+wm)*t)+c*cos((2*wc-wm)*t)+d*cos(2*wc*t)
 

SLK001

Joined Nov 29, 2011
1,549
Your circuit isn't a demodulator. It is a simple mixer. The carrier will always appear at the output of this circuit.

What you have is: (cos α)⋅(cos β)

That, of course, results in ½ cos(α+β) + ½ cos(α-β).
 

Thread Starter

Felipe Kalinski Ferreira

Joined Jul 28, 2016
5
I'm having a hard time understanding how your circuit multiplies. Rather, I see a superposition (adding) of the two incoming signals.
I'm having trouble with this too.. but its a aproximation of a multiplication, acctualy if you take the equation of a current of a JFET we will see
that

Id=Idss*(1-Vgs/Vp)²

Where the Vp is constant and Vgs=Vg-Vs. In my circuit Vg=(carrier+AM). Vs is also a constant, then Vout would be something like

Vout=ampgain*k*(ctevalue - Vgs)² = ampgain*k*(ctevalue - carrier+AM)^2) = k*(carrier² + a*carrier*AM + AM² + ctevalue²)

Look! Is not a perfect product, but in the middle of the last equation we can see a product of the 2 signals in the Gate. But if we expand de cossines² the harmonic component of carrier should not appear. I built it in my protoboard and the carrier frequency component always appear..
 

Thread Starter

Felipe Kalinski Ferreira

Joined Jul 28, 2016
5
Your circuit isn't a demodulator. It is a simple mixer. The carrier will always appear at the output of this circuit.

What you have is: (cos α)⋅(cos β)

That, of course, results in ½ cos(α+β) + ½ cos(α-β).
Yeah, I wanted to multiply that analog signals, I thought because the properties of a current in the drain of a JFET we can take that multiplication, I built one gilbert cell to solve that problem, but I'm having the same problems... Do you Know how can I multiply those signals??
 

KL7AJ

Joined Nov 4, 2008
2,229
Hello my friends, I'm new here and I hope someone may can help with this mixer.
I'm trying to do a synchonous demodulator, and multplying the AM signal with the local carrier I'm hope that the carrier frequency does not appear in the result of this multiplication. The expected would be that the information would appear in the baseband and the other harmonic component would be around twice the frequency of the carrier. I'm using a single JFET mixer (in the atached file) and the frequency component of the carrier ALWAYS appear in the Vout. Please can someone help me, I'm just trying to understand why this happened. Thank you!

View attachment 114384
A synchronous AM detector will indeed reproduce the carrier. It is multiplied with itself (autocorrelation). If you want to reject the carrier, you must null it either with filtering or with a balanced modulator/demodulator.
 
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