Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Math for balanced mixer...
- Thread starter Lewish
- Start date
1) Sorry Nbucska I spelled your name wrong in my last post.
2) Lewish: I found this is an old scribbler of course notes I have. Assume cosA is the input and cosB is the local oscillator.
Single Balanced Mixer (2 diodes)
The simple mathematical explanation is the input cosA is multiplied by a square wave of frequency B. Obviously it won't exactly be a square wave but it will be a periodic signal with a dc offset due to only two diodes (so there is an average value in the fourier series). Assuming a square wave the mixer outputs are cosA plus DSBSC of cosA at B, 3B, 5B etc with decreasing amplitudes according to the fourier series of a square wave.
Double balanced mixer (4 diodes)
Same cosA input multiplied by same pseudo-square wave of frequency B except this time the pseudo-square wave has no dc offset (average =0) due to 4 diodes. So the mixer outputs are DSBSC of cosA at B, 3B, 5B etc.
The difference between the two is the inherent suppression of the unwanted cosA term in the double balanced version. Obviously filtering is required to extract the desired signal, bandpass for the single balanced and low pass for the double balanced.
Note I used a square wave to make the math simple. Suffice to say that the signal multiplied by cosA will be a periodic function which will have a dc offset for the single balanced case and none for the double balanced case. So the real fourier series of that signal may contain even harmonics. It just means DSBSC of cosA at those frequencies as well. Either way they are unwanted signals.
Hope this helps.
2) Lewish: I found this is an old scribbler of course notes I have. Assume cosA is the input and cosB is the local oscillator.
Single Balanced Mixer (2 diodes)
The simple mathematical explanation is the input cosA is multiplied by a square wave of frequency B. Obviously it won't exactly be a square wave but it will be a periodic signal with a dc offset due to only two diodes (so there is an average value in the fourier series). Assuming a square wave the mixer outputs are cosA plus DSBSC of cosA at B, 3B, 5B etc with decreasing amplitudes according to the fourier series of a square wave.
Double balanced mixer (4 diodes)
Same cosA input multiplied by same pseudo-square wave of frequency B except this time the pseudo-square wave has no dc offset (average =0) due to 4 diodes. So the mixer outputs are DSBSC of cosA at B, 3B, 5B etc.
The difference between the two is the inherent suppression of the unwanted cosA term in the double balanced version. Obviously filtering is required to extract the desired signal, bandpass for the single balanced and low pass for the double balanced.
Note I used a square wave to make the math simple. Suffice to say that the signal multiplied by cosA will be a periodic function which will have a dc offset for the single balanced case and none for the double balanced case. So the real fourier series of that signal may contain even harmonics. It just means DSBSC of cosA at those frequencies as well. Either way they are unwanted signals.
Hope this helps.
- Thread starter
- #22
Update Time.
Here is what I have done so far. I created a one million sample file of a 55KHz signal in cosine form. Call it Cos(A). I did a FFT of this file in Matlab and verified that I have only one signal. Then I created a second file of the same signal but with a 3 Hz modulation on it. Again I did a FFT and verified that only 2 signals existed. Call this Cos(B). Then using Matlab, I did 2*Cos(A)*Cos(B) to create the mixer output file. When I did a FFT of this file I see 3 signals - one at 3 Hz, one at 55KHz and one at 110.003KHz.
From this I conclude my reference book is correct that the equation 2*Cos(A)*Cos(B) is indeed the equation for a simple mixer.
So, what is the equation for a balanced mixer where there is no carrier term in the output?
Please no replies about modeling diodes or transistors or other non-linear devices. I don't have the resources in the DSP to do that, and I don't think I should need to do it.
Here is what I have done so far. I created a one million sample file of a 55KHz signal in cosine form. Call it Cos(A). I did a FFT of this file in Matlab and verified that I have only one signal. Then I created a second file of the same signal but with a 3 Hz modulation on it. Again I did a FFT and verified that only 2 signals existed. Call this Cos(B). Then using Matlab, I did 2*Cos(A)*Cos(B) to create the mixer output file. When I did a FFT of this file I see 3 signals - one at 3 Hz, one at 55KHz and one at 110.003KHz.
From this I conclude my reference book is correct that the equation 2*Cos(A)*Cos(B) is indeed the equation for a simple mixer.
So, what is the equation for a balanced mixer where there is no carrier term in the output?
Please no replies about modeling diodes or transistors or other non-linear devices. I don't have the resources in the DSP to do that, and I don't think I should need to do it.
"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"
What I get from the above is you have a 55kHz carrier with two sidebands and you want to extract the sidebands and ditch the carrier.
You last post indicated that when you modulated the 55kHz carrier with 3Hz, the FFT gave two signals - I assume at 54.997kHz and 55.003kHz. So the amplitude modulation you did in Matlab was suppressed carrier - just a straight mixing of 3Hz and 55kHz -is that right? Otherwise I would have expected 3 terms, i.e a 55 kHz carrier as indicated above.
Assuming signals at 54.997 kHz and 55.003 kHz and then mixing these (mathematically) with 55kHz I get 109.997kHz, 110.003KHz and 3 Hz. I don't get anything at 55kHz.
By the way when I did the same thing assuming a non supressed carrier I get 3Hz, 109.997 kHz, 110kHz and 110.003kHz - still no 55kHz
Are you sure you aren't getting any artifacts from the FFT?
This is probably not helping with your problem or your patience but I'd just like to make sure we're talking about the same thing.
What I get from the above is you have a 55kHz carrier with two sidebands and you want to extract the sidebands and ditch the carrier.
You last post indicated that when you modulated the 55kHz carrier with 3Hz, the FFT gave two signals - I assume at 54.997kHz and 55.003kHz. So the amplitude modulation you did in Matlab was suppressed carrier - just a straight mixing of 3Hz and 55kHz -is that right? Otherwise I would have expected 3 terms, i.e a 55 kHz carrier as indicated above.
Assuming signals at 54.997 kHz and 55.003 kHz and then mixing these (mathematically) with 55kHz I get 109.997kHz, 110.003KHz and 3 Hz. I don't get anything at 55kHz.
By the way when I did the same thing assuming a non supressed carrier I get 3Hz, 109.997 kHz, 110kHz and 110.003kHz - still no 55kHz
Are you sure you aren't getting any artifacts from the FFT?
This is probably not helping with your problem or your patience but I'd just like to make sure we're talking about the same thing.
- Thread starter
- #24
Hi brennaj, thanks for the reply. A staight modulated carrier should contain 2 terms only. In my case 3 Hz and 55KHz. That is what I see in the FFT in Matlab. Maybe I am using the term modulation incorrectly here. What I have in the real world is an oscillator putting out a very pure 55KHz sinewave. In the setup, a 3 Hz signal is being impressed onto the 55KHz carrier. They are not being mixed, so I don't expect to see the 2 sidebands you refer to. My FFT extends out to 1MHz, and I only see these 2 signals on the sampled waveform.
Since the amplitude of the 55KHz is variable, not due to modulation, just random noise, I want to run the signal thru a mixer to strip off the 55KHz and leave the 3Hz signal and some other signal which will be removed by filtering.
Any additional thoughts would be appreciated.
Since the amplitude of the 55KHz is variable, not due to modulation, just random noise, I want to run the signal thru a mixer to strip off the 55KHz and leave the 3Hz signal and some other signal which will be removed by filtering.
Any additional thoughts would be appreciated.
Give me a bit more info. How is the 3Hz being modulated on the 55KHz or are they being modulated at all, perhaps just added? Is the '3 Hz modulation' being done on purpose or is this the problem you are dealing with?
Maybe you could post the mathematical equation of the 'signal' - ie 55kHz plus the 3Hz 'modulation'
I know this isn't getting you closer to your answer but I'd like to understand the problem you're trying to solve.
Maybe you could post the mathematical equation of the 'signal' - ie 55kHz plus the 3Hz 'modulation'
I know this isn't getting you closer to your answer but I'd like to understand the problem you're trying to solve.
- Thread starter
- #26
Hi brennaj, OK, lets see if this helps any. The 55KHz is a differential signal impressed across a human body. As blood flows, it changes the impedance presented to the drive signal. This changes the voltage across the body, which is then seen by a differential amplifier acting as the receiver. The body's impedance as we are connected is typically about 3Kohms. This value will change by about 3 ohms as blood is pumped. From this it is possible to estimate the volumetic efficiency of the heart. This is an oversimplified explanation. The "modulating signal will of course change with the heart rate. One Hz being a more typical signal. Because of the tiny signal levels involved, I am not sure whether simple addition or AM modulation is occuring. That is part of what I am trying to resolve. But either way, I think a balanced mixer should strip off the 55KHz carrier. Yes - No??
Thanks for you input.
Thanks for you input.
Actually not... What you are describing is conventional heterodyne AM detection, which is mixing the AM signal with the local oscillator at the carrier frequency, which results in the baseband signal and the AM signal shifted up by the LO frequency. You then have to run through a lowpass filter to get the baseband by itself.
from your description, the process is essentially AM modulation x(t)*cos(2*pi*f*t), except that x(t) is 1/Z(t), where Z(t) is the impedance function.
The following article describes AM demodulation:
TTFN
from your description, the process is essentially AM modulation x(t)*cos(2*pi*f*t), except that x(t) is 1/Z(t), where Z(t) is the impedance function.
The following article describes AM demodulation:
TTFN
- Thread starter
- #28
Lewish:
Based on your explanation of the system I agree with IRStuff's logic that the 55kHz signal is amplitude modulated by the 1 to 3Hz "tone". The spectrum of that is a 55kHz tone and two sidebands. Assuming you want to recover the 1 to 3Hz tone the simplest method is an envelope (peak) detector - can you implement that in a DSP?
The balanced mixer(with a synchronous LO) would work as well with a low pass filter as both you an IRStuff said but from an electronic standpoint the peak detector is far easier. From a DSP standpoint - I couldn't say as I'm no expert
Based on your explanation of the system I agree with IRStuff's logic that the 55kHz signal is amplitude modulated by the 1 to 3Hz "tone". The spectrum of that is a 55kHz tone and two sidebands. Assuming you want to recover the 1 to 3Hz tone the simplest method is an envelope (peak) detector - can you implement that in a DSP?
The balanced mixer(with a synchronous LO) would work as well with a low pass filter as both you an IRStuff said but from an electronic standpoint the peak detector is far easier. From a DSP standpoint - I couldn't say as I'm no expert
- Status
Similar threads
- Locked
- Question
- Locked
- Question
- Locked
- Question