Demodulating a signal buried in noise

Suppose I have a signal that is buried in noise two or three orders of
magnitude larger. The desired signal is coherent with a reference
waveform in the low audio range. 

What will be the difference between mixing the incoming (noisy)
signal, say x= (sin(w*t+alpha) + noise(t)) by multiplying by the
reference y= sin(w*t+alpha) as opposed to using sgn(y) =
sgn(sin(w*t+alpha) (IOW a gain of +1 for sin(..) > 0 and -1 for
sin(...) < 0 ? 



Best regards, 
Spehro Pefhany
-- 
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Spehro Pefhany wrote:

> Suppose I have a signal that is buried in noise two or three orders of
> magnitude larger. The desired signal is coherent with a reference
> waveform in the low audio range. 
> 
> What will be the difference between mixing the incoming (noisy)
> signal, say x= (sin(w*t+alpha) + noise(t)) by multiplying by the
> reference y= sin(w*t+alpha) as opposed to using sgn(y) =
> sgn(sin(w*t+alpha) (IOW a gain of +1 for sin(..) > 0 and -1 for
> sin(...) < 0 ? 

Asumming the noise is white, the difference is 3 dB.


Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

Spehro Pefhany  <speffSNIP@interlogDOTyou.knowwhat> wrote:

>Suppose I have a signal that is buried in noise two or three orders of
>magnitude larger. The desired signal is coherent with a reference
>waveform in the low audio range. 

>What will be the difference between mixing the incoming (noisy)
>signal, say x= (sin(w*t+alpha) + noise(t)) by multiplying by the
>reference y= sin(w*t+alpha) as opposed to using sgn(y) =
>sgn(sin(w*t+alpha) (IOW a gain of +1 for sin(..) > 0 and -1 for
>sin(...) < 0 ? 

I would say one difference is that that sgn(sin(w*t)) has non-zero 
correlation with sin(2*w*t), sin(3*w*t), etc. so you are
going to be getting noise components from these frequencies
in your result.  But the significance of this effect would be
reduced if there is narrowband filtering ahead of your mixer.

Steve

On Sun, 28 Sep 2008 17:13:56 -0500, Vladimir Vassilevsky wrote:

> Spehro Pefhany wrote:
> 
>> Suppose I have a signal that is buried in noise two or three orders of
>> magnitude larger. The desired signal is coherent with a reference
>> waveform in the low audio range.
>> 
>> What will be the difference between mixing the incoming (noisy) signal,
>> say x= (sin(w*t+alpha) + noise(t)) by multiplying by the reference y=
>> sin(w*t+alpha) as opposed to using sgn(y) = sgn(sin(w*t+alpha) (IOW a
>> gain of +1 for sin(..) > 0 and -1 for sin(...) < 0 ?
> 
> Assuming the noise is white, the difference is 3 dB.
> 

... because in multiplying by a square wave you're letting in all of the 
noise energy associated with the odd harmonics of your carrier 
('reference waveform').

But it's more like 1dB, because your desired signal is gained up by 4/pi 
(+2dB), but corrupted by a bunch of noise that rides in on the edges of 
the square wave (-3dB).  So the end effect is that you lose a dB.

Realistically, you may lose more if the noise tends to be worse at high 
frequencies, or if there's noise associated with the carrier frequency 
not quite matching up to the sampling rate.  But maybe it's not too bad.

-- 
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott  <tim@seemywebsite.com> wrote:

>On Sun, 28 Sep 2008 17:13:56 -0500, Vladimir Vassilevsky wrote:

>> Assuming the noise is white, the difference is 3 dB.

>... because in multiplying by a square wave you're letting in all of the 
>noise energy associated with the odd harmonics of your carrier 
>('reference waveform').

>But it's more like 1dB, because your desired signal is gained up by 4/pi 
>(+2dB), but corrupted by a bunch of noise that rides in on the edges 
> of the square wave (-3dB).  

I'm not sure I agree with this.  The fourier expansion of a 
square wave has coefficients:

        1 ,  1/3, 1/5, 1/7 ....

And so the 3dB value would be true only if

        1^2    =   (sum over k = 3, 5, 7...) ( 1 / (k^2))

which I don't think is the case, as the RHS of this expression
is bounded from above by 1/4.

I may, of course, have miscalculated somewhere.

Steve

"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:7KednZUQk82I233VnZ2dnUVZ_obinZ2d@web-ster.com...
> On Sun, 28 Sep 2008 17:13:56 -0500, Vladimir Vassilevsky wrote:
>
> > Spehro Pefhany wrote:
> >
> >> Suppose I have a signal that is buried in noise two or three orders of
> >> magnitude larger. The desired signal is coherent with a reference
> >> waveform in the low audio range.
> >>
> >> What will be the difference between mixing the incoming (noisy) signal,
> >> say x= (sin(w*t+alpha) + noise(t)) by multiplying by the reference y=
> >> sin(w*t+alpha) as opposed to using sgn(y) = sgn(sin(w*t+alpha) (IOW a
> >> gain of +1 for sin(..) > 0 and -1 for sin(...) < 0 ?
> >
> > Assuming the noise is white, the difference is 3 dB.
> >
>
> ... because in multiplying by a square wave you're letting in all of the
> noise energy associated with the odd harmonics of your carrier
> ('reference waveform').
>
> But it's more like 1dB, because your desired signal is gained up by 4/pi
> (+2dB), but corrupted by a bunch of noise that rides in on the edges of
> the square wave (-3dB).  So the end effect is that you lose a dB.

Tim,

I did the math. Taking the signal gained up, it is 3dB.

VLV

>What will be the difference between mixing the incoming (noisy)
>signal, say x= (sin(w*t+alpha) + noise(t)) by multiplying by the
>reference y= sin(w*t+alpha) as opposed to using sgn(y) =
>sgn(sin(w*t+alpha) (IOW a gain of +1 for sin(..) > 0 and -1 for
>sin(...) < 0 ? 

Interesting question!

If the filter following this operation is fairly sharp (with cut-off
frequency less than 3w) then there is no difference. (Steve pointed this
out. But as Tim pointed, there will only be odd harmonics in your case.)  

If an integrator follows this operation, then the loss in using the square
wave multiplier is only 0.06 dB. 

If no filtering is applied, additional noise you will collect from all the
harmonics (3w, 5w, ...) causes 0.91 dB higher (white) noise power. (Though
I can't think of a good reason for not using a filter in sequel.)  At any
rate, 0.91 dB is the upper bound to the amount of extra noise; it should
not get any worse than that given the noise is white. (Bravo, Tim, for your
offhand precision.)

Emre

>If the filter following this operation is fairly sharp (with cut-off
>frequency less than 3w) then there is no difference. (Steve pointed this
>out. But as Tim pointed, there will only be odd harmonics in your case.) 


Sorry, the filtering needs to be before this operation (as Steve said), if
you want to eliminate the effect. 

>If an integrator follows this operation, then the loss in using the
square
>wave multiplier is only 0.06 dB.

Never mind this. An integrator can't precede this operation anyway (it
would destroy the signal), and it doesn't help afterwards.
 
>If no filtering is applied, additional noise you will collect from all
the
>harmonics (3w, 5w, ...) causes 0.91 dB higher (white) noise power.

The answer still remains the same when there is no pre-filtering. Roughly
1 dB. If anyone is wondering, this is how I got the answer in Matlab:  
n = 1:2:1000;  
10*log10(sum(1./n.^2))
ans = 0.9103

as the Fourier series coefficients of the pulse train given by
sgn(sin(wt)) decay with 1/n for odd n, and are zero for even n.

Emre

emre <eguven@ece.neu.edu> wrote:

>The answer still remains the same when there is no pre-filtering. Roughly
>1 dB. If anyone is wondering, this is how I got the answer in Matlab:  
>n = 1:2:1000;  
>10*log10(sum(1./n.^2))
>ans = 0.9103

I agree with this method, and got a number just around 0.9 dB
with the formula given in my last post.

Tangentially: I once saw a 2.4 GHz direct-conversion radio
that drove the mixer with a 800 MHz square wave.  This way
the third harmonic of the signal injected into the mixer
was the actual component used.  It was cheap, and (with
a ceramic RF filter) it actually worked.

Steve

I had trouble desiphering teh question but you mean using a
purecoherent  sine way as opposed to  a coherent square wave! I wish
people would use English! In a word - bugger all.


Hardy



Spehro Pefhany wrote:

> Suppose I have a signal that is buried in noise two or three orders of
> magnitude larger. The desired signal is coherent with a reference
> waveform in the low audio range.
>
> What will be the difference between mixing the incoming (noisy)
> signal, say x= (sin(w*t+alpha) + noise(t)) by multiplying by the
> reference y= sin(w*t+alpha) as opposed to using sgn(y) =
> sgn(sin(w*t+alpha) (IOW a gain of +1 for sin(..) > 0 and -1 for
> sin(...) < 0 ?
>
>
>
> Best regards,
> Spehro Pefhany
> --
> "it's the network..."                          "The Journey is the reward"
> speff@interlog.com             Info for manufacturers: http://www.trexon.com
> Embedded software/hardware/analog  Info for designers:  http://www.speff.com