Result of AM anti-phase waveforms

Why don't you do the math?

And, by "modulate", do you mean...

sin(2*pi*f*t)*sin(2*pi*f*t)

or...

(1+m*sin(2*pi*f*t))*sin(2*pi*f*t)

or what? ...Jim Thompson

AM is just multiplying the carrier by the modulation signal, with enough of a DC offset that the resulting envelope never goes negative.

Assuming 100% AM, and letting u = omega*t, your example would be

f(t) = sin(u)(1+sin(u+phi))

= sin(u) + sin(u)(sin(u)cos(phi)-cos(u)sin(phi))

= sin(u) + sin^2(u)cos(phi) - sin(u)cos(u)sin(phi)

= sin(u) + 0.5*(1-cos(2u))cos(phi) -0.5*sin(2u)sin(phi)

= sin(u) + 0.5*cos(phi)- 0.5*(cos(phi)*cos(2u)+sin(phi)sin(2u))

= sin(u) +0.5*cos(phi) - cos(2u-phi)

So assuming I remembered my trig identities correctly, it's the difference of a signal at omega and 2 omega, with a DC offset depending on the phase shift between the modulation and the carrier.

Cheers

Phil Hobbs

"Ken Jacobs"

If locked in phase, then the result is a sine squared wave:

sin squared X = - 0.5 ( cos 2X -1 )

This looks like a sine wave of double frequency and half amplitude shifted positive so it never goes below zero - it has a DC component of 0.5.

If some phase shift exists, then the DC level changes.

... Phil

The DC null at 90 degree's is the basis of lock-in detection. (As you know.)

Say, speaking of finding signals in noise. I've been playing around with looking for a signal, when you don't know the 'carrier' frequency. I guess this is a bit like the SETA search. So with signals bigger than noise you can trigger your scope and average. With DSO's you can use this trick with A S/N of 0.1 (if you trigger at the tippy-top of the noise) Below that you need bigger guns. With a spectrum analyzer you can get down to the 0.01 level, in a reasonable time. (You've got to limit the band width and average for a while.) I'm pretty sure the exact numbers will follow from Dickes' noise in the noise equation.

So below that is there some 'lock-in' approach? A slowly swept LO. It's limited by the same equation, but you can make really narrow 'filters'.

George H.

(I'm sure there's some previous thread about this if someone knows a good search term.)

snip

Possibly related:

Or this one:

Yes, from the thread title, there IS a 180 degree phase offset between the two signals. Frequencies are the same.

Sorry, I should have reiterated that point in the body of the question.

And, if one signal is inverted from the other, would I be left with a simple rising and falling DC level at their common frequency?

Ken

Maybe you should have said his idea went pi-f(f)t(t)...(eg: piff..).

Quit P-f-f-f-f-f-artin' around and get on with it.

"Ken Jacobs" "Phil Allison"

Read what you just wrote .

... Phil

It depends on the relative PSD of the signal and the noise. If the signal PSD is below the noise PSD, then narrowing the bandwidth stops helping, and you have to start relying on known correlation properties of the signal, e.g. for a direct-sequence spread-spectrum (DSSS), you have to know which PRBS polynomial is used and then look for the proper phase.

If the signal PSD is above the noise PSD in some bandwidth, then you can often use a sufficiently-slowly scanned PLL to lock to it. The maximum scan rate (df/dt) has to be well below (BW)**2, where BW is the PLL loop bandwidth. That means that the scan has to slow down quadratically as you reduce the bandwidth.

One of my favourite tricks for this, which I came up with about 30 years ago when I worked for a couple of years before grad school, is to build an integrating loop with a little bit of positive DC feedback wrapped around the integrator. That makes it sweep back and forth slowly until it gets within pull-in range, at which point the feedback becomes negative and the loop locks.

Cheers

Phil Hobbs

ed

No, you will be left with zero.

"Fred Bloggs" (Ken Jacobs)

No, you will be left with zero.

Care to explain what the OP means by

" .. rising and falling DC level at their common frequency" ???

.... Phil

well cos(2A)=3Dcos(A)^2-sin(A)^2=3D 1-2 x sin(A)^2 or sin(A)^2=3D(1-cos(2A)= )/

"Fred Bloggs" "Phil Allison" :

well cos(2A)=cos(A)^2-sin(A)^2= 1-2 x sin(A)^2 or sin(A)^2=(1-cos(2A))/

** So not "zero" as you wrongly posted.

... Phil

A))/

My original impression was that he was adding the two signals, obviously, otherwise why care about antiphase.

"Fred Bloggs" , "Phil Allison" <

My original impression was that he was adding the two signals,

obviously, otherwise why care about antiphase.

But not the tiniest clue.

As usual.

... Phil

gh

i)

i)sin(2u))

Hmm Thanks Phil, I don't think my question was very well formed. I was thinking of a single frequency. (I'm not sure what the PSD of a pure sine wave is, I guess it depends on how long I measure it.)

I'll have to think about it some more.

George H.

t -

If you're looking for a pure sine wave, then its PSD is a Dirac delta function(*), so you can always narrow the PLL bandwidth sufficiently to bring it up out of the noise. The limitation then is how long you're prepared to wait for lock acquisition.

Cheers

Phil Hobbs

(*) For SED's purists, yes of course its finite duration sets a minimum bandwidth--I dream of a signal source pure enough for that to be the actual limitation.