Author Topic: Help understanding AM  (Read 3535 times)

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Offline Dan MoosTopic starter

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Help understanding AM
« on: December 05, 2016, 02:35:16 am »
Ok, hopefully this isn't a breach of forum etiquette. I posted this in the RF section, but got crickets. What follows is a direct cut and paste of my original post.

Ok, this is a pretty noob question.

I am a long time electronics hobbyist, but just dipping my feet into radio. Something has always confused me about AM. All the texts I see show the positive and negative halves of the modulated signal as being 180 degrees out of phase with each other. This always seemed strange, since to my mind, it would seem that basically modulating the DC operating point of a sine wave would result in having the top and bottom halves be in phase.

Ok, so I built  (and Spice simmed) a circuit to investigate. I made a simple, low gain BJT common emitter stage, with the carrier injected at the base. I used two resistors from emitter to ground, and injected my modulating signal through a coupling  cap to the junction between   the two resistors. Thus, instead of the emitter being grounded, it was modulated by my signal. AM, right?

I got the result my thinking expected, instead of the text book signal. My new modulated signal's top and bottom halves were in phase. What's more, it seemes to me that that should still be a valid way to do AM. Since the detector dumps half of it anyway, phase between the two halves should be irrelevant, right?

And so I experimented. I tuned a transistor radio to my carrier frequency (about 600kHz), and was able to hear a clean tone. Carrier is from an opamp based relaxation oscillator. Modulation from my sig gen at 1k.

So what am I missing? Is the textbook version wrong? I doubt it, as its what you see in virtually every text I read. Or am I doing AM wrong? Also seems odd, since, by experimentation, i found my way to get a proper result.

Any thoughts?
 

Online helius

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Re: Help understanding AM
« Reply #1 on: December 05, 2016, 02:51:10 am »
"180 degrees out of phase" is equivalent to inverted polarity for primarily sinusoidal signals. You can't have any phase difference between "positive and negative halves" of a signal: that's mixing up the time axis with the f(t) axis.

AM is not "modulating the DC operating point". It is modulating the amplitude, which is most often symbolized by a direct multiplication of the carrier with the signal. By superimposing your signal with a capacitor, you've essentially done adding instead of multiplying. The results appear similar for SMALL modulation indexes. When you make the modulation index LARGE they are very different and I doubt you will cleanly recover the signal anymore.
 

Offline CatalinaWOW

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Re: Help understanding AM
« Reply #2 on: December 05, 2016, 02:55:51 am »
What you did is not amplitude modulation.  Amplitude modulation mathematically involves a product of the modulating signal with the carrier.  Something like (1+a*sin(modulation signal))*sin(carrier signal).  a is the fraction modulation.    The multiplication is why the upper and lower halves are in phase. What you have done is a summation of the two signals.

If you looked at your signal in the frequency domain you would see one spike at the modulation frequency and another at the carrier frequency.  An AM modulation signal would show a spike at the carrier frequency, with two other spikes just above and below the carrier, spaced off by the modulation frequency.

Your receiver may have produced a response to this signal for a number of reasons.  The most likely is that a non-linearity in the input stages actually produced AM modulation which then worked in later stages.
 

Offline Dan MoosTopic starter

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Re: Help understanding AM
« Reply #3 on: December 05, 2016, 03:15:55 am »
Ok, I sort of get it now. I'm basically taking a carrier of constant P to P amplitude, and modulating the center point, rather than actually modulating the amplitude.

Two questions. First, why isn't this a valid method? If you draw a horizontal line  at the average mid point of my  modulated signal, the two halves individually look like a demodulated, but yet to be filtered signal. I realize it's not the correct way, but the end result seems right. To my eyes, a detector should get the same result as real AM.

Second, correct me if I'm wrong, but instead of modulating the operating point, I want some sort of voltage controlled amplifier, right?
 

Offline hamster_nz

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Re: Help understanding AM
« Reply #4 on: December 05, 2016, 04:06:26 am »
I'm doing some thinking and it all comes down to this:

  cos(x)/2+cos(-x)/2 = cos(x)

and

  sin(x)/2+sin(-x)/2 = 0

Because your modulation signal lacks the quadrature information it is acting as half the sine of  positive frequency summed with half the sine of negative frequency, which gives a constant quadrature value of 0.

When you scale (multiply) your carrier by the modulation signal you end up with the final three signals in the output.

- a proportion of the carrier speed up by f_modulation
- an equal proportion of the carrier slowed down by f_modulation
- the reminder being the rest of the carrier

The balance between the three depends on the modulation depth, at 100% modulation there will be no carrier left in the final signal, at 0% modulation it will be all carrier :)
 
« Last Edit: December 05, 2016, 04:14:57 am by hamster_nz »
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Offline CatalinaWOW

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Re: Help understanding AM
« Reply #5 on: December 05, 2016, 04:19:21 am »
Ok, I sort of get it now. I'm basically taking a carrier of constant P to P amplitude, and modulating the center point, rather than actually modulating the amplitude.

Two questions. First, why isn't this a valid method? If you draw a horizontal line  at the average mid point of my  modulated signal, the two halves individually look like a demodulated, but yet to be filtered signal. I realize it's not the correct way, but the end result seems right. To my eyes, a detector should get the same result as real AM.


You could use this.  Your radio accidentally works, and you could design a detector which would capture this modulation on purpose.  But.

In a radio system a high pass filter is used to eliminate the noise from DC to just below the carrier frequency. And another to get rid of higher frequency noise.   This would eliminate the modulation you are using.  If you don't use this filter the signal to noise ratio will be far lower.  The noise is proportional to the square root of the bandwidth.  In your example of a 600 kHz carrier with a 1 kHz modulation you could get good results with a filter of just over 2 kHz bandwidth when using AM modulation.  Using your modulation the bandwidth would be just over 600 kHz and would capture roughly 17 times more noise.   Not a problem crossing a workbench, but for long range work a no go.


Second, correct me if I'm wrong, but instead of modulating the operating point, I want some sort of voltage controlled amplifier, right?

That is certainly one way to achieve the desired effect.  Once you feel like you understand this method you can move on to other ways of getting there.  Don't feel bad about not understanding them all at first.  A lot of really bright people spent the first three decades or more of radio figuring this stuff out. 

 

Offline Dan MoosTopic starter

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Re: Help understanding AM
« Reply #6 on: December 05, 2016, 04:50:30 am »
Thanks for the help! The net seems to be devoid of well explained fundamental rf stuff. It 's particularly disappointing that the Art of Electronics says so little on the matter. Strange gap in an otherwise great text if you ask me.

My goal is to build a decent receiver that doesn't depend on modern digital stuff. Towards that goal, I figured I needed a weak transmitter to test it with. I am close to purchasing a better signal generator (siglent dsg2042x) that will do AM, but even then, I want to roll my own for the learning experience.

What would be a good method to amplitude modulate a signal that doesn't require more than jelly bean parts?
 

Offline jh15

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Re: Help understanding AM
« Reply #7 on: December 05, 2016, 05:10:47 am »
Decent receiver? germanium diode or galena cat whisker, coils on oatmeal box, black high impedance headphones.
Tek 575 curve trcr top shape, Tek 535, Tek 465. Tek 545 Hickok clone, Tesla Model S,  Ohio Scientific c24P SBC, c-64's from club days, Giant electric bicycle, Rigol stuff, Heathkit AR-15's. Heathkit ET- 3400a trainer&interface. Starlink pizza.
 

Offline Dan MoosTopic starter

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Re: Help understanding AM
« Reply #8 on: December 05, 2016, 05:40:50 am »
Ha! No, I am going for a a super het type.
 

Offline AG6QR

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Re: Help understanding AM
« Reply #9 on: December 05, 2016, 06:18:18 am »
What would be a good method to amplitude modulate a signal that doesn't require more than jelly bean parts?

Simplest: Take one of those standard four-pin crystal oscillators in a can, and instead of feeding it steady DC power, feed it a varying DC signal.

The result will be low-power, crude, and full of harmonics (use a low-pass filter to clean them up if that's important to you, and please don't try to transmit the signal at any significant power), but it can be detected nearby with a simple AM radio tuned to the frequency of the crystal.


As an alternative, use an ordinary signal generator, and modulate its output by combining it with the modulating signal using an Analog Devices AD633 analog multiplier IC.  The data sheet has an example circuit to do linear amplitude modulation.  That's probably more expensive, and it requires that you have some sort of RF oscillator already, but you aren't restricted to using a fixed frequency crystal controlled carrier.
 

Offline DimitriP

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Re: Help understanding AM
« Reply #10 on: December 05, 2016, 08:43:55 am »
Quote
It 's particularly disappointing that the Art of Electronics says so little on the matter.

Time for another couple of books:
ARRL Radio Handbook and Experimental methods in RF design

As far as designing with discrete components go take a look here: http://www.n5ese.com/2n2222.htm




   If three 100  Ohm resistors are connected in parallel, and in series with a 200 Ohm resistor, how many resistors do you have? 
 

Offline C

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Re: Help understanding AM
« Reply #11 on: December 05, 2016, 02:18:27 pm »
One old method to create AM is to use an audio transformers secondary as power source to RF amplifier.

Connect audio transformer secondary as follows,
One connection to DC Level.
Second connection to RF amp's DC source.
You now have a DC level that changes +-audio level.
100% modulation is when the audio input level is such that the lowest DC voltage peak just turns off RF output.

One trick I have seen is using a second secondary winding of audio transformer as a power source to boost to the DC source. The result is a carrier of x watts with no modulation with transmitter power increasing to a higher wattage output at 100%.

 

Offline radiogeek381

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Re: Help understanding AM
« Reply #12 on: December 06, 2016, 01:42:53 pm »

The reason your radio detected an AM signal is that you actually produced
a very slightly modulated AM signal, in addition to adding the low frequency
offset.

If I understand your circuit correctly, you have two resistors in series between
the emitter and ground.  You are driving the node between the two resistors
with your AF generator.  We'll call this node E1, and the emitter E. (Base
and collector will be B and C).

If the AF generator impedance is low WRT the emitter resistors, then E1 will
wiggle up and down.  But the voltage at B will, largely, be affected only by
the RF generator, as its impedance is surely lower than the effective impedance
looking into the base terminal.  That means that the voltage between B and
E1 will be modulated by the AF generator.  This voltage determines the base-emitter
current and, because it is a BJT, the collector-emitter current. 

Another way of saying this is, if you raise the voltage at E1 very slowly (relative
to the signal at the base) then you move the operating point of the transistor
and thus reduce the gain. 

If you were to take the output of your "modulating amplifier" and pass it through a
.01uF cap that is then shunted on the output by a 1K resistor, you'll remove most
of the low frequency "DC" offset from the output and you might even see a
(small?) amount of modulation on a carefully synchronized oscilloscope.  This will
look like the textbook AM envelope.

 

Offline TimFox

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Re: Help understanding AM
« Reply #13 on: December 06, 2016, 09:16:25 pm »
It is important to note that the audio signal that is multiplied by the carrier sinusoidal waveform is of the form M = 1 + A(t) where A(t) > -1 for all time.  Therefore, M > 0 for all time.  The simplest modulation would be an audio tone A(t) = 1 x cos (2 pi f t), where f is an audio frequency.  When A(t) = 0 (between words), the RF output is an unmodulated carrier, which is otherwise a waste of power but is needed to demodulate the AM signal in a simple detector.
 


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