I suspect both ideas are correct for a crystal radio, though I also think the more correct explanation is the sidebands are heterodyned with the RF carrier in the setup.

The arguments I make for the transmitter aren't nearly as strong for a crystal receiver, though they do work. You can also use a heterodyne mixer with a synthesized carrier to pull the audio out. It is typically used with suppressed carrier, but it will work on a conventional AM. The frequency must must be exact though. I have seen more than one AM receiver based on this principle.

 

Hello,

At the recieving side we have also the Direct-Conversion-Reciever:
http://en.wikipedia.org/wiki/Direct-conversion_receiver

This will have a local oscillator on (or very close to) the carrier of the AM signal.

Bertus

 

Ah! I had a hunch I had gone wrong somewhere in the linearity of the Fourier transformation. You are correct, it isn't applicable to any non-linear system. Thanks.

Also, I just thought about a case where the Fourier representation doesn't provide us with enough tools whereas the time-domain is more useful:
The simple demodulation circuit uses a germanium diode to "filter" the signal envelope and do away with the carrier frequency.
As far as I know, the non-linear diode can't be easily coded into a frequency-response system.
Am I right in this or both ways work here?

I think you are basically correct.

When you are dealing with a non-linear element and want to talk about it's behavior in the frequency domain, I think the normal way to go about it is to model it in the time domain and then see what effect it has one the spectra and then create a non-linear frequency domain block that is simply defined to behave that way. This is what we do with a mixer. We draw a box with a circuit in then simply define that the output is a signal that is the half the product of the input amplitudes at frequencies that are the sum and difference of the input frequencies. There is no transfer function (H(s) or H(jw)) that we multiply the Fourier transform of the input signal by to get the Fourier signal of the output.

So you would take your envelope detector and analyze it in the time-domain and then come up with a frequency-dome processing block that applies a specific rule to transform the input to the output. Once you have that block, you can then pretty easily fool yourself into thinking that you have a complete frequency-domain model of the system that is not reliant at all on the time-domain descriptions. That will be particularly easy to have happen when you are talking about a student that may or may not have had this process pointed out explicitly and may or may not have walked through the development even once.

 

I suspect both ideas are correct for a crystal radio, though I also think the more correct explanation is the sidebands are heterodyned with the RF carrier in the setup.

Perhaps this is just due to using the same term for two different things, but the sidebands don't exist until after the baseband is heterodyned with the RF carrier. Using the example that we have been working with most of the time, 1MHz is the carrier, 1.001MHz (and 0.999MHz) are the sidebands, and 1kHz is the baseband. At least that is the way I am trying to consistently use those terms.

If, somehow, I have the sidebands, then I don't have to heterodyne them with the carrier at all -- I merely have to add them. In fact, I don't want to heterodyne them with the carrier because then I will be generating signals at 2.001MHz and 1kHz (the sum and difference frequencies for 1000kHz and 1001kHz, so ignoring the lower sideband).

Note that two waveforms (say 1000kHz and 1001kHz sinusoids) constructively and destructively interfering with each other is NOT, in general, a nonlinear process. With few exceptions, it is linear. Consequently, adding the two produces NO content at 1kHz. Which is why the receiver HAS to use a nonlinear process to extract anything at that frequency.

 

Hello,

At the recieving side we have also the Direct-Conversion-Reciever:
http://en.wikipedia.org/wiki/Direct-conversion_receiver

This will have a local oscillator on (or very close to) the carrier of the AM signal.

Bertus

Actually that was what I was describing in the previous post.

<snip> You can also use a heterodyne mixer with a synthesized carrier to pull the audio out. It is typically used with suppressed carrier, but it will work on a conventional AM. The frequency must must be exact though. I have seen more than one AM receiver based on this principle.

 

In the early days of AM there was great controversy over the question of whether there were really sidebands. Some people contended that since it was just the amplitude of the carrier that was varied while its frequency remained constant, a high Q filter could tune the carrier (and reject static and other interference) without losing the modulated audio.

This notion was soon dispelled however, and by 1918 the telephone company (AT&T) had single sideband working on its lines.

See this early paper from the Bell Systems Technical Journal:

http://www.alcatel-lucent.com/bstj/vol02-1923/articles/bstj2-2-90.pdf

So, even though an AM signal consists of a carrier and sidebands, the easiest method to produce it (without using modern DSP methods) is to use circuitry that does in fact vary the amplitude of the carrier, which is a multiplicative (non-linear) process. It would be entirely possible to produce the desired sidebands by some other process and then add (rather than multiply) to the carrier, but this would be more complicated.

One could produce a square wave by adding large numbers of pure sine waves in the proper amplitude and phase together, but why bother with such a complicated method when much simpler methods are available? But the well known harmonic structure of a square wave must be present no matter how it's produced.

The time domain point of view and the frequency domain point of view are both useful aids to a full understanding of AM.

 

This came out of a thread in a different sub-forum, but can someone elaborate on what's the wrong (but pedagogically useful) model of AM modulation, and what's the right one?

This has got me wondering if I was taught the wrong model, and if I'm still in the dark about the right one...

M

I wouldn't say it's taught wrong....there are two very different models, however. When you actually do the number crunching, the answer is the same, whether you do vector summation of the sidebands or do a Fourier analysis of the envelope shape!

Where I have a real interesting time is in explaining what a diode detector does! The old analogy simply describes it as a rectifier, so that you "lop off" the redundant half of the amplitude. The "modern" translation treats the diode as a mixer which gives you the difference frequency between the sidebands and the carrier. BOTH analogies, again....are fully compatible!

Eric

 

I'm confused... why is it called FM and AM... whats the difference between the two?

 

I'm confused... why is it called FM and AM... whats the difference between the two?

Imagine I have a carrier waveform, c(t), at a frequency of 1MHz and an amplitude of 1V. So

c(t) = 1V*sin(wt) where w = 2pi(1MHz)

Now imagine I have a signal waveform, s(t), that I want to impress onto this carrier. The signal waveform might be the signal from a microphone.

In AM modulation, the signal that is applied to the antenna is the combination, v(t), of c(t) and s(t) such that, ideally,

v(t) = (1V+ßs(t))*sin(wt). (ß is the coefficient that translates signal voltage into an amplitude deviation).

So at any moment in time, the actual signal, v(t), is basically the value that the carrier would have had if the amplitude of the carrier were (1V+ßs(t)) instead of a fixed 1V. Now, in this example, s(t) can be both positive and negative, but it can never be less than -1V because we want the amplitude of the signal to always be positive.

For FM modulation, the signal that is applied to the antenna is the combination, v(t), of c(t) and s(t) such that, ideally,

v(t) = 1V*sin((w+ßs(t))t). (ß is the coefficient that translates signal voltage into a frequency deviation).

So at any moment in time, the actual signal, v(t), is basically the value that the carrier would have had if the frequency of the carrier were w+ßs(t) instead of a fixed w. Again, in this example, s(t) can be both positive and negative.

Perhaps a mental picture might help. Let's imagine two signals, s1(t) and s2(t), in which both are is constant (or very, very slowly changing) but s1(t) is different, say considerably smaller, than s2(t). Now, you might be thinking along the lines, for a mental picture, that s1(t) is the output of the microphone when someone is humming real softly and steadily while s2(t) is the output when they are humming real loudly and steadily. But it's important that you realize that these are NOT constant signals, even if the amplitudes in both cases are rock steady. If the humming is at a tone of 1hHz (and we'll assume our hummer can vocalize a pure, single tone somehow), then the signal is changing through a complete period from a maximum positive value, through zero, to a maximum negative value, back through zero and back to the maximum positive value a thousand times a second. This is NOT a constant or slowly changing signal.

Instead, think of the output signal coming from a thermometer that takes many minutes, at least, to produce a noticeable change in the output. So s1(t) might be coming from a thermometer in the shade while s2(t) might be coming from a thermometer in full sunlight. For simplicity, let's say that for the AM case our the combination of ß and s(t) is such that the amplitude deviation for the first thermometer is 0.25V while it is 0.5V for the second. If you were to look at the signal sent to the antenna from the first thermometer, you would see a 1MHz sine wave that had an amplitude of 1.25V while looking at the antenna for the second signal you would see a 1MHz sine wave that had an amplitude of 1.50V.

In the FM case, let's say that our combination of ß and s(t) is such that the frequency deviation for the first thermometer is 1kHz while it is 2kHz for the second. If you were to look at the signal sent to the antenna from the first thermometer, you would see a 1.001MHz sine wave that had an amplitude of 1V while looking at the antenna for the second signal you would see a 1.002MHz sine wave that had an amplitude of 1V.

As the temperature slowly changes, you would see the amplitude of the AM signal slowly changing while you would see the frequency of the FM signal slowly changing.

 

The simple answer is AM varies amplitude, FM varies frequency, which is why the initials, AM (Amplitude Modulation) and FM (Frequency Modulation). When you look at the sidebands it gets much more complex, but the beginning concepts are simple enough.

 

I wouldn't say it's taught wrong....there are two very different models, however. When you actually do the number crunching, the answer is the same, whether you do vector summation of the sidebands or do a Fourier analysis of the envelope shape!

Where I have a real interesting time is in explaining what a diode detector does! The old analogy simply describes it as a rectifier, so that you "lop off" the redundant half of the amplitude. The "modern" translation treats the diode as a mixer which gives you the difference frequency between the sidebands and the carrier. BOTH analogies, again....are fully compatible!

Eric

So how are you going to figure out the sidebands from an oscilloscope picture? This is one of the problems I have with it. I may be wrong, but I don't think you can. This is why I think using time domain on a oscope is flawed.

Even with sidebands, there are phase relationships that are not clear cut, at least to me.

Ultimately, you have to teach the sideband model for completeness, the oscope version leaves too many holes.

 

Both models have holes.

If I showed you two (appropriately scaled) scope captures, one with an AM signal and one with an FM signal, could you quickly identify which is the AM and which is the FM?

If I showed you spectragraphs instead, could you do so? I doubt I could, at least not without a lot of consideration. Could you even tell from looking at a spectragraph whether the signal is contant envelope or not? I can pretty much guarantee I couldn't, not without transforming it back into the time domain.

While either view, in theory, is complete, neither view is really adequate to tell us all the things we are generally interested in. In theory, yes, those things can be determined from either view, but in practice some things are trivial to determine in the time domain and extremely difficult to determine in the frequency domain, while other things are the reverse.

Then there are things that neither view captures well but that a hybrid view makes trivial to see. For instance, consider a frequency-hop spread spectrum system. By combining the frequency and time domains to produce a waterfall display, you can immediately see the essence of the signal and its primary behavior whereas this would be quite difficult in either domain alone.

 

So how are you going to figure out the sidebands from an oscilloscope picture? This is one of the problems I have with it. I may be wrong, but I don't think you can. This is why I think using time domain on a oscope is flawed.

Even with sidebands, there are phase relationships that are not clear cut, at least to me.

Ultimately, you have to teach the sideband model for completeness, the oscope version leaves too many holes.

Hi Bill:

NO argument there....you just have to remember that oscilloscopes were around a few decades before spectrum analyzers!

HOWEVER...with a narrowband receiver, it's easy to detect the individual sidebands of an A.M. signal. I can do that with my 1940 Hallicrafters Super Defiant!

Eric