AM modulation and sidebands

hello,
An AM modulated signal is composed of an unmodulated carrier and two modulated sidebands.
I am struggling to understand how it works.
If I mix a 1KHz audio signal with an RF carrier, the carrier plus two 1KHz apart from it sidebands will occur. This is clear.

However if I have an unmodulated carrier and I manage to switch it on and off at 1KHz range, will two sidebands appear that are 1KHz apart from the carrier?

There is no mixing in the second case, just switching the carrier on/off.

Hi,

switching ON and OF is like mixing. You will see the sidebands.

Not only at 1kHz apart, but - because square wave includes overtones - you will also see 3kHz, 5kHz, 7kHz...

***
You can do an experiment: (even excel is good for this)
generate your carrier frequency. and generate a second "sideband frequency" (Carrier + 1kHz) then add both. What is the result?


Klaus

Thank you. This has been very helpful.

Let us say sin(w*t) is a carrier frequency. You want to modulate this with a low frequency signal sin(a*t). The modulated carrier will be sin(a*t)*sin(w*t) (we are ignoring some const factors for the sake of simplicity).

Now we know that 2*sin(a*t)sin(w*t)=cos((w+a)*t) + cos(w-a)*t)

You see that both sum and difference frequencies appear


The modulation function can be any function provided it has a Fourier decomposition. You will get so many frequencies....

Both situations are mixing. Both will produce sidebands, but may differ in how strong the sidebands are relative to the carrier in the overall output spectrum. The first case sounds like the carrier will be suppressed entirely, while in the second case the carrier will remain.

a mixer is very much like an on/off switch

I am not fully aware how it works, but I see that there are switching mixers out there, in contrast to the non-linear mixing process that occurs in diodes or semiconductors.

Mixing can be ambiguous unless the kind of 'mixing' involved is clearly stated or defined.

In AM theory, they often start everybody off with 'multiplication' ..... pure mathematical multiplication. So here, mixing would mean 'multiplying'. cos(w1.t)xcos(w2.t) ..... if you mathematically expand it out using the cos(A)cos(B) expansion, you end up with cos([w1+w2]t) + cos([w1-w2]t), which is mathematically two sinusoids, one of them with frequency w1+w2, and the other frequency at w1-w2. This isn't 'AM' as such. It is called DSB-SC....aka double sideband suppressed carrier. If w1 is the carrier frequency, then the carrier mathematically interacts with the message to yield the two 'sidebands'.

You only get 'AM' if you add some DC voltage or constant voltage to the message .... ie.. instead of cos(w2.t) being the message, you add some DC ...such as 1V DC.... to get a message signal of 1 + cos(w2.t). So if you do this, then the product between the carrier and the message (containing DC), will yield something similar to before.... ie two sidebands, but you also get the cos(w1t) in there as well. This results in a 'carrier' component between the two sidebands. This is called 'AM'.

But, even for DSB-SC, with apparently no visible carrier..... the original carrier signal is actually responsible for the two side-bands, as it is the whole interaction that puts those side-bands there.

Now, for a single carrier that is switched on and off at some frequency. This is multiplication as well, but it's multiplying a sinewave with a higher frequency square-wave, where the square wave has levels of +1 and -1. The result is a waveform that looks like time-domain DSB-SC, except the edges are vertical. However, the signal can turn into something that really looks like DSB-SC (when it goes through a transmission medium that has limited bandwidth).

On-off switching (OOK modulation) is a non-linear operation. As said it's the equivalent to 100% AM with a square wave modulation signal.

Thank you very much for the explanation!

You may want to take a look under "intermodulation products" also. You should find explanations there too.

Both you and me (#5) are guilty of oversimplification. Mixing should be considered as a convolution or faltung (original term) of two functions. This will appear as a simple product in the frequency domain (Fourier transform). When the two frequencies are far, this can be approximately considered as an envelope function.

If you just add a constant term (you add some DC ...such as 1V DC), this will simply appear as unmodulated carrier.

I believe (not so sure!) that the term mixing came up because the two functions are allowed to interact for some time (tau)- if the interaction time (tau) is zero, there will be no (AM) modulation.

The sidebands are a fact of life.

Please note that "convolution or faltung (original term)" and "simple product" are interchanged in the above post (#12).

I don't see what this mean. The two functions are needed at all time in order to form the output.

My first words in my previous post was - the kind of mixing we want to talk about needs to be defined. Clearly defined. So your statement about 'both of us' oversimplying things isn't right.

Mathematically, a product of two time domain functions is equivalent to a convolution in the frequency domain. However, there is also non-linear mixing, associated with non-linear devices - like diodes.

Mixing, for two functions f(t) and g(t) is defined as the integral (f(t-tau).g(tau)).d(tau)

The two signals are allowed to mix for time tau (kind of running average). Under certain conditions, this will become the envelope function (AM signal). If tau is zero, you will get the result mentioned in #8; sidebands plus unmodulated carrier.

Sorry; that is completely wrong.
The above-mentioned integral is not mixing f(t) with g(t), but their convolution. Is what we get if we apply f(t) to a filter whose impulse reponse is g(t), or conversely. Mixing is k*f(t)*g(t), possibly plus other terms that can be filered out.

Time and frequency domains must not be confused.
As it was said, mixing two signals is associated with obtaining a term that is the their multiplication in time domain. It corresponds to convolution in frequency domain.

Most welcome. I can also mention that when I wrote carrier switched 'on and off' should have really been saying sinusoidal-type carrier having a very small section of its waveform directly passed from input to output, while the next very small section gets passed through (but multiplied by -1), and the next tiny section after that gets directly passed through to output, and the following section gets multiplied by -1 again, and so on. The resulting waveform is basically a chopped up sine-wave with alternating portions inverted. This kind of waveform is the same as for a hypothetical (ideal) double-balanced mixer. Due to the perfectly vertical edges of this kind of hypothetical waveform, the expected bandwidth is very large (or infinite). However, for practical circuits, the bandwidth will be limited, so we can end up with a signal that looks like DSB-SC. And DSB-SC will basically have 2 sidebands only (and no visible carrier component). But also, due to some non-linearity in double-balanced mixers, we expect to see some other components other than the 2 sidebands, but the sideband components are expected to be dominant, and those other unwanted components may be filtered out.

Multiplying with +/- 1 is a possible mixing scenario, but different from the OOK modulation described in post #1.

OOK spectrum includes carrier frequency and multiple side bands at +/- the fundamental and odd harmonics of the modulation square wave, +/- 1 kHz, +/- 3 kHz, +/- 5 kHz and so on.

As previously stated , switching as well as multiplication in time domain (operation of an ideal amplitude modulator) is already a non-linear operation, both in terms of mathematics and electronics. But there's in fact an impact of using non-ideal diode mixers instead of ideal multipliers. You get additional higher order intermodulation products that would be avoided by an ideal modulator.

I expect the OP is well confused by now.

How about someone drawing a simple spectrum caused by switching a 1khz tone on and off?
(and mentioning the result will depend on what equipment is used and how good the filtering is
and what level the spectrum analyzer is resolving)
And maybe pointing out the internationally acceptable levels for transmission that exist?
I know at least 2 of you here are better qualified than me to do this.
Just KISS it.

Thanks for pointing this out.
But why are side bands and multiple side bands generated in that process.
Is it because of the "starting" and "ending" points of the carrier, that may not be at zero voltage at the point it is switched on/off?
I am telling this because the "chopping" proccess assimes an already running oscillator is chopped on/off. One cannot instantly switch on/off an oscillator by itself I guess.



It will be very interesting for all of us to see this.
You see how things considered obvious by many of us, are actually more complex than thought.
Hopefully few that have an understanding in depth, can guide others. That is why I like this forum!