# Does an AM radio wave make use of a lower side band?

I believe my problem is I am having some confusion on the meaning of "side bands". I read the wiki but I am wondering that since an AM signal is rectified doesn't that eliminate the lower part of the wave?

On the other hand what would the difference be between just calling the upper side band just amplitude modulation since after the rectification that is all you have left anyway.

Needless to say the terminology side band is confusing to me so maybe someone can explain in the context of amplitude modulation.

• Side bands are not the upper and lower parts of the waveform as displayed on an oscilloscope. Dec 29, 2022 at 23:16

We need to distinguish between different ways of looking at a signal. On an oscilloscope, a sine wave will look like this:

But if we look at that on a spectrum analyzer, it'll look something like this:

...a little noise (mostly from sampling and such) but mostly one sharp peak at the sine wave's frequency. If we synthesize data, run it through an FFT, and graph the input and output, the result is more "perfect"--a single spike at one precise frequency, with none of the noise:

Here the axis/waveform is green, and the FFT is that single, vertical black line.

When we amplitude modulate the sine wave, we mix the audio signal with the radio signal. In the process of mixing the audio signal with the carrier, we not only modulate the carrier, but also produce two sidebands: one at the carrier frequency + the audio frequency, and one at the carrier - the audio frequency.

So, we look at this on the spectrum analyzer, we get something like this:

The tallest spike at the center is the carrier frequency, and the two others are the side bands (the one to the left, at the lower frequency, is the lower side band, and the one to the right at the higher frequency is the upper side band).

If we synthesize a modulated wave form by simply generating 3 perfect sine waves at slightly different frequencies, and then run them through an FFT, we get a rather cleaner display:

Either way, what we have is three sine waves all being transmitted together--but each is still just a normal sine wave like that shown in the first picture above. The only thing unusual involved is that they're at three slightly different frequencies (and as we see in the amplitude plot, the result no longer looks like a sine wave).

Putting them through a diode won't change that basic fact. We'll just have three sine waves at different frequencies, but with all the negative (or all the positive, depending on the diode's orientation) removed. So each of the sine waves will look like this:

...but we'll still have the upper halves of the sine waves at all three frequencies, so on the spectrum analyzer it'll still look roughly the same as it did without the diode. It does change a little bit (because rectifying it obviously changes the shape of the waveform) but not drastically.

Oh, we could also use a full wave rectifier, giving an output roughly like this:

If we run a modulated wave form through a diode, then an FFT, we do see some artifacts from the rectification. For example, with three sine waves at relative frequencies of 0.7, 1.0, and 1.3, we get a result like this:

So at this point, we get a fundamental, two sidebands, and some artifacts from the rectification--the sudden change of direction where it's dropping to zero, then starts to rise again acts like some other frequency of signal for the duration of that transition (and the apparent frequency of an artifact will depend on the rate of change on each side).

Getting to the exact question you asked: it depends on what you mean by AM radio. If you mean the commercial AM broadcast band (540 kHz up to 1700 kHz), then the answer is that all signals there were and are broadcast with the carrier and both sidebands.

On the other hand, the carrier uses a lot of power, that carries essentially no information. If you use a narrow band-pass filter right after the modulator, you can filter out the carrier and one of the side bands. By transmitting only one side band, a given amount of total output power can result in considerably greater range. You can also optimize bandwidth usage, by allowing receivers to tune in only a lower side band or an upper side band at one time, so each can act as essentially a separate channel.

Although (as stated above) this hasn't been done in commercial AM radio, it has (for one example) been done in CB radios.

• Amplitude modulation preserves information about the phases of the components in the modulated signal, which is not preserved in SSB modulation. Further, an amplitude-modulated signal whose carrier frequency is known approximately will convey precise information about the frequencies of the modulated components. By contrast, when using SSB, any uncertainty of the carrier frequency or tuning reference will yield uncertainty in the frequencies of the signals that were modulated. Dec 31, 2022 at 2:03
• ..it took a while for me to read it and understand but I learned a lot from reading this. Thank you Dec 31, 2022 at 14:59
• "By contrast, when using SSB, any uncertainty of the carrier frequency or tuning reference will yield uncertainty in the frequencies of the signals that were modulated". Yes. Anyone who has operated an SSB ham radio of short wave receiver knows that if you don't have the LO set correctly, you get a horrendous squeal in the audio that decreases as the LO gets closes to the carrier frequency. Jan 5, 2023 at 22:50

since an AM signal is rectified

Yes, but it could be half wave or full-wave rectified.

Regardless, that has nothing to do with sidebands. An AM receiver uses both sidebands equally, regardless of whether it uses half-or full-wave rectification.

You are conflating two unrelated concepts:

• Instantaneous negative and positive voltages (with respect to 0 V) of a RF carrier in the time domain
• Lower and higher frequencies (with respect to the carrier frequency) in the frequency domain

Rectification only deals with the first one.

• so there is another signal riding on the same carrier wave maybe? Dec 29, 2022 at 22:57
• Whenever you modulate the carrier, sidebands are produced. This all works back into fundamental mathematics so you can predict these sidebands. Dec 29, 2022 at 23:06
• @Sedumjoy You are thinking backwards. We don't set out to create the sidebands. We start with a single frequency carrier, then we multiply this carrier with the audio signal (that's what AM means). If we do this, and then analyze the frequencies in the multiplied signal, we find that it has the carrier frequency and two symmetrical sidebands. That's just what the frequency spectrum of a multiplied signal happens to look like. It's not like each sideband has a purpose. The multiplication is what has a purpose. The sidebands are coincidental. Dec 30, 2022 at 18:06
• @Sedumjoy and then it is quite logical to ask "well, couldn't we do the same but with only one sideband, so we aren't using twice as many frequencies as we need to?" and the answer is yes, but if you do that then it's not AM any more - it's something more complicated, and it's more complicated than simple multiplication. Dec 30, 2022 at 18:08
• @user253751 I am glad you clarified this. Excellent. Some of the "how to" manuals don't really explain this very well, so this helps me a lot.thank you Dec 30, 2022 at 18:33

RussellH has given a very fine answer, showing the math.

Shown below is a LTspice simulation showing a 100,000 Hz carrier modulated by a 1000 Hz sine wave.

• Green waveform shows 100% AM modulation, with both sidebands present along with carrier.
• Purple waveform shows the same with lower sideband missing (full carrier with USB -6dBc)

The simulation is imperfect. Had it been perfect, the purple waveform would have reached |0.5| where the green waveform reached |1.0V|.
And where the green waveform dropped to |0V|, purple waveform would have bottomed out at |0.25V|

Where only one sideband is present, it appears similar to 50% modulation, versus 100% modulation of the green waveform.

Below, the 1kHz modulator has been modified so that carrier at 100kHz is suppressed:

• V(in) waveform in green has LSB at 99 kHz, and USB at 101 kHz. NO CARRIER at 100 kHz. Both USB, LSB have same amplitude.
• V(out) waveform in purple is USB (101kHz), with LSB (99 kHz attenuated.

This last method was commonly used to generate single-sideband for radio transmission. You can use a double-balanced mixer, or a four-quadrant multiplier for the modulator (both can suppress carrier). The RF filter to pass one sideband while rejecting the other sideband usually required a multipole crystal filter.

Station CHU actually transmits the top waveform (full carrier, only one sideband) at a few different frequencies: 3.33 MHz, 7.85 MHz and 14.67 MHz. Only one sideband (upper sideband) is transmitted, while lower sideband is attenuated.

• Since the full carrier is also transmitted, an AM receiver can properly demodulate this waveform.
• A receiver set for single-sideband (upper sideband) can also receive CHU properly.
• A receiver set for single-sideband (lower sideband) hears nothing.

If CHU were to transmit both sidebands, an AM receiver's audio would be twice amplitude.

• These presentations are exteremly well written and I am grateful to have them for reading. Dec 30, 2022 at 18:28
• How does the modulation depth affect the existence of the sidebands? Referring to RusselH's last equation, changing Am does not eliminate one cosine term and keep the other. I thought single sideband modulation required quadrature modulation/demodulation. Jan 5, 2023 at 17:03
• @DavidG25 Hope I haven't misled you - the AM-with-LSB-missing described here is NOT upper-sideband-SSB. In proper SSB, no carrier is transmitted - it is artificially reproduced in the receiver. So "modulation depth" doesn't apply to true SSB. Quadrature sampling is very often used in a digital SSB detector, but additional phase shifting is also needed if you want to reject the unwanted sideband while enhancing the wanted sideband. The amplitude-independent phase shift consumes considerable software resources. Analog phase shifters require lots of circuits too, especially for high bandwidth Jan 5, 2023 at 17:37
• Wikipedia says there is SSB and suppressed carrier SSB—seems that the presence of a carrier is not the determining factor of whether an AM technique is SSB or not. Regardless, how does the purple trace have USB only? Can you add spectrum plots for both the green and purple waveforms to show that the LSB is missing in the purple one? Jan 5, 2023 at 18:01
• @DavidG25 I did verify that LSB was attenuated about 40 dB, while USB was 6dB below carrier amplitude (by using LTspice FFT tool). Also added waveforms showing suppressed-carrier modulator, with LSB attenuated - not surprising, it's output is a constant amplitude sinewave at 101 kHz. Jan 5, 2023 at 21:42

The sidebands are frequency domain regions on either side of the carrier frequency, $$\f_{c}\$$. Each sideband carries the same information of the modulating signal, call it $$\m(t)\$$ which could be audio from a microphone. The carrier signal contains no information apart from its own frequency.

If we let a sinusoid represent the audio, then $$m(t)=A_m\cos(2\pi f_mt )$$ The carrier is written as: $$c(t)=A_c\cos(2\pi f_ct )$$

By setting the amplitude of the carrier equal to the modulating signal and adding in the unmodulated carries

The AM signal is then $$f_{AM}(t)=c(t)+m(t)\cos(2\pi f_ct )$$

The second term contains both the upper and lower sidebands.

$$m(t)\cos(2\pi f_ct )=A_m\cos(2\pi f_mt )\cos(2\pi f_ct )$$

By using the trigonometric identity: $$\\cos(u)cos(v)=\frac{1}{2}cos(u-v)+\frac{1}{2}cos(u+v)\$$

$$m(t)\cos(2\pi f_ct )=\frac{A_m}{2}\left(\cos(2\pi (f_c-f_m)t )+\cos(2\pi (f_c+f_m)t )\right)$$

This demonstrates the trigonometry of sidebands. Here is a link for more explanation.

• ..this is elegant ...thank and the link is wonderful. Kudos. Dec 30, 2022 at 15:16

In AM modulation, we start with a carrier wave which is a pure sine wave, and then we multiply the carrier wave with the audio signal.

When we analyze the frequencies in the carrier wave, using a spectrum analyzer (which can be analog, or a computer algorithm called FFT) we find that it has only one frequency which is the carrier frequency - let's say 800 kHz. No big surprises.

If we multiply this with a 1 kHz audio wave, we get another wave. If we analyze the frequencies in this wave, we find a lot of 800 kHz, but also a little bit of the frequencies 799 kHz and 801 kHz. Those are the sidebands.

In reality, the audio wave isn't just a sine wave, either. If we multiply our carrier wave with an audio wave that has a mix of frequencies 0-5kHz, and then analyze the frequencies in this multiplied wave, we find a lot of 800 kHz, and also a mixture of frequencies 795-800kHz and a mixture of frequencies 800-805kHz. Those are the sidebands.

Sidebands are something you discover once you analyze the frequency spectrum of your modulated signal, with a spectrum analyzer. They aren't the point, they are just a side effect of the multiplication which does the modulation. You might naively think that if you AM-modulate an 800kHz carrier and then analyze the frequencies, you only see 800kHz, but it turns out that in reality it has these sidebands. It can be proven mathematically, but I assume you don't need to see that.

It turns out the frequency spectrum of a signal is interchangeable with the signal itself - but in many cases, it's a lot easier to analyze and calculate things with the frequency spectrum. So much easier, that a lot of engineers only think about the frequency spectrum. They talk about things like sidebands and assume you are also thinking about the frequency spectrum, but you are thinking about the signal itself (the so-called "time domain") so what they are saying makes no sense to you.

With that in mind, sidebands are a frequency thing and rectification is a time domain thing. They are completely separate, they are not really related at all. Of course there's nothing stopping you from rectifying a signal and analyzing its frequency spectrum, but it's hard to predict exactly what that will do to the frequency spectrum. It certainly won't just cut off the frequencies below 800kHz.

There's also nothing stopping you from using a high-pass filter to cut off the frequencies below 800kHz, and then seeing what the signal looks like in the time domain. It certainly won't look anything like a rectified signal, and it probably won't look like AM either.

You might think that AM is wasteful since it uses twice the frequencies it needs to, and design a modulation system which only uses one sideband. In fact that does exist, and it's called (drumroll) single sideband modulation. But it's not AM, and it's more complicated than just multiplying. The advantage of AM is that it's really simple to use, not that it's efficient with frequencies. You barely need any signal processing knowledge to use AM, because on one side it's just multiplication, and on the other hand it's just rectification, really simple. If you use single-sideband, it's more complicated.

• @ user253751 this stackexchange site is wonderful. This is so helpful. I wish it was presented this way right up front in the "how to" manual. Dec 31, 2022 at 14:03
• I think “If we multiply … find a lot of 800 kHz” is not correct. There will only be 799 and 801 unless you add the carrier back in. Jan 5, 2023 at 18:46
• @DavidG25 AM modulation is done by offsetting the audio signal so it doesn't go below 0 which is the same as adding a bunch of carrier in. Jan 5, 2023 at 18:49