# Amplitude modulation

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#### GW8LJJ

##### New Member
GW8LJJ
1/1/09

Gentlemen.
An RF carrier has an amplitude of say, 50 watts (output power). If it is modulated with an AM signal of say, 100% what will the power output be (at peaks) and why?

It is argued that it is 400 watts PEP. If the am receiver chops off the lower sideband during detection, is it then considered to be 200 watts transmitted power?

How does it increase the carrier strength so that it is greater than 50 watts on peaks?

A few questions there but all related. I would appreciate a 'plain language' answer. Thanks, regards, Happy new year, Eric.

#### audioguru

##### Well-Known Member
If the unmodulated carrier is 50W, then its voltage and its current are doubled at the peaks of full modulation. Doubling the voltage and the current causes the instantaneous output power to be 4 times as high. The nulls of a fully modulated AM transmitter are at zero W.

#### Nigel Goodwin

##### Super Moderator
GW8LJJ
1/1/09

Gentlemen.
An RF carrier has an amplitude of say, 50 watts (output power). If it is modulated with an AM signal of say, 100% what will the power output be (at peaks) and why?

It is argued that it is 400 watts PEP. If the am receiver chops off the lower sideband during detection, is it then considered to be 200 watts transmitted power?

How does it increase the carrier strength so that it is greater than 50 watts on peaks?

A few questions there but all related. I would appreciate a 'plain language' answer. Thanks, regards, Happy new year, Eric.
'Plain answer' - you don't measure AM in PEP - it doesn't apply. An AM radio also doesn't chop off one sideband, you seem to be mixing your modulations.

If I remember back long ago to my RAE, 150W DC input at AM is the equivalent of 400W PEP SSB.

#### GW8LJJ

##### New Member
Hi,

Yes I understand the theory. However, I would like to know how that is the case (in prtactice). How does the RF increase in power 4 fold? what creates this increase?

If the power output is only capable of providing say 50 watts of carrier where does this extra power come from?

Is the carrier measured as RMS or peak or pek to peak?

Regards, Eric.

#### Nigel Goodwin

##### Super Moderator
Hi,

Yes I understand the theory. However, I would like to know how that is the case (in prtactice). How does the RF increase in power 4 fold? what creates this increase?

If the power output is only capable of providing say 50 watts of carrier where does this extra power come from?

Is the carrier measured as RMS or peak or pek to peak?

Regards, Eric.
Like I said, it's measured as DC input to the output stage - it's a question of what can be measured easily and sensibly.

Draw yourself a diagram, 10MHz, 50W carrier, with AM modulation of 50% at 1kHz sinewave - you get three peaks on the spectrum - what frequencies are they at, and what amplitude are they?.

You took the RAE back when I did (G8MMV), have you forgotten it all?

#### GW8LJJ

##### New Member
Ha! yes your are right. I did but not forgotten the theory.

I need to be able to pass on this information in an understandable way.

I shall leave it now until tomorrow to come back with a proper reply to you.

Eric

#### JimB

##### Super Moderator
Lets try this explanation:

Assume a transmitter giving 50w RF output, if we look at the output using an oscilloscope we see a sinewave output at the carrier frequency.
If we slow down the scope timebase we will see an "envelope" containing this sinewave.
The amplitude of the envelope is 50 volts, assuming a 50Ω load on the transmitter.
Power = voltage²/resistance
Power = 50 x50 / 50 = 50 watts

If we amplitude modulate the carrier to 100%, we will see that the carrier envelope is now varying between 0 and 100 volts.

We know that the power in a circuit is proportional to the square of the voltage across that circuit.
So now at the modulation peaks the voltage is 100volts.

Power = voltage²/resistance
Power = 100 x100 / 50 = 200 watts

Be aware that this is the Peak Power, the Peak Envelope Power.

The average power, (averaged over 1 cycle of the modulation waveform) will be 75watts.

Where does this extra power come from?
In a classic anode modulated transmitter, the power comes from the modulator.
In more recent AM transmitter, where the modulation is done at low level, the extra power comes from the transmitter PA stage which must be running in a linear mode, and have sufficient capacity to provide the 75w peak power, not running flat out at 50watts.

JimB (GM3ZMA)

#### GW8LJJ

##### New Member
Gentlemen of the air!
Thank you all for the fast response.

Nigel: Yes I remember my RAE days of the 1970s and I had just finished my apprenticeship ten years earlier as a radio and T/V engineer so was quite well up on the theory then! I still engage in it and a very active constructor of radio gear (mainly valves!)

The waveform on the scope shows a voltage representation of the modulated waveform (many signals superimposed) so I can see the modulation waveform created by the 'audio' signal. I am also aware that the carrier itself does not vary only the sidebands and these are created by the audio. I can also see that the waveform varies, when modulated at 100%, by a double amount,as you say Jim. I understand all that ... BUT.... what I had forgotten, and you guys have now cleared it up, is that I am looking on the scope and spectrum analyser, at the VOLTAGE increase. As you say it doubles (on peaks) and as this is 6dbs, then at power ratings 6dbs is X4. Yes Jim, the formulae Power = V²/R. so it is X4

I realse that this is peak power and the average, of course, would be lower.

So the questions have ben answered. Of course this increase, as you say comes from the modulator (at high level mod) will only apply with say plate/screen as any other mod will only reach carrier output (grid, controlled carrier etc).

Well, many thanks for that, and I can now confidently pass on that information with a full understanding of how it works.

Thanks to Audioguro, I understand what you say about doubling the voltage and current (power) provides the X4.

And thanks to Nigel and JimB

73's Eric GW8LJJ

#### Nigel Goodwin

##### Super Moderator
what I had forgotten, and you guys have now cleared it up, is that I am looking on the scope and spectrum analyser, at the VOLTAGE increase.
Bear in mind, this is ONLY for high level (anode) modulation - modern SSB based AM don't increase the supply voltage, nor did cathode modulation methods etc.

#### GW8LJJ

##### New Member
No, that's right I understand that.
As I made the comment on my last reply. It does not happen when grid mod is used or grid controlled carrier (screen grid mod).

Also, when I said after the signal is rectified (detected), one half is chopped off. What I meant by that is the lower sideband is not required (as it contains the same information) and by detection it is removed (or the other sideband if the diode is reversed). I now know however, that this is no longer relavent. I was assuming, wrongly, voltage amplitude and not power which is the square of the voltage.

Anyway, all good fun and it's surprising how the basics are sometimes overlooked but taken for granted.

Thanks again, 73's Eric

#### audioguru

##### Well-Known Member
A diode detector rectifies the RF waveform and has nothing to do with a sideband.
The rectification produces a varying DC voltage that increases when the amplitude of the AM signal increases when it has a peak from modulation and decreases when the amplitude of the AM signal decreases when it has a null from modulation. The detector detects and smooths the envelope of the rectified waveform.

An upper sideband is at an upper frequency, it is not the upper half of the waveform seen on a scope.

The polarity of the diode detector cannot select an upper or a lower sideband. The sideband is selected with a frequency filter.

##### Well-Known Member
Audioguru is correct. I have heard academic theoreticians explain this another way, like this. One can also characterize the diode detector as a mixer followed by a low pass filter. The mixer mixes the incoming carrier plus two sidebands and generates sum and difference frequencies. the sum frequencies are filtered out by the low pass filter. In the case of the difference frequencies, the carrier subtracts from itself and is zeroed out, while the low side and high side sidebands (whose magnitudes are exactly the same) both exist at the same time at baseband and sum together to form the audio.

Because the mixer is acting on the incoming carrier and sidebands, you can imagine that it doesn't work anymore if you take away the carrier only. This is a way of demonstrating that the simple diode demodulator doesn't work for double sideband with no carrier, nor for single sideband. In these cases, the BFO is used to inject a carrier to make the mixer work once again.

Strangely, I have met many university students who understand this explanation much more readily than the practical one repeated by audioguru. Its a disturbing symptom of the theoretician who has studied the science thoroughly but would not know a resistor if he met one in the street, so to speak.

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#### audioguru

##### Well-Known Member
I didn't memorize radio theory. I just use common sense.

#### ericgibbs

##### Well-Known Member
I didn't memorize radio theory. I just use common sense.
Nicely put, I like it.

#### Hero999

##### Banned
I don't fully understand Ron's explanation but I've known how diode detection works since the age of 10.

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