DSB and SSB Radio: Explaining the Confusion

[Averagesupernova]

So how is 25 watts a quarter of 150 watts? THAT is my point.

 

[TurtleMeister]

So how is 25 watts a quarter of 150 watts? THAT is my point.

It's not a quarter of the total power, it's a quarter of the carrier power.

 

[Averagesupernova]

Maybe so, I am not above a mistake in that area. Where I AM above a mistake is this quote: This means that each sideband is just a quarter of the total power. In other words for a transmitter with a 100 watt carrier, the total sideband power would be 50 watts and each individual sideband would be 25 watts.
TOTAL power and CARRIER power are not the same thing. For the record I would not believe much off a site that gets that wrong.

 

[TurtleMeister]

Maybe so, I am not above a mistake in that area. Where I AM above a mistake is this quote: This means that each sideband is just a quarter of the total power. In other words for a transmitter with a 100 watt carrier, the total sideband power would be 50 watts and each individual sideband would be 25 watts.
TOTAL power and CARRIER power are not the same thing. For the record I would not believe much off a site that gets that wrong.

Wow, that is an error. Not mine, but from the web site. I didn't even notice that. It should read "This means that each sideband is just a quarter of the carrier power. In other words for a transmitter with a 100 watt carrier, the total sideband power would be 50 watts and each individual sideband would be 25 watts." Apologies, I see why you were confused.

 

[Averagesupernova]

I can't take the time to do the math right now but I was REALLY CERTAIN that the total power doubles from 0% to 100%. The link does not support that no matter how you look at it. I will check in here later.

 

[TurtleMeister]

I'm pretty sure it's right except for that one thing. But like you, I don't have much time right now. I will check it more carefully later.

 

[jim hardy]

It was 1962-63 i studied AM radio and my RAM is increasingly volatile with the passing years.

I think my earlier mistake came from my partially remembering receiving circuits, specifically vacuum tube mixers (6BE6)

but the thread doesn't seem to have been hurt .

Thanks guys

old jim

 

[Merlin3189]

I see you guys have been having fun while I was away.
And I think you have finally got it sorted. I'll have to join Jim in the "Ooops! There I go again." club.
Even after TurtleMeister pointed out my error, and I did think about it again, I still got it wrong. I might plead, I was mislead by his first quote, but I think my error was really in adding the two sideband amplitudes before squaring it. I should have worked out the power in each sideband, by squaring the amplitude, before I added them.
Then you do get the power ratio carrier:usb:lsb = $1^2:(\frac{1}{2})^2:(\frac{1}{2})^2\ =\ 1:0.25:0.25\ =\ 4:1:1$
and carrier power : sideband power = 4:2 , total power : carrier power = 6:4 , total power : total sb power = 6:2 , etc.

I just hope Weightofananvil is not led to think this is a difficult topic, but rather is encouraged to see that he is not alone. Even people who have been playing with this stuff for half a century can get confused at times.

 

[Averagesupernova]

http://www.careerride.com/view.aspx?id=21847
-
The link tells me that 100% modulation gets you double the total power compared to no modulation. So 1/4 of the total power at 100% modulation is in each sideband. A quick back of the envelope drawing tells me the same thing. Typically the area under the curve will give you the average power. Hasn't failed me yet anyway. Imagine a 100 watt load into 50 ohms. Turns out to be 70.7 Vrms. Apply the modulation and the voltage goes to 141.4 Vrms at the peak. Draw the waveform for the power. The line drawn will be all positive of course peaking at 400 and then just touching zero. Average out the area and you get 200.
-Edit:
Hmmmm. Forgot to divide by 2. Power will only go to 1.5 times. Must be overlooking something in the quick drawing I made too.

 

[Merlin3189]

What I think you may be forgetting, is the change in wave shape.
The 100W carrier into 50Ω is a simple sinewave and your calculations for power and V_RMS can apply.
When you modulate it, the peak voltage does double, but the waveform is no longer a simple sinewave and you can't do the same calculations for V_RMS and power.
You may be stuck with resolving the waveform into three pure sinewaves (carrier, lsb and usb) to which you can separately apply your calculations of power, then adding the results. (Though I wouldn't put it past these maths wallahs to integrate the combined function over a cycle of the modulation and calculate it directly. Maybe something like, $P=\frac{1}{T_m}\int _0 ^{T_m} \frac{V^2_A_M(t)}{R} \, dt$ with V_AM(t) =sin(ω_ct )(1 + Msin(ω_mt)) , T_m= 2π/ω_m and M = modulation index. )
That probably makes no sense, 1 because LaTex doesn't seem to be working, 2 my maths is getting v.poor.

 

[Averagesupernova]

I was assuming that the envelope that is created in the power graph is a solid line but of course it is not. A quick program with a bunch of samples would tell the story on that. I seem to recall that sidebands are 6 db down on the spectrum from the carrier which would indicate what we seem to have all agreed on now. For some reason I seem to recall the power meter doubling upon applying modulation. But in recent years I have learned not to trust what I think I remember.

 

[sophiecentaur]

http://www.careerride.com/view.aspx?id=21847
-
The link tells me that 100% modulation gets you double the total power compared to no modulation. So 1/4 of the total power at 100% modulation is in each sideband. A quick back of the envelope drawing tells me the same thing. Typically the area under the curve will give you the average power. Hasn't failed me yet anyway. Imagine a 100 watt load into 50 ohms. Turns out to be 70.7 Vrms. Apply the modulation and the voltage goes to 141.4 Vrms at the peak. Draw the waveform for the power. The line drawn will be all positive of course peaking at 400 and then just touching zero. Average out the area and you get 200.
-Edit:
Hmmmm. Forgot to divide by 2. Power will only go to 1.5 times. Must be overlooking something in the quick drawing I made too.

Yep
1 unit of carrier power plus two units of quarter carrier power = 1.5 times unmodulated carrier power. Which is why SSB is a very neighbourly modulating system to use in a crowded spectrum. The interference put out is 0.25/1.5, which is 17% of that of DSB (nearly 8dB less) and it leaves one sideband gap totally unused, ideally.
Problem is that an SSB transmitter is more complicated (and definitely doesn't use 8dB less mains power!) and so is an SSB receiver (particularly if you don't like all your announcers to sound like Donald Duck).