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Old March 20th 12, 01:01 PM posted to rec.radio.cb
JWG JWG is offline
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Default AM vs. SSB technical question

In short, how does a 5-watt CB produce 12W PEP in the SSB mode?

Once the carrier and one sideband are filtered out of the signal to
produce a SSB signal, and this is fed to the CB's RF amplifier (which
would generate 4-5W of AM), why wouldn't the result just be 4-5W of
SSB?

It is often described as the power being focused into one sideband,
and the increased bandwidth efficiency is clear, but: if the power
input to the final amplifier is the same as with AM, where do the
"extra watts" come from? Or is the amplification somehow applied in a
different way to the SSB signal? Not sure how to conceptualize this.

Thanks!
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Old March 21st 12, 12:45 AM posted to rec.radio.cb
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Default AM vs. SSB technical question

JWG wrote:
In short, how does a 5-watt CB produce 12W PEP in the SSB mode?

Once the carrier and one sideband are filtered out of the signal to
produce a SSB signal, and this is fed to the CB's RF amplifier (which
would generate 4-5W of AM), why wouldn't the result just be 4-5W of
SSB?

It is often described as the power being focused into one sideband,
and the increased bandwidth efficiency is clear, but: if the power
input to the final amplifier is the same as with AM, where do the
"extra watts" come from? Or is the amplification somehow applied in a
different way to the SSB signal? Not sure how to conceptualize this.

Thanks!



A 4 watt AM signal sends out 4 watts when there is no modulation (A0)
To fully modulate an AM signal to 100%, you mix 50% more to it, in this
case, your audio. This creates 2 identical mirror image sidebands, each
being 1 watt. Added together, this would show 6 watts on an RMS (average)
reading meter. Ohms law being what it is, and power being related to
ohms law, when you increase the voltage 50% to a fixed resistance (your
antenna), the current also increases by 50%, and now your PEAK TO PEAK
power is now 16 watts. The allowed increase from 4 to 12 comes from the
fact that the FCC allows the power of the selected sideband to be the
amount of peak power that 80% AM modulation would put out. That is part
of the increase. The other part of the increase is that the receiver is
more sensitive when it only has to listen to half the bandwidth of an AM
signal.

All told, sideband is a much more efficient use of power and radio spectrum.
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Old March 21st 12, 01:00 AM
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Quote:
Originally Posted by JWG View Post
In short, how does a 5-watt CB produce 12W PEP in the SSB mode?

Once the carrier and one sideband are filtered out of the signal to
produce a SSB signal, and this is fed to the CB's RF amplifier (which
would generate 4-5W of AM), why wouldn't the result just be 4-5W of
SSB?

It is often described as the power being focused into one sideband,
and the increased bandwidth efficiency is clear, but: if the power
input to the final amplifier is the same as with AM, where do the
"extra watts" come from? Or is the amplification somehow applied in a
different way to the SSB signal? Not sure how to conceptualize this.

Thanks!
Lets break it down into a easier way to explain it.
If we could see one individual radio wave, we would be able to draw a line ( carrier ) though the center and the upper side band ( the radio wave above the carrier ) would be above the carrier and the lower side band ( the radio wave below the carrier ) would be below the carrier.

When we subtract one of the carriers, the forward power is transferred into the other band.
Now the carrier is a power hog and probably wastes 60% of the transmitted power.
So at that rate - if both side bands combined had 5 watts and you removed one of the side bands - you would effectively double the transmit power.

However, you have to account for losses, the loss of the connectors, the loss in the coax, the values of all the components in the radio, to be conservative lets say it is 12 watts PEP when it is new.

PEP stands for Peak Envelope Power.
http://en.wikipedia.org/wiki/Peak_envelope_power

The PEP output of an AM transmitter at full modulation is four times its carrier PEP; in other words, a sold-state, 100-watt amateur transceiver is usually rated for no more than 25 watts carrier output when operating in AM

When you turn on a vac cleaner, at the initial start up, you have a power spike which is needed to start the motor turning, that can be considered a peak power level. When manufactures sells a vac cleaner, they include the PEP power level to make the motor seem more impressive.

On the other hand, a VOM meter reads Root Mean Square - RMS power.
http://en.wikipedia.org/wiki/Root_mean_square

Because of their usefulness in carrying out power calculations, listed voltages for power outlets, e.g. 120 V (USA) or 230 V (Europe), are almost always quoted in RMS values, and not peak values.
Peak values can be calculated from RMS values from the above formula, which implies Vp = VRMS × √2, assuming the source is a pure sine wave.
Thus the peak value of the mains voltage in the USA is about 120 × √2, or about 170 volts.
The peak-to-peak voltage, being twice this, is about 340 volts.
A similar calculation indicates that the peak-to-peak mains voltage in Europe is about 650 volts.
It is also possible to calculate the RMS power of a signal. By analogy with RMS voltage and RMS current, RMS power is the square root of the mean of the square of the power over some specified time period.
This quantity, which would be expressed in units of watts (RMS), has no physical significance.
However, the term "RMS power" is sometimes used in the audio industry as a synonym for "mean power" or "average power".
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Old March 21st 12, 03:07 PM posted to rec.radio.cb
JWG JWG is offline
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Default AM vs. SSB technical question

THANKS Scott and Channel Jumper, I really appreciate the clear and
thoughtful replies. Extremely helpful! What I could find in texts and
online has either been too technical or just glosses over this without
stating "why" the increased wattage level is achieved on SSB.

JG
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Old March 21st 12, 11:47 PM
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You are welcome.

One thing I did not mention was the fact that since AC - alternating current - is harder to measure then DC current / so when we talk about RMS power.
( Root Mean Square ) we are talking about the amount of power necessary to produce a certain amount of heat into a resistor of a known value at X Volts DC.

This type of measurement is what keeps us from mistaking one countries voltage - at the wall outlet, from another.
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