November 1960 Electronics World
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from
Electronics World, published May 1959
 December 1971. All copyrights hereby acknowledged.

Although this article discusses
audio waveform measurements, the lesson learned applies equally well to any waveform
frequency. In the RF realm, we are accustomed to injecting two sinewaves at equal
amplitudes into a unit under test (UUT) and reading the relative output powers of
the two input signals and the norder intermodulation (IM) signals. It is usually a very
simple test with simple to interpret results handily shown on the display of a spectrum
analyzer. The task is made a bit more difficult when injecting signals of unequal
strengths and especially when measuring in units of voltage as a viewed on an oscilloscope
display. I dare say most of us need to do some head scratching and looking up of
formulas to pull off such a measurement.
Power Ratings & IM Tests
By L. W. Born
Be careful of what power rating you are quoting when checking
hifi units for IM distortion.
We had sold the amplifier kit, a well known 60watt basic amplifier, only a few
days before and we now had a pretty unhappy owner on our hands.
"This amplifier won't come even close to putting out 60 watts without terrible
distortion," he said with disgust. "There must be something loused up with it!"
"How do you know it's not putting out its rated power?" we asked. "Does it sound
bad or overload easily?"
"Well, no, it seems to sound pretty good. But a friend of mine came over with
his IM distortion analyzer and measured it for me. About the most it would do was
40 watts, and that's a far cry from the 60 watts it's supposed to deliver:"
Further questioning revealed that he had, in fact, been quite pleased with the
sound quality produced by his new amplifier, but since it had been measured and
found wanting, it no longer seemed to sound quite right. We set up the amplifier
on the test bench, loaded it into our 300watt noninductive load, clipped the scope
across the output terminals for visual observation of the waveform, and proceeded
to check it out.
Everything seemed in order. We touched up the bias adjustment slightly and, with
rated output, measured considerably less than one percent intermodulation distortion.
Our "friend" kept his eyes glued to every move that was made. "It's doing a real
good job," we said. "It's running about sixtenths of one percent distortion at
60 watts."
"Wait a minute, now; what scale are you. reading on the wattmeter?" he inquired
with a skeptical glance.
"Read the power on the 150watt scale," we replied.
"But that's not showing 60 watts; that indicates only about 40 watts, just as
I said!"
"True," we agreed, "but you must remember that we are not measuring a simple
sine wave. We are using a complex wave consisting of two sine waves: one at a frequency
of 60 cycles per second, and the other at a frequency of 7000 cycles mixed together
in a voltage ratio of fourtoone respectively. With such a complex signal, the
wattmeter reading must be multiplied by a factor of 1.47 to obtain the equivalent
single sinewave power."
It was apparent that we were not getting through. "Look," we said, "let's feed
a singlefrequency sine wave into the amplifier and adjust the input until the wattmeter
reads 60 watts." The scope showed a beautiful waveform, and the amplifier delivered
60 watts easily. "Now, notice the vertical deflection on the scope. We adjust the
gain of the scope until the pattern covers exactly ten divisions. But see what happens
when we add the 7000cycle signal which was preset to equal just onefourth the
value of the 60cycle signal: the amplifier is obviously overloading. Next we reduce
the combined input signal until the scope shows the same tendivision deflection
that we saw corresponded to a power output of 60 watts. Then we are still driving
the amplifier to its maximum peakpower output, but because we are no longer measuring
a simple sine wave, and because the wattmeter happens to be an averagepower instrument,
the wattmeter shows the true average power output of 40.8 watts."
"You mean that a sixtywatt amplifier can overload even when it's producing only
40 watts output?" he asked incredulously.
"Absolutely! As a matter of fact, it can overload when it is putting out much
less than 40 watts average power, depending on the waveform it is handling. With
some very complex waveforms generated by many kinds of music, it is entirely possible
that an average power output of only a few watts could demand peak powers in excess
of the capacity of the 60watt amplifier:
Fig. 1  With a singlefrequency sinewave signal applied
to the amplifier, these are the output power relations.
"In many ways, it would be better if we used the peakpower output rating for
amplifiers rather than averagepower ratings, because when an amplifier is producing
its maximum peakpower output, this is its limit no matter what the averagepower
output may be. With a single sinewave signal, the peakpower output is exactly
twice the value of the averagepower output. In other words, when your amplifier
was delivering 60 watts as measured by the wattmeter when using the single sinewave
signal, it was necessarily and mathematically producing 120 watts peakpower output.
This simple twotoone ratio is true only for a simple sinewave input. With complex
waveforms, the ratio can be many times as great. With the 4to1 ratio signals used
in intermodulation distortion measurements, the ratio of peaktoaverage power is
2.94 to 1, and as the wave becomes more complex the ratio in general becomes even
larger.
"When your friend measured the distortion at an indicated power of 40.8 watts,
he was actually measuring the distortion with the amplifier delivering 120 watts
peak power with the complex wave and that's all the amplifier can deliver."
"Then the amplifier is really all right?"
"Yes, sir. It's doing even better than the specifications call for and we're
certain you will be completely satisfied with its performance. And, by the way,
ask your friend with the analyzer to come in for a chat, will you?"
Improper Power Ratings
Misunderstandings and dissatisfaction occur repeatedly in connection with IM
distortion measurements caused by improper evaluation of amplifier output when using
complex waveforms, despite words of caution from the manufacturers of the test equipment.
Let's look into the problem a little more deeply.
Let's assume we have an amplifier set up and adjusted in such a manner that with
an input signal of 60 cycles per second measuring one volt on a meter indicating
r.m.s. value, an average power output of ten watts is produced in a 16ohm load.
Elementary a.c. theory tells us that the peak value of the input voltage is 1.414
times the r.m.s. value or, in this instance, 1.414 volts peak.
Looking at the output side of the amplifier, we would measure an r.m.s. voltage
across the load of 12.65 volts with a corresponding peak value of 1.414 x 12.65
= 17.89 volts, as in Fig. 1.
Using the wellknown relationship:
Power = E^{2}/R we find average power
is = 10 watts peak
power is = 20 watts.
Here again, we find the situation where the peak power is exactly twice the average
power, using a single sine wave.
Now, if we remove the 60cycle, 1volt signal and substitute a 7000cycle, 1/4volt
signal with all other adjustments unchanged, we find the peak value of the input
signal is 1.414 x 0.25 = 0.3535 volt. At the load we now find an r.m.s. voltage
of 3.1625 volts with its corresponding peak value of 4.4725 volts. With the 1/4volt
input signal we find average power is
= 0.625 watt
peak power is = 1.25
watts and again, since a single sine wave is involved, the ratio of peakpower
output to averagepower output is twotoone.
Fig. 2  When two frequencies are applied for an intermodulation
distortion test, the picture changes considerably.
In intermodulation distortion measurements, these two above signals are mixed
or added together, and the higher frequency component "rides" on the lower frequency
signal as shown in Fig. 2, producing a combinedsignal peak voltage that is
1.414 plus 0.3535 = 1.7675 volts at the amplifier input. At the output, assuming
no overloading we find a peak voltage equal to the sum of the two separate peak
values or 17.89 plus 4.4725 = 22.36 volts.
This value of peak voltage indicates a peak power of
peak watts.
Yet the average power with the combined signals is actually the same as if each
of the two signals operated independently, or 10 plus 0.625 = 10.625 watts. Hence
it is seen that adding the highfrequency signal to the larger lowfrequency signal
adds only 0.625 watt to the average power output while at the same time, it increases
the peakpower output from 20 watts to 31.24 watts.
A single sine wave that would produce a peakpower output of 31.24 watts would
have an averagepower output of onehalf this amount or 15.62 watts. Thus we say
the above complex signal produces an equivalent sinewave power of 15.62 watts.
The ratio of the equivalent sinewave power to the actual average power is therefore
15.62/10.625 or 1.47. Hence, if the wattmeter used to measure power responds to
average power, it will be necessary to multiply the indicated reading by the factor
of 1.47 to obtain the equivalent sinewave power. Likewise, if a voltmeter is used
to measure the output voltage across the load, and power output computed from this
measured voltage, and if the voltmeter responds to the r.m.s. value of voltage,
the power output computed from the meter reading must be multiplied by the same
1.47 factor.
Other Instruments
If the wattmeter or voltmeter used to measure power responds to the peak value,
no correction factor should be used since the meters will be indicating equivalent
sinewave power or voltage. Different types of instruments respond Differently to
the same a.c. wave. Thus in addition to those that respond to peak values, simple
rectifier or diode types produce a deflection in an d'Arsonval meter movement proportional
to the average value of the rectified wave; thermocouple, electrodynamometer, movingiron,
and electrostatic type instruments produce a deflection proportional to the r.m.s.
value of any waveform. Vacuumtube voltmeters may produce a deflection which range
from proportionality to the average value up to the peak value, depending upon the
design of the instrument.
Where available, the oscilloscope may be used to adjust the output of the amplifier
under test to have the same peak output with the complex wave as it had with the
single sine wave of known power output. Then the amplifier will have the same equivalent
sinewave output when using the twofrequency test signal.
Should there be some uncertainty as to just what the wattmeter or voltmeter is
responding to, a simple test may be performed to determine "watts what." Adjust
the amplifier under test to produce a power output of about half its rated power
output, using a 60cycle signal of known voltage at the input. Introduce the 7000cycle
signal of a value just onefourth that of the 60cycle voltage. The indicated power
output should then increase 56.2% and the indicated output voltage should increase
25%. Should these percentages not be obtained, it is a simple matter to determine
from the observed increases, how much correction factor is required for the particular
meter in question to obtain the necessary percentage increases. It must be remembered
that the correction factors so determined are valid only for twofrequency signals
mixed In a fourtoone ratio for "the particular meter in question.
Posted March 22, 2022 (updated from original post on 1/10/2014)
