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<p>Hi John,</p>
<p>The transmitter limitation is peak power. The tone magnitude
values in the code below are the peak I and Q values which can be
viewed as "voltages." With each tone set for peak values of 0.5,
when their peaks coincide, the peak "voltage" is 1.0 which reaches
the maximum voltage and consequently maximum power supported by
the transmitter. Now, what about the AVERAGE power of a single
tone peaking at 1.0 voltage versus two equal-amplitude tones with
a combined peak voltage of 1.0: the answer is that the single
tone has double the AVERAGE power of the two-tones, even though
the peak power is the same. Here's a reference that goes through
the math:</p>
<p><a class="moz-txt-link-rfc2396E" href="http://www.eetimes.com/document.asp?doc_id=1275672"><http://www.eetimes.com/document.asp?doc_id=1275672></a> <br>
</p>
<p>Hence, to stay within the peak power limitations of the
transmitter, I believe the factors are correct.<br>
</p>
<p>73,</p>
<p>Warren NR0V<br>
</p>
<br>
<div class="moz-cite-prefix">On 11/22/2016 8:15 PM,
<a class="moz-txt-link-abbreviated" href="mailto:john_eckert@keysight.com">john_eckert@keysight.com</a> wrote:<br>
</div>
<blockquote
cite="mid:888FC1A46449D04EA4E340F7BC2621112736BD98@wcosexch01k.cos.is.keysight.com"
type="cite">
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<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">There is an error in the calculation of the
levels for two tone testing. The result is that the total
output power is half of what is expected. I believe the
author intended to reduce the power of each tone by half with
respect to a single CW tone. This would have resulted in
exactly the same output power. Unfortunately, the voltage
levels are being reduced by one half resulting in each tone
dropping to one quarter the power. The result is that the Two
Tone power is at one half the desired power. <o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">With the current code IMD testing can’t be
made at full power and current tests are at one half the
indicated power.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Problem routine:<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"> private void
chkTestIMD_CheckedChanged(object sender, System.EventArgs e)<o:p></o:p></p>
<p class="MsoNormal"> {<o:p></o:p></p>
<p class="MsoNormal"> if (chkTestIMD.Checked)<o:p></o:p></p>
<p class="MsoNormal"> {<o:p></o:p></p>
<p class="MsoNormal"> udTestIMDFreq1.Enabled =
false;<o:p></o:p></p>
<p class="MsoNormal"> udTestIMDFreq2.Enabled =
false;<o:p></o:p></p>
<p class="MsoNormal"> udTwoToneLevel.Enabled =
false;<o:p></o:p></p>
<p class="MsoNormal"> udTestIMDPower.Enabled =
false;<o:p></o:p></p>
<p class="MsoNormal"> chkInvertTones.Enabled =
false;<o:p></o:p></p>
<p class="MsoNormal"> double ttfreq1 =
(double)udTestIMDFreq1.Value;<o:p></o:p></p>
<p class="MsoNormal"> double ttfreq2 =
(double)udTestIMDFreq2.Value;<o:p></o:p></p>
<p class="MsoNormal"> double ttmag =
(double)udTwoToneLevel.Value;<o:p></o:p></p>
<p class="MsoNormal"> double ttmag1, ttmag2;<o:p></o:p></p>
<p class="MsoNormal"> ttmag1 = ttmag2 = 0.49999 *
Math.Pow(10.0, ttmag / 20.0); // 11/22/16 k2ox
should be 0.707107<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Hopefully a maintainer will change the
0.49999 constant to 0.707107.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">73,<o:p></o:p></p>
<p class="MsoNormal">John<o:p></o:p></p>
<p class="MsoNormal">K2ox<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
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