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Re: [Amps] How intermod limits your PEP

To: Tony King - W4ZT <amps080605@w4zt.com>
Subject: Re: [Amps] How intermod limits your PEP
From: "Dr. David Kirkby" <david.kirkby@onetel.net>
Date: Thu, 18 May 2006 02:17:54 +0100
List-post: <mailto:amps@contesting.com>
Tony King - W4ZT wrote:
> I would like to see the solution to the math problem.  However, I think 
> that it is important for us to remember that Gary's point is valid. 

<snip>

> 73, Tony W4ZT

I don't guarantee this is right, but here is my attempt at the maths of 
this. It looks reasonable, but I have not checked it carefully.

Assume the amp has at the output two wanted signal voltage V1 and V2 
with a *peak* voltage of A volts.

V1=A Sin(2 pi f1 t)
V2=A Sin(2 pi f2 t + phi2)

where f1 and f2 and the frequencies in Hz, t is time in seconds, and 
phi2 is the phase difference between the two signals at t=0.

So V1 and V2 are the two wanted signals at the output of the amp.

There are two unwanted intermodulation products with a *peak* voltage of 
B. (B will be a lot less than A in any half-decent amp).

V3 = B Sin( 2 Pi f3 t + phi3)
V4 = B Sin( 2 Pu f4 t + phi4)

Assume a load resistance of R.

The powers P1, P2, P3 and P4 generated by the 4 signals will be P1, P2, 
P3 and P4, where these are given by:

P1 = A^2 / (2 R)
P2 = A^2 / (2 R)
P3 = B^2 / (2 R)
P4 = B^2 / (2 R)

Define the IMD (in dB) as

10 Log_base_10(P1/P3)

Since P is proportional to V^2, it follows that

IMD = 10 log_base_10(V1^2 / V1^2)
= 20 log_base 10(V1/V2)

The PEP, which is the average power at the peak of the waveform occurs 
when the 4 voltages V1, V2, V3 and V4 are all in phase.

The PEP is (2A + 2B)^2 / 2 R.

As someone said, these need to be solved as a set of simultaneous 
equations. Rather than work that lot out, I'll trust Mathematica

http://www.wolfram.com/products/mathematica/index.html

to do it. Note, Mathematica is expensive, but you can get a free 14 day 
demo which works like the real products, but does not have the ability 
to save results).

I'm using 'Solve[]' in Mathematica, solving to find the voltages A and B.

The 'Solve[]' command is documented at

http://documents.wolfram.com/v5/TheMathematicaBook/APracticalIntroductionToMathematica/SymbolicMathematics/1.5.7.html

for anyone interested, but basically I solve for the voltages A and B, 
subject to these two equations

PEP == (2 A + 2 B)^2 / (2 R)
IMD == 20 Log[10, A/B]

Let Mathematica crunch it.

sparrow /export/home/drkirkby % math
Mathematica 5.2 for Sun Solaris (UltraSPARC)
Copyright 1988-2005 Wolfram Research, Inc.
  -- Motif graphics initialized --

In[1]:=  Solve[{PEP == (2 A + 2 B)^2 / (2 R), IMD == 20 Log[10, A/B]},{A,B}]

                              IMD/20
                            10       Sqrt[PEP] Sqrt[R]
Out[1]= {{A -> ---------------------------------------------------,
                          2 + IMD/20  IMD/20    1 + IMD/10  IMD/10
                Sqrt[2 + 2           5       + 2           5      ]

                             Sqrt[PEP] Sqrt[R]
 >     B -> ---------------------------------------------------},
                      2 + IMD/20  IMD/20    1 + IMD/10  IMD/10
            Sqrt[2 + 2           5       + 2           5      ]

               -(1/2) + IMD/20  IMD/20
              2                5       Sqrt[PEP] Sqrt[R]
 >    {A -> -(------------------------------------------),
                         1 + IMD/20  IMD/20     IMD/10
               Sqrt[1 + 2           5       + 10      ]

                               Sqrt[PEP] Sqrt[R]
 >     B -> -(---------------------------------------------------)}}
                        2 + IMD/20  IMD/20    1 + IMD/10  IMD/10
              Sqrt[2 + 2           5       + 2           5      ]


That looks complex, but it basically says there are two pairs of 
solutions. It so happens the two pairs are the negative of one another, 
so A can be 100 V or - 100 V for example. Nothing too hard really, 
despite it looks a bit messy.

Now, lets stick some numbers in it. I'll use the same as Gary.

First, we will use a PEP of 1000 W, R = 50 Ohms and and IMD of 20 dB.

In[2]:= N[%1] /. {PEP->1000, R->50, IMD->20}

Out[2]= {{A -> 143.74, B -> 14.374}, {A -> -143.74, B -> -14.374}}

So taking the positive solutions, we have the main signals having a 
voltage of 143.74 V peak and the interference having a voltage of 14.374 
peak.

The powers in these will be as follows - I divide by 2, since these are 
peak values and I need to get RMS to compute the powers. Since they need 
to be squared, I divide by Sqrt(2) twice, which is the same as dividing 
by 2.

P1 = 143.74^2 / (2 * 50) = 206.612 W  and
P2 = 14.374^2 / (2 * 50) = 2.06612 W,

so the intermodulation products are 1/100th of the main signals, which 
is to be expected, as I chose an IMD of 20 dB or 1/100.

Now do it for the same 1000 W PEP, but this time make the 
intermodulation signals 30 dB down.

In[5]:=  N[%1] /. {PEP->1000, R->50, IMD->30}

Out[5]= {{A -> 153.267, B -> 4.84673}, {A -> -153.267, B -> -4.84673}}


Now, lets make the amplifier really clean, and make the IMDs 100 dB down.

In[6]:=  N[%1] /. {PEP->1000, R->50, IMD->100}

Out[6]= {{A -> 158.112, B -> 0.00158112}, {A -> -158.112, B -> -0.00158112}}

So that says when the IMD is very very small, the peak voltages are 
158.112 for the main signal, and next to nothing for the interference. 
Does that make sense?

Lets compute the power in these two signals.

P1=P3=158.112%2/ (50 * 2) = 249.994 W

So we find virtually 250 W is in each sideband,  at a PEP of 1000 W. 
That is to be expected. There are basically just two carriers in this 
case (as the IM products are so small). In which case the PEP should be 
4x the power in any one carrier, which is what I found.

I then produced a table, so to get some results quickly. I used IMD of 
0, 5, 10, 15, ...50 dB down. The voltages are given below for A and B.

In[8]:=  N[%1] /. {PEP->1000, R->50, IMD->Table[IMD,{IMD,0,50,5}]}

Out[8]= {{A -> {79.0569, 101.203, 120.127, 134.242, 143.74, 149.696, 
153.267,

 >       155.351, 156.548, 157.23, 157.615},

 >     B -> {79.0569, 56.9107, 37.9873, 23.872, 14.374, 8.41802, 4.84673,

 >       2.76258, 1.56548, 0.884168, 0.498424}},

 >    {A -> {-79.0569, -101.203, -120.127, -134.242, -143.74, -149.696,

 >       -153.267, -155.351, -156.548, -157.23, -157.615},

 >     B -> {-79.0569, -56.9107, -37.9873, -23.872, -14.374, -8.41802,

 >       -4.84673, -2.76258, -1.56548, -0.884168, -0.498424}}}


So this means:

IMD = 0 dB, A=79.0569 V, B = 79.0569 V
IMD = 5 dB, A=101.203 V, B = 56.9107 V
IMD = 10 dB,A=120.127 V, B = 37.9873 V
and so on.


As I was saying before, as the IMD decreases, so the signal must go up 
but the IMD voltage must go down.

These numbers are quite different to what Gary got. I think his analysis 
is flawed, but I don't think it is so badly flawed that we should get 
quite different numbers. Anyway, perhaps someone else will look at it.

-- 
Dr. David Kirkby BSc MSc PhD CEng MIEE
Chartered Engineer
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