SM 5 BSZ - Intermodulation in receivers. Math details.
(Sept 18 2003)
X(t), a voltage that varies with time, is the input to a receiver stage that is well characterized by the following equation which gives the output Y(t+d), a voltage that varies with time in a similar fashion where d is a time delay that is not interesting in the current context.
Y(t+d) = k1X(t)+ k2{X(t)}2 + k3{X(t)}3 + _______________(1)

We want to find the output Y(t+d) for the following input function:
X(t)=A*{ sin(2*PI*f1) + sin(2*PI*f2) } _______________(2)

Substitute the trigonometric functions
u=sin(2*PI*f1)=sin(w1) _______________(3)
v=sin(2*PI*f2)=sin(w2) _______________(4)

Y(t+d) = A*k1*(u+v)+ A2* k2*(u+v)2 + A3*k3*(u+v)3 + _______________(5)

Split the result in two parts Yo for odd terms and Ye for even terms:
Y(t+d) = Yo + Ye _______________(6)

This means that Yo and Ye look like this:
Yo = A*k1(u+v)+ A3*k3*(u+v)3 + A5*k5*(u+v)5 + _______________(7)
Ye = A2*k2*(u+v)2 + A4*k4*(u+v)4 + _______________(8)

Rewrite Yo and Ye:
Yo = (u+v)* { A*k1 + A2*(u+v)2 * {k3 + A2*(u+v)2 * {k5 + }}} _______________(9)
Ye =A2*(u+v)2 * {k2 + A2*(u+v)2 * {k4 + }} _______________(10)

Find out what the square of the sum of u and v squared means by use of (3) and (4):
(u+v)2= u2 + 2 * u * v + v2=
= sin(w1) 2 + 2 * sin(w1) * sin(w2) + sin(w2) 2 =
= 1 - {cos(2*w1) + cos(2*w2)} / 2 + cos(w1 - w2) - cos(w1 + w2) _____(11)

We now see that (u+v)2 is composed of four signals. A dc voltage and the overtones 2*f1 and 2*f2 with an amplitude of 1.0 plus the sum and difference frequencies (f1+f2) and (f1-f2) with an amplitude of 0.5. It is easier to understand the result if we write is as sums and differences only. The following frequencies are generated. All with an amplitude of 0.5:

These frequencies contribute to Ye multiplied by A2*k2. It should be obvious that k4 will give contributions that can be written as a combination of four frequencies and that Ye is composed of frequencies that can be written as a combination of an even number of frequencies only.

In the current context, receiver dynamic range measurements, we are interested in the third order frequencies only. They can be written f1+f1-f2 and f2+f2-f1 and it should be clear by now that they can not be generated among the contributions in Ye.

Multiply (11) with (u+v) using (3) and (4)
[1 - {cos(2*w1) + cos(2*w2)} / 2 + cos(w1 - w2)) - cos(w1 + w2)] * [sin(w1) + sin(w2)]=
= sin(w1) + sin(w2) -sin(w1)*cos(2*w1)/2 -sin(w2)*cos(2*w1)/2 -sin(w1)*cos(2*w2)/2 -sin(w2)*cos(2*w2)/2+ +sin(w1)*cos(w1 - w2)) +sin(w2)*cos(w1 - w2)) -sin(w1)*cos(w1 + w2)) -sin(w2)*cos(w1 + w2))=

= sin(w1) + sin(w2) + { - sin(3*w1) - sin(-w1) - sin(w2+2*w1)) - sin(w2-2*w1)) - sin(w1+2*w2)) - sin(w1-2*w2)) - sin(3*w2) - sin(-w2)}/4 + {sin(2*w1-w2)) + sin(w2) + sin(w1) + sin(2*w2-w1)) - sin(2*w1 + w2) - sin(-w2) - sin(2*w2 + w1) - sin(-w1) }/2

= 2.25*{sin(w1) + sin(w2)}+ 0.25*{sin(-3*w1) + sin(-3*w2)} + 0.75*{sin(-w2-2*w1)+ sin(-w1-2*w2)+ sin(2*w1-w2) + sin(2*w2-w1)} _______________(12)

We now see that the first contribution to the third order intermodulation arise from the k3term:
Yo=0.75*A3*k3*{sin(2*w1-w2) + sin(2*w2-w1)} + _______________(13)

This is not the only contribution to Yo however. All the higher odd terms contribute. We get the fifth order terms by multiplying (11) and (12):
[1 - {cos(2*w1) + cos(2*w2)} / 2 + cos(w1 - w2)) - cos(w1 + w2)) ] * [1.75*{sin(w1) + sin(w2)}+ 0.25*{sin(-3*w1) + sin(-3*w2)} + 0.5*{sin(-w2-2*w1)+ sin(-w1-2*w2)+ sin(2*w1-w2) + sin(2*w2-w1)}]

The number of terms is becoming unpleasant....... There must be text-books that give the full treatment - the matematicians have methods to find the formulae for sums like this.........

The contributions to the third order frequencies sum up to a constant. Rather than doing the algebra let us simply call it c5. The exact value is not important to the discussion of receiver dynamic ranges, it should be clear from the above that the output at the third order intermodulation frequencies can be written like this:
IM3=0.5*A3*k3 + c5*A5*k5 + c7*A7*k7 _______________(13)

To get an idea of the size of these numbers, we make the amplitude A expressed as a percentage of the voltage at the 1 dB compression point. Let us also make k1 = 1, for unity gain and calculate the signal and the IM3 amplitudes:
IM3 = 0.5*0.13*k3 _______________(15)

It is common knowledge that IP3 is in the order of 15 dB higher than the 1 dB compression point. A signal with an amplitude A=0.1 is 20 dB below the 1 dB compression point according to our definition of the amplitude scale. This means that such a signal is in the order of 35 dB below IP3. As a consequence, 'common knowledge' tells us that IM3 should be about 70 dB (0.000316 times) below the signal at this amplitude, which leads to:
0.1 * 0.000316 = 0.5*0.13*k3 _______________(16)

This means that 'common knowledge' is that:
k3 = 0.06 _______________(17)

It is a trivial thing to add a third order term of this size, full scale times 0.6, to the input data stream from the A/D converter in Linrad and locate the input signal level that gives IM3 at -70 dB. To get the same level of IM3 from the k5 term only, one has to make k3 = 2. This means that:

0.5 * k3*0.13 = c5 * 2 * 0.15 _______________(18)

which leads to:
c5 = 1.5 _______________(19)

In the same way, by a numerical experiment, we find:
c7 = 3 _______________(20)

As a final result we find the level of the third order intermodulation products for a receiver stage that is well described by (1) to be:
IM3=0.5*k3*A3+ 1.5*k5*A5 + 3*k7A7 + _______________(21)

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