# Creating harmonics

#### apr 2003

Harmonics are created by non-linear devices. Amplifiers run in their "linear" region will still create harmonics at a low level. An amplifier can be parameterized at V= X0 +X1*V(0) + X2V(0)^2 + X3V(0)^3  where the coefficients Xn tell how much of each term is created.

#### Two tones:

Suppose we have 2 tones Acos(at) and Bcos(bt). Plugging them into the above equation gives the strengths of the various frequency components generated by the device. All of the amplitudes should be multiplied by the corresponding Xn.
• Linear: (Acos(at) + Bcos(bt))

• The individual freq terms are then:
• a: A
• b: B
• square:[Acos(at) + Bcos(bt)]^2:
[Acos(at)]^2 +[Bcos(bt)]^2 + 2ABcos(at)cos(bt)
or
A^2(cos(2at) -1)/2
B^2(cos(2bt)-1)/2
AB(cos(a+b) + cos(a-b))
The individual freq terms are then:
• (2a): A^2
• (2b): B^2
• (a+b) : AB
• (a-b) : AB
• cube:[Acos(at) + Bcos(bt)]^3:

• [Acos(at)]^3  + [Bcos(bt)]^3 +
Acos(at)[Bcos(bt)]^2 + 2Bcos(bt)[Acos(at)]^2 +
Bcos(bt)[Acos(at)]^2 + 2Acos(at)[Bcos(bt)]^2
or
[Acos(at)]^3  + [Bcos(bt)]^3 +
(AB^2 + 2AB^2)cos(at)[cos(bt)]^2 +
(BA^2 + 2BA^2)cos(bt)[cos(at)]^2  The terms are then:

• A^3cos(at)*[cos2at - 1)/2=A^3([cos(3at) + cos(at)]/2 - cos(at)]/2=A^3(cos(3at) -cos(at))/4
• B^3cos(bt)*[cos2bt - 1)/2=B^3([cos(3bt) + cos(bt)]/2 - cos(bt)]/2=B^3(cos(3bt) -cos(bt))/4

• let C2= 3AB^2 and C3=3BA^2
• C2cos(at)[cos(2bt)-1]/2=C2[{cos((2b-a)t)+cos((2b+a)t)}/2+cos(at)]/2
• C3cos(bt)[cos(2at)-1]/2=C3[{cos((2a-b)t)+cos((2a+b)t)}/2+cos(bt)]/2

The individual freq terms are then:

• (3a): A^3/4
• (3b):B^3/4
• (2b-a):3AB^2/4
• (2b+a):3AB^2/4
• (2a-b):3BA^2/4
• (2a+b):3BA^2/4
• a:A(3B^2/2 -A^2/4)
• b:B(3A^2/2 - B^2/4)
Each of these needs to be multiplied by the Xi coef of the amplifier/device.

#### Single tone:

If there is just a single tone, then the values generated are the same as the first line of each section above:
1. Linear:Acos(at)

2. (a):A
3. Square:A^2(cos(2at) -1)/2

4. (2a): A^2/2
(DC): -A^2/2
5. cubed:A^3(cos(3at) -cos(at))/4

6. (3a):A^3/4
(a):-A^3/4