Power Series Equation for Amplifier and Harmonics

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 2K views
Natalie Johnson
Messages
38
Reaction score
0
Hi,

I keep reading in multiple sources that amplifier output can be given by

Vout = a0 + a1v(t) + a2v2(t) + a3v3(t) + ... + anvn(t)

I've checked in three of my textbooks and there is not a clear definition (its often just stated) why this equation is used and why it works. I am not looking for an explanation of what each term is, but why it needs to be a power series and not something else?

I've searched on google and my textbooks, I'm looking for some mathematical vigour of how harmonics form

Please can someone advise who knows or point me in the direction of a good explanation?
 
Engineering news on Phys.org
I am not sure if this is what you are looking for, but here are my two cents:
For harmonics, the best expansion would be in terms of trigonometric functions, not a power series. That being said, here is a brief description of the power series.

a0 + a1x + a2x2 + a3x3 + ... + anxn is a Taylor series approximation. Since it has powers of (x-x0), where x0=0, it is expanded around x0=0 and is called a Maclaurin series. The first term, a0, is the function value at x=0. The second term, a1x, adjusts for the slope (first derivative) of the function at x=0. The third term, a2x3, adjusts for the curvature (second derivative) of the function at x=0. For well behaved functions, more terms give better approximations of the function farther away from the central point, x0 = 0. Since you are interested in harmonics, the expansion of sin(x) and cos(x) will be of special interest. Here is a figure showing Taylor series expansions of f(x)=cos(x) at x0 = 0 with more and more terms. The function cos(x) is an even function, so the coefficients of the odd powers of x are all 0. g(x)=1-x2/2 looks ok very near x=0, but the errors get large away from x=0. As more and more terms are added, the functions, g(x), h(x), p(x), q(x), r(x), and s(x) get more and more accurate farther from x=0. They follow the higher derivatives of f(x) better. The final function, t(x), shows the error between s(x) and f(x). You can see that the error is fairly small out to about x=5 and then grows rapidly.
cosTaylorSeries.png
 

Attachments

  • cosTaylorSeries.png
    cosTaylorSeries.png
    35.6 KB · Views: 479
Last edited:
  • Like
Likes   Reactions: anorlunda