Sinusoidal Output of Electronic Oscillators

Date: 04/28/2005 at 09:51:19
From: Rajan
Subject: electronic oscillators 

I am a second year engineering student and doing signals and systems.
I just wanted to know the explanation for electronic oscillators
oscillating only in a sinusoidal wave shape when electronic devices
work in a linear region.

I can understand electric generators giving sinusoidal output because 
the rotor in the generator goes through a circular motion and the flux 
contributing to generation of voltage has a sine component.  But in 
oscillators there is no such motion, so how does this also give a 
sinusoidal output?

Date: 04/28/2005 at 11:41:14
From: Doctor Douglas
Subject: Re:  electronic oscillators

Hi Rajan.

You must be careful to distinguish what you mean when you use the word 
"linear" to describe a system.  The behavior of a system on its 
parameters may be linear, even though the actual time-dependence of a 
particular variable may have some complicated variation with time, 
such as sinusoidal or exponential.

Suppose you have a differential equation:

  dx
  -- = Ax             (A = a constant)
  dt 

This is a LINEAR differential equation, because all of the 
coefficients of the derivatives and the source terms are constants
(1 and A) and the equation is linear in x and its derivatives.

You probably know that the solution for the above differential
equation is

  x(t) = X0*exp(A*t)

where X0 is a constant that has to be determined from the initial
boundary conditions (such as x(t=0) = X0).

The exponential function certainly isn't a linear function--it varies 
exponentially as a function of time, of course.  It is not too hard to 
generalize this to sinusoidally varying functions by allowing the 
constant A to take on complex values, or by going to the second-order 
linear differential equation (DEQ):

  d^2 x
  ----- = -w^2 x
   dt^2

Again, note that this differential equation is linear.  The way that
this linearity comes out in the system is that if you have two
solutions of this equation, then their linear combination is also
a solution:

   x1(t) = cos(wt)
   x2(t) = sin(wt)

   x(t) = A*x1(t) + B*x2(t)           will also satisfy the DEQ
        = A*cos(wt) + B*sin(wt)       check it!

You might also be interested in checking out the following answers in 
our archives:

  Math in Electrical Engineering
    http://mathforum.org/library/drmath/view/62937.html 

  Non-homogeneous Differential Equation Solutions
    http://mathforum.org/library/drmath/view/52128.html 

  Homogeneous Linear Differential Equations
    http://mathforum.org/library/drmath/view/52131.html 

  What is nonlinear math?
    http://mathforum.org/library/drmath/view/53603.html 

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/