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/
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