Superposition Theorem

Thread Starter

steeve_wai

Joined Sep 13, 2007
47
why can this theorem be applied only to linear functions such as voltage and current,and not power which is non-linear.please explain with equations.thank you.:confused:
 

recca02

Joined Apr 2, 2007
1,214
its not ' non linear functions' that they r not applicable to IIRC (correct me if i m wrong).
its not applicable to non linear circuits.
 
If you model the current entering/exiting a node as a function, then you can say that \(f(\sum_{k=1}^n{I_k}) = \sum_{k=1}^n{f(I_k)}\) iff f is linear, where \(I_k\) is the current contributed by source k. This is easy to prove if one writes \(f(x) = ax + b\). It's also easy to extend it to voltages.

How to prove it the other way eludes me at the moment.

\(f(E) = {1 \over R} E^2\), which is for power, certainly doesn't obey the above relationship.
 

BlackBox

Joined Apr 22, 2007
20
Sorry to appoint it Eduard, but you are wrong: \( f(x) = a x + b[\tex] is not a linear function if you want to speak of the superposition theorem.

A linear function is a function between vector fields (a scalar field can be a vector field too).
A function is linear if it follows the following, simple, rules:
\( f(ax)= a \cdot() f (x)[\tex]
\( f(x_1+x_2)=f(x_1)+f(x_2)[\tex]
given x in a vector field

Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

\( f(x) = x^2 [\tex]
\( f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
\( f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
\( f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
Q.E.D.

It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

Holla\)\)\)\)\)\)\)
 
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