Feedback

Given single-input single-output block, what happens when the output is connected to the input?

A determinate system has at most one behavior. With the feedback arrangement, either the arrangement is invalid, or has a unique behavior. The unique behavior would be the fixed-point for the system.

If x denotes the signal on the arc connecting the output to the input, then the fixed-point for the system would be

f(x) = x

That is, if x is input into f, then the result would be x.

Fixed-point solutions

Fixed-point solutions to equations arise frequently in mathematics. Consider developing an algorithm that finds a solution to the following equation for x where x is a real value:

2 x^7 - x = 1

One approach would be to isolate x and iterate its computation until the computed value converged:

                  7    
x    = f(x ) = 2 x  - 1
 i+1      i       i

At the fixed point, x = f(x). In this case, the fixed point is x = 1. If the initial guess for x is between [-1, 1], then the iteration will converge; otherwise, it will not.

From real analysis, a fixed-point iteration will be guaranteed to converge if | f'(x) | < 1 over all possible values of x for the problem. For the example problem, it is better to run the iteration using the 7th root of x + 1:

                 1             1/7
x    = f(x ) = ( - ( x  + 1 ) )   
 i+1      i      2    i

which converges for all x. Examples of sequences for i = 0, 1, 2, ...:

For x₀ = 10, the sequence is 10, 1.2758, 1.0186, 1.0013, ...., which converges to 1.
For x₀ = 1/2, the sequence is 0.5, 0.9597, 0.9971, ..., which converges to 1.
For x₀ = 0, the sequence is 0, 0.9057, 0.9931, 0.9995, ...., which converges to 1.
For x₀ = -1/2, the sequence is -0.5, 0.8203, 0.9866, 0.9990, ..., which converges to 1.
For x₀ = -1, the sequence is -1, 0, 0.9057, 0.9931, 0.9995, ...., which converges to 1.

Updated 02/16/04.