In a previous post, I showed what a recursive feedback loop looks like and how it affects the transfer function of a first order system. Now, I’m looking at what this recursive feedback loop with a first order system inside does. What is it good for?
In the limit, when you apply a unit step to it, the output of this system approximates the inverse of the Golden Ratio, φ-1 = 2/(1 + √ 5 ) = 0.618 (roughly).
Recall that the transfer function for the first level of recursion is
And that from this level of recursion on, all transfer functions has one pole and one zero. In general, the transfer function has the form, (I’ve switched to calling the loop gain K, instead of a1 in the previous post on this subject — I think it’s clearer that way)
for the i-th level of recursion, where Fi is the i-th number in the Fibonacci sequence, Fi-1 the one before it, and so on…
The difference equation described by this transfer function is easily found with some rearranging and taking the inverse z-transform,
So we can draw a generic block diagram that holds regardless of how many levels of recursion we have,
This is a lot easier than drawing block diagrams of infinitely nested feedback loops (check out the block diagram of the first level of recursion for an example) — and they are functionally equivalent! From a hardware perspective, you could actually build something (in this case, a discrete-time filter) with a finite (and very small) number of math elements and interconnects that emulates recursion.
The difference equation and the block diagram show how the output of the recursive feedback loop would be determined for any arbitrary input sequence, x[n]. Say the input sequence was a unit step, that is, x[n] = 1 indefinitely. The step response of having one level of recursion could look like this.
This is the step response for H1(z). To calculate this, K can no longer be an abstract variable, we need to choose a value for it. K = 0.5 is a responsible choice, since it is less than 1, which means our feedback loop will be stable — that is, it will not increase nor decrease without bound (in the limit of levels of recursion, the choice of the loop gain constant K, doesn’t matter much).
Notice that the step response for H1(z) settles out to about 0.6 for an input of 1.0 and peaks at 0.63 before doing so. It overshoots it’s final value by 0.03, or 4% of it’s final value. It doesn’t take long for it to settle to its final value, only 4 samples.
Before the levels of recursion get too deep, i.e., in the first few iterations, the resulting transfer function has one pole and one zero that actually do something — they influence the response to a time-varying input.
The step responses of the first four levels of recursion are plotted on top of each other in the figure below. The general trend is to converge more quickly to φ-1 for increasing levels of recursion.
Remember that the ratio of two consecutive numbers in the Fibonacci sequence is the equal to φ.
So in the recursion limit, the recursive feedback loop does a good job of approximating φ-1 with minimal and rudimentary hardware.
