$L_\infty$ norm minimization

Question

I write supplemental explanation to render my aim of optimization.

First Consider discrete time linear system below:

\begin{equation} G(z)=\frac{1-e^{-0.3T}}{z^{100}(z-e^{-0.3T})} \end{equation}

where $z = e^{j\omega}$ and $\omega$ is frequency, $\omega \in [-\pi,\pi)$, and $T$ is sampling period, assume that $T = 1$. It's obvious that the absolute value of $z$ is equal to $1$ for all $\omega$ between $-\pi$ and $\pi$.

When one is talking about maximum absolute value of $G(z)$, in fact he/she consider below function :

\begin{equation} \max_{\omega} |G(e^{j\omega})|=|\frac{1-e^{-0.3T}}{e^{j\omega}-e^{-0.3T}}| . \end{equation}

note that $|z^{100}| = |z|^{100} = 1.$

On the other hand consider another discrete time linear system :

\begin{equation} C(z) = \frac{(x+\frac{T}{2}y)z+(\frac{T}{2}y-x)}{z-1} \end{equation}

Consider these two linear system connected serially.Then the closed loop system transfer function (with unity feedback) is:

\begin{equation} f(x,y,z) = \frac{G(z)C(z)}{1+G(z)C(z)}\end{equation}

or

\begin{equation} f(x,y,e^{j\omega}) = \frac{G(e^{j\omega})C(e^{j\omega})}{1+G(e^{j\omega})C(e^{j\omega})}\end{equation}

My aim is to minimize the maximum absolute value of $f$ (respect to $\omega$) over $x $ and $ y$ as mathematically described bellow:

\begin{equation} \min_{x, y} \max_{z \; or \; \omega} |f(x,y,z)|\end{equation}

This is equal to minimizing the $L_\infty$ norm of $f$ (respect to $\omega$) over $x$ and $y$ (right?) In fact:

\begin{equation}\min_{x, y} \max_{\omega} |f(x,y,\omega)| \equiv \min_{x, y} ||f(x,y,\omega)||_{\infty , \;\omega}\end{equation}

How can I do the optimization with Mathematica?

---

I try the code below for defining the function and maximizing $f$ respect to $z$:

T = 0.01
z = Exp[I w]
f = (((x + (T/2) y) z + ((T/2) y - x)) (1 - Exp[-0.3 T]))/(
      (z^100) (z - Exp[-0.3 T]) (z - 1) + ((x + (T/2) y) z + ((T/2) y - x)) (1 - Exp[-0.3 T]))
maxp = First[ NMaximize[{f[x, y, w], 0 <= w <= 10}, w, Method -> DifferentialEvolution]]

But when I run code, the following error appears:

NMaximize::nnum: The function value 
-((0.0029955 (-x + (0.89724 + 0.441543 I) (x + 0.005 y) + 0.005 y))/((0.120832 -0.165872 I)
 + 0.0029955 ( -x + (0.89724 + <<20>> I) (<<1>>)+0.005 y)))[x,y,0.457318] 
is not a number at {w} = {0.457318}. >>

I try different intervals for $w$ but nothing changed.

What's wrong with it?

Answer 1

This is along the right lines, I think. But I had to change the minimization to use a log, in order for the scaling not to be too much of a problem for FindMinimum. I also removed the complex exponential definition of z. You might want to exponentiate the result. I also used an Abs to make the logarithm real valued, so that too could have consequences.

t = 0.01;
f[x_, y_, z_] := 
 Log[Abs[(((x + (t/2) y) z + ((t/2) y - x)) (1 - 
        Exp[-0.3 t]))/((z^100) (z - Exp[-0.3 t]) (z - 
         1) + ((x + (t/2) y) z + ((t/2) y - x)) (1 - Exp[-0.3 t]))]]

fzmin[z_?NumberQ] := Module[{x, y, res},
  res = FindMinimum[f[x, y, z], {x, -10, 10}, {y, -10, 10}];
  If[! ListQ[res], -1000000., res[[1]]]]

max = 
 NMaximize[{fzmin[z], 0 <= z <= 10}, {z, 0, 10}, 
  Method -> "DifferentialEvolution"]

(* Out[317]= {-1.11022302463*10^-16, {z -> 0.416231991971}} *)

--- edit ---

Based on updates to the question, the code below should be appropriate. I remark that it is not hugely modified from the prior attempt.

T = 0.01;
f[x_, y_, w_] := (((x + (T/2) y) Exp[I w] + ((T/2) y - x)) (1 - 
       Exp[-0.3 T]))/((Exp[I w]^100) (Exp[I w] - 
        Exp[-0.3 T]) (Exp[I w] - 1) + 
        ((x + (T/2) y) Exp[I w] + ((T/2) y - x)) (1 - Exp[-0.3 T]));

g[x_?NumberQ, y_?NumberQ] := 
 NMaximize[{Evaluate[Abs[f[x, y, w]]], 0 <= w <= 10}, {w, 0, 10}][[1]]

NMinimize[g[x, y], {x, y}]

{0.777743387928, {x -> -6.56647441199, y -> -2.87876723563}}

--- end edit ---