An inverse system is a system such that $x \to \fbox{$P(\omega)$} \to \fbox{$P(\omega)^{-1}$} \to y = x$.
%\begin{center}
%\includegraphics{images/inverses/invdiag2.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-4)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.5,0){$k$} 
\rput(3.75,-2){$P(\omega)$}
\rput(1.5,-2.5){$H$}

% plus or minus for adder
\rput(2.1,-0.5){$-$}
\rput(1.45,0.35){$+$}

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\psline[linewidth=1.25 pt](3,-2)(1.75,-2)
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Given the system in the figure, find the frequency response. Note that the plant is a multiplier. After inspection, we find that 
$$ H(\omega) = \frac{k}{1+kP(\omega)}  $$

Thus, if we let $k$ be large enough, then $kP(\omega) \gt\gt 1$ and $kP(\omega)$ dominates the denominator. Using this information, we can make a good approximation:

$$ H(\omega) = \frac{k}{1+kP(\omega)} \approx  \frac{k}{kP(\omega)} =  \frac{1}{P(\omega)}$$  
This works well as long as $kP(\omega) \gt\gt 1$ works for our frequency range of interest. Lets look at a system where the controller is a multiplier:

%\begin{center}
%\includegraphics{images/inverses/invdiag.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-4)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.75,0){$P(\omega)$}
\rput(4,-2){$k$}
\rput(1.5,-2.5){$H$}

% plus or minus for adder
\rput(2.1,-0.5){$-$}
\rput(1.45,0.35){$+$}

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\psframe(3,-0.5)(4.5,0.5)

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\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.75,-2)(1.75,-0.25)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(4.5,0)(7,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(5.5,-2)(4.5,-2)
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We have a similar situation. If we let $k$ be large enough, then $kP(\omega) \gt\gt 1$ and we have 

$$ H(\omega) =  \frac{P(\omega)}{1+kP(\omega)}  \approx  \frac{P(\omega)}{kP(\omega)} = \frac{1}{k} $$

It is important to note that letting $k$ become large introduces some consequences. We lose magnitude for flatness over a frequency range. $1/k$ is not much gain, but a flat signal, whereas $P(\omega)$ may be a signal with large amplitude but not flat. 


Inverse Systems