Difference between revisions of "User:DukeEgr93/MagOrder"

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== Comparison ==
 
== Comparison ==
These two items will have the same magnitude (though potentially different signs) if
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These two calculations will thus have the same magnitude (though potentially different signs) if
  
 
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Latest revision as of 19:38, 20 November 2009

Question: for a transfer function

\( \begin{align} \Bbb{H}(j\omega)&=H(\omega)e^{j\theta(\omega)} \end{align} \)

what is the relationship between

\( \begin{align} \frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right) \end{align} \)

and

\( \begin{align} \left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right| \end{align} \)

Magnitude First

\( \begin{align} \frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right)&= \frac{d}{d\omega}\left(\left|H(\omega)e^{j\theta(\omega)}\right|\right)\\ ~&=\frac{d}{d\omega}\left(H(\omega)\right) \end{align} \)

which makes sense since \(H\) is the magnitude of \(\Bbb{H}\).

Derivative First

\( \begin{align} \left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|&= \left|\frac{d}{d\omega}\left( H(\omega)e^{j\theta(\omega)} \right)\right|\\ ~&=\left| \left( \frac{d}{d\omega}H(\omega)e^{j\theta(\omega)}+ H(\omega)e^{j\theta(\omega)} j\frac{d}{d\omega}\theta(\omega) \right)\right|\\ ~&=\left| \frac{d}{d\omega}H(\omega)+ H(\omega) j\frac{d}{d\omega}\theta(\omega) \right| \end{align} \)

Comparison

These two calculations will thus have the same magnitude (though potentially different signs) if

\( \begin{align} H(\omega)&=0 & \mbox{o}&\mbox{r} & \frac{d}{d\omega}\theta(\omega)&=0 \end{align} \)

otherwise, due to the orthogonality of the terms, they must have different magnitudes.