User:DukeEgr93/MagOrder
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\(
\begin{align}
\Bbb{H}(j\omega)&=H(\omega)e^{j\theta(\omega)}
\end{align}
\)
\(
\begin{align}
\frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right)
\end{align}
\)
\(
\begin{align}
\left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|
\end{align}
\)
\(
\begin{align}
\frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right)&=
\frac{d}{d\omega}\left(\left|H(\omega)e^{j\theta(\omega)}\right|\right)\\
~&=\frac{d}{d\omega}\left(H(\omega)\right)
\end{align}
\)
\(
\begin{align}
\left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|&=
\left|\frac{d}{d\omega}\left( H(\omega)e^{j\theta(\omega)} \right)\right|\\
~&=\left| \left(
\frac{d}{d\omega}H(\omega)e^{j\theta(\omega)}+
H(\omega)e^{j\theta(\omega)} j\frac{d}{d\omega}\theta(\omega)
\right)\right|\\
~&=\left|
\frac{d}{d\omega}H(\omega)+
H(\omega) j\frac{d}{d\omega}\theta(\omega)
\right|
\end{align}
\)
\(
\begin{align}
H(\omega)&=0 &
\mbox{o}&\mbox{r} &
\frac{d}{d\omega}\theta(\omega)&=0
\end{align}
\)
Question: for a transfer function
what is the relationship between
and
Magnitude First
which makes sense since \(H\) is the magnitude of \(\Bbb{H}\).
Derivative First
Comparison
These two calculations will thus have the same magnitude (though potentially different signs) if
otherwise, due to the orthogonality of the terms, they must have different magnitudes.