Difference between revisions of "User:DukeEgr93/MagOrder"
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\(
\begin{align}
\Bbb{H}(j\omega)&=H(\omega)e^{j\theta(\omega)}
\end{align}
\)
\(
\begin{align}
\frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right)
\end{align}
\)
\(
\begin{align}
\left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|
\end{align}
\)
\(
\begin{align}
\frac{d}{d\omega}\left(\left|\Bbb{H}(j\omega)\right|\right)&=
\frac{d}{d\omega}\left(\left|H(\omega)e^{j\theta(\omega)}\right|\right)\\
~&=\frac{d}{d\omega}\left(H(\omega)\right)
\end{align}
\)
\(
\begin{align}
\left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|&=
\left|\frac{d}{d\omega}\left( H(\omega)e^{j\theta(\omega)} \right)\right|\\
~&=\left| \left(
\frac{d}{d\omega}H(\omega)e^{j\theta(\omega)}+
H(\omega)e^{j\theta(\omega)} j\frac{d}{d\omega}\theta(\omega)
\right)\right|\\
~&=\left|
\frac{d}{d\omega}H(\omega)+
H(\omega) j\frac{d}{d\omega}\theta(\omega)
\right|
\end{align}
\)
\(
\begin{align}
H(\omega)&=0 &
\mbox{o}&\mbox{r} &
\frac{d}{d\omega}\theta(\omega)&=0
\end{align}
\)
| (One intermediate revision by the same user not shown) | |||
| Line 33: | Line 33: | ||
\begin{align} | \begin{align} | ||
\left|\frac{d}{d\omega}\left(\Bbb{H}(j\omega)\right)\right|&= | \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|\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} | \end{align} | ||
</math></center> | </math></center> | ||
| + | |||
| + | == Comparison == | ||
| + | These two calculations will thus have the same magnitude (though potentially different signs) if | ||
| + | |||
| + | <center><math> | ||
| + | \begin{align} | ||
| + | H(\omega)&=0 & | ||
| + | \mbox{o}&\mbox{r} & | ||
| + | \frac{d}{d\omega}\theta(\omega)&=0 | ||
| + | \end{align} | ||
| + | </math></center> | ||
| + | |||
| + | otherwise, due to the orthogonality of the terms, they must have different magnitudes. | ||
Latest revision as of 19:38, 20 November 2009
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.
