= Wien bridge band-pass response
{c}
{title2=$H=-[3+i(u-u^{-1})]^{-1}$}
Use the equal-component <Wien bridge> with upper-left parallel arm and lower-left series arm. Let $A$ be the left arm junction, $C$ the upper input node, and $V_i$ the upper-minus-lower input <electric potential difference>. With $u=\omega RC$, the unloaded transfer ratio $H=(V_A-V_C)/V_i$ is the displayed expression. Its <band-pass filter> magnitude is $[9+(u-u^{-1})^2]^{-1/2}$, maximal at $u=1$ with $H=-1/3$. One continuous <wave phase> is $\phi=-\pi-\arctan[(u-u^{-1})/3]$, tending from $-\pi/2$ at low positive <frequency> to $-3\pi/2$ at high <frequency>. Reversing the measured output polarity multiplies $H$ by $-1$ and shifts its phase by $\pi$ without changing its magnitude.
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