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Past exam of the physics 2 paper of the University of Oxford

2018

The output is measured across the <capacitor>. The <voltage divider> gives the <RC low-pass filter> transfer function
$$
H(i\omega)=\frac{Z_C}{R+Z_C}=\frac1{1+i\omega RC}.
$$
Therefore
$$
\boxed{V_{\mathrm{out}}(t)=\frac{V_0}{\sqrt{1+(\omega RC)^2}}\cos\bigl(\omega t-\arctan(\omega RC)\bigr).}
$$
The amplitude ratio tends to $1$ and the <phase shift> to zero as $\omega\to0$. For $\omega\to\infty$, the amplitude ratio is asymptotic to $1/(\omega RC)$ and the <phase shift> tends to $-\pi/2$.

\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2018-rc-frequency-and-square-response.png]
{title=The <RC low-pass filter> amplitude and phase response, and periodic responses to a symmetric square-wave input at three values of $\omega RC$}

Solution

Past exam of the physics 2 paper of the University of Oxford / 2018 / 9 / a / Solution

Past exam of the physics 2 paper of the University of Oxford / 2018 / 9 / a / Solution

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Past exam of the physics 2 paper of the University of Oxford / 2018 / 9 / a / Solution

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