Past exam of the physics 2 paper of the University of Oxford / 2019 resit / 5

Table of contents
- Solution 5

Solution

Let

v_{C}

denote the capacitor voltage with the same polarity as the applied voltage. By Kirchhoff's voltage law, Ohm's law and

I = C d v_{C} / d t

,

RC \frac{d v _{C}}{d t} + v_{C} = V, v_{C} (0) = 0.

Solving this first-order differential equation gives the RC low-pass step response

v_{C} (t) = V (1 - e^{- t / τ}), τ = RC = 0.050 s .

The capacitor voltage is continuous at switching because an instantaneous finite jump would require an impulsive electric current.

For the square-wave response of an RC low-pass filter, take equal positive and negative half-periods. Let

A = 10 V

,

T = 1/ f

,

h = T /2

and

α = e^{- h / τ}

. After the transient, symmetry makes the capacitor voltage alternate between

- v_{p}

and

+ v_{p}

. Over the positive half-period,

v_{C} (t) = A - (A + v_{p}) e^{- t / τ}, 0 \leq t < h .

Requiring

v_{C} (h) = v_{p}

gives

v_{p} = A \frac{1 - α}{1 + α} = A tanh (\frac{T}{4 τ}) .

The next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:

At $1 kHz$ , $h / τ = 0.01$ . The voltage barely changes in each half-period, so the output is nearly triangular about zero, with extrema about $\pm 0.0500 V$ . Its small-signal slope is approximately $\pm A / τ = \pm 200 V s^{- 1}$ .
At $10 Hz$ , $h = τ$ . Clearly curved exponential rises and falls reach only $\pm 4.621 V$ ; the output never settles to either input level.
At $0.1 Hz$ , $h = 100 τ$ . The output closely follows the input plateaux at $\pm 10 V$ , joined by continuous exponential transitions with time constant $0.050 s$ .

Figure 1.
Steady periodic capacitor voltages for three square-wave frequencies
. The blue curves use the exact periodic RC low-pass filter solution with $RC = 0.050 s$ . Dashed grey curves show the applied square wave in the lower panels; the first panel uses an expanded output scale. Time is measured in input periods. Details.

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