RC circuit

A couple of simple observations:

1) The voltage across the capacitor cannot change discontinuously. Since the source voltage is equal to the capacitor voltage plus the resistor (output) voltage, if the source voltage changes discontinuously, the resistor (output) voltage must also. This means the answer is either (b) or (d).

2) In DC steady state, a capacitor is an open circuit (very high impedance). 5 time constants ( $5 \tau)$ is about the time it takes to get there. The source pulse width is ( $3 \tau)$ , which gets us close enough to DC steady state that the capacitor eats up nearly the entire source voltage before the polarity flips again, leaving almost nothing for the resistor. Choice (b) exhibits this behavior, but (d) does not. Hence, the answer is (b).

The circuit shown is a high-pass filter that attenuates the low frequencies and acts as a differentiating element. This follows from the equations for the capacitor and the resistance: $\begin{aligned} & & C (U_\text{in} - U_\text{out}) &= Q \\ \Rightarrow & & U_\text{out} &= RI = R \frac{dQ}{dt} = \tau \frac{d}{dt} (U_\text{in} - U_\text{out}) \end{aligned}$ with the time constant $\tau = RC$ . For a piecewise constant input signal $U_\text{in}$ , its derivative is zero almost everywhere, so that the differential equation is reduced to $U_\text{out} + \tau \frac{d}{dt} U_\text{out} = 0 \quad \Rightarrow \quad U_\text{out} = U_0 e^{-t/\tau}$ This exponential decay can be found in Figures b) and d). But the width of the square pulses is $\Delta t = 3 \tau$ , so that the exponential function must have dropped to $e^{-3} \approx 5\,\text{\%}$ at the end of the pulse, which corresponds to the image b).