The study the response of high pass RC circuits to a square wave input; numerical examples with graphs of voltages are presented.
An online calculator and grapher on high pass RC circuit response to a square wave is also included.

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Problem with Solution

Find and graph the voltages across the capacitor \( R \) as function of time in the high pass \( RC \) ciruit below

given that the input voltage is \( v_i(t) \) is a square wave as shown in the graph below.
Solution to the Problem
In the study of the low pass RC circuit response to a square wave, it was found that the voltage across the capacitor is given by
\( \displaystyle v_C(t) = \displaystyle V_0 \sum_{n=0}^{n=\infty} \left \{ u(t-nT) \; \left(1 - e^{- \dfrac{t - n \; T}{R \;C} } \right)
\\\\
\quad \quad \quad \quad
- u(t-(n+1/2)T) \; \left(1 - e^{-\dfrac{ t - (n + 1/2) T}{\; R \; C} } \right) \right\} \)
when the input \( v_i(t) \) voltage is a square wave modelled by a sum of positive and negative step unit functions of the form

\( \displaystyle v_i(t) = V_0 \sum_{n=0}^{n=\infty} \left\{ u(t - n\;T)- u (t-(n+1/2)\;T) \right\} \)
In this study we need to find the voltage \( v_R(t) \) cross the resistor which is given by
\( v_R(t) = v_i(t) - v_C(t)\)
When \( v_i(t) \) and \( v_C(t) \) are substituted by their expression given above, we can simplify \( v_R(t) \) to
\( \displaystyle v_R(t) = \displaystyle V_0 \sum_{n=0}^{n=\infty} \left \{ u(t-nT) \; \left(e^{- \dfrac{t - n \; T}{R \;C} } \right)
\\\\ \quad \quad \quad
- u(t-(n+1/2)T) \; \left(e^{-\dfrac{ t - (n + 1/2) T}{\; R \; C} } \right) \right\} \)

Numerical Applications
Let \( V_0 = 10 \) V , \( R = 200 \; \Omega \) and \( C = 5 \) mF.
\( R\;C = 200 \times 5 \times 10^{-3} = 1 \) s (seconds)
Below are shown the graphs of the input \(v_i(t) \) as a square wave defined above as a sum of shifted step functions and the voltage \( v_R(t) \) across the resistor also given above. There are four graphs for different values of the period \( T \) of the input square wave.
a) \( T = 15 RC = 15 \) s

b) \( T = 10 RC = 10 \) s
c) \( T = 5 RC = 5 \) s
d) \( T = 2 RC = 2 \) s