Find and graph the voltages across the capacitor \( C \) and the resistor \( R \) and the current \( i \) as functions of time in the ciruit below
Fig.1 - Low Pass RC circuit
given that the input voltage is \( v_i(t) \) is a square wave as shown in the graph below.
Fig.2 - Square Wave as Input to the RC CircuitSolution to the Problem
The equation, in the \( s \) domain
, relating the voltage \( V_C(s) \) across the capacitor and the input voltage \( V_i(s) \) in an RC circuit has already been determined.
\( V_i(s) - R \; C \; s \; V_C(s) - V_C(s) = 0 \) (I)
We now need to determine the Laplace transform \( V_i(s) \) of the square wave \( v_i(t) \).
We first express the square wave as a sum of shifted unit step functions as follows
\( \displaystyle v_i(t) = V_0 \sum_{n=0}^{n=\infty} \left\{ u(t - n\;T)- u (t-(n+1/2)\;T) \right\} \)
We take the Laplace transform of both sides of the above and use the property of linerity of the Laplace transform to write
\( \displaystyle \mathscr{L} \{ v_i (t) \} = V_0 \sum_{n=0}^{n=\infty} \left\{ \mathscr{L} \{ u(t - n\;T) \} - \mathscr{L} \{ u (t-(n+1/2)\;T) \} \right\} \)
The Laplace transform of a shifted unit step function of the form \( u(t - \alpha) \) is given by
\( \dfrac{e^{-\alpha s }}{s} \)
We use the above to write the Laplace transform \( V_i(s) = \mathscr{L} \{ v_i (t) \} \) as
\( \displaystyle V_i(s) = V_0 \sum_{n=0}^{n=\infty} \left\{ \dfrac{ e^{-n\;T\;s}}{s} - \dfrac {e^{-(n+1/2)\;T \; s}}{s} \right\} \)
Substitute \( V_i(s) \) in equation (i) by the above expression and solve for \( V_C(s) \) and put all terms with \( V_C(s) \) on the right
\( \displaystyle V_0 \sum_{n=0}^{n=\infty} \left\{ \dfrac{ e^{-n\;T\;s}}{s} - \dfrac {e^{-(n+1/2)\;T \; s}}{s} \right\} = R \; C \; s \; V_C(s) + V_C(s) \)
Solve for \( V_C(s) \)
\( \displaystyle V_C(s) = \dfrac{V_0}{s(R\;C\;s + 1)} \sum_{n=0}^{n=\infty} \left\{ e^{-n\;T\;s} - e^{-(n+1/2)\;T s} \right\} \) (II)
Decompose the rational expression \( \dfrac{V_0}{s(R\;C\;s + 1)} \) into partial fractions ( see Appendix - A ) and rewrite as
\( \dfrac{V_0}{s(R\;C s + 1)} = \dfrac{V_0}{s} - \dfrac{R\;C V_0}{R\;C s + 1} \)
Divide numerator and denominator in the term on the right side by \( R\;C \) and factor \( V_0 \) out and rewrite it as
\( \dfrac{V_0}{s(R\;C s + 1)} = V_0 \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \)
Substitute the above in the expression of \( V_C(s) \) given in (II) to write \( V_C(s) \) as
\( \displaystyle V_C(s) = V_0 \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \sum_{n=0}^{n=\infty} \left\{ e^{-n\;T\;s} - e^{-(n+1/2)\;T \; s} \right\} \)
The above may be written as
We now use the formulas and properties of Laplace transform to find the inverse Laplace transform \( v_C(t) \) (time domain) of \( V_C(s) \)
We need to apply the inverse Laplace transformation to find \( v_C(t) \) from \( V_C(s) \)
\( \displaystyle v_C(t) = \mathscr{L^{-1}} \left\{ V_c(s) \right\} \)
\( \displaystyle = V_0 \sum_{n=0}^{n=\infty} \left\{ \mathscr{L^{-1}} \left\{ e^{-n\;T\;s} \left (\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \right\}
\\\\
\quad \quad \quad \quad - \mathscr{L^{-1}} \left\{ e^{-(n+1/2)\;T \; s} \left (\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right) \right\} \right\} \)
The two main terms within the curly brackets may be written as
\( \mathscr{L^{-1}} ( e^{-\tau s} F(s) ) \)
Property 2 in properties of Laplace transform may be written as
\( \mathscr{L^{-1}} ( e^{-\tau s} F(s) ) = u(t- \tau) f(t - \tau) \) , where \( f(t) \) is the inverse Laplace transform of \( F(s) \)
We now use the formulas of Laplace transform to evaluate
\( \displaystyle \mathscr{L^{-1}} \left\{ \dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right\} \)
\( \displaystyle = \mathscr{L^{-1}} \left\{ \dfrac{1}{s} \right\} - \mathscr{L^{-1}} \left\{ \dfrac{1}{s + \dfrac{1}{R\;C}} \right\} \)
\( = u(t) - u(t) e^{-\frac{t}{R\;C}} = u(t)(1 - e^{-\frac{t}{R\;C}} ) \)
Using all the above, we now write \( v_C(t) \) as
\( \displaystyle v_C(t) = \displaystyle V_0 \sum_{n=0}^{n=\infty} \left \{ u(t-nT) \; \left(1 - e^{- \dfrac{t - n \; T}{R \;C} } \right)
\\\\
\quad \quad \quad \quad
- u(t-(n+1/2)T) \; \left(1 - e^{-\dfrac{ t - (n + 1/2) T}{\; R \; C} } \right) \right\} \)
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 \( v_C(t) \) across the capacitor 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
Fig.3 - Graphs the input square wave and the voltage v_C(t) across the capacitor for a period T = 15 RC
b) \( T = 10 RC = 10 \) s
Fig.4 - Graphs the input square wave and the voltage v_C(t) across the capacitor for a period T = 10 RC
c) \( T = 5 RC = 5 \) s
Fig.5 - Graphs the input square wave and the voltage v_C(t) across the capacitor for a period T = 5 RC
d) \( T = 2 RC = 2 \) s
Fig.6 - Graphs the input square wave and the voltage v_C(t) across the capacitor for a period T = 2 RC
