Question 4.7: Response of an RC lowpass filter to a square wave using supe......
Response of an RC lowpass filter to a square wave using superposition
Use the principle of superposition to find the response of an RC lowpass filter to a square wave that is turned on at time t = 0 . Let the RC time constant be 1 ms, let the time from one rising edge of the square wave to the next be 2 ms, and let the amplitude of the square wave be 1 V (Figure 4.24).
We have no formula for the response of the RC lowpass filter to a square wave, but we do know how it responds to a unit step. A square wave can be represented by the sum of some positive and negative time-shifted unit steps. So x(t) can be expressed analytically as
x(t) = x_0 (t) + x_1(t) + x_2(t) + x_3(t) + \cdot \cdot \cdotx(t) = u(t) − u(t − 0.001) + u(t − 0.002) − u(t − 0.003) + \cdot \cdot \cdot
The RC lowpass filter is a linear, time-invariant system. Therefore, the response of the filter is the sum of the responses to the individual unit steps. The response to one unshifted positive unit step is y_0 (t) = (1− e^{−1000t} )u(t). Invoking time invariance,
y_1(t)= −(1−e^{−1000(t−0.001)})u(t−0.001)\\ y_2(t)= (1−e^{−1000(t−0.002)})u(t−0.002)\\ y_3(t)= −(1−e^{−1000(t−0.003)})u(t−0.003)\\ \quad \vdots \quad\quad\quad\quad\quad \vdots \quad\quad\quad\quad\quad \vdotsThen, invoking linearity and superposition,
y(t) = y_0 (t) + y_1(t) + y_2(t) + y_3(t) + \cdot \cdot \cdot\\(see Figure 4.26).
