Question 12.6: Integrating a Square Wave Determine the output voltage for t...
Integrating a Square Wave
Determine the output voltage for the integrator circuit of Figure 12.31 if the input is a square wave of amplitude \pm A and period T.
Known Quantities: Feedback and source impedances; input waveform characteristics.
Find: v_{\text {out }}(t).
Schematics, Diagrams, Circuits, and Given Data: T=10 \mathrm{~ms} ; C_{F}=1 \mu \mathrm{F} ; R_{S}=10 \mathrm{k} \Omega.
Assumptions: Assume ideal op-amp. The square wave starts at t=0 and therefore v_{\text {out }}(0)=0.
Analysis: Following equation 12.67,
v_{\mathrm{out}}(t)=-\frac{1}{R_S C_F} \int_{-\infty}^t v_S\left(t^{\prime}\right) d t^{\prime} (12.67)
we write the expression for the output of the integrator:
\begin{aligned}v_{\text {out }}(t) &=-\frac{1}{R_{F} C_{S}} \int_{-\infty}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}=-\frac{1}{R_{F} C_{S}}\left(\int_{-\infty}^{0} v_{S}\left(t^{\prime}\right) d t^{\prime}+\int_{0}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}\right) \\&=-\frac{1}{R_{F} C_{S}}\left(v_{\text {out }}(0)+\int_{0}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}\right)\end{aligned}
Next, we note that we can integrate the square wave in a piecewise fashion by observing that v_{S}(t)=A for 0 \leq t<T / 2 and v_{S}(t)=-A for T / 2 \leq t<T. We consider the first half of the waveform:
\begin{aligned}v_{\text {out }}(t) &=-\frac{1}{R_{F} C_{S}}\left(v_{\text {out }}(0)+\int_{0}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}\right)=-100\left(0+\int_{0}^{t} A d t^{\prime}\right) \\&=-100 A t \quad 0 \leq t<\frac{T}{2} \\v_{\text {out }}(t) &=v_{\text {out }}\left(\frac{T}{2}\right)-\frac{1}{R_{F} C_{S}} \int_{T / 2}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}=-100 A \frac{T}{2}-100 \int_{T / 2}^{t}(-A) d t^{\prime} \\&=-100 A \frac{T}{2}+100 A\left(t-\frac{T}{2}\right)=-100 A(T-t) \quad \frac{T}{2} \leq t<T\end{aligned}
Since the waveform is periodic, the above result will repeat with period T, as shown in Figure 12.32.
Comments: The integral of a square wave is thus a triangular wave. This is a useful fact to remember. Note that the effect of the initial condition is very important, since it determines the starting point of the triangular wave.
