Signals and Systems: RC as an Integrator

Thursday, July 25, 2013

RC as an INTEGRATOR

Integrator

Consider the output across the capacitor at high frequency i.e.

$\omega \gg \frac{1}{RC}$ .

This means that the capacitor has insufficient time to charge up and so its voltage is very small. Thus the input voltage approximately equals the voltage across the resistor. To see this, consider the expression for $I$ given above:

$I = \frac{V_{in}}{R+1/j\omega C}$

but note that the frequency condition described means that

$\omega C \gg \frac{1}{R}$

$I \approx \frac{V_{in}}{R}$ which is just Ohm's Law.

Now,

$V_C = \frac{1}{C}\int_{0}^{t}Idt$

$V_C \approx \frac{1}{RC}\int_{0}^{t}V_{in}dt$ ,

Here RC=1u*1k=1ms. It means capacitor needs approximately 5RC=5m to charge completely.However input pulse width is 2ms. so the waveform will be integrated version of input . SO RC acts as integrator.

IF c=0.5u RC=0.5m.So RC=2.5m.So capacitor will be charged almost completely as shown below

SIMULATION SOFTWARE :PSPICE .