Signals and Systems: RC as a differentiator

Monday, July 29, 2013

RC as a differentiator

Differentiator

Consider the output across the resistor at low frequency i.e.,

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

This means that the capacitor has time to charge up until its voltage is almost equal to the source's voltage. Considering the expression for $I$ again, when

$R \ll \frac{1}{\omega C}$ ,

$I \approx \frac{V_{in}}{1/j\omega C}$

$V_{in} \approx \frac{I}{j\omega C} = V_C$

Now,

$V_R = IR = C\frac{dV_C}{dt}R$

$V_R \approx RC\frac{dV_{in}}{dt}$

Circuit

output

For C=0.1u R=1k

Since RC=0.1ms . 5RC=0.5ms which is less compared to pulse width of 2ms. The resistor will lose its voltage very soon. Initially V(R)=2v because capacitor acts as short. As RC is very small.Capacitor gets 2v at the end of 0.5ms.

Undesirable output

For C=1u R=1k

SEE Video here

Monday, July 29, 2013

RC as a differentiator

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