So I have a circuit like this:

With the output equation as: \$ V_{out} = -(\frac{R_2}{R_1})\frac{s}{s+\frac{1}{R_1C}} V_{in} \$

So the question is to derive the time-domain equation in terms of the input and also to show that the circuit performs the function of a differentiator.

So here's my work:
\[ V_{out} = -(\frac{R_2}{R_1})\frac{s}{s+\frac{1}{R_1C}} V_{in} \] \[ \mathcal{L^{-1}}(V_{out}) = -(R_2C)\mathcal{L^{-1}}(\frac{s}{sR_1C+1} V_{in}) \] If we assume that \$R_1=0\$ then we have \[ \mathcal{L^{-1}}(V_{out}) = -(R_2C)\mathcal{L^{-1}}(s V_{in}) \] The \$sV_{in}\$ implies a differentiator since \$ \mathcal{L^{-1}}(s) = \delta'(t) \$

But I'm pretty sure that is either totally wrong or mostly wrong. Where am I going wrong on this derivation for the time-domain equation?


1 Answer 1


The most straightforward way to work this with R1 = 0 is to see that the time-domain current through the capacitor is just:

\$i_C = C \dfrac{dv_c}{dt} = C\dfrac{dv_{in}}{dt} \$

due to the virtual ground at pin 2 of the (ideal) op amp.

It follows that:

\$v_{out} = -R_2C\dfrac{dv_{in}}{dt} \$

Now, the circuit pictured is not a differentiator; it is a high-pass filter. However, when the frequency is much lower than \$\frac{1}{R_1C} \$, the frequency-domain output is approximately:

\$V_{out} = -R_2CsV_{in} \$

But differentiation in the time domain is multiplication by s in the complex frequency domain. So, the circuit is an approximation of a differentiator when the input frequencies of interest are much lower than \$\frac{1}{R_1C} \$.

  • \$\begingroup\$ A real op-amp will of course have a pole in it's open-loop response and this interacts with R2 & C to form a bandpass filter with very low damping & consequent ringing. R1 helps reduce this effect in practical circuits. \$\endgroup\$
    – MikeJ-UK
    Commented Feb 22, 2013 at 9:02

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