I'm trying to analyse the following circuit to find the differential equation describing the relationship between \$V_{in}(t)\$ and \$V(t)\$:

Circuit Diagram

Where \$C_1 = 2uF\$, \$C_2= 1uF\$ and \$R_1=R_2=1k\Omega\$.

I understand that:


\$I_c = C \dfrac{dv}{dt}\$

and \$X_C = \dfrac{1}{2\pi f C}\$

So from this, I can work out that current through \$R_1\$ and \$C_1\$ is given by:

(1) \$I_{c_1} = 2\dfrac{dv}{dt} \times 10^{-6}A\$

And I get 3 KVL equations:

(2) \$V_{in}(t) - V_{R_1} - V_{C_1} - V_{R_2} = 0\$

(3) \$V_{in}(t) - V_{R_1} - V_{C_1} - V_{C_2} = 0\$

(4) \$V(t) - R_{C_2} + R_{R_2} = 0\$

This shows that: \$V_{C_2} = V_{R_2} = V(t)\$ which is obvious from the diagram.

This can then be substituted into either (2) or (3) to give:

(5) \$V_{in}(t)-V_{R_1}-V_{C_1}=V(t)\$

It's at this juncture I get stuck and I'm not sure how to proceed. I don't need a full solution, just some advice on what to try next.


2 Answers 2


Here you can take two approaches. In the second approach you can see where your confusion lies..

Approach 1 - s-domain. This approach leads to an equation containing Laplace terms, which, additionally, requires you to take Laplace inverse to find the equation in the time domain. That's ok if you're familiar with the Laplace transform. For example, if the initial conditions are zero, then taking Laplace inverse of s^2V(s) gives V''(t), that of sV(s) gives V'(t), and of V(t) gives V(s).

Approach 2 - time domain. This sometimes might be tricky to get around but still easy to handle for this problem. Assuming initial conditions are zero, what you need here is to first find the current through the circuit, which is simply $$ i(t)=\frac{V(t)}{R}+C_2\frac{dV(t)}{dt}\ \ (*)$$

Now a KVL leads to

$$V_{in}(t)=i(t)R_1+V_{C_1}+V(t) \ \ (**)$$ where that comes your confusion. You just need to put \$V_{C_1}=\frac{1}{C_2}\int i(t)dt\$ and get the expression as \$V_{in}(t)=i(t)R_1+\frac{1}{C_2}\int i(t)dt+V(t)\$. Now to get rid of the integration try to differentiate from both sides w.r.t time,

$$ \frac{V_{in}(t)}{dt}=R_1\frac{di(t)}{dt}+\frac{i(t)}{C_2}+\frac{dV(t)}{dt} \ \ (***)$$

Now simply substitute \$i(t)\$ from (*) into (***) and you'll get the desired differential equation. As an extra exercise for you, try to compare the resultant equation to the one you obtained from s-domain. You must get pretty the same equations..

  • \$\begingroup\$ Thanks, to confirm, that would be \$R_2\dfrac{dv}{dt}\times 10^{-6}=V(t)\$? \$\endgroup\$
    – user159015
    Commented Feb 20, 2019 at 11:21
  • \$\begingroup\$ Which would be: \$V(t) = \dfrac{dv}{dt}\times10^{-3}\$ \$\endgroup\$
    – user159015
    Commented Feb 20, 2019 at 11:33
  • \$\begingroup\$ No, no, how did you arrive at this eq.? Another way of saying above is that think of series R1 and C1 as Z1 and parallel C2 and R2 as Z2. Now can you tell me what V is in terms of V(in)? \$\endgroup\$
    – dirac16
    Commented Feb 20, 2019 at 11:46
  • 1
    \$\begingroup\$ Okay, I see where I've gone wrong there... so if I let those equal Z1 and Z2, when I apply KCL (\$ I=\dfrac{V}{Z}\$) I get \$ \dfrac{V_{in}(t) - V(t)}{Z_1}=\dfrac{V(t)}{Z_2}\$ which I believe can be rearranged to: \$ V(t)=\dfrac{V_{in}(t)}{\dfrac{Z_1}{Z_2}+1}\$ and then onto \$ \dfrac{V_{in}(t)Z_2}{Z_1+Z_2}\$. Which looks like a voltage divider. So if I work out the reactance of the caps and combine them with the resistors I should get \$Z_1\$ and \$Z_2\$? \$\endgroup\$
    – user159015
    Commented Feb 20, 2019 at 12:04
  • 1
    \$\begingroup\$ Thanks @dirac16, I make \$Z_1 =1000+\dfrac{1}{2j\omega\times10^{-6}}\$ and \$\dfrac{1}{Z_2}=\dfrac{1}{1000}+j\omega\times10^{-6} \implies Z_2 = \dfrac{1,000,000}{1000j\omega}\$ I'm just trying to substitute that into the equation for \$V(t)\$ and simplify \$\endgroup\$
    – user159015
    Commented Feb 20, 2019 at 13:31

Well, we know that:


Using the values that are given:


For sine functions we can set \$\text{s}=\text{j}\omega\$, then we get:

$$\mathcal{H}\left(\text{j}\omega\right)=\Re\left(\mathcal{H}\left(\text{j}\omega\right)\right)+\text{j}\cdot\Im\left(\mathcal{H}\left(\text{j}\omega\right)\right)=$$ $$\frac{2500000\omega^2}{6250000\omega^2+\left(\omega^2-500000\right)^2}-\frac{1000\omega\left(\omega^2-500000\right)}{250000000000+5250000\omega^2+\omega^4}\cdot\text{j}\tag3$$


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