I'm trying to analyse the following circuit to find the differential equation describing the relationship between \$V_{in}(t)\$ and \$V(t)\$:
Where \$C_1 = 2uF\$, \$C_2= 1uF\$ and \$R_1=R_2=1k\Omega\$.
I understand that:
\$V=IZ\$
\$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.
. In general, if you have 2 circuits connected together, the first one's output to the second one's input, you do not want the second one to attenuate the signal that passes through the first one. Therefore, you must make sure that the second circuit's input impedance (represented by R2 in the voltage divider) is larger than your first circuit's output impedance. How much larger? The rule of thumb is 10x larger.
Best Answer
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..