# Step response and time constant of RC circuit with multiple capacitors

I need to find the step response and the time constant of the given RC circuit: :

Can anybody please share the method to solve this circuit?

Also, if R1=R2=R and C1=C2=C; then is the step response a step function( with Vmax= Vi/2) ? I tried solving it using frequency domain analysis;

Vo(s)/Vin(s) = (R||1/Cs)/( R|| 1/Cs + R|| 1/Cs)

         = 1/2


since transfer function is a constant and the step input can be modeled byout Vmax/s;

Vout(s) = Vmax/2s; Vout(t) = Vmax/2 * u(t)

which shows that the voltage across capacitor increases immediately. Is this correct? If so, how do we justify the immediate rise in the voltage across capacitor?

• There's a strong smell of homework from this question. You need to show some effort before you'll get help. Show your calculations and where you got stuck. After all, you might be competing for my job when you qualify and if you can't solve this ... ;^) Aug 13 '16 at 22:02
• I tried; i have mentioned my approach for R1= R2 and C1=C2; what i am not able to justify is the immediate change in capacitor potential which is valid for all other cases too. Aug 14 '16 at 3:57
• Your Laplace TF equation is correct for the case of equal resistor and equal capacitor values. Now solve with the appropriate numerical subscripts in-place.
– Chu
Aug 14 '16 at 9:14
• @Chu So, the voltage across the capacitor jumps immediately, right? Aug 14 '16 at 18:29
• Use conservation of charge to derive the voltage across the capacitors at t=0+. Q = C1*V1 = C2*V2 and Q = (C1||C2) * Vi. With the assumption that charges (integration of current) going through R1 or R2 are essentially 0 for time duration of 0. Aug 14 '16 at 20:43

This is indeed a interesting question, I have to write a very long explanation for this. So bear with me if I made some mistakes.

My approach is based more on basics than Laplace etc etc.

When you hear something like:

"The voltage across a capacitor cannot change immediately" or

"The current in a inductor cannot change immediately".

Always understand the assumptions behind these statements. They are true yes, but with some assumptions. We generally tend to miss that small asterisk*.. Anyways :)

First let me give you two simple scenario in which both of the above statements can go for a toss. Look at the figure below.

Can you see that the voltage across the capacitor changed instantly at t=0 ? Likewise, the current in the inductor.

Where do these two statements come from ?? Lets take a step back and understand this.

For a capacitor : $$I(t)=C\frac{dV(t)}{dt}$$ $$V(t)=\frac{1}{C}\int_{-inf}^{t}I(t)dt$$

1. lets calculate V(t) at certain time t0

$$V(t0)=\frac{1}{C}\int_{-inf}^{t0}I(t)dt$$ -- Eq(1)

1. Using the same equation, lets now calculate V(t) at certain time $t0+\Delta{t}$, where the delta is a very small time instant.

$$V(t0+\Delta{t})=\frac{1}{C}\int_{-inf}^{t0+\Delta{t}}I(t)dt$$ -- Eq(2)

1. The change in voltage between these two instants is the difference between Eq(2) and Eq(1)

$$V(t0+\Delta{t})-V(t0)=\frac{1}{C}\int_{-inf}^{t0+\Delta{t}}I(t)dt-\frac{1}{C}\int_{-inf}^{t0}I(t)dt$$

$$\Delta{V}=\frac{1}{C}\int_{t0}^{t0+\Delta{t}}I(t)dt$$ Now here you have to observe carefully, as the delta t tends to 0. Let I(t) be any function which occurs in our day to day life ( exponential or sinusoidal or ramp etc ). No matter what the function I(t) is, the above integral tends to zero.

$$(lim \Delta{t}->0)\Delta{V}=\frac{1}{C}\int_{t0}^{t0+\Delta{t}}I(t)dt =0$$

Hence the statement "The voltage across a capacitor cannot change immediately" Yes, this statement is absolutely true in day to day life.

1. However, when you bring special functions like dirac-delta or impulse function etc ( with infinite amplitude at an instant etc etc.. but finite area under that curve ), which actually do not occur in reality but occur in textbooks and can be mimicked in simulators etc. The above statement doesn't hold true.

$$(lim \Delta{t}->0)\Delta{V}=\frac{1}{C}\int_{t0}^{t0+\Delta{t}}\delta(t)dt \neq 0$$

Hence when we allow for impulse currents it can be easily seen that the voltage can indeed change instantly across a capacitor.

A similar line of equations can be written for inductors as well and shown that when we allow for impulse voltages, its current can change instantly.

1. Now, think of all capacitors as short circuits for a very short time whenever you applied a instant change. In the example I have quoted, you can see that the capacitor acting as a short initially, leads to a huge instant current ( which is the impulse current we introduced unknowingly ), leading to a instant change in voltage.

2. If we change the circuit slightly.

Now even though the capacitor is initially a short, your current is still limited because you have a 1 ohm resistor. Hence in this case the voltage on your cap can never change instantly.

1. In your case, you are unkowingly forming a short to ground

This is leading to a instant impulse current, which charges the two caps instantly ( like the resistors do not exist ).

Therefore you see a instant change in voltage across the caps. - Cap1 at t=0, will charge to viC2/(C1+C2). - Cap2 at t=0, will charge to viC1/(C1+C2).

The final voltages across the cap however are determined by the resistor - Cap1 at t=inf, will be charged to viR1/(R1+R2). - Cap2 at t=inf, will be charged to viR2/(R1+R2).

They have to go from that initial voltage to that final voltage with a time constant of (R1||R2)*(C1+C2) ---- Explanation for this can be found anywhere.

The step response is composed of two exponentials, of the forms: $A(1-e^{-t/\tau})$ and $Be^{-t/\tau}$, so the response at $\small t=0$ is non-zero.

The TF for this circuit has one pole and one zero and, as is often the case, the zero makes the problem look more difficult than it actually is. There is a useful 'trick' for time domain analysis for systems with one or more zeros, based on the fact that differentiation in the time domain is equivalent to multiplying by 's' in the Laplace domain.

To illustrate, take the TF: $$G(s)=\dfrac{1+as}{1+bs}$$

This has step response: $$R(s)=\dfrac{1}{s}.\dfrac{1+as}{1+bs}$$

We can decompose the TF: $$R(s)=\dfrac{1}{s}.\left ( \dfrac{1}{1+bs}+\dfrac{as}{1+bs}\right)$$

The step response of the first term is simply: $$R_1(s)=\dfrac{1}{s}.\dfrac{1}{1+bs}\rightarrow r_1(t)=1-e^{-t/b}$$

and the step response of the second term is the derivative of $r_1(t)$, multiplied by $a$, thus: $$r(t)= r_1(t) +a\:\dfrac{d}{dt}r_1(t)=1-e^{-t/b}+a\dfrac{d}{dt}(1-e^{-t/b})=1-e^{-t/b}+\dfrac{a}{b} \:e^{-t/b}$$ which gives $r(t)=\dfrac{a}{b}$ at $t=0$.

This approach helps conceptually as well as being useful analytically. For instance it demonstrates that numerator zeros have no influence on absolute stability (unless they're open-loop zeros!) but only augment the natural response. This is because the derivative of a stable exponential is, essentially, the same stable exponential.

Additionally, it can be seen that a zero in the right-half of the s-plane (i.e. a negative value of $a$) can give rise to an initial negative-going step response.

which shows that the voltage across capacitor increases immediately. Is this correct? If so, how do we justify the immediate rise in the voltage across capacitor?

This shows you're thinking about it, not just blindly plugging in numbers. That's good, but you need to think a little further.

Pretend for a moment that the resistors aren't there. You have a voltage divider made of only two capacitors. Ideally, the input voltage appears across each cap, scaled inversely proportional to its capacitance. So yes, a theoretical step input yields a step across each cap.

What you are probably struggling with is that this requires infinite current for a infinitely short time. This again shows you're thinking. Keep in mind this is only in theory, where you can have things like perfect step functions and Dirac delta functions. In the real world this can't happen because of the infinite current. Any attempt to change the voltage across a cap too quickly requires so much current that the voltage can't be changed so fast. Also, at that level secondary effects like the ESR (equivalent series resistance) of the cap, the resistance and inductance of the wire, and the like, start to matter.

In theory, practice is like theory. In practice, it's not.