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Course: Electrical engineering > Unit 2
Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations
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RC step response solve (2 of 3)
We find the RC step response using the method of Natural + Forced response = Total response. Created by Willy McAllister.
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Why not use just regular methods for solving ODE rather than going through all that mathematical acrobatics?(8 votes)
Please don't be selfish, and let others know what methods you would have used to solve the ODE? This type of math isn't really geared towards mathematicians, the same is seen in physics lectures where they often just "guess" the solution.
The reason for the "acrobatics" is to gain an understanding of different concepts as you go through the problem.(8 votes)
I posted this first as a tip, but think it will be seen more here... is this helpful?
I was initially confused by this lesson, because the instructor is purporting to solve the total response through a summation of the natural and forced responses. However, the terminology used here is a bit loose. Each response consists of two parts, a transient portion and a permanent portion.
For the forced response, the transient portion is the charging of the capacitor until the voltage across the capacitor is equal to Vs. The permanent portion of the forced response is the steady input voltage of Vs that is applied to the circuit.
For the natural response, the transient portion is the drop in voltage of the capacitor as it discharges to 0 volts. The permanent portion of the natural response is nullified, because its steady-state is 0.
Now, what the instructor refers to as the "natural response" portion of the total response is actually a sum of the two transient portions of both the natural and forced responses. What he refers to as the "forced response" portion of the total response is actually just the permanent portion of the forced response.
However, the instructor's solution will always work out correctly, because he is still adding up all the pieces, just not in the intuitive order described. Also, the transient portions of the responses have the same form of A*e^(-t/RC), so when we solve for that constant 'A', we should get the correct value for the total response solution, as long as we do this solving under the constraints of the total response, and not just those of the natural response (which would give us a value of A = V_0).
This link should help clarify some of the confusion on this lesson (read the top-rated answer by Felipe_Ribas): https://electronics.stackexchange.com/questions/93061/difference-between-natural-response-and-forced-response(7 votes)- Hi,
I think the guessing that the solution of the equation
dv/dt + v/RC = Vs/RC is a constant was not very successful
The question of Chambers Wong was legitimate, the natural response is as you said in a previous video V0 e^-t/RC
and there is no reason to pick an other constant because a time t=0 the voltage across the capacitor is V0
The question of Learner was also legitimate the voltage across the capacitor can't be Vs (at least at the beginning).
All this inconstancy is because the guess was not successful.
Consider the mathematical solution for dv/dt + v/RC = Vs/RC which is Vs(1-e^-t/RC) [you can verify that's a valid solution].
This time if we consider superposition we get:
1) V0e^-t/RC
2) Vs(1-e^-t/RC)
and adding these give us the solution (V0-Vs)e^-t/RC + Vs(4 votes) 
In previous video, it says natural response equation has a fixed constant ratio Vo. and in this video, you replace it with Kn. Since we already have presumed the initial i and v in natural response condition, then why not a Vo? thank you for your help.(2 votes)
In the previous RC Natural Response video, we made a guess at a solution. This is where we introduced a K as a placeholder for the amplitude term in our guess. In that problem, it turned out K = Vo.
In this video on RC Step Response we start with a different circuit that includes a step voltage source. There is no justification for bringing forward the solution from the previous just-RC circuit. In this step response problem we use the same guessing technique to solve a different differential equation. Again, we introduce K as a placeholder for the amplitude term. K turns out to be Vo - VS in this case.(1 vote)
For step response, is it valid to derive the step response formula from the natural response formula by changing the reference point so that Vs becomes ground (0)?
The step response graph looked really similar to a natural response graph that was flipped over the x-axis and translated up.
Natural response equation: V(t)=(V0)e^(-t/RC)
Transform the graph so that V becomes V - Vs,
and Vs becomes Vs - Vs = 0,
and V0 becomes V0 - Vs,
and V(t) becomes V(t) - Vs,
so that we get the same formula without any calculus:
V(t)=(V0)e^(-t/RC) becomes V(t)-Vs=(V0-Vs)e^(-t/RC)
and we end up with the same V(t)=(V0-Vs)e^(-t/Rc)+Vs
This looks too simple to be true. Is there anything wrong with this reasoning?(1 vote)- There's nothing wrong with your reasoning, but I think perhaps you are taking full advantage of hindsight since you know what the solution should be. I don't know if you could take this path if you started from scratch with just the natural response solution. It might be a leap to justify Vs --> 0.
But you are correct, the step response is an 'upside down' version of the natural response. It makes intuitive sense at the end. For me the surprise is how twisty the math is compared to how intuitively obvious the answer comes out.(2 votes)
- I have a conceptual question: the forced voltage on the capacitor is equal to the voltage from the voltage source according to the video (the yellow box Vf = Vs at8:01). Why isn't there a voltage drop across the resistor? In other words, I do not understand why putting a capacitor after a resistor changes what a resistor does.(1 vote)
- At8:01, the forced response v_f turned out to be equal to the step voltage, V_S. It may seem strange that the forced response is a constant for all time. But remember, the forced response is not the whole answer, it's only a portion of the answer. We have to add the forced response to the natural response to get the total (the actual) response. So all by itself, the forced response is only part of the whole story.
This method of forced + natural is a mathematical technique we use to break a hard problem into two easier problems. While the math works great, the intermediate result (the forced response) may seem a little strange.
One way to interpret the forced response is to say it is the solution to the circuit after a long time has passed (this is true because any natural response of any circuit always dies out to zero, so the only thing happening after a long time is the forced response). In this step response circuit, the step happens and the capacitor charges up. After a long time, the capacitor voltage grows until it matches the source voltage, which means the voltage difference across the resistor becomes 0, and current stops flowing.(1 vote)
- I have two question. Question 1. At5:13, why dVs/dt=0, can it be dVs/dt=Dirac(t)?
Question 2, how if input Vs=Diract(t)+V0. I want to calculate the impulse response, but based on the above analysis in the video, V=Kn*exp(-t/RC), I cannot solve Kn.(1 vote)- Q1: Why is dVs/dt = 0 as opposed to dVs/dt = impulse.
A1: We are solving the step response using the method of natural+forced response. The point is to break the problem into two simpler parts rather than trying to solve everything in a single analysis. We can separate the circuit problem like this because the principle of Superposition holds, which allows us to add natural + forced = total response.
As the video says, we figure out the forced response after t=0, not for all time. For t > 0 the input voltage is constant at Vs, so the derivative with respect to time is 0.
It might seem a little fishy that the forced response ignores what happened prior to t = 0. The forced response only deals with after t = 0. Dealing with before t = 0 is the job of the natural response part of the problem. When we add them together, the entire answer emerges.(1 vote)
- How do we get vsubn = Ksubn e^t/RC from s=-1/RC?(1 vote)
This equation appears at2:00minutes of this video. This result is not derived here, but rather it is taken from a previous video on the RC Natural Response.(1 vote)
@7:29at t=0 , Vt=Vo , is Vt=Vo before t equal to zero ?(1 vote)- Yes. Vt = V_0 before the step happens. Prior to the input voltage step the value of the voltage source is V_0. Let's let V_0=1V just to have something to talk about. Assuming it has been at V_0 for a "long time", what is the state of the circuit?
The capacitor will be charged up to 1V. That makes the voltage across resistor R to be 1V - 1V = 0. With 0V across the resistor you know the current is 0, (I = V/R = 0/R = 0). So some charge is sitting on C sufficient to make the capacitor voltage = 1V. The circuit is just sitting there waiting for the step to come along. The step is the 'forcing function" that makes the circuit change state.(1 vote)
At5:52, I saw you move the screen for space.How do you do that?(1 vote)- It's pretty simple. I draw in Autodesk Sketchbook Pro. I make the page size really tall, like 2 screens high. The video is captured using Camtasia, which does a continuous screen grab of a rectangular section of my monitor. To move the drawing up it's just a matter of scrolling up the Sketchbook window while Camtasia is running.(1 vote)
8:01Video transcript
- [Voiceover] In the last
video on step response, we set up the differential equation that describes our
circuit, and we found that it was a non-homogeneous equation, and now we're gonna follow through on the strategy of solving it with a forced response plus a natural response. So here's two copies of our circuit, and we're gonna, on the top one here, we're gonna solve for
the natural response, and on the bottom one, we'll use that one for the forced response. Alright, let's go to work on the natural. Now, to do a natural response, what we do is we set
the initial conditions to whatever they were
for the original circuit, so that involves some q and a v zero, a v naught on the capacitor. So let's just draw in
quick, let's draw in our q. There's some q here, that'll
be a plus or minus v naught on the capacitor. Now the other thing we
do, is for the inputs, we set the inputs to zero
to do the natural response. And how you do that, our input here is a voltage source, and we're gonna suppress
the voltage source, that's how we do it with super position. When you suppress a voltage source, what you do is you set
the voltage to zero. And that means that's the same as creating a short circuit for this. So, for the natural response we're gonna take out our voltage source and replace it with a
short circuit, like this. So, this circuit now is identical to the circuit that we used
for the natural response. So, if you haven't seen that video on how to solve for the natural response, this would be a good
time to go look at that. What I'm gonna do now is just write down the answer from that, this
will be plus or minus, we'll call this v natural, and that equals some k natural times e to the minus t over r c. This is the natural
response of an r c circuit, so I'll put a square around
that so we can remember it. And I've left the constant in here, and we'll work out what this
constant is a little bit later. Okay, so now let's move on and do the forced
response for this circuit. And for the forced response, we remember the initial
conditions are set to zero. So that means that q equals zero here, and that means that initial v is zero. I'll just write in zero. So that's what it means to set the initial conditions to zero. And the inputs are equal to, we use to inputs this time, the input is equal to v s. And in particular, the
input is equal to v s, we're gonna solve the forced response after time is equal to
zero, so that means that the input is v s, capital v s. And as a reminder, what
we're trying to solve here is the differential equation
from the previous screen, and that is c times d v d t, I'll put a forced in here, plus one over r, times v forced equals one over r times v s, and I can plug in capital v s here because we're trying to solve this for this initial condition. So this is our differential equation for the forced response, and we're gonna take a strategy here that's like we did with
the natural response, is we're gonna guess at an answer for v f, and then plug it into this
differential equation, this non-homogeneous
differential equation, and we're gonna see if it works. And a good guess here
is to guess something that looks like the input. So a good guess is gonna be some function that looks like v s. And v s looks like what? V s looks like, over here,
v s looks like a constant. So we're gonna make a guess here that the forced response
looks like some constant, we'll call that k f. The way we test our guess is
to plug it back into here. Plug it back into the
differential equation. So I'm gonna do that, alright so we get c times the derivative of k f, with respect to time, plus one over r, k sub f, equals one over r times v s. Alright, now here's the interesting
thing that happens next. What's the derivative of a
constant with respect to time? That's That goes to zero. So this first leading term here
of our differential equation goes to zero, and now I'm left with k f times one over r equals v s times one over r. So that makes k f equal to v s. So I'll tuck this in here, our forced response, v f equals v s. So now I have our natural response right here, and in this square we
have our forced response, the forced response is
just a constant v s. So now we're ready to come
up with our total response, and we'll call that v capital t, the total response is equal
to the natural response plus the forced response. So now we're using our
principal of super position. Okay, v t equals k some constant times e to the minus t over r c. And let's add the forced response, and the forced response is right here, it's v s, and we're getting close, the only thing we have left is
to figure out this value now, we have to figure out the gain factor in front of the exponential
term of the natural response. Now the way we go after that, is we would know this
if we knew what v s is, and we do, it's called v s. If we knew v t at some time, we could plug in two values here. We could plug in a v t and a t, and one of the most
convenient times to know this, if we set t equal to zero, and we know that the total
at time equals zero is what? It's basically, let me roll
it up here a little bit, let's go backwards a little bit. Let me go back here and
use this diagram again, this was the forced response, so we're not gonna use that right now. We're gonna use, this
is now the total circuit all assembled together,
and we have to figure out what is the actual initial charge on here, and if we recall, the
initial charge on here was v naught, it was this value here, just before time equals zero, this was the value of the
voltage on that capacitor. So I can fill that in here, v naught. Alright, let's go back
to our total solution and plug in these two values, time is zero and v t equals v naught. So that looks like v naught equals natural constant times e to the minus zero over r c plus v s. And let's solve for k n k n equals, this term is e to the zero is one, so this term goes to one, that says that it's k n plus v s, is this side, and I can write down here v naught minus v s. So now I've solved for k, and we can finish our total response, we can say the total
response is v t equals k sub n, which is v naught minus v s, e to the minus t over r c and plus v s. And that is the total response of our circuit, and we saw that in two steps, first we did the natural
and then we did the forced and we added them together and
worked out the last constant, whatever the constant was,
and there's our answer. So in the next video we'll
do an explicit example with values for r and c in the step and we'll see what it actually looks like.