We use the method of natural plus forced response to solve the challenging non-homogeneous differential equation that models the $\text R\text C$ step circuit.

The $\text{RC}$ step response is the most important analog circuit. In analog systems it is the building block for filters and signal processing. In digital systems it sets the speed limit for how fast the system runs—the $\text{RC}$ step happens billions of times every second inside computers and other digital devices.

There is a fair amount of new vocabulary introduced in this derivation. Relax, let it soak in.

Written by Willy McAllister.

Contents

Before diving into the step response you may want to review two articles,

To gain an intuitive understanding of the step circuit see

The step response of an $\text R\text C$ circuit is,

$v(t) = \text V_\text S + (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}}$

where $v(t)$ is the voltage on the capacitor, $\text V_0$ is the beginning voltage of the step, and $\text V_\text S$ is the ending voltage of the step.

Apply an abrupt step to a resistor-capacitor $(\text{RC})$ circuit and watch the voltage across the capacitor. A step input is a common way to give a circuit a little “kick” to see what it does. The step response tells us quite a lot about the properties of the $\text{RC}$ circuit.

This is the circuit we will study,

The input is an abrupt voltage step that starts at $\text V_0$ and instantly jumps to $\text V_{\text S}$ at $t = 0$.
We want to find the voltage on the capacitor, $\goldC{v(t)}$, as a function of time.

instantaneous step

A step function is a mathematical idea. The voltage has only two values, $\text V_0$ when $t < 0$ and $\text V_{\text S}$ when $t > 0$. At precisely $t = 0$ the step function is not defined. There are no in-between values. When we draw a step with an orange vertical line at $t = 0$ it’s simply a graphical connection between the top and bottom horizontal lines. The vertical line isn’t meant to suggest intermediate voltage values exist at $t = 0$.

Technically, the step function does not meet the definition of a mathematical function, since there’s this discontinuous weirdness at $t=0$. You will see how we deal with that when we determine the initial conditions.

In the real world step functions always have some finite slope. We still call it a step if the slope is really steep relative to the response of the $\text{RC}$ circuit. It’s close enough to an ideal step if it looks like a step on the time scale we are interested in.

It is awkward to express a mathematical model of a step source. There is an equivalent way to draw this circuit. We change the source to a constant voltage $\text V_{\text S}$ and add a switch to cause the step. The initial voltage, $\text V_0$, is placed directly on the capacitor,

The starting voltage on the capacitor is the initial condition, also called the boundary condition. The moment of flipping the switch establishes a boundary.

How did $\text V_0$ get there?

If you just have to know how the initial charge got onto $\text C$, here’s a circuit that does that,

Both switches are thrown at the exact same moment. The switch on the right opens, and the one on the left closes at $t = 0$. Prior to $t = 0$ the voltage source on the right charges the capacitor up to $\text V_0$.

Model the circuit

We start by modeling the circuit with Kirchhoff’s Current Law for the two currents flowing out of the top right node. Replace each current with the proper $i$-$v$ expression,

$\begin{array}{cccc} i_\text R & + & i_\text C &= 0 \\ \\ \dfrac{v - \text V_\text S}{\text R} & + & \text C\,\dfrac{dv}{dt}&= 0 \end{array}$

Rearrange the terms a little to format it like a differential equation. Collect the $v$ terms on the left side and put the non-$v$ terms on the right side. We like to have a coefficient of $1$ on the leading $dv/dt$ term,

$\dfrac{v}{\text R} - \dfrac{\text V_\text S}{\text R} + \text C\,\dfrac{dv}{dt} = 0$

$\text C\,\dfrac{dv}{dt} + \dfrac{v}{\text R} = \dfrac{\text V_\text S}{\text R}$

$\dfrac{dv}{dt} + \dfrac{v}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}\qquad$ with initial condition: $v(0) = \text V_0$

This is the differential equation we have to solve.

It is hard to solve

There’s an important difference between this equation and the one we developed for the $\text{RC}$ natural response shown here,

$\dfrac{dv}{dt} + \dfrac{v}{\text{RC}} = 0\qquad$ with initial condition: $v(0) = \text V_0$

The natural response circuit has no voltage source, so the right side of the differential equation came out $0$. The natural response equation is homogeneous. That means all terms involve the independent variable $v$ or a derivative of $v$.

For the step response the expression on the right side is not $0$,

$\dfrac{dv}{dt} + \dfrac{v}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}\qquad$ with initial condition: $v(0) = \text V_0$

The term $\dfrac{\text V_\text S}{\text{RC}}$ is not related to $v$ or a derivative of $v$. That makes this equation non-homogeneous.

Solving a non-homogeneous differential equation is not the simplest thing in the world. In fact it is usually a pain in the neck. Fortunately, this particular non-homogeneous equation is solvable without too much torture. Lucky us.

The electrical reason the step response is challenging is because of the two energy sources. Energy comes from 1. the input voltage source and 2. the initial charge on the capacitor. The trouble is the two energy sources have no relationship to each other. This complication makes it difficult to solve.

Solve a driven circuit

We find the step function using the method of homogeneous and particular solutions. If you have taken a differential equation class, what we are doing is the method of undetermined coefficients.

Our derivation is based on this non-obvious theorem of differential equations,

The total solution to a non-homogeneous differential equation can be found by adding the general solution of the circuit’s natural response to any particular response (found without regard for the initial conditions), followed by applying the initial conditions to resolve unknown constants.

This is a pretty bold theorem to state without proof. It is central to finding the step response. For the moment please take it on faith. Save the proof for later.

In mathematical vocabulary the the theorem says,

Complete solution = homogeneous solution + particular solution

$v_c = v_h + v_p\qquad$ subject to initial conditions

The subscripts $c$, $h$, and $p$ stand for complete solution, homogeneous solution, and a particular solution.

In circuit vocabulary the notation becomes,

Complete response = Natural response + forced response (steady-state)

$v_{tot} = v_n + v_f\qquad$ subject to initial conditions

where $tot$, $n$, and $f$ stand for total response, natural response, and forced response.

The theorem says we can pick any particular response—we could choose any of the myriad of particular solutions. However… Ingenious part: We hunt for one particular particular response. It is relatively easy to find and has a nice real-world significance. Somewhere in the family of curves represented by the general solution there is always one particular response that corresponds to the long-term steady-state behavior of the circuit—the place the circuit ends up after the natural response dies out. We call that the forced response.

What is meant by, “… we can pick any particular response”?

Here’s an example based on the general solution to the $\text{RC}$ natural response.

$v = K_n\,e^{-t/\text{RC}}$

The general solution is an infinite family of functions that all make the differential equation true. We plot this equation with different values of $K_n$. Each curve represents a different particular solution,

The family of particular solutions represented by the general solution: $K_n\,e^{-t/\text{RC}}$

The theorem says we could select any of these. But there’s a special one. Find the particular response that’s a straight line along the time axis, the line $v = 0$ (bold orange). This particular response is interesting—out of all possible particular responses this one matches the long-term steady-state response.

The point of this example is to show how the long-term steady-state response appears somewhere in the general solution. We will use this little nugget in a little while.

When we change from natural response to a step function the steady-state response is no longer zero—it will be some other value. Nevertheless, the long-term steady-state response will still appear somewhere in the general solution to the step response.

Strategy

The steps to solve a driven circuit,

1. Find the general form of the natural response. Do this by suppressing the driven input (set the input to $0$). Ignore the initial conditions for now.
2. Find the forced response. The forced response is the particular solution that looks like a scaled version of the input forcing function. Continue ignoring the initial conditions. (Sometimes the particular solution might need to include the input function plus its derivatives.)
3. Superimpose (add) the natural response to the forced response to get the general form of the total response.
4. Last, apply the initial conditions to the total response and resolve the unknown constants.

Holding off the initial conditions until the last step allows the natural and forced response to blend together to create the transition from initial state to steady state.

Let’s work our strategy and see what happens.

Natural response

To find the natural response $v_n$ we suppress (turn off, set to zero, short) the input,

Find the natural response with the input suppressed. To suppress a voltage source, replace it with a short. (For more on suppressing sources see Superposition.)

We already worked through the $\text{RC}$ natural response in great detail. This is a quick repeat of the derivation.

Write a KCL equation for the upper right node. Model $\text R$ and $\text C$ by their $i$-$v$ equations,

$\begin{array}{cccc} i_\text C & + & i_\text R &= 0 \\ \\ \text C\,\dfrac{dv}{dt} & + & \dfrac{v_n}{\text R}&= 0 \end{array}$

to get this differential equation,

$\dfrac{d}{dt}v_n + \dfrac{v_n}{\text{RC}} = 0$

Notice: When we suppress the input, the right side of the differential equation comes out $0$, so it is a homogeneous differential equation.

Propose a solution in the form of an exponential with two adjustable parameters, $K_n$ and $s$,

$v_n = K_n\,e^{st}$

Test the proposed solution to see if it makes the differential equation true.
Plug the proposed $v_n$ into the homogeneous equation,

$\dfrac{d}{dt}K_n\,e^{st} + \dfrac{1}{\text{RC}}K_n\,e^{st} = 0$

Perform the derivative in the first term,

$sK_n\,e^{st} + \dfrac{1}{\text{RC}}K_n\,e^{st} = 0$

Factor out the common $K_n\,e^{st}$ term,

$K_n\,e^{st}\,\left (s + \dfrac{1}{\text{RC}} \right ) = 0$

To make this equation true, one of the three terms on the left side has to be $0$. We could set $K_n$ to $0$, which means there was no initial energy stored in the capacitor. The circuit just sits there and does nothing. The term $e^{st}$ becomes zero if $s$ is negative and we wait for $t$ to go to infinity. All natural responses approach $0$ if you wait long enough. Something interesting happens if we make the third term $0$,

$s + \dfrac{1}{\text{RC}} = 0$

This is the characteristic equation of the $\text{RC}$ natural response. Solve the characteristic equation for $s$ (find the roots of the characteristic equation).

$s=-\dfrac{1}{\text{RC}}$

Put this value of $s$ back into the proposed solution and we get the general form of the natural response for $t > 0$,

$v_n = K_n\,e^{-t/\text{RC}}$

General solution to the $\text{RC}$ natural response. $v_n = K_n\, e^{-t/\text{RC}}$

This general solution is an infinite family of curves based on all possible values of $K_n$. The natural response is an exponential curve whose rate of descent is determined by the product $\text{RC}$. If you set $t = 0$ the exponential term becomes $1$ and the equation is telling us $K_n$ corresponds to the initial voltage at $t = 0$. We hold off figuring out a specific value of $K_n$ until after we assemble the total response.

Forced response

Our special theorem says we can pick any particular solution to the differential equation. However, we won’t go for any particular solution, but rather for a special one, the particular solution that corresponds to the long-term steady-state of the circuit. This particular solution gets a special name, the forced response.

$\dfrac{d}{dt}v + \dfrac{v}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}$

The term on the right side of the equation represents the input forcing function. The input will be driving the circuit long after the natural response dies out. It is definitely going to have a strong influence on the long-term behavior of the output. If we hope to make both sides equal, $v$ should bear some resemblance to the input—so let’s pick a solution that resembles the right side. Since the input is a specific constant let’s propose an arbitrary constant,

$v_f = K_f$

Pop this into the non-homogeneous equation and see what happens. Does this $v_f$ makes the equation true? That’s the same as asking: Is this $v_f$ a particular solution?

$\dfrac{d}{dt} K_f + \dfrac{K_f}{\text{RC}} \stackrel{?}{=} \dfrac{\text V_\text S}{\text{RC}}$

In the first term, the derivative of a constant is always $0$,

$0 + \dfrac{K_f}{\text{RC}} \stackrel{?}{=} \dfrac{\text V_\text S}{\text{RC}}$

The equation comes true when what? When,

$K_f = \text V_\text S$

This resolves the unknown constant $K_f$, which means $v_f$ is a particular solution to the non-homogeneous differential equation,

$v_f = \text V_\text S$

A particular solution for the $\text{RC}$ step is $v_f(t) = \text V_\text S$, also known as the forced response.

We call this particular response the forced response because our proposed solution was derived from the forced input.

Things to notice about the forced response

Notice: The forced response is exactly the input step final value. That doesn’t always happen for every circuit, but it did this time.

Notice: The force response solves the equation but completely misses the initial condition, $v(0) = \text V_0$. That’s okay. The natural response will take care of that when we merge them together.

Notice: The forced response is plotted for all time after $t=0$ bit you can’t observe it with an oscilloscope. The same holds for the natural response component. When we find the total response we can observe that with an oscilloscope.

What happens if the guess is wrong?

What if you guess a forced response and it doesn’t work? Let’s see what that looks like. Here’s an example of a guess that doesn’t work.

Suppose you guess the solution is a constant $\times$ time, $v_f = K_f\,t$.
When you plug $v_f$ into the differential equation,

$\dfrac{d}{dt}K_f\,t + \dfrac{K_f\,t}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}$

You get,

$K_f + \dfrac{K_f\,t}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}$

When you solve this for $K_f$ no amount of algebra will get rid of the $t$ term. This means $K_f$ depends on time, so it’s not constant. That’s a fail for this proposed solution.

Natural plus forced

The total step response is the sum of the natural and forced responses,

$\begin{array}{cccc} v_{tot} = & v_n &+& v_f \\ \\ v_{tot} = & K_n\,e^{-t/\text{RC}} &+& \text V_\text S \end{array}$

This is the general solution to the step response because it has one remaining undetermined constant. The general solution is the infinite family of all possible particular solutions. Here are a few of them,

General solution to the $\text{RC}$ step response. $v_{tot} = K_n\, e^{-t/\text{RC}} + \text V_\text S$

Look at the plot and find the particular solution that is a horizontal straight line at $v = \text V_\text S$ (hint: the fifth one from the top). That’s the particular solution we found above—the forced response, equivalent to the long-term steady state of the system.

Apply the initial conditions

Now, finally, we apply the initial conditions to resolve the last unknown, $K_n$.

We know the voltage at a particular point in time: At $t = 0$ the voltage on the capacitor is $\text V_0$. We find a $K_n$ such that the total response—not just the natural response—matches this boundary condition.

In the total response expression set $t$ to $0$ and replace $v_{tot}$ with $\text V_0$,

$\text V_0 = K_n\,e^{-0/\text{RC}} + \text V_\text S$

The exponential expression becomes $1$ and we are left with,

$\text V_0 = K_n + \text V_\text S$

Solving for $K_n$,

$K_n = \text V_0 - \text V_\text S$

Step response

Replace $K_n$ in the total response and we have the step response,

$\boxed{v_{tot} = (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}} + \text V_\text S}$

The total response looks like this,

The $\text{RC}$ step total response. $v_{tot} = (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}} + \text V_\text S$

We did it! This is the step response of an $\text{RC}$ network.

$K_n$ turned out to be the height of the step, the difference between the starting and ending voltages, $\text V_0 - \text V_\text S$. This is just the right value to flip the natural response term over and make it fit perfectly between start and end.

Special case: No charge on the capacitor

If the capacitor starts with no initial charge then $\text V_0 = 0$, and the total response expression gets a little simpler,

$v_{tot} = (0 - \text V_\text S)\,e^{-t/\text{RC}} + \text V_\text S$

$v_{tot} = - \text V_\text S\,e^{-t/\text{RC}} + \text V_\text S$

or

$\boxed{v_{tot} = \text V_\text S \left (1 - e^{-t/\text{RC}}\right )}$

If the capacitor voltage starts at $0$ the $\text{RC}$ step total response is $v_{tot} = \text V_\text S \, \left (1 - e^{-t/\text{RC}}\right )$.

What to remember

You don’t have to memorize the equation for the step response, but you should remember what it means,

• The voltage starts at some initial value prior to the step, $\text V_0$.
• When the step arrives the voltage heads toward the destination voltage, $\text V_\text S$, rising (or falling) with a smooth exponential shape controlled by the time constant, $\text{RC}$.
• The output approaches $\text V_\text S$, the steady-state value forced by the input.

Concept check

Suppose the input voltage makes another step in the downward direction a little later, from $\text V_{\text S}$ back down to $\text V_0$. Assume the original $\text V_0 = 2\,\text V$ and $\text V_{\text S} = 6\,\text V$.

Let $\text R = 3\,\text K\Omega$, $\text C = 0.4\,\mu\text F$.

How does the capacitor voltage respond?

Hint: Reverse the roles of the high and low voltages, so you can think of it this way,
$\text V_0$ (starting voltage) $= 6\,\text V$ and $\text V_{\text S}$ (ending voltage) $= 2\,\text V$.

$v(t) =$ ________

The moment of interest is the downward step, so reassign $t=0$ to be the time of the downward step. Repositioning $t=0$ lets us use the step response expression above.

$v = \text V_\text S + (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}}$

$v = 2 + (6 - 2)\,e^{-t/(3\,\text k \,\cdot \, 0.4 \,\mu)}$

$v(t) = 2 + 4\,e^{-t/1.2\,\text{msec}}$

Explore further

Run this simulation model.

1. Click on DC in the top menu bar to perform a static DC analysis.
• What is the initial voltage on the capacitor?
2. Click on TRAN to run a transient analysis to observe the $\text{RC}$ step response voltage.
• What are the starting and ending voltages of the step input (red)?
• What is the final voltage on the capacitor (cyan)?
3. Double click on the voltage source. Change the start and end voltages and see what happens.
• What happens if the start voltage is higher than the end voltage?
4. Add a current probe anywhere in the loop and give it a distinct color. TRAN again.
• What is the current at the start?
• What is the current at the end?
• Describe what happens in between.
5. Change the voltage source from step to square. Return the low and high voltages to 2V and 6V. Set the frequency to $50\,\text{Hz}$. Change the TRAN time to $29\,\text{msec}$.
• Compare the down step to an up step.
• What is different?
• What is the same?
6. Change the square wave frequency to $200\,\text{Hz}$ and TRAN again.
• Describe what happens.
7. Zoom out the schematic page and build a copy of the circuit next to the original. Try different values for $\text R$ and $\text C$ in the new circuit. Add a voltage probe to the new circuit with a new color. Simulate.
• Compare both voltages on the same graph.

Summary

We solved an $\text{RC}$ network driven by a step voltage. We started with Kirchhoff’s Current Law to create a non-homogeneous differential equation representing the circuit. We found the homogeneous solution (the natural response) and a special particular solution called the forced response. Then we added them together and applied the initial conditions to get the total response.

• The natural response is what the circuit does with its initial energy, with the input suppressed. The natural response always fades to $0$.

• The forced response is the particular response corresponding to the long-term steady-state circuit response.

• The total response is the natural response plus the forced response.

The step response of an $\text R\text C$ network is,

$v = \text V_\text S + (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}}$

where $\text V_\text S$ is the step voltage and $\text V_0$ is the starting voltage on the capacitor.

If the capacitor voltage starts at $0$ the step response expression is a bit simpler,

$v = \text V_\text S \, \left (1 - e^{-t/\text{RC}}\right )$

Appendix - Separable differential equation

$\dfrac{dv}{dt} + \dfrac{v}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}$

It is possible to solve this non-homogeneous differential equation directly without decomposing the problem into natural and forced response.

This turns out to be a separable differential equation. If you learned this technique in a differential equations class you can solve both the $\text{RC}$ and $\text{RL}$ steps function with this method.

A differential equation is separable if you can sort all the voltage terms, $v$’s and $dv$’s, into a product on one side and sort the time terms, $dt$’s, into a product on the other side. That happens with this equation.

This isn’t the usual way to teach this circuit because the method doesn’t generalize to other forcing inputs, and it does not provide insight into the natural and forced components of the solution.

The non-homogeneous differential equation is,

$\dfrac{dv}{dt} + \dfrac{v}{\text{RC}} = \dfrac{\text V_\text S}{\text{RC}}\qquad$ with initial condition: $v(0) = \text V_0$

Rearrange terms to isolate the derivative,

$\dfrac{dv}{dt} = -\dfrac{v - \text V_\text S}{\text{RC}}$

Rearrange again to get the $v$ and $dv$ terms to one side and $dt$ term to the other,

$\dfrac{dv}{v - \text V_\text S} = - \dfrac{dt}{\text{RC}}$

Integrate both sides and apply the initial condition,

\begin{aligned} \displaystyle \int_{\text V_0}^{v(t)}{\dfrac{dv}{v - \text V_\text S}} &= - \int_0^t{\dfrac{dt}{\text{RC}}} \\ \\ \ln(v - \text V_\text S) \,\bigg \vert_{V_0}^{v(t)} &= - \dfrac{t}{\text{RC}}\,\bigg \vert_0^{t} \\ \\ \ln(v(t) - \text V_\text S) - \ln(\text V_0 - \text V_\text S) &= -\dfrac{t}{\text{RC}} + 0 \\ \\ \ln \dfrac{v(t) - \text V_\text S}{\text V_0 - \text V_\text S} &= -\dfrac{t}{\text{RC}} \end{aligned}

Clear the natural log by taking the exponential of both sides,

$\dfrac{v(t) - \text V_\text S}{\text V_0 - \text V_\text S} = e^{-t/\text{RC}}$

Rearrange to isolate $v(t)$ on the left side and we get the same result we came up with in the main article,

$v(t) = \text V_\text S + (\text V_0 - \text V_\text S)\,e^{-t/\text{RC}}$

Sal has a sequence of videos where he covers this type of separable differential equation. He does a worked example to solve a population growth problem.

The derivation in this appendix is from Alexander and Sadiku, Fundamentals of Electric Circuits, 5th Edition, p. 274.

Appendix - Vocabulary

There are so many terms used in math and engineering related to this type of differential equation. Don’t try to memorize all this.

• Natural response is the homogeneous solution or the force-free solution or the complementary solution (because it complements the particular solution). In control systems you might see it called the free response.

• Forced response is a particular solution chosen from the general solution. Not just any particular solution but the one specifically derived from the the long term behavior of the circuit, the steady-state response.

• Total response is also called the complete solution.

Even more,

Our differential equation is a mouthful, it is a
non-homogeneous first-order constant-coefficient ordinary differential equation.

What does all this mean?

• Homogeneous means the equation contains $v$ and derivatives of $v$, and nothing else. The natural response equation is homogeneous.
• Non-homogeneous or inhomogeneous means there is some term that’s not $v$ or one of its derivatives. Our step response differential equation includes a $\text V_{\text S}$ term not related to $v$, so it is non-homogeneous.
• First-order means the highest derivative is the first derivative $dv/dt$. If there was a second derivative, $dv^2/d^2v$, it would be a second-order equation. $\text{LC}$ and $\text{RLC}$ circuits produces second-order equations.
• Constant coefficient means the values of the components $\text R$ and $\text C$ are constant and do not change as time goes by. This is also referred to as time invariant. You may see a system described by the acronym LTI for linear time invariant. LTI means if you run the circuit today and tomorrow with the same initial conditions it will do the same thing both days. It’s a nice property.
• Ordinary means there is just one independent variable, $t$.