how to get the equation for i= V0/W0*e−αt*t in crirically damped?

Jonathan is correct. The expression for current should be i = V0/L t e^(-at). This has been corrected in the article. The article still just states the result, without derivation. Inspired by Jonathan's solution in this clarification, I wrote up this more complete derivation of the Critically Damped case: http://spinningnumbers.org/a/rlc-natural-response-variations.html#critically-damped This website is a continuation of the EE contributions I made at KA.

why do you get barely any questions, tips, or thanks on your electrical engineering section? on other sections, i see over 100 questions but on this section, i hardly ever see 10

There are lots of questions scattered throughout the EE subject area. You can see them if you add a /d at the end of the url, like this: https://www.khanacademy.org/science/electrical-engineering/d

If I add a SPDT switch in the circuit, I can charge the capacitor. But if there is one way to charge without the switch. (For exmaple, I add a battery and it will work and form same-frequncy wave until battery does not have any energy)

Providing an initial charge to a capacitor or initial current to an inductor is a tricky bit of work in a circuit simulator. It is difficult to implement a mechanical switch. They are usually not part of the simulator's primitive elements. It's possible to mimic a mechanical switch using very large transistors. I show an example of this on my web site, spinningnumbers.org. This is a follow-on to my work at KA where all the articles are updated. If you navigate to the Natural and Forced Response topic, find the article on RLC Variations. The simulation models have giant MOSFETS acting like mechanical switches to initialize the voltage. Remove the spaces and try this URL, https : //spinningnumbers . org/a/rlc-natural-response-variations.html#under-damped (Scroll down to the end of the Under Damped chapter to find the simulation model.)

Main content

Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response

RLC natural response - variations

Google Classroom

The RLC natural response falls into three categories: overdamped, critically damped, and underdamped. Written by Willy McAllister.

Introduction

The natural response of a resistor-inductor-capacitor circuit

(RLC)

can take on three different forms, depending on the specific component values.

In two prior articles, we covered an intuitive description of how the

RLC

behaves, and did a formal derivation where we modeled the circuit with a

2

nd-order differential equation and solved a specific example circuit. In this article, we look closely at the characteristic equation and give names to the various solutions.

What we're building to

The

RLC

characteristic equation is:

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

We will solve for the roots of the characteristic equation using the quadratic formula:

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

By substituting variables

α

and

ω_{o}

we can write

s

a little simpler as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

where,

α = \frac{R}{2 L},

ω_{o} = \frac{1}{\sqrt{LC}}

α

is called the damping factor and

ω_{o}

is the resonant frequency.

Depending on the relative size of

α

and

ω_{o}

, there will be three different forms of the solution for

i (t)

overdamped, $α > ω_{0}$ ‍, leads to the sum of two decaying exponentials
critically damped, $α = ω_{0}$ ‍, gives $t$ ‍ times a decaying exponential
underdamped, $α < ω_{0}$ ‍, leads to a decaying sine

Modeling and solving the circuit - review

In a previous article we created and solved a

2

nd-order differential equation modeling the

RLC

circuit. That equation looks like this:

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 0

We proposed a solution with an exponential form (which worked out really nicely for us), and came up with what is called the characteristic equation with this form:

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

We solved for

s

, the roots of the

RLC

characteristic equation, using the quadratic formula,

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

By substituting variables

α

and

ω_{o}

we wrote

s

a little simpler as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

where

α = \frac{R}{2 L},

and

ω_{o} = \frac{1}{\sqrt{LC}}

α

is called the damping factor, and

ω_{o}

is called the resonant frequency.

We revised our proposed solution to have this form,

i = K_{1} e^{s_{1} t} + K_{2} e^{s_{2} t}

We now take a close look at the expression for

s

, the roots of the

RLC

characteristic equation, and the impact it has on the solution for

i

Exact solution

If we want an exact answer for particular values of

R

L

, and

C

, we perform a computation like the one we did in the previous article for the example circuit. Alternatively, we can enter the circuit into a circuit simulator to help us find a result.

Overdamped, critically damped, underdamped

We can get an impression of the full richness of the natural response by looking three possible outcomes in a qualitative sense.

The solution for

s

depends on the sign of the subtraction that happens under the square root term in the equation:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

How the roots turn out:

relation	sign of $α^{2} - ω^{2}$ ‍	nickname	$s$ ‍
$α > ω_{o}$ ‍	$+$ ‍	overdamped	2 real roots
$α = ω_{o}$ ‍	$0$ ‍	critically damped	2 repeated roots
$α < ω_{o}$ ‍	$-$ ‍	underdamped	2 complex roots

How the response turns out,

i (t)

relation	sign of $α^{2} - ω^{2}$ ‍	nickname	$i (t)$ ‍
$α > ω_{o}$ ‍	$+$ ‍	overdamped	2 decaying exponentials
$α = ω_{o}$ ‍	$0$ ‍	critically damped	$t \cdot$ ‍ decaying exponential
$α < ω_{o}$ ‍	$-$ ‍	underdamped	decaying sine

If your engineering studies take you into the area of Control Theory, these terms are used to describe how dynamic systems act. For example, the motion of a robot's arm can be described by a second-order differential equation. If your ask your robot to quickly reach for an object, you can describe how its hand moves using these words.

Let's take a look at the three possible outcomes in a bit more detail.

$α^{2} - ω^{2} > 0$ ‍ overdamped

Under this condition, the

ω_{o}^{2}

term is small relative to

α^{2}

, so we know the expression inside the square root will be positive. We also know the square root expression will be smaller than

α

. This means

s

will be two real numbers, both negative.

s_{1, 2} = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

s_{1} = - {real number}_{1}

and

s_{2} = - {real number}_{2}

(Convince yourself that

s_{1}

and

s_{2}

will both be negative.)

The current will be the superposition of two real exponentials that both decay to zero.

i = K_{1} e^{- {real}_{1} t} + K_{2} e^{- {real}_{2} t}

The circuit is said to be overdamped because two superimposed exponentials are both driving the the current to zero. A circuit will be overdamped if the resistance is high relative to the resonant frequency.

Let

R = 10 Ω, L = 1 H

, and

C = 1 / 9 F

.
Let the initial conditions be

v_{c} (0) = 10 V

and

i_{L} = 0

Find $i (t)$ ‍.

Work out the damping factor and resonant frequency:

α = \frac{R}{2 L}

α = \frac{10}{2 \cdot 1} = 5

ω_{o} = \frac{1}{\sqrt{LC}}

ω_{o} = \frac{1}{\sqrt{1 \cdot 1 / 9}} = 3

Find the (negative real) roots of the characteristic equation:

s_{1, 2} = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

s_{1, 2} = - 5 \pm \sqrt{5^{2} - 3^{2}} = - 5 \pm \sqrt{16} = - 1, - 9

Assemble a proposed solution:

i = K_{1} e^{s_{1} t} + K_{2} e^{s_{2} t}

i = K_{1} e^{- t} + K_{2} e^{- 9 t}

Use the two initial conditions (same as the previous worked example):

i (0^{+}) = 0

\frac{d i}{d t} (0^{+}) = 10

Write expressions for

K_{1}

and

K_{2}

t = 0

. Using the first initial condition:

i (0) = 0 = K_{1} e^{- 0} + K_{2} e^{0}

0 = K_{1} + K_{2}

Now we take the derivative of the proposed current and use the second initial condition:

\frac{d i}{d t} = \frac{d}{d t} [K_{1} e^{- t} + K_{2} e^{- 9 t}] = - K_{1} e^{- t} - 9 K_{2} e^{- 9 t}

\frac{d i}{d t} (0) = 10 = - K_{1} e^{- 0} - 9 K_{2} e^{0}

10 = - K_{1} - 9 K_{2}

Now we solve the two initial condition equations together. From the first initial condition:

K_{1} = - K_{2}

Substituting into the second initial condition:

10 = - (- K_{2}) - 9 K_{2}

10 = - 8 K_{2}

K_{2} = - 1.25,

and therefore

K_{1} = + 1.25

And the solution for current is:

i = 1.25 e^{- t} - 1.25 e^{- 9 t}

$α^{2} - ω^{2} = 0$ ‍ critically damped

The boundary between underdamped and overdamped is when

α = ω_{o}

. The damping factor and the resonant frequency are in balance, and the terms under the square root subtract to

0

. The roots of the characteristic equation,

s

, are two identical real numbers, called repeated roots:

s_{1, 2} = - α \pm {\sqrt{α^{2} - ω_{o}^{2}}}^{0}

s_{1, 2} = - α

Solving a

2

nd-order differential equation with repeated roots is a bit tricky. I'm not going to do the derivation here, but instead I refer you to a great video Sal did on solving repeated roots. Welcome back ... With repeated roots, the answer is an exponential term multiplied by

t

i = \frac{V_{0}}{L} t e^{- α t}

This response is said to be critically damped.

Let

R = 4 Ω

L = 1 H

, and

C = 1 / 4 F

.
Let the initial conditions be

v_{c} (0) = 10 V

and

i_{L} = 0

Find $i (t)$ ‍.

The differential equation modeling the circuit is,

\frac{d^{2} i}{d t^{2}} + 4 \frac{d i}{d t} + 4 i = 0

which leads to the corresponding characteristic equation,

s^{2} + 4 s + 4 = 0

We can factor this directly (no need to use the quadratic formula),

(s + 2) (s + 2) = 0

giving us repeated real roots,

s_{1} = - 2

and

s_{2} = - 2

With repeated real roots the solution is,

i = \frac{V_{0}}{L} t e^{- α t}

i = \frac{10}{1} t e^{- 2 t}

i = 10 t e^{- 2 t}

The critically damped response looks a lot like an over damped response. It's hard to tell the difference by eye. The critical response is the boundary of all possible decaying responses.

$α^{2} - ω^{2} < 0$ ‍ underdamped

When

α

is smaller than

ω_{o}

, the square root term has a negative number inside, and

s

comes out as two complex conjugate numbers, with real and imaginary parts. The example circuit we worked out in the RLC natural response derivation article is an underdamped system.

The current looks like a sine wave that diminishes over time. Think of the sound a bell makes when you strike it. The bell's note rings out and fades over time. That is an underdamped second-order mechanical system. For second-order electrical circuits, we borrow the term and say the underdamped system "rings" at a frequency of approximately

ω_{o} = \frac{1}{\sqrt{LC}} .

If we let the resistance get really small and eventually go to

0

, then

α = R / 2 L

goes to zero and

s_{1, 2}

becomes

ω_{o}

. The circuit becomes a pure

LC

configuration. When we analyzed the natural response of the LC circuit, we came up with a sine wave that lasted forever. (In real life,

R

is never really

0

, so there is always some energy lost. A bell does not ring forever.)

The first example circuit we worked through earlier in this article had

R = 2 Ω, L = 1 H,

and

C = 1 / 5 F .

We're not going to repeat the solution, but here are a few observations using the

α

and

ω_{o}

notation.

The damping factor

α

α = \frac{R}{2 L} = \frac{2}{2 \cdot 1} = 1

The resonant frequency,

ω_{o}

ω_{o} = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \cdot 1 / 5}} = \sqrt{5}

Looking at the terms under the square root:

α^{2} - ω^{2} = 1^{2} - \sqrt{5}^{2} = - 4 =

a negative number, which we saw led to a decaying sine solution. Therefore, we would describe the example circuit as an underdamped system.

Summary

The

RLC

circuit is the electronic equivalent of a swinging pendulum with friction. The circuit can be modeled by this

2

nd-order linear differential equation:

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 0

The resulting characteristic equation is:

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

We solved for the roots of the characteristic equation using the quadratic formula:

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

By substituting variables

α

and

ω_{o}

we wrote

s

a little simpler as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

where

α = \frac{R}{2 L},

and

ω_{o} = \frac{1}{\sqrt{LC}}

Depending on the relative size of

α

and

ω_{o}

, we came up with three different forms of the solution:

overdamped, $α > ω_{0}$ ‍, leads to the sum of two decaying exponentials
critically damped, $α = ω_{0}$ ‍, leads to $t$ ‍ times decaying exponential
underdamped, $α < ω_{0}$ ‍, leads to a decaying sine

Want to join the conversation?

Sort by:

315137771xy
Posted 8 years ago. Direct link to 315137771xy's post “how to get the equation f...”
how to get the equation for i=
V0/W0*e−αt*t in crirically damped?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Jonathan is correct. The ...”
  From the author:Jonathan is correct. The expression for current should be i = V0/L t e^(-at). This has been corrected in the article. The article still just states the result, without derivation.
  
  Inspired by Jonathan's solution in this clarification, I wrote up this more complete derivation of the Critically Damped case:
  http://spinningnumbers.org/a/rlc-natural-response-variations.html#critically-damped
  This website is a continuation of the EE contributions I made at KA.
  Button navigates to signup page
  (2 votes)
seanwallawallaofficial
Posted 7 years ago. Direct link to seanwallawallaofficial's post “why do you get barely any...”
why do you get barely any questions, tips, or thanks on your electrical engineering section?
on other sections, i see over 100 questions but on this section, i hardly ever see 10
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “There are lots of questio...”
  There are lots of questions scattered throughout the EE subject area. You can see them if you add a /d at the end of the url, like this: https://www.khanacademy.org/science/electrical-engineering/d
  Button navigates to signup page
  (3 votes)
浩然费
Posted 2 years ago. Direct link to 浩然费's post “If I add a SPDT switch in...”
If I add a SPDT switch in the circuit, I can charge the capacitor. But if there is one way to charge without the switch. (For exmaple, I add a battery and it will work and form same-frequncy wave until battery does not have any energy)
Button navigates to signup pageComment on 浩然费's post “If I add a SPDT switch in...”
(1 vote)
Answer
- Willy McAllister
  Posted 2 years ago. Direct link to Willy McAllister's post “Providing an initial char...”
  Providing an initial charge to a capacitor or initial current to an inductor is a tricky bit of work in a circuit simulator. It is difficult to implement a mechanical switch. They are usually not part of the simulator's primitive elements.
  
  It's possible to mimic a mechanical switch using very large transistors. I show an example of this on my web site, spinningnumbers.org. This is a follow-on to my work at KA where all the articles are updated. If you navigate to the Natural and Forced Response topic, find the article on RLC Variations. The simulation models have giant MOSFETS acting like mechanical switches to initialize the voltage. Remove the spaces and try this URL,
  
  https : //spinningnumbers . org/a/rlc-natural-response-variations.html#under-damped
  
  (Scroll down to the end of the Under Damped chapter to find the simulation model.)
  Comment on Willy McAllister's post “Providing an initial char...”
  (2 votes)
Deepak Gautam
Posted 6 years ago. Direct link to Deepak Gautam's post “The article begins by tel...”
The article begins by telling:
( α > ωo ) implies overdamped circuit. Or,
( R/2L > ωo ) implies overdamped circuit..................... (1)

But at a later point, the article also tells:
"A circuit will be overdamped if the resistance is high relative to the resonant frequency."
This means,
( R > ωo ) implies overdamped circuit..................... (2)

If ( R/2L > ωo ) and ( R > ωo ), both implies an overdamped circuit, this means both are equivalent statements.

How can ( R/2L > ωo ) means the same thing as ( R > ωo )?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Equation (1) is correct. ...”
  Equation (1) is correct.
  For equation (2) you are taking a very qualitative statement and trying to turn it into an equation, but this doesn't work. The units are all wrong... ohms > radians/sec. I'm just trying to say you end up with an over damped circuit if the resistance is high. In technical terms, a circuit is over damped when equation (1) is true.
  Comment on Willy McAllister's post “Equation (1) is correct. ...”
  (2 votes)
ronen peri
Posted 3 years ago. Direct link to ronen peri's post “In the hidden section abo...”
In the hidden section about the underdamped natural frequency approximation, it states that we can only approximate the frequency as w0 because the damping factor under the radical shifts the frequency over time (the wavelength changes).

I think this is misleading and not true. The natural frequency does not change; only the envelope changes.
lets begin with the end solution for the differential eqn:
i = <some function of t> * <a sinusoidal>. in our RLC case:
i = 5e^-t * sin2t.
We can clearly see that the argument of the sine function depends linearly on t. Nothing else. The multiplier (or Eigenvalue) acts only as an envelope function which does nothing in term of frequency.

The natural frequency is determined by the roots of the differential equation, which in turn are characterized by the "damping factor" alpha and the "frequency" omega. But these names are only naming convention after the case that the damping factor equals 0 (the only case we can state that w is the natural frequency, as it is the only factor which determines the roots). Otherwise, the frequency is a "combination" of alpha and w. And it stays constant! The name "damping factor" arise from the fact that when it is not zero, the sinusoidal decays.

The only way to change the system's FIXED natural frequency is by changing the boundary conditions.
Button navigates to signup pageComment on ronen peri's post “In the hidden section abo...”
(1 vote)
Answer
Erkan Giray Arat
Posted 4 years ago. Direct link to Erkan Giray Arat's post “I used a program to graph...”
I used a program to graph the overdamped current solution, and I am pretty sure it never goes up to two amperes as it is shown. The equation 1.25*(exp(-x)-exp(-9x)) has a zero derivative at (-ln(1/9))/8=0.2747 which makes the function value 0.8443, never rises above this.
Button navigates to signup pageComment on Erkan Giray Arat's post “I used a program to graph...”
(1 vote)
Answer
- Willy McAllister
  Posted 4 years ago. Direct link to Willy McAllister's post “https://spinningnumbers.o...”
  https://spinningnumbers.org/a/rlc-natural-response-variations.html#over-damped
  
  image: https://spinningnumbers.org/i/rlc_overdamped_current.svg
  Button navigates to signup page
  (1 vote)
muhammad atif
Posted 6 years ago. Direct link to muhammad atif's post “Suppose a parallel RLC ci...”
Suppose a parallel RLC circuit is critically damped at the natural frequency of 100 Hz. I have values for R and L. How do I calculate a value of C such that it is critically damped. I know it can be calculated using 1/2RC = 1/sqrt(LC) but how do I incorporate an element of 100 Hz from this formula? This formula doesn't give me the critical damped circuit with the natural frequency of 100 Hz. Please tell me.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
wmburtis
Posted 5 years ago. Direct link to wmburtis's post “I don't really have a goo...”
I don't really have a good feel for what the i-t curves would look like for these three variations of the RLC natural response. Why don't you give specific examples with their graphs?

A more general criticism of the entire EE course -- there are not nearly enough problems to try to solve to test the reader's understanding. Please provide problems for each module!!
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “The currents are fully wo...”
  The currents are fully worked out and plotted. Search in this article for "Example of over damping" and "Example of critical damping". These are hints that open up to reveal the plots.
  
  If you follow this link to the derivation of the under damped RLC you will find a plot of the current for that variation, https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/w/ee/a/ee-rlc-natural-response-derivation
  Button navigates to signup page
  (1 vote)

Electrical engineering

Course: Electrical engineering > Unit 2

Introduction

What we're building to

Modeling and solving the circuit - review

Exact solution

Overdamped, critically damped, underdamped

α2−ω2>0‍ overdamped

α2−ω2=0‍ critically damped

α2−ω2<0‍ underdamped

Summary

Want to join the conversation?

$α^{2} - ω^{2} > 0$ ‍ overdamped

$α^{2} - ω^{2} = 0$ ‍ critically damped

$α^{2} - ω^{2} < 0$ ‍ underdamped