EGR 2201 Unit 10 Second-Order Circuits

EGR 2201 Unit 10 Second-Order Circuits
Read Alexander & Sadiku, Sections 8.1 through 8.4. Homework #10 and Lab #10 due next week. Quiz next week.

Review: Four Kinds of First-Order Circuits
The circuits we studied last week are called first-order circuits because they are described mathematically by first-order differential equations. We studied four kinds of first-order circuits: Source-free RC circuits Source-free RL circuits RC circuits with sources RL circuits with sources

Review: A General Approach for First-Order Circuits (1 of 3)
General approach for most of the problems we studied last week: Find the quantity’s initial value 𝑥(0). Find the quantity’s final value 𝑥(∞). Find the time constant: 𝜏=𝑅𝐶 for an RC circuit. 𝜏=𝐿/𝑅 for an RL circuit. Once you know these items, solution is : 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏

The equation from the previous slide, 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏 graphs as either: A decaying exponential curve if the initial value x(0) is greater than the final value x(). Or a saturating exponential curve if the initial value x(0) is less than the final value x().

Recall also that we can think of the complete response as being the sum of a transient response that dies away with time and a steady-state response that is constant and remains after the transient has died away: 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏 Steady-state response Transient response

Transient Analysis with Multisim
The textbook’s Sections 7.8 and 8.9 discuss using PSpice simulation software to perform transient analysis of first-order and second-order circuits. We can also do this with Multisim, as shown here. The steps are summarized in Lab 10. Walk them through the simple analysis shown here.

Our Goal: A General Approach for Second-Order Circuits
Next we will develop a general approach for analyzing more complicated circuits called second-order circuits. Unfortunately the general approach for second-order circuits is quite a bit more complicated than the one for first-order circuits.

Second-Order Circuits
The circuits we’ll study are called second-order circuits because they are described mathematically by second-order differential equations. Whereas first-order circuits contain a single energy-storing element (capacitor or inductor), second-order circuits contain two energy-storing elements. These two elements could both be capacitors or both be inductors, but we’ll focus on circuits containing one capacitor and one inductor.

Four Kinds of Second-Order Circuits
The book treats four kinds of second-order circuits: Source-free series RLC circuits Source-free parallel RLC circuits Series RLC circuits with sources (We won’t cover these.) Parallel RLC circuits with sources (We won’t cover these.)

Natural Response and Step Response
Recall that the term natural response refers to the behavior of source-free circuits. And the term step response refers to the behavior of circuits in which a source is applied at some time. So the goal of this chapter in the book is to understand the natural response of source-free RLC circuits, and to understand the step response of RLC circuits with sources. We’ll only have time for the natural response of source-free circuits.

Redraw, Redraw, Redraw! Our procedure will usually require us to find values of voltages or currents at the following three times: At t = 0, just before a switch is opened or closed. At t = 0+, just after a switch is opened or closed. As t  , a long time after a switch is opened or closed. Usually the circuit looks different at these three times, so you’ll want to redraw the circuit for each of these times.

Finding Initial Values
To completely solve a first-order differential equation, you need one initial condition, usually either: An initial inductor current i(0+), or An initial capacitor voltage v(0+). To completely solve a second-order differential equation, you need two initial conditions, usually either: An initial inductor current i(0+) and its derivative di(0+)/dt, or An initial capacitor voltage v(0+) and its derivative dv(0+)/dt.

Finding Initial Derivative Values
To find initial derivative values such as dv(0+)/dt, we’ll rely on the basic relationships for capacitors and inductors: For example, if we know a capacitor’s initial current i(0+), then we can use the left-hand equation above to find the initial derivative of that capacitor’s voltage, dv(0+)/dt. 𝑖=𝐶 𝑑𝑣 𝑑𝑡 𝑣=𝐿 𝑑𝑖 𝑑𝑡

Quantities that Cannot Change Abruptly
We’ll also rely on the fact that a capacitor’s voltage and an inductor’s current cannot change abruptly. Example: In this circuit, i(0+) must be equal to i(0), and v(0+) must be equal to v(0). Since these values must be equal, we don’t really need to distinguish between their values at time t = 0 and at time t = 0+. So we could just write i(0) instead of i(0+) and i(0).

Caution: Some Quantities Can Change Abruptly
Don’t assume that every quantity has the same value at times t = 0 and t = 0+. Example: In the same circuit, iC(t) changes abruptly from 0 A to 2 A at time t = 0. So we must distinguish between iC(0) and iC(0+): iC(0) = 0 A iC(0+) = 2 A iC(0) is undefined. Do practice question 1 parts a, b, c.

Finding Final Values Our procedure will sometimes also require us to find final or “steady-state” values, such as: A final inductor current i() A final capacitor voltage v(). Usually these final values are easier to find than initial values, because: We don’t have to worry about abrupt changes as t  , so we never need to distinguish between t   and t  +. We don’t have to find derivatives of currents or voltages as t  . Do practice question 1 part d and practice question 2.

Natural Response of Source-Free Series RLC Circuit (1 of 2)
Consider the circuit shown. Assume that at time t=0, the inductor has initial current I0, and the capacitor has initial voltage V0. As time passes, the initial energy in the capacitor and inductor will dissipate as current flows through the resistor. This results in changing current i(t), which we wish to calculate.

Natural Response of Source-Free Series RLC Circuit (2 of 2)
Applying KVL, 𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 + 1 𝐶 −∞ 𝑡 𝑖 𝜏 𝑑𝜏 =0 A standard trick for such integro-differential equations is to take the derivative of both sides: 𝑅 𝑑𝑖 𝑑𝑡 +𝐿 𝑑 2 𝑖 𝑑 𝑡 𝐶 𝑖=0 Now divide by L and rearrange terms: 𝑑 2 𝑖 𝑑 𝑡 2 + 𝑅 𝐿 𝑑𝑖 𝑑𝑡 + 1 𝐿𝐶 𝑖=0

A Closer Look at Our Differential Equation
Our equation, 𝑑 2 𝑖 𝑑 𝑡 2 + 𝑅 𝐿 𝑑𝑖 𝑑𝑡 + 1 𝐿𝐶 𝑖=0, is a second-order linear differential equation with constant coefficients. Such equations have been studied extensively. The following slides outline the standard solution.

Solving Our Differential Equation (1 of 4)
To solve 𝑑 2 𝑖 𝑑 𝑡 2 + 𝑅 𝐿 𝑑𝑖 𝑑𝑡 + 1 𝐿𝐶 𝑖=0, first assume that the solution is of the form 𝑖 𝑡 =𝐴 𝑒 𝑠𝑡 where A and s are constants to be found. Plugging this into the equation yields: 𝐴 𝑠 2 𝑒 𝑠𝑡 + 𝑅 𝐿 𝐴𝑠 𝑒 𝑠𝑡 + 1 𝐿𝐶 𝐴 𝑒 𝑠𝑡 =0 Factoring out the common term, 𝐴 𝑒 𝑠𝑡 (𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶 )=0

From 𝐴 𝑒 𝑠𝑡 (𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶 )=0 it follows that: 𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶 =0 This is called the characteristic equation of our original differential equation. Note that it is purely algebraic with one variable (s). It has no derivatives, no integrals, no exponentials. Now we can use the quadratic formula to solve for s. But first….

For convenience we introduce two new symbols: 𝛼= 𝑅 2𝐿 and 𝜔 0 = 1 𝐿𝐶 We call  the neper frequency, and we call 0 the undamped natural frequency. Then we can rewrite 𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶 =0 as: 𝑠 2 +2𝛼𝑠+ 𝜔 0 2 =0 Now use the quadratic formula…. Do practice question 3 parts a, b.

Applying the quadratic formula to 𝑠 2 +2𝛼𝑠+ 𝜔 0 2 =0 gives two solutions for s, which we call the natural frequencies: 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 Because of the square root, we must distinguish three cases:  > 0, called the overdamped case.  = 0, called the critically damped case.  < 0, called the underdamped case. Do practice question 3 parts c, d.

The Overdamped Case ( > 0)
If  > 0, our two solutions for s are distinct negative real numbers: 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 In this case, the solution to our differential equation is 𝑖 𝑡 = 𝐴 1 𝑒 𝑠 1 𝑡 + 𝐴 2 𝑒 𝑠 2 𝑡 The initial conditions determine A1 and A2: 𝐴 1 + 𝐴 2 =𝑖( 0 + ), 𝑠 1 𝐴 1 + 𝑠 2 𝐴 2 = 𝑑𝑖( 0 + ) 𝑑𝑡 Real number

The Critically Damped Case ( = 0)
If  = 0, our two solutions for s are equal to each other and to : 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 0 2 =−𝛼 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 =−𝛼 In this case, the solution to our differential equation is 𝑖 𝑡 =( 𝐴 2 + 𝐴 1 𝑡) 𝑒 −𝛼𝑡 The initial conditions determine A1 and A2: 𝐴 2 =𝑖( 0 + ), 𝐴 1 = 𝑑𝑖( 0 + ) 𝑑𝑡 +𝛼 𝐴 2 Zero

The Underdamped Case ( < 0)
If  < 0, our two solutions for s are complex numbers: 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 0 2 =−𝛼+𝑗 𝜔 𝑑 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 =−𝛼−𝑗 𝜔 𝑑 Here j is the imaginary unit, 𝑗= −1 , and d is called the damped natural frequency, 𝜔 𝑑 = 𝜔 0 2 − 𝛼 2 . In this case, the solution to our differential equation is 𝑖 𝑡 = 𝑒 −𝛼𝑡 ( 𝐵 1 cos 𝜔 𝑑 𝑡 + 𝐵 2 sin 𝜔 𝑑 𝑡 ) The initial conditions determine B1 and B2: 𝐵 1 =𝑖( 0 + ), 𝐵 2 = 𝑑𝑖( 0 + ) 𝑑𝑡 + 𝛼 𝐵 1 𝜔 𝑑 Imaginary number

Graphs of the Three Cases
Do practice question 4. Details will differ based on initial conditions and element values, but the shapes shown here are typical. Note the oscillation in the underdamped case.

Typing Equations in Word 2013
Select Insert > Equation on Word’s menu bar. Use the toolbar’s Structures section to create fractions, exponents, square roots, and more. Use the toolbar’s Symbols section to insert basic math symbols, Greek letters, special operators, and more.

Oscilloscope Looking ahead, we’ll use an oscilloscope to display and measure fast-changing voltages, including transients.

Oscilloscope Challenge Game
The oscilloscope is a complex instrument that you must learn to use. To learn the basics, play my Oscilloscope Challenge game at

Natural Response of Source-Free Parallel RLC Circuit (1 of 2)
Preview: The math for a source-free parallel circuit is almost the same as the math on the previous slides, except that: Now the variable in our differential equation is v(t) instead of i(t). We define  (the neper frequency) to be equal to 1 2𝑅𝐶 instead of 𝑅 2𝐿 .

Natural Response of Source-Free Parallel RLC Circuit (2 of 2)
By applying KCL and some algebra, we get: 𝑑 2 𝑣 𝑑 𝑡 𝑅𝐶 𝑑𝑣 𝑑𝑡 + 1 𝐿𝐶 𝑣=0 By assuming a solution of the form 𝑣 𝑡 =𝐴 𝑒 𝑠𝑡 we can derive the characteristic equation 𝑠 𝑅𝐶 𝑠+ 1 𝐿𝐶 =0

Solving Our Differential Equation
For convenience we define: 𝛼= 1 2𝑅𝐶 and 𝜔 0 = 1 𝐿𝐶 Note that 0 (the undamped natural frequency) is the same as for series RLC circuits, but  (the neper frequency) is not. Therefore, just as for series circuits, 𝑠 2 +2𝛼𝑠+ 𝜔 0 2 =0 We get the same three cases as before (overdamped, critically damped, and underdamped)….

The Overdamped Case ( > 0)
If  > 0, our two solutions for s are distinct negative real numbers 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 In this case, the solution to our differential equation is 𝑣 𝑡 = 𝐴 1 𝑒 𝑠 1 𝑡 + 𝐴 2 𝑒 𝑠 2 𝑡 The initial conditions determine A1 and A2: 𝐴 1 + 𝐴 2 =𝑣( 0 + ), 𝑠 1 𝐴 1 + 𝑠 2 𝐴 2 = 𝑑𝑣( 0 + ) 𝑑𝑡

The Critically Damped Case ( = 0)
If  = 0, our two solutions for s are equal to each other and to : 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 0 2 =−𝛼 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 =−𝛼 In this case, the solution to our differential equation is 𝑣 𝑡 =( 𝐴 2 + 𝐴 1 𝑡) 𝑒 −𝛼𝑡 The initial conditions determine A1 and A2: 𝐴 2 =𝑣( 0 + ), 𝐴 1 = 𝑑𝑣( 0 + ) 𝑑𝑡 +𝛼 𝐴 2

The Underdamped Case ( < 0)
If  < 0, our two solutions for s are complex numbers: 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 0 2 =−𝛼+𝑗 𝜔 𝑑 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 =−𝛼−𝑗 𝜔 𝑑 Here j is the imaginary unit, 𝑗= −1 , and d is called the damped natural frequency, 𝜔 𝑑 = 𝜔 0 2 − 𝛼 2 . In this case, the solution to our differential equation is 𝑣 𝑡 = 𝑒 −𝛼𝑡 ( 𝐵 1 cos 𝜔 𝑑 𝑡 + 𝐵 2 sin 𝜔 𝑑 𝑡 ) The initial conditions determine B1 and B2: 𝐵 1 =𝑣( 0 + ), 𝐵 2 = 𝑑𝑣( 0 + ) 𝑑𝑡 + 𝛼 𝐵 1 𝜔 𝑑

Graphs of the Three Cases
Do practice question 5. Details will differ based on initial conditions and element values, but the shapes shown here are typical. Note the oscillation in the underdamped case.

A General Approach for Source-Free Series or Parallel RLC Circuits (1 of 3)
To find x(t) in a source-free series or parallel RLC circuit, where x could be any current or voltage: Find the quantity’s initial value 𝑥( 0 + ). Find the quantity’s initial derivative 𝑑𝑥( 0 + ) 𝑑𝑡 . Find the neper frequency: 𝛼= 𝑅 2𝐿 for a series RLC circuit. 𝛼= 1 2𝑅𝐶 for a parallel RLC circuit. Find the undamped natural frequency: 𝜔 0 = 1 𝐿𝐶 Compare  to 0 to see whether circuit is overdamped, critically damped, or underdamped…

If it’s overdamped ( > 0), then: 𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 , 𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2 Solve for A1 and A2: 𝐴 1 + 𝐴 2 =𝑥( 0 + ), 𝑠 1 𝐴 1 + 𝑠 2 𝐴 2 = 𝑑𝑥( 0 + ) 𝑑𝑡 𝑥 𝑡 = 𝐴 1 𝑒 𝑠 1 𝑡 + 𝐴 2 𝑒 𝑠 2 𝑡 If it’s critically damped ( = 0), then: 𝐴 2 =𝑥( 0 + ), 𝐴 1 = 𝑑𝑥( 0 + ) 𝑑𝑡 +𝛼 𝐴 2 𝑥 𝑡 =( 𝐴 2 + 𝐴 1 𝑡) 𝑒 −𝛼𝑡

If it’s underdamped ( < 0), then: 𝜔 𝑑 = 𝜔 0 2 − 𝛼 2 𝐵 1 =𝑥( 0 + ), 𝐵 2 = 𝑑𝑥( 0 + ) 𝑑𝑡 + 𝛼 𝐵 1 𝜔 𝑑 𝑥 𝑡 = 𝑒 −𝛼𝑡 ( 𝐵 1 cos 𝜔 𝑑 𝑡 + 𝐵 2 sin 𝜔 𝑑 𝑡 )

EGR 2201 Unit 10 Second-Order Circuits

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EGR 2201 Unit 10 Second-Order Circuits

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