Download presentation
Presentation is loading. Please wait.
1
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.
2
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
3
Review: A General Approach for First-Order Circuits (1 of 3)
General approach for most of the problems we studied last week: Find the quantityโs initial value ๐ฅ(0). Find the quantityโs final value ๐ฅ(โ). Find the time constant: ๐=๐
๐ถ for an RC circuit. ๐=๐ฟ/๐
for an RL circuit. Once you know these items, solution is : ๐ฅ ๐ก =๐ฅ(โ)+(๐ฅ(0)โ๐ฅ(โ)) ๐ โ ๐ก ๐
4
Review: A General Approach for First-Order Circuits (2 of 3)
The equation from the previous slide, ๐ฅ ๐ก =๐ฅ(โ)+(๐ฅ(0)โ๐ฅ(โ)) ๐ โ ๐ก ๐ graphs as either: A decaying exponential curve if the initial value x(0) is greater than the final value x(๏ฅ). Or a saturating exponential curve if the initial value x(0) is less than the final value x(๏ฅ).
5
Review: A General Approach for First-Order Circuits (3 of 3)
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: ๐ฅ ๐ก =๐ฅ(โ)+(๐ฅ(0)โ๐ฅ(โ)) ๐ โ ๐ก ๐ Steady-state response Transient response
6
Transient Analysis with Multisim
The textbookโs 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.
7
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.
8
Second-Order Circuits
The circuits weโll 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 weโll focus on circuits containing one capacitor and one inductor.
9
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 wonโt cover these.) Parallel RLC circuits with sources (We wonโt cover these.)
10
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. Weโll only have time for the natural response of source-free circuits.
11
Redraw, Redraw, Redraw! Our procedure will usually require us to find values of voltages or currents at the following three times: At t = 0๏ญ, just before a switch is opened or closed. At t = 0+, just after a switch is opened or closed. As t ๏ฎ ๏ฅ, a long time after a switch is opened or closed. Usually the circuit looks different at these three times, so youโll want to redraw the circuit for each of these times.
12
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.
13
Finding Initial Derivative Values
To find initial derivative values such as dv(0+)/dt, weโll rely on the basic relationships for capacitors and inductors: For example, if we know a capacitorโs initial current i(0+), then we can use the left-hand equation above to find the initial derivative of that capacitorโs voltage, dv(0+)/dt. ๐=๐ถ ๐๐ฃ ๐๐ก ๐ฃ=๐ฟ ๐๐ ๐๐ก
14
Quantities that Cannot Change Abruptly
Weโll also rely on the fact that a capacitorโs voltage and an inductorโs current cannot change abruptly. Example: In this circuit, i(0+) must be equal to i(0๏ญ), and v(0+) must be equal to v(0๏ญ). Since these values must be equal, we donโt really need to distinguish between their values at time t = 0๏ญ and at time t = 0+. So we could just write i(0) instead of i(0+) and i(0๏ญ).
15
Caution: Some Quantities Can Change Abruptly
Donโt assume that every quantity has the same value at times t = 0๏ญ 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๏ญ) and iC(0+): iC(0๏ญ) = 0 A iC(0+) = 2 A iC(0) is undefined. Do practice question 1 parts a, b, c.
16
Finding Final Values Our procedure will sometimes also require us to find final or โsteady-stateโ values, such as: A final inductor current i(๏ฅ) A final capacitor voltage v(๏ฅ). Usually these final values are easier to find than initial values, because: We donโt have to worry about abrupt changes as t ๏ฎ ๏ฅ, so we never need to distinguish between t ๏ฎ ๏ฅ๏ญ and t ๏ฎ ๏ฅ+. We donโt have to find derivatives of currents or voltages as t ๏ฎ ๏ฅ. Do practice question 1 part d and practice question 2.
17
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.
18
Natural Response of Source-Free Series RLC Circuit (2 of 2)
Applying KVL, ๐
๐+๐ฟ ๐๐ ๐๐ก + 1 ๐ถ โโ ๐ก ๐ ๐ ๐๐ =0 A standard trick for such integro-differential equations is to take the derivative of both sides: ๐
๐๐ ๐๐ก +๐ฟ ๐ 2 ๐ ๐ ๐ก ๐ถ ๐=0 Now divide by L and rearrange terms: ๐ 2 ๐ ๐ ๐ก 2 + ๐
๐ฟ ๐๐ ๐๐ก + 1 ๐ฟ๐ถ ๐=0
19
A Closer Look at Our Differential Equation
Our equation, ๐ 2 ๐ ๐ ๐ก 2 + ๐
๐ฟ ๐๐ ๐๐ก + 1 ๐ฟ๐ถ ๐=0, is a second-order linear differential equation with constant coefficients. Such equations have been studied extensively. The following slides outline the standard solution.
20
Solving Our Differential Equation (1 of 4)
To solve ๐ 2 ๐ ๐ ๐ก 2 + ๐
๐ฟ ๐๐ ๐๐ก + 1 ๐ฟ๐ถ ๐=0, first assume that the solution is of the form ๐ ๐ก =๐ด ๐ ๐ ๐ก where A and s are constants to be found. Plugging this into the equation yields: ๐ด ๐ 2 ๐ ๐ ๐ก + ๐
๐ฟ ๐ด๐ ๐ ๐ ๐ก + 1 ๐ฟ๐ถ ๐ด ๐ ๐ ๐ก =0 Factoring out the common term, ๐ด ๐ ๐ ๐ก (๐ 2 + ๐
๐ฟ ๐ + 1 ๐ฟ๐ถ )=0
21
Solving Our Differential Equation (2 of 4)
From ๐ด ๐ ๐ ๐ก (๐ 2 + ๐
๐ฟ ๐ + 1 ๐ฟ๐ถ )=0 it follows that: ๐ 2 + ๐
๐ฟ ๐ + 1 ๐ฟ๐ถ =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 firstโฆ.
22
Solving Our Differential Equation (3 of 4)
For convenience we introduce two new symbols: ๐ผ= ๐
2๐ฟ and ๐ 0 = 1 ๐ฟ๐ถ We call ๏ก the neper frequency, and we call ๏ท0 the undamped natural frequency. Then we can rewrite ๐ 2 + ๐
๐ฟ ๐ + 1 ๐ฟ๐ถ =0 as: ๐ 2 +2๐ผ๐ + ๐ 0 2 =0 Now use the quadratic formulaโฆ. Do practice question 3 parts a, b.
23
Solving Our Differential Equation (4 of 4)
Applying the quadratic formula to ๐ 2 +2๐ผ๐ + ๐ 0 2 =0 gives two solutions for s, which we call the natural frequencies: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 Because of the square root, we must distinguish three cases: ๏ก > ๏ท0, called the overdamped case. ๏ก = ๏ท0, called the critically damped case. ๏ก < ๏ท0, called the underdamped case. Do practice question 3 parts c, d.
24
The Overdamped Case (๏ก > ๏ท0)
If ๏ก > ๏ท0, our two solutions for s are distinct negative real numbers: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 In this case, the solution to our differential equation is ๐ ๐ก = ๐ด 1 ๐ ๐ 1 ๐ก + ๐ด 2 ๐ ๐ 2 ๐ก The initial conditions determine A1 and A2: ๐ด 1 + ๐ด 2 =๐( 0 + ), ๐ 1 ๐ด 1 + ๐ 2 ๐ด 2 = ๐๐( 0 + ) ๐๐ก Real number
25
The Critically Damped Case (๏ก = ๏ท0)
If ๏ก = ๏ท0, our two solutions for s are equal to each other and to ๏ญ๏ก: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ 0 2 =โ๐ผ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 =โ๐ผ In this case, the solution to our differential equation is ๐ ๐ก =( ๐ด 2 + ๐ด 1 ๐ก) ๐ โ๐ผ๐ก The initial conditions determine A1 and A2: ๐ด 2 =๐( 0 + ), ๐ด 1 = ๐๐( 0 + ) ๐๐ก +๐ผ ๐ด 2 Zero
26
The Underdamped Case (๏ก < ๏ท0)
If ๏ก < ๏ท0, our two solutions for s are complex numbers: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ 0 2 =โ๐ผ+๐ ๐ ๐ ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 =โ๐ผโ๐ ๐ ๐ Here j is the imaginary unit, ๐= โ1 , and ๏ทd is called the damped natural frequency, ๐ ๐ = ๐ 0 2 โ ๐ผ 2 . In this case, the solution to our differential equation is ๐ ๐ก = ๐ โ๐ผ๐ก ( ๐ต 1 cos ๐ ๐ ๐ก + ๐ต 2 sin ๐ ๐ ๐ก ) The initial conditions determine B1 and B2: ๐ต 1 =๐( 0 + ), ๐ต 2 = ๐๐( 0 + ) ๐๐ก + ๐ผ ๐ต 1 ๐ ๐ Imaginary number
27
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.
28
Typing Equations in Word 2013
Select Insert > Equation on Wordโs menu bar. Use the toolbarโs Structures section to create fractions, exponents, square roots, and more. Use the toolbarโs Symbols section to insert basic math symbols, Greek letters, special operators, and more.
29
Oscilloscope Looking ahead, weโll use an oscilloscope to display and measure fast-changing voltages, including transients.
30
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
31
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 ๏ก (the neper frequency) to be equal to 1 2๐
๐ถ instead of ๐
2๐ฟ .
32
Natural Response of Source-Free Parallel RLC Circuit (2 of 2)
By applying KCL and some algebra, we get: ๐ 2 ๐ฃ ๐ ๐ก ๐
๐ถ ๐๐ฃ ๐๐ก + 1 ๐ฟ๐ถ ๐ฃ=0 By assuming a solution of the form ๐ฃ ๐ก =๐ด ๐ ๐ ๐ก we can derive the characteristic equation ๐ ๐
๐ถ ๐ + 1 ๐ฟ๐ถ =0
33
Solving Our Differential Equation
For convenience we define: ๐ผ= 1 2๐
๐ถ and ๐ 0 = 1 ๐ฟ๐ถ Note that ๏ท0 (the undamped natural frequency) is the same as for series RLC circuits, but ๏ก (the neper frequency) is not. Therefore, just as for series circuits, ๐ 2 +2๐ผ๐ + ๐ 0 2 =0 We get the same three cases as before (overdamped, critically damped, and underdamped)โฆ.
34
The Overdamped Case (๏ก > ๏ท0)
If ๏ก > ๏ท0, our two solutions for s are distinct negative real numbers ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 In this case, the solution to our differential equation is ๐ฃ ๐ก = ๐ด 1 ๐ ๐ 1 ๐ก + ๐ด 2 ๐ ๐ 2 ๐ก The initial conditions determine A1 and A2: ๐ด 1 + ๐ด 2 =๐ฃ( 0 + ), ๐ 1 ๐ด 1 + ๐ 2 ๐ด 2 = ๐๐ฃ( 0 + ) ๐๐ก
35
The Critically Damped Case (๏ก = ๏ท0)
If ๏ก = ๏ท0, our two solutions for s are equal to each other and to ๏ญ๏ก: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ 0 2 =โ๐ผ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 =โ๐ผ In this case, the solution to our differential equation is ๐ฃ ๐ก =( ๐ด 2 + ๐ด 1 ๐ก) ๐ โ๐ผ๐ก The initial conditions determine A1 and A2: ๐ด 2 =๐ฃ( 0 + ), ๐ด 1 = ๐๐ฃ( 0 + ) ๐๐ก +๐ผ ๐ด 2
36
The Underdamped Case (๏ก < ๏ท0)
If ๏ก < ๏ท0, our two solutions for s are complex numbers: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ 0 2 =โ๐ผ+๐ ๐ ๐ ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 =โ๐ผโ๐ ๐ ๐ Here j is the imaginary unit, ๐= โ1 , and ๏ทd is called the damped natural frequency, ๐ ๐ = ๐ 0 2 โ ๐ผ 2 . In this case, the solution to our differential equation is ๐ฃ ๐ก = ๐ โ๐ผ๐ก ( ๐ต 1 cos ๐ ๐ ๐ก + ๐ต 2 sin ๐ ๐ ๐ก ) The initial conditions determine B1 and B2: ๐ต 1 =๐ฃ( 0 + ), ๐ต 2 = ๐๐ฃ( 0 + ) ๐๐ก + ๐ผ ๐ต 1 ๐ ๐
37
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.
38
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 quantityโs initial value ๐ฅ( 0 + ). Find the quantityโs initial derivative ๐๐ฅ( 0 + ) ๐๐ก . Find the neper frequency: ๐ผ= ๐
2๐ฟ for a series RLC circuit. ๐ผ= 1 2๐
๐ถ for a parallel RLC circuit. Find the undamped natural frequency: ๐ 0 = 1 ๐ฟ๐ถ Compare ๏ก to ๏ท0 to see whether circuit is overdamped, critically damped, or underdampedโฆ
39
A General Approach for Source-Free Series or Parallel RLC Circuits (2 of 3)
If itโs overdamped (๏ก > ๏ท0), then: ๐ 1 =โ๐ผ+ ๐ผ 2 โ ๐ , ๐ 2 =โ๐ผโ ๐ผ 2 โ ๐ 0 2 Solve for A1 and A2: ๐ด 1 + ๐ด 2 =๐ฅ( 0 + ), ๐ 1 ๐ด 1 + ๐ 2 ๐ด 2 = ๐๐ฅ( 0 + ) ๐๐ก ๐ฅ ๐ก = ๐ด 1 ๐ ๐ 1 ๐ก + ๐ด 2 ๐ ๐ 2 ๐ก If itโs critically damped (๏ก = ๏ท0), then: ๐ด 2 =๐ฅ( 0 + ), ๐ด 1 = ๐๐ฅ( 0 + ) ๐๐ก +๐ผ ๐ด 2 ๐ฅ ๐ก =( ๐ด 2 + ๐ด 1 ๐ก) ๐ โ๐ผ๐ก
40
A General Approach for Source-Free Series or Parallel RLC Circuits (3 of 3)
If itโs underdamped (๏ก < ๏ท0), then: ๐ ๐ = ๐ 0 2 โ ๐ผ 2 ๐ต 1 =๐ฅ( 0 + ), ๐ต 2 = ๐๐ฅ( 0 + ) ๐๐ก + ๐ผ ๐ต 1 ๐ ๐ ๐ฅ ๐ก = ๐ โ๐ผ๐ก ( ๐ต 1 cos ๐ ๐ ๐ก + ๐ต 2 sin ๐ ๐ ๐ก )
Similar presentations
© 2025 SlidePlayer.com Inc.
All rights reserved.