1. 0 ECE 222 Electric Circuit Analysis II Chapter 9 Natural Response in Parallel RLC Circuits Herbert G. Mayer, PSU Status 5/13/2016 For use at CCUT Spring 2016

2. 1 Syllabus Definitions Initial State RLC Circuit Characteristic Equation v(t) in Parallel RLC Damping Example 1 Bibliography

3. 2 Definitions Natural Response is mode of creating voltage in a parallel RLC circuit by releasing energy stored in the circuit’s capacitor and inductor, without any further energy sources In a First-Oder Differential Equations (DE) the characteristic variable x = x(t) occurs solely in the first derivate x’(t) or dx/dt, and in original form x. E.g. for variable x and constants a, b, c: a * x + b * dx/dt + c = 0 In a Second-Order DE, the characteristic variable x also occurs also as a first- as well as second derivative; Example for variable x with constants a, b, c, d: a * x + b * dx/dt + c * d 2 x/dt 2 + d = 0

4. 3 Initial State At t = 0 the capacitor C holds an initial charge in its electric field between the 2 plates No current flowing at t = 0 or before, i.e. t < 0 At t = 0 the inductor L has a magnetic field built up, caused by a prior current flowing at t < 0 through its windings There is residual energy in L and C that starts discharging at time t = 0 Then and at time t = 0 the source is removed, the circuit can follow its natural response

5. 4 Parallel RLC Circuit (Taken from text [1], pages 266 ff.)

6. 5 RLC Circuit Circuit below is under analysis for second-order DE Assume current I 0 in L flowing at t = 0 And V 0 is voltage at terminals of capacitor C

7. 6 RLC Circuit – Recall from First-Order SI unit of [H] = [V s A -1 ] SI unit of [H] = [V s A -1 ] SI unit of [F] = [A s V -1 ] SI unit of [F] = [A s V -1 ] KVL for Natural Response in RL circuit with single R, L (i.e. first-order), and current i(t) = i: KVL for Natural Response in RL circuit with single R, L (i.e. first-order), and current i(t) = i: L di / dt + R i = 0 KCL for Natural Response in RC circuit with single R, C (i.e. first-order), and voltage v(t) = v: KCL for Natural Response in RC circuit with single R, C (i.e. first-order), and voltage v(t) = v: C dv / dt + v / R = 0

8. 7 RLC Circuit – Second-Order KCL for Natural Response in parallel R L C circuit with single R, L, C and voltage v(t) = v, and initial current I 0 : KCL for Natural Response in parallel R L C circuit with single R, L, C and voltage v(t) = v, and initial current I 0 : v / R + 1/L v dt + C dv / dt + I 0 = 0 Find voltage v first, since v applies to all 3 parallel elements (RLC); then 3 currents can be computed Find voltage v first, since v applies to all 3 parallel elements (RLC); then 3 currents can be computed Eliminate integral by differentiating, const I 0 drops out: Eliminate integral by differentiating, const I 0 drops out: dv/dt * 1/R + v/L + C d 2 v/dt 2 + 0 = 0 Divide by C, re-order by powers of derivative, yields second-order DE for v. We’ll seek a solution to this: Divide by C, re-order by powers of derivative, yields second-order DE for v. We’ll seek a solution to this: d 2 v / dt 2 + 1 / (RC) dv / dt + v / (LC) = 0 v’’ + v’ / RC + v / LC = 0

9. 8 RLC Circuit – Characteristic Equation The method to find solution for v(t) used in first order DE does not work for second order DE The method to find solution for v(t) used in first order DE does not work for second order DE We know d 2 v / dt 2 + 1 / (RC) dv / dt + v / (LC) = 0 must always yields 0 We know d 2 v / dt 2 + 1 / (RC) dv / dt + v / (LC) = 0 must always yields 0 Since this involves 1. and 2. derivatives, the Σ = 0 for all values of t can only occur if the derivatives have the same general form as the actual equation Since this involves 1. and 2. derivatives, the Σ = 0 for all values of t can only occur if the derivatives have the same general form as the actual equation Only the e-function fulfills this unusual constraint Only the e-function fulfills this unusual constraint Plausible to assume (and later demonstrate the assumption) that v(t) is in fact an e-function Plausible to assume (and later demonstrate the assumption) that v(t) is in fact an e-function Also we observed exponentials with first-order DE, so use of an e-function sounds plausible Also we observed exponentials with first-order DE, so use of an e-function sounds plausible

10. 9 RLC Circuit – Characteristic Equation To try an e-function for v(t), e.g. we guess To try an e-function for v(t), e.g. we guess v(t) = A e st Where A and s are constants to be derived Where A and s are constants to be derived A is the initial circuit voltage A is the initial circuit voltage Constants A and s are yet to be computed Constants A and s are yet to be computed If guess v = A e st is right, it has to satisfy the second order DE d 2 v / dt 2 + 1 / (RC) dv / dt + v / (LC) = 0 If guess v = A e st is right, it has to satisfy the second order DE d 2 v / dt 2 + 1 / (RC) dv / dt + v / (LC) = 0 Reminder: this is KCL for 3 currents in RLC circuit, computing currents as function of voltages and electric components in the RLC circuit Reminder: this is KCL for 3 currents in RLC circuit, computing currents as function of voltages and electric components in the RLC circuit If the guess is right, we should be able to substitute that guessed equation’s R.H.S. into the DE If the guess is right, we should be able to substitute that guessed equation’s R.H.S. into the DE First a quick digression into verifying the SI units: First a quick digression into verifying the SI units:

11. 10 RLC Circuit – SI Units Verify the SI units (dimension) for KCL equation Verify the SI units (dimension) for KCL equation v / R + 1/L v dt + C dv/dt + I 0 = 0 All summands must have unit [A]. Confirm this: All summands must have unit [A]. Confirm this: SI units of v / R= [V A V -1 ]= [A] SI units of v / R= [V A V -1 ]= [A] SI units of 1/L v dt= [V -1 s -1 A V s]= [A] SI units of 1/L v dt= [V -1 s -1 A V s]= [A] SI units of C dv / dt= [A s V -1 V s -1 ]= [A] SI units of C dv / dt= [A s V -1 V s -1 ]= [A] Confirmed! Confirmed!

12. 11 v(t) in Parallel RLC Circuit

13. 12 v(t) in Parallel RLC Circuit Solve: d 2 v / dt 2 + 1 / (R C) * dv / dt + v / (LC) = 0 using: v(t) = A e st Where A is constant, a function of initial values and of RLC And s similarly is a function of RLC, but constant! Goal: find solutions for A and s to solve our guessed function for v(t) Since we have a quadratic (second order) equation, we must find 2 solutions for A and s Name them A 1, A 2, and s 1, s 2 Where A 1 and A 2 are initial values defined by the circuit under analysis

14. 13 v(t) in Parallel RLC Circuit If our guess v(t) = v = A e st holds, then: v’ = s A e st v’’ = s 2 A e st Substituting v = A e st and the first 2 derivatives into d 2 v / dt 2 + 1 / (R C) * dv / dt + v / (LC) = 0 We finally get: s 2 A e st + s A e st / RC + A e st / LC = 0 And with A e st factored out: A e st ( s 2 + s / RC + 1 / LC ) = 0 Equation must hold for all values of t

15. 14 v(t) in Parallel RLC Circuit A e st ( s 2 + s / RC + 1 / LC ) = 0 holds, if either A e st is 0, or the parenthesized expression is 0 But A e st cannot be zero for all times of t, thus the parenthesized factor must be 0: s 2 + s / RC + 1 / LC = 0 This is the characteristic equation CE!! CE is a simple quadratic equation, yielding 2 roots for s, named s 1 and s 2 Roots s 1 and s 2 determine (characterize) the physical behavior of the voltage over time, or the mathematical character of v(t) Now solve the simple quadratic equation to find roots s 1 and s 2

16. 15 Remember Quadratic Equation a x 2 + b x + c= 0 x 2 + x b/a + c/a= 0 x 2 + x b/a= -c/a x 2 + x b/a + b 2 /4a 2 = -c/a + b 2 /4a 2 ( x + b/2a ) 2 = -c/a + b 2 /4a 2 x + b/2a= ±( -c/a + b 2 /4a 2 ) ) ½ x 1 = -b/2a +( -c/a + b 2 /4a 2 ) ) ½ x 2 = -b/2a -( -c/a + b 2 /4a 2 ) ) ½

17. 16 v(t) in Parallel RLC Circuit – Solve s 1, s 2 s 2 + s / RC + 1 / LC = 0 s 2 + 2s / (2 RC) = - 1/LC s 2 + 2s / (2 RC) + 1/(2 RC) 2 = - 1/LC + 1/(2 RC) 2 ( s + 1/(2RC) ) 2 = 1/(2RC) 2 - 1/LC s + ½ RC = +-( ( 1/(2RC) ) 2 - 1/LC ) ½ s 1 = -½ RC + ( 1/(2RC) 2 - 1/LC ) ½ s 2 = -½ RC - ( 1/(2RC) 2 - 1/LC ) ½

18. 17 v(t) in Parallel RLC Circuit – Solve s 1, s 2 Substituting either s 1 or s 2 into v(t) = A e st will by definition satisfy the DE! And A 1 and A 2 are initial electric sizes/units defined for any specific circuit under analysis: v 1 = A 1 e s1 t v 2 = A 2 e s2 t v(t) = v = v 1 + v 2 = A 1 e s1 t + A 2 e s2 t

19. 18 v(t) in Parallel RLC Circuit With v(t) and derivatives v’ and v’’ v(t) = v = v 1 + v 2 = A 1 e s1 t + A 2 e s2 t v’ = dv/dt = A 1 s 1 e s1 t + A 2 s 2 e s2 t v’’ = d 2 v/dt 2 = A 1 s 1 2 e s1 t + A 2 s 2 2 e s2 t Substitute v, v’, and v’’: A 1 s 1 2 e s1 t + A 2 s 2 2 e s2 t + A 1 s 1 e s1 t / RC + A 2 s 2 e s2 t / RC + A 1 e s1 t / LC + A 2 e s2 t / LC = 0 A 1 e s1 t ( s 1 2 + s 1 / RC + 1/ LC ) + A 2 e s2 t ( s 2 2 + s 2 / RC + 1/ LC ) = 0 Both blue parenthesized expressions ( ) have to be 0 for the overall function to yield = 0! We know s 1 and s 2 are roots of CE, and are ≠ 0 And s 1 and s 2 cannot be 0 for finite t

20. 19 Some Abbreviations For Parallel RLC Circuit

21. 20 v(t) in Parallel RLC Circuit – α and ω 0 Rewrite certain subexpressions, to generate a simple, standardized notation: s 1 = -1 / (2RC) + ( 1 / (2RC) 2 - 1/LC ) ½ s 2 = -1 / (2RC) - ( 1 / (2RC) 2 - 1/LC ) ½ Using abbreviations α (Neper frequency) and ω 0 (resonant radian frequency) we write: α = 1 / 2RC ω 0 = 1 / ( LC ) ½ s 1 = -α + ( α 2 - ω 0 2 ) ½ s 2 = -α - ( α 2 - ω 0 2 ) ½

22. 21 v(t) in Parallel RLC Circuit – EE Terms Solution for v: v = A 1 e s1 t + A 2 e s2 t Where s 1 and s 2 have been determined, A 1 and A 2 are initial values in circuit at t = 0 Common EE terms: s 1 and s 2 - characteristic roots, or complex frequencies α- Neper frequency ω 0 - Resonance Radian frequency ω 0 2 < α 2 - roots are real, different; voltage over damped ω 0 2 > α 2 - s 1 and s 2 are complex, different; voltage under damped ω 0 2 = α 2 - s 1 and s 2 are real, equal; voltage critically damped

23. 22 Damping Classes

24. 23 v(t) in Parallel RLC Circuit - Dimensions The exponents of e must be dimensionless But we see t in the exponent Hence factors s 1 and s 2 must be frequency, or t -1 Thus α and ω 0 better have same SI unit, namely t -1 [α]= [ 1 / 2RC ] = [ 1 / V A -1 A s V -1 ] = [s -1 ] [ω 0 ]= [ 1 / ( LC ) ½ ]= [ 1 / ( A s V -1 A -1 s V ) ½ ] = [s -1 ] Indeed α and ω 0 have frequency dimension [s -1 ]

25. 24 Damping in Parallel Electric Circuit

26. 25 Damping Analogy from Mechanical Engineering Imagine a restaurant’s kitchen door swinging into and out of the adjacent kitchen, driven by a door-spring Waiters kick the door to open, go through, and the door swings back and forth, until mechanical damping of air + hinge friction gradually bring stand still How long and how wide does the door oscillate? As wide as initially “kicked”; rest depends on: Mass of door, analogous to inductivity L: Resists push, but when moved, holds kinetic energy to swing Spring, analogous to capacitor C: able to compress and expand in both directions, accelerating or damping the door’s movement = oscillation Air- and mechanical friction, analogous to resistor R: Slows down oscillating movement due to energy spent

27. 26 Damping Analogy from Mechanical Engineering If there were no friction (or there were no R in electric circuit), the system could oscillate undamped forever But generally there is air to cause friction for the door, R to cause resistance; we assume real system, with friction and resistance If all else remained unmodified, an increasing door mass would slow down the oscillation frequency, but enable it to oscillate for a longer time, due to large energy injected against friction If all else remained unmodified, a weaker spring stiffness would slow down the oscillation frequency

28. 27 Damping

29. 28 Remember: Damping Classes

30. 29 Overdamped 1.) Root types:real, distinct roots s 1, s 2 2.) α vs. ω 0 : α 2 > ω 0 2 3.) Voltage v(t):v(t) = A 1 e s1 t + A 2 e s2 t 4.) Root Functions s1, s2: s 1 = -α + ( α 2 - ω 0 2 ) 1/2 s 2 = -α - ( α 2 - ω 0 2 ) 1/2

31. 30 Underdamped 1.) Root types:complex, distinct roots s 1, s 2 2.) α vs. ω 0 : α 2 < ω 0 2 3.) Voltage v(t): v(t) = B 1 e -α t cos( ω d t ) + B 2 e -α t sin( ω d t ) 4.) Root Functions s1, s2: s 1 = -α + ( α 2 - ω 0 2 ) 1/2 s 2 = -a - ( α 2 - ω 0 2 ) 1/2 s 1 = -α + j ω d s 2 = -α - j ω d ω d = ( ω 0 2 - α 2 ) 1/2

32. 31 Critically Damped 1.) Root types:real, equal roots s 1, s 2 2.) α vs. ω 0 : α 2 = ω 0 2 3.) Voltage v(t):v(t) = D 1 t e -α t + D 2 e -α t 4.) Root Functions s1, s2: s 1 = -α s 2 = -α

33. 32 Example 1 Parallel RLC Circuit (See also Example 2 in Series RLC Circuit)

34. 33 Example 1 – RLC Circuit Given the circuit for Example 1 below, we compute the following parts of the characteristic equations: 1. Neper frequency α 2. Square of Neper frequency α 2 3. Square of Resonant Radian frequency ω 0 2 4. Root s 1 5. Root s 2 6. Determine, whether the circuit is overdamped, underdamped, or critically damped 7. Find value for R such that the circuit will be critically damped

35. 34 Example 1 – RLC Circuit Parallel RLC circuit below has components: R = 200 Ω, C = 0.2 μF, and L = 50 mH

36. 35 Example 1 – RLC Circuit 1. Compute Neper α (see page 19): α = 1 / 2RC α = 1 / ( 2 * 200 * 0.2 * 10 -6 ) = 1 / ( 0.8 * 10 -4 ) α = 1.25 10 4 -- [Radians / s] 2. Compute α 2 : α 2 = 1 / ( 2RC ) 2 α 2 = 1.5625 10 8 -- [Radians 2 / s 2 ] 3. Compute square of Resonant Radian frequency ω 0 2 : ω 0 2 = 1 / LC= 1 / ( 50 * 10 -3 * 0.2 * 10 -6 ) ω 0 2 = 10 8 -- [Radians 2 / s 2 ]

37. 36 Example 1 – RLC Circuit 4. Compute root s 1 : s 1 = -α + ( α 2 + ω 0 2 ) ½ s 1 = -1/(2RC) + (1/(2RC) 2 - 1/LC ) ½ s 1 = -α + ( α 2 + ω 0 2 ) ½ s 1 = - 1.25 10 4 + 7,500 s 1 = - 5,000 -- [Radians / s] 5. Compute root s 2 : s 2 = -α - ( α 2 + ω 0 2 ) ½ s 2 = -1/(2RC) - (1/(2RC) 2 - 1/LC ) ½ s 1 = - 20,000 -- [Radians / s]

38. 37 Example 1 – RLC Circuit 6. Determine the class of damping: We classified as follows ω 0 2 < a 2 - overdamped ω 0 2 > a 2 - underdamped ω 0 2 = a 2 - critically damped ω 0 2 = 10 8 -- [Radians 2 / s 2 ] α 2 = 1.5625 10 8 Hence this circuit is overdamped

39. 38 Example 1 – RLC Circuit 7. Find the value for R such that the circuit will be critically damped, which means: such that ω 0 2 = a 2 Given ω 0 2 = 1 / ( LC ) α 2 = 1 / ( 2RC ) 2 1/LC=1 / 4R 2 L 2 R 2 =LC/(4C 2 )=L / 4C R 2 =50*10 -2 / ( 4 * 0.2 * 10 -6 ) = 50/0.8 * 10 3 R=250 Ω

40. 39 Bibliography  Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13: 9780-13-376003-3. Most samples here are copied directly from the text book  Differentiation rules: http://www.codeproject.com/KB/recipes/Differentiati on.aspx  Euler’s Identities: https://en.wikipedia.org/wiki/Euler%27s_identity  Table of integrals: http://integral- table.com/downloads/single-page-integral-table.pdf  Damping: https://en.wikipedia.org/wiki/Damping