1. ECE 222 Electric Circuit Analysis II Chapter 10 Natural Response
in Series RLC Circuits
Herbert G. Mayer, PSU
Status 4/19/2016
For use at CCUT Spring 2016

2. Syllabus Definitions Initial State RLC Circuit i(t) in Series RLC
Example 2
Bibliography

3. a * x + b * dx/dt + c * d2x/dt2 + d = 0
Definitions
Natural Response is current discharged in a series RLC circuit by releasing energy stored in the circuit’s capacitor and inductor
In a First-Oder Differential Equation (DE), the characteristic variable x = x(t) would occur only in its simple form x, and as first derivate x’(t) or dx/dt. 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 occurs in original form x, and first derivative, as well as a second derivative
Example for variable x with constants a, b, c, d:
a * x + b * dx/dt + c * d2x/dt2 + d = 0

4. Initial State
In a series circuit at time t < 0 an electric source (to be removed at t = 0) may charge components C and L
If so, there exists residual energy in the series circuit
At t = 0 capacitor C in the series circuit holds an initial charge, creating its electric field between the 2 plates
And at t = 0 inductor L holds a magnetic field between its terminals
This energy discharges, when the circuit is closed and the source is removed exactly at time t = 0

5. Series RLC Circuit

6. RLC Circuit Below is a series circuit, analyzed for second-order DE
Assume V0 across C in the circuit, no current at t = 0

7. RLC Circuit – Recall from First-Order
SI unit of [H] = [V s A-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:
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:
C dv / dt + v / R = 0

8. RLC Circuit – Second-Order
KVL for Natural Response in series R L C circuit with single R, L, C, with i = i(t) at time t > 0:
R i + 1/C i dt + L di / dt + V0 = 0
Once we have current i, we can compute the voltages in all 3 series elements (RLC)
Eliminate integral by differentiating:
R di/dt + i/C + L d2i/dt2 + 0 = 0
Divide by L, re-order by powers of derivative, and we have a second-order DE for i(t) to solve (a.):
d2i / dt2 + R / L * di / dt + i / (LC) = 0
i’’ + i’ R / L + i / LC = 0

9. RLC Circuit – Characteristic Equation
Rewrite as characteristic equation for a series circuit:
s2 + s R / L + 1 / (LC) = 0
The roots for the characteristic equation are:
s1/2 = -R/2L ± ( (R/2L)2 - 1/LC )½
Abbreviate with α and ω0 as before:
s1 = -α + ( α2 - ω02 )½
s2 = -α - ( α2 - ω02 ) ½
α = R / ( 2L )
ω0 = ( LC )-½
Verify SI units [V] for integral-form: R i + 1/C i dt + L di / dt + V0 = 0

10. RLC Circuit – Second-Order
Verify the SI units (dimension) for KCL equation
R i + 1/C i dt + L di / dt + V0 = 0
All summands must be of units [V] in the end
SI units of R i = [V A-1 A] = [V]
SI units of 1/C i dt = [A-1 s-1 V A s] = [V]
SI units of L di / dt = [V s A-1 A s-1] = [V]
Confirmed!

11. Series Electric Circuit
Damping in
Series Electric Circuit

12. Damping, Like for Parallel RLC

13. Functions for Damping Classes

14. Overdamped 1.) Root types: real, distinct roots s1, s2
2.) α vs. ω0: α2 > ω02
3.) Voltage v(t): i(t) = A1 es1 t + A2 es2 t
4.) Root Functions s1, s2:
s1 = -α + ( α2 - ω02 )1/2
s2 = -α - ( α2 - ω02 )1/2

15. Underdamped 1.) Root types: complex, distinct roots s1, s2
2.) α vs. ω0: α2 < ω02
3.) Voltage v(t):
i(t) = B1 e-α t cos( ωd t ) + B2 e-α t sin( ωd t )
4.) Root Functions s1, s2:
s1 = -α + ( α2 - ω02 )1/2
s2 = -a - ( α2 - ω02 )1/2
s1 = -α + j ωd
s2 = -α - j ωd
ωd = ( ω02 - α2 )1/2

16. Critically Damped 1.) Root types: real, equal roots s1, s2
2.) α vs. ω0: α2 = ω02
3.) Voltage v(t): i(t) = D1 t e-α t + D2 e-α t
4.) Root Functions s1, s2:
s1 = -α
s2 = -α

17. RLC Circuit – Second-Order
Once current i(t) is found in the series circuit, all voltages vRLC(t) across the RLC components can be computed
Given that we know the units for R, L, and C

18. (See also Example 1 in Parallel RLC Circuit)
Series RLC Circuit
(See also Example 1 in Parallel RLC Circuit)

19. Example 2 – Series RLC Circuit
Given the circuit for Example 2 below, we compute the following parts of the characteristic equations:
Neper frequency α
Square of Neper frequency α2
Square of Resonant Radian frequency ω02
Determine, whether the circuit is overdamped, underdamped, or critically damped
If applicable , find ωd2
Find the current equation i(t)

20. Example 2 – Series RLC Circuit
Series RLC circuit below has components:
R = 560 Ω, C = 0.1 μF, and L = 100 mH
C is charged with 100 V at t<=0

21. Example 2 – Series RLC Circuit
1. Compute Neper frequency α:
α = R / 2L
α = 560 / ( 2 * 100 * 10-3 ) = 5600 / 2
α = [Radians / s]
2. Compute α2:
α2 = [Radians2 / s2]
3. Compute square of Resonant Radian frequency ω02:
ω02 = 1 / LC = 1 / ( 100 * 10-3 * 0.1 * 10-6 )
ω02 = [Radians2 / s2]

22. Example 2 – Series RLC Circuit
4. Determine the class of damping:
We classify as follows
ω02 < a2 - overdamped
ω02 > a2 - underdamped
ω02 = a2 - critically damped
ω02 = [Radians2 / s2]
α2 = =
We see that ω02 > α2
Hence this circuit is underdamped

23. Example 2 – Series RLC Circuit
5. Compute root ωd:
ωd = ( ω02 - α2 )½
ωd = ( 108 – )½
ωd = 0.96 * [Radians / s]
6. Compute current equation i(t):
i(t) = B1 e -α t cos( ωd t ) + B2 e -α t sin( ωd t )

24. Bibliography
Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13:
Differentiation rules:
Euler’s Identities:
Table of integrals: