Overview
Second Order
Circuits: Series
RLC Circuit
• Recap 1st order circuits
• Introduction to 2nd order circuit solutions
o Two handouts
• Natural response for series RLC circuits
• Complete solution method
• Form of solution: “damping” of the circuit
response
o Expectations
• Circuit solutions
EGR 220, Chapter 8
March 10, 2016
o Source-free response
o Complete response: Step response
2
1st
1st Order Circuit Review
Order Circuit Review
• Find i(t) for t > 0
o Expectations for i(t)?
• Graph and math expression
o Quantities to find (time period for
each)
4
5
1
Recap:
Circuit Response: R Network
Input, Sources è
System
è
Circuit Response: 1st Order
Input, Sources è
System
è
Output, Circuit Behavior
Output, Circuit Behavior
20V
è
$"
# = xn (t) + x f (t)
x f (t) = X S (1− e−t/τ )$%
xn (t) = X 0 e−t/τ
x1 (t) = Xo e
• DC, static circuits: Chapters 1 through 4
−t
τ
(
+ X s 1− e
−t
τ
è
)
x2 (t) = x(∞) +[x(0) − x(∞)]e
−t
τ
= x1 (t)
8
7
Circuit Response: 2nd Order
Input, Sources è
System
è
2nd Order System Mechanics Analogy
• A spring-mass system:
Output, Circuit Behavior
d2x
mg − kx = ma = m 2
dt
x = A ⋅ f (sin ωt)
è
ωo =
9
k
m
10
€
2
Energy Storage Elements
Capacitor
Second Order Circuit Analysis
• Second order differential equations
describe the circuit behavior
• Contain:
Inductor
In DC conditions,
behaves as:
o _________ (how many) storage elements?
The Constituent
Relationship is:
• The time periods of interest:
o
The differential in
the Constituent
Relationship
reveals:
• Information we need:
o
12
13
Series RLC
Natural Response Expression
Series RLC Natural Response Handout
d 2i R di 1
+
+
i=0
dt 2 L dt LC
• Let Vs=0 for natural response (assume fully charged)
we assume that i = Ae st , and plug this in
AR st A st
As e +
se +
e =0
L
LC
R
1
Ae st (s 2 + s +
)=0
L
LC
either i = Ae st is zero or (.) is zero...
• Write KVL
2 st
• Substitute in for each V
(s 2 +
• Take the derivative of the equation
R
1
s+
)=0
L
LC
is solved by finding the roots 's' s.t. (.) = 0
2
s1,2 = −
α≡
•
14
" R%
R
1
± $ ' −
# 2L & LC
2L
=
− α ± α 2 − ω 02
R
1
, ω0 ≡
2L
LC
Note that having two roots means that we need two
constants of integration and so two initial conditions
15
3
Typical Responses (h.o.)
Form of Response
• Overdamped α > ω0
o The roots, ‘s,’ are negative, unequal and real
• What are the options for the
values s1 and s2?
x(t) = A1e s1t + A2 e s 2 t
• Critically damped α = ω0
o The roots, ‘s,’ are equal and real
€
x(t) = (A1 + A2 t)e −αt
• Underdamped α < ω0
• What is the relationship
of α and ωo for each type
of response?
o The roots, ‘s,’ are complex
€
x(t) = e −αt ( A1 cosω d t + A2 sinω d t)
16
17
€
Solution Methodology
For Lab 7 – RLC Circuit
1.
2.
3.
4.
Find initial and final conditions
Find s1 & s2 and α & ωo (instead of τ)
Identify type of damping
Construct complete response
expression
5. Solve for constants (after constructing
complete solution)
6. Write out solution equation, and be
able to graph this final solution
• Check the discussion on pp 323-324,
for series RLC circuits
— Role of R in
damping
— SeRling time as an
indicator of type
of damping
— Note critical
damping – not
always with
overvshoot
18
19
4
Text Example 8.2
Text Example 8.2
• Find the initial and final conditions –
carefully drawing the circuit at each time
stage
• For t = 0–
20
21
Text Example 8.2
Text Example 8.2
• For t = 0+
• For t à ∞
22
23
5
Beyond Text Example 8.2
Summary
• Find i(t) for t > 0
• Concepts for 2nd order circuits
o Behavior of 2 energy storage elements
with dissipating element(s) and sources
o Types of circuit response expected
• Methods for determining specific
response
o Familiarity with solving and interpreting
• Differential equations
• Characteristic equation
• Taking derivatives
24
25
26
27
Lab After Break
• Back to Knowledge Building, to
continue through the end of the
semester
6