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lecture 18 2
2nd Order Circuits
• Any circuit with a single capacitor, a single inductor, an arbitrary number of sources, and an arbitrary number of resistors is a circuit of order 2.
• Any voltage or current in such a circuit is the solution to a 2nd order differential equation.
lecture 18 3
Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping ratio
lecture 18 4
A 2nd Order RLC Circuit
• The source and resistor may be equivalent to a circuit with many resistors and sources.
R
Cvs(t)
i (t)
L
+–
lecture 18 5
Applications Modeled by a 2nd Order RLC Circuit
• Filters
– A lowpass filter with a sharper cutoff than can be obtained with an RC circuit.
lecture 18 6
The Differential Equation
KVL around the loop:
vr(t) + vc(t) + vl(t) = vs(t)
R
Cvs(t)
+
–
vc(t)
+ –vr(t)
L
+– vl(t)
i (t)
+–
lecture 18 7
Differential Equation
)()(1)(
)( tvdxxiCdt
tdiLtRi s
t
dt
tdv
Lti
LCdt
tdi
L
R
dt
tid s )(1)(
1)()(2
2
lecture 18 8
The Differential Equation
Most circuits with one capacitor and inductor are not as easy to analyze as the previous circuit. However, every voltage and current in such a circuit is the solution to a differential equation of the following form:
)()()(
2)( 2
002
2
tftidt
tdi
dt
tid
lecture 18 9
Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping ratio
lecture 18 10
The Particular Solution
• The particular (or forced) solution ip(t) is usually a weighted sum of f(t) and its first and second derivatives.
• If f(t) is constant, then ip(t) is constant.
• If f(t) is sinusoidal, then ip(t) is sinusoidal.
lecture 18 11
The Complementary Solution
The complementary (homogeneous) solution has the following form:
K is a constant determined by initial conditions.
s is a constant determined by the coefficients of the differential equation.
stc Keti )(
lecture 18 12
Complementary Solution
02 2002
2
ststst
Kedt
dKe
dt
Ked
02 200
2 ststst KesKeKes
02 200
2 ss
lecture 18 13
Characteristic Equation
• To find the complementary solution, we need to solve the characteristic equation:
• The characteristic equation has two roots-call them s1 and s2.
02 200
2 ss
lecture 18 14
Complementary Solution
• Each root (s1 and s2) contributes a term to the complementary solution.
• The complementary solution is (usually)
tstsc eKeKti 21
21)(
lecture 18 15
Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping ratio
lecture 18 16
Damping Ratio () andNatural Frequency (0)
• The damping ratio is .
• The damping ratio determines what type of solution we will get:
– Exponentially decreasing ( >1)
– Exponentially decreasing sinusoid ( < 1)
• The natural frequency is 0
– It determines how fast sinusoids wiggle.
lecture 18 17
Roots of the Characteristic Equation
The roots of the characteristic equation determine whether the complementary solution wiggles.
12001 s
12002 s
lecture 18 18
Real Unequal Roots
• If > 1, s1 and s2 are real and not equal.
• This solution is overdamped.
tt
c eKeKti
1
2
1
1
200
200
)(
lecture 18 19
Overdamped
0
0.2
0.4
0.6
0.8
1
-1.00E-06
t
i(t)
-0.2
0
0.2
0.4
0.6
0.8
-1.00E-06
ti(t)
lecture 18 20
Complex Roots
• If < 1, s1 and s2 are complex.
• Define the following constants:
• This solution is underdamped.
tAtAeti ddt
c sincos)( 21
0 2
0 1 d
lecture 18 21
Underdamped
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1.00E-05 1.00E-05 3.00E-05
t
i(t)
lecture 18 22
Real Equal Roots
• If = 1, s1 and s2 are real and equal.
• This solution is critically damped.
ttc teKeKti 00
21)(
lecture 18 23
Example
• This is one possible implementation of the filter portion of the IF amplifier.
10
769pFvs(t)
i (t)
159H
+–
lecture 18 24
More of the Example
dt
tdv
Lti
LCdt
tdi
L
R
dt
tid s )(1)(
1)()(2
2
)()()(
2)( 2
002
2
tftidt
tdi
dt
tid
For the example, what are and 0?
lecture 18 25
Example continued
• = 0.011
• 0 = 2455000
• Is this system over damped, under damped, or critically damped?
• What will the current look like?
lecture 18 26
Example (cont.)
• The shape of the current depends on the initial capacitor voltage and inductor current.
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1.00E-05 1.00E-05 3.00E-05
t
i(t)
lecture 18 27
Slightly Different Example
• Increase the resistor to 1k• What are and 0?
1k
769pFvs(t)
i (t)
159H
+–
lecture 18 28
Example cont.
• = 2.2
• 0 = 2455000
• Is this system over damped, under damped, or critically damped?
• What will the current look like?
lecture 18 29
Example (cont.)
• The shape of the current depends on the initial capacitor voltage and inductor current.
0
0.2
0.4
0.6
0.8
1
-1.00E-06
t
i(t)
lecture 18 30
Damping Summary
ζ Roots (s1, s2) Damping
ζ>1 Real and unequal Overdamped
ζ=1 Real and equal Critically damped
0<ζ<1 Complex Underdamped