Transcript Second-Order Circuits (7.3)

Second-Order Circuits (7.3)
Dr. Holbert
April 19, 2006
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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.
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Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping
ratio
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A 2nd Order RLC Circuit
i (t)
R
vs(t)
+
–
C
L
• The source and resistor may be equivalent
to a circuit with many resistors and sources.
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Applications Modeled by a 2nd
Order RLC Circuit
• Filters
– A lowpass filter with a sharper cutoff
than can be obtained with an RC circuit.
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The Differential Equation
i (t) + vr(t) –
R
vs(t)
+
–
+
vc(t)
C
– vl(t) +
–
L
KVL around the loop:
vr(t) + vc(t) + vl(t) = vs(t)
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Differential Equation
t
di (t ) 1
Ri (t )  L
  i ( x)dx  vs (t )
dt
C 
2
d i (t ) R di (t ) 1
1 dvs (t )


i(t ) 
2
dt
L dt
LC
L dt
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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
d i (t )
di (t )
2
 2 0
  0 i(t )  f (t )
2
dt
dt
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Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping
ratio
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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.
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The Complementary Solution
The complementary (homogeneous) solution
has the following form:
ic (t )  Ke
st
K is a constant determined by initial conditions.
s is a constant determined by the coefficients of
the differential equation.
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Complementary Solution
2
st
st
d Ke
dKe
2
st
 2 0
  0 Ke  0
2
dt
dt
s Ke  2 0 sKe   Ke  0
2
st
st
2
0
st
s  2 0 s    0
2
2
0
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Characteristic Equation
• To find the complementary solution, we
need to solve the characteristic equation:
s  2 0 s    0
2
2
0
• The characteristic equation has two rootscall them s1 and s2.
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Complementary Solution
• Each root (s1 and s2) contributes a term to
the complementary solution.
• The complementary solution is (usually)
ic (t )  K1e  K 2 e
s1t
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s2t
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Important Concepts
• The differential equation
• Forced and homogeneous solutions
• The natural frequency and the damping
ratio
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Damping Ratio () and
Natural 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.
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Roots of the Characteristic
Equation
The roots of the characteristic equation
determine whether the complementary
solution wiggles.
s1   0   0   1
2
s2   0   0   1
2
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Real Unequal Roots
• If  > 1, s1 and s2 are real and not equal.
ic (t )  K1e
    2 1  t
0
0


 K 2e
    2 1  t
0
0


• This solution is overdamped.
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1
0.8
0.8
0.6
0.6
i(t)
i(t)
Overdamped
0.4
0.4
0.2
0.2
0
-1.00E-06
-0.2
0
-1.00E-06
t
t
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Complex Roots
• If  < 1, s1 and s2 are complex.
• Define the following constants:
   0
d  0 1  
ic (t )  e
 t
2
 A1 cos  d t  A2 sin  d t 
• This solution is underdamped.
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Underdamped
1
0.8
0.6
i(t)
0.4
0.2
0
-1.00E-05
-0.2
1.00E-05
3.00E-05
-0.4
-0.6
-0.8
-1
t
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Real Equal Roots
• If  = 1, s1 and s2 are real and equal.
ic (t )  K1e
 0t
 K 2 te
 0t
• This solution is critically damped.
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Example
i (t)
10W
vs(t)
+
–
769pF
159mH
• This is one possible implementation of the
filter portion of the IF amplifier.
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More of the Example
2
d i (t ) R di (t ) 1
1 dvs (t )


i(t ) 
2
dt
L dt
LC
L dt
2
d i (t )
di (t )
2
 2 0
  0 i(t )  f (t )
2
dt
dt
For the example, what are  and 0?
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Even More Example
•  = 0.011
• 0 = 2p455000
• Is this system over damped, under damped,
or critically damped?
• What will the current look like?
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Example (cont.)
• The shape of the current depends on the
initial capacitor voltage and inductor
current.
1
0.8
0.6
i(t)
0.4
0.2
0
-1.00E-05
-0.2
1.00E-05
3.00E-05
-0.4
-0.6
-0.8
-1
t
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Slightly Different Example
i (t)
1kW
vs(t)
+
–
769pF
159mH
• Increase the resistor to 1kW
• What are  and 0?
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More Different Example
•  = 2.2
• 0 = 2p455000
• Is this system over damped, under damped,
or critically damped?
• What will the current look like?
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Example (cont.)
• The shape of the current depends on the
initial capacitor voltage and inductor
current.
1
i(t)
0.8
0.6
0.4
0.2
0
-1.00E-06
t
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Damping Summary
ζ
Roots (s1, s2)
Damping
ζ>1
Real and unequal Overdamped
ζ=1
Real and equal
0<ζ<1 Complex
Critically damped
Underdamped
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Class Example
• Learning Extension E7.9
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