EECS 216 2nd -ORDER CIRCUITS: SINUSOIDAL SOURCES
NATURAL RESPONSE (NO SOURCE PRESENT)
Given:
Initial:
t=0:
Goal:
Series RLC circuit (resistor+inductor+capacitor connected together).
Capacitor is charged up to v(0) and inductor is “juiced” up to i(0).
Close or throw the switch at t = 0. Recall i(t) and v(t) don’t jump.
Compute current through inductor (and everything else) i(t) for t > 0.
t
1
di
1 t
Devices: i = C dv
→
v(t)
=
i(t
)dt
and
v
=
L
→
i(t)
=
dt
C −∞
dt
L −∞ v(t )dt .
t
1 d
d2 i
R di
1
KVL: C1 −∞ i(t )dt + L di
dt + iR = 0. Take L dt → dt2 + L dt + LC i = 0.
Note: Any circuit variable satisfies this differential equation. So use the following procedure to compute inductor current iL (t) or capacitor voltage vC (t) (usually easiest), then compute other quantities from those.
Soln: The trial solution i(t) = Iest solves the above differential equation if
[s2 +
R
Ls
+
1
st
LC ]Ie
= 0 → s2 +
R
Ls
+
1
LC
= 0 (characteristic equation)
So: i(t) = C1 es1 t + C2 es2 t for some constants C1 , C2 , s1 , s2 (Math 216 idea)
where: C1 and C2 determined by initial conditions v(0) and i(0)
and: s1 and s2 are the two roots of the characteristic equation.
Three different cases: Overdamped,
underdamped, critically damped:
R 2
) >
1. ( 2L
R 2
) <
2. ( 2L
where: α =
Note:
1
α
=
R
2L
L
2R
R 2
) =
3. ( 2L
1
LC
R
→ s1,2 = − 2L
±
R 2
1
( 2L
) − LC
< 0 (so it always decays)
1
R 2
± j LC
− ( 2L
) = −α ± jωd
R
→ s1,2 = − 2L
1
R 2
and ωd = LC
− ( 2L
) = the damped natural frequency.
1
LC
acts like a time constant, affecting amplitude of sinusoids.
1
LC
→ i(t) = C1 e−αt + C2 te−αt (called critical damping).
COMPLETE RESPONSE (NOW INCLUDE SOURCE)
Formula: i(t) =
iSS (t)
steady−state
+ C1 es1 t + C2 es2 t for t > 0.
transient decays to 0
where: C1 and C2 are determined by the initial conditions v(0) and i(0).
Note: Follows from linearity of differential equation; C1 and C2 are deterdi +
mined by the initial conditions, used to determine i(0+ ) and dt
(0 ).
EECS 216
2nd -ORDER CIRCUITS: EXAMPLES
Given: Circuit shown at right:
✤✜
Initially: i(0) = 0 and v(0) = −1
❅ ❅
+
4Ω
2cos(3t)
Then: Throw switch at t = 0.
–
✣✢
Goal: Compute i(t) for t > 0
Char. poly.: s2 +
R
Ls
+
1
LC
= s2 + 41 s +
1
1( 13 )
1H
I
1
3F
✲ +
V
–
= 0 → s1 = −1;
s2 = −3.
1
Steady Phasors: KVL → 2ej0 − 4I − j(3)(1)I − j(3)(1/3)
I = 2 − (4 + j2)I = 0
o
2
= 0.447e−j27 → iSS (t) = 0.447 cos(3t − 27o ).
State: → I = 4+j2
Complete: i(t) = C1 e−t + C2 e−3t + 0.447 cos(3t − 27o ). To compute C1 and C2 :
• 0 = i(0− ) = i(0+ ) = C1 + C2 + 0.447 cos(−27o ) → C1 + C2 = −0.4.
+
)
di +
• Using KVL, we get dt
(0 ) = vL (0
= 2−0−(−1)
= 3 since i(0+ ) = 0.
L
1
3 = −C1 − 3C2 − 0.447(3) sin(−27o ) → C1 + 3C2 = −2.4.
Solve: Both C1 + C2 = −0.4 and C1 + 3C2 = −2.4 → C1 = 0.6 and C2 = −1.
Sol’n: i(t) = 0.447 cos(3t − 27o ) + 0.6e−t − e−3t for t > 0. i(0) = 0 checks.
steady−state
Given:
Initially:
Then:
Goal:
The circuit shown at right:
Switch closed for long time
Switch is opened at t = 0.
Compute vC (t) for t > 0.
transient decays to 0
✤✜
+
10V
–
✣✢
t = 0− : C → open and L → short → iL (0− ) =
10
50
❅ ❅
50Ω
0.02F
10Ω
✆
5H
✆
✆
= 0.2 and vC (0− ) = 0.
t = 0+ : iL (0+ ) = iL (0− ) = 0.2 and vC (0+ ) = vC (0− ) = 0 (no jumps) →
dvC
1
−0.2
+
+
+
+
dt (0 ) = C ic (0 ) = 0.02 = −10 since ic (0 ) = −iL (0 ) (1oop).
1
2
s2 + 10
s2 = −1 − j3.
5 s + (5)(.02) = s + 2s + 10 = 0 → s1 = −1 + j3;
Char.:
SS: Steady-state response=0 snce no sources once switch is opened at t = 0.
Complete: vC (t) = Ce(−1+j3)t +C ∗ e(−1−j3)t = C1 e−t cos(3t)+C2 e−t sin(3t)where:
Solve: 0 = vC (0+ ) = C1 and −10 = dvdtC (0+ ) = 3C2 → C2 = −3.33.
Sol’n: vC (t) = −3.33e−t sin(3t) for t > 0. Plug this in and confirm it works.
Note: Since vC (0+ ) = 0, can see that there will be no cosine term in vC (t).
1