1. Linear second-order circuits
1.1. Series RLC circuit
Dynamic circuits containing two capacitors or two inductors or one inductor and one
capacitor are called the second order circuits. At first we consider a special class of the
second-order circuits, namely a series connection of a resistor, an inductor and a capacitor
driven by a voltage source (see Fig. 1.1).
R
i
L
vR
vL
vC
C
vs(t)
Fig. 1.1
We describe this circuit using the capacitor voltage vC as a variable. Therefore we express
voltages vR and vL in terms of vC
vR = Ri = RC
vL = L
dvC
,
dt
diL
d ⎛ dv ⎞
d 2v
= L ⎜ C C ⎟ = LC 2C .
dt
dt ⎝ dt ⎠
dt
(1.1)
(1.2)
Next we write KVL equation
vR + vL + vC − vs = 0
(1.3)
and substitute into this equation relationships (1.1) and (1.2)
RC
dvC
d 2v
+ LC 2C + vC = vs .
dt
dt
After simple rearrangements we obtain
d 2vC R dvC
1
1
+
+
vC =
vs .
2
dt
L dt
LC
LC
For convenience in manipulation let us define two parameters
α=
R
,
2L
ω0 =
1
1
.
LC
(1.4)
The parameter α is called the damping constant and the parameter ω0 the angular resonant
frequency.
Using these parameters we rewrite equation (1.4) in the form
d 2vC
dv
+ 2α C + ω02vC = ω02vs .
2
dt
dt
(1.5)
The initial conditions are:
vC (0) = V0
and
iL (0 ) = I 0 .
The second initial condition can be expressed as follows
C
dvC
(0) = iL (0) = I 0
dt
or
dvC
(0) = 1 I 0 .
dt
C
Equation (1.5) that describes the circuit of Fig. 1.1 is a linear second order, nonhomogenous
differential equation with constant coefficients. As we know from mathematics the solution of
this equation consists of two terms vn and v f
vC = vn + v f ,
(1.6)
where vn is a general solution of the homogenous equation
d 2 vn
dv
+ 2α n + ω02vn = 0 ,
dt
dt
(1.7)
where v f is a particular solution of the original (nonhomogenous) equation (1.5).
At first we take into account the homogenous equation (1.7). To solve this equation we write
the characteristic equation
s 2 + 2αs + ω02 = 0 .
The zeros of this equation, called the natural frequences of the circuit, are
s1 = −α + α 2 − ω02 ,
(1.8)
s2 = −α − α 2 − ω02 .
(1.9)
The form of the solution of equation (1.7) depends on the relative values of α and ω0 .
Therefore we consider the following cases.
2
Case 1
α > ω0 (overdamped case)
On the basis of (1.8) and (1.9) we state that s1 and s2 are real and negative. The solution of
the equation (1.6) is as follows
vn = k1e s1t + k2e s2t ,
(1.10)
where k1 and k 2 are constants that depend on the initial conditions. A typical plot of vn (t ) is
depicted in Fig. 1.2.
vn (t)
0
t
Fig. 1.2
Case 2
α = ω0 (critically damped)
Using (1.8) and (1.9) we obtain
s1 = s2 = −α ,
that is the characteristic equation has a double zero being real and negative. The solution of
(1.7) is
vn = (k1 + k 2t )e −αt .
(1.11)
The plot of vn (t ) has similar shape as previously. The critically damped case is rather
theoretical one because in practical application the equality α = ω0 cannot be satisfied.
Case 3
α < ω0 (underdamped case)
Since α < ω0 implicates
3
α 2 − ω02 < 0 ,
the expression α 2 − ω02 is an imaginary number as follows
(
α 2 − ω02 = − ω02 − α 2
)
= jω d ,
where ωd = ω02 − α 2 . Hence, we have
s1 = −α + jωd ,
s2 = −α − jωd .
Thus, both natural frequencies are complex conjugate. The solution of equation (1.7) has the
form
vn = ke −αt cos (ω d t + θ ) ,
(1.12)
where k and θ are constants that depend upon the initial conditions. A typical plot of the
waveform vn (t ) is shown in Fig. 1.3.
vn (t)
0
t
Fig. 1.3.
If in particular α = 0 , or equivalently R = 0 , we have
ω d = ω02 − α 2 = ω0
and equation (1.12) reduces to
vn = k cos (ω0t + θ ) .
We now consider the nonhomogenous equation (1.6). The response v f , called a forced
response, depends on the source voltage vs (t ) . It can be shown that it is a constant if vs is a
4
DC voltage and a sinusoid if vs is a sinusoidal voltage. We consider only a situation when vs
is a constant voltage, i.e. vs (t ) = Vs . Observe that v f = Vs satisfies equation (1.6)
d 2Vs
dV
+ 2α s + ω02Vs = ω02Vs .
2
dt
dt
(1.13)
Thus, v f = Vs is a particular solution of the nonhomogenous equation and using (1.6) we
obtain
vC = vn + Vs ,
(1.14)
where, as shown above vn depends on the relative values of α and ω0 .
We wish to find the solution vC (t ) so that it satisfies equation (1.6) and the initial conditions.
We will derive the solution in all the discussed above cases.
1.2.
Overdamped case
(α > ω0 )
Using (1.10) and (1.14) we obtain
vC = k1e s1t + k 2 e s2t + Vs .
(1.15)
Since vC includes two constants k1 and k 2 that must be determined, we need another
equation. To form this equation we compute current i
i=C
dvC
= Ck1 s1 e s1t + Ck 2 s2 e s2t .
dt
Then we write (1.15) and (1.16) at t = 0
vC (0) = V0 = k1 + k 2 + Vs ,
i (0) = iL (0) = I 0 = C s1 k1 + C s2 k 2
and solve this set of equations for k1 and k 2 . We use the substituting method:
k 2 = V0 − Vs − k1 ,
C s1 k1 + C s2 (V0 − Vs − k1 ) = I 0 .
Hence, we obtain
I0
− s2 (V0 − Vs )
,
k1 = C
s1 − s2
I0
− s2 (V0 − Vs )
.
k 2 = V0 − Vs − C
s1 − s2
5
(1.16)
and substitute into equation (1.15)
I0
I0
⎛
⎞
− s2 (V0 − Vs )
− s2 (V0 − Vs ) ⎟
⎜
⎟e s2t + Vs .
vC (t ) = C
e s1t + ⎜V0 − Vs − C
s1 − s2
s1 − s2
⎜
⎟
⎜
⎟
⎝
⎠
1.3.
(1.17)
(α = ω0 )
Critically damped case
Using (1.6) and (1.11) yields
vC (t ) = (k1 + k2t )e −αt + Vs .
(1.18)
d vC
= C (k2 − α (k1 + k2t ))e −αt .
dt
(1.19)
Hence, we obtain
i=C
Let us write equations (1.18) and (1.19) at t = 0 and substitute i (0 ) = I 0 , vC (0) = V0
V0 = k1 + Vs ,
(1.20)
I 0 = C (k2 − α k1 ) .
(1.21)
We solve the set of equation (1.20) and (1.21) for k1 and k 2
k1 = V0 − Vs ,
(1.22)
I0
+ α (V0 − Vs ) .
C
(1.23)
k2 =
Next we substitute (1.22) and (1.23) into (1.18)
⎛
⎛I
⎞ ⎞
vC (t ) = ⎜⎜V0 − Vs + ⎜ 0 + α (V0 − Vs )⎟ t ⎟⎟ e −αt + Vs .
⎝C
⎠ ⎠
⎝
(1.24)
(α < ω0 )
1.4. Underdamped case
Setting (1.12) into (1.6) we obtain
vC = ke −αt cos (ω d t + θ ) + Vs .
(1.25)
Hence, it holds
i=C
(
)
dvC
= C − αke −αt cos (ωd t + θ ) − ω d ke −αt sin (ω d t + θ ) .
dt
6
(1.26)
Note that the same current traverses inductor L and capacitor C . Consequently,
i (0) − iL (0) = I 0 . Let us write equations (1.25) and (1.26) at t = 0
vC (0) = V0 = kcosθ + Vs ,
(1.27)
i (0) = I 0 = −C (αkcosθ + ω d ksinθ ) .
(1.28)
We solve this set of equation for k and θ as follows. Equation (1.27) leads to
kcosθ = V0 − Vs .
(1.29)
Setting (1.29) into (1.28) we obtain, after simple manipulation,
ksinθ = −
I0
α
(V0 − Vs ) .
−
Cω d ω d
(1.30)
On the basis of (1.29) and (1.30) we find
−
tanθ =
I0
α
(V0 − Vs )
−
Cω d ω d
V0 − Vs
and next
I
α
⎛
(V0 − Vs ) ⎞⎟
⎜− 0 −
ω
ω
C
d
d
⎟ .
θ = tan -1 ⎜
⎜
⎟
V0 − Vs
⎜
⎟
⎝
⎠
Having determined θ and used (1.29) to find k
k=
V0 − Vs
.
cosθ
Now we substitute θ and k into (1.25) and obtain the solution vC (t ) that satisfies both
equation (1.5) describing the circuit and the initial conditions.
Example 1
Let us consider the circuit shown in Fig. 1.1 where vs (t ) = Vs = 2 V , R = 10 Ω , L = 1H ,
1
1
C = F , vC (0 ) = V0 = 3 V , iL (0 ) = I 0 = A . We wish to determine vC (t ) for t ∈ [0 , ∞ ) .
9
3
At first we compute α and ω0
R
α=
=5 ,
2L
ω0 =
1
=3 .
LC
7
Since α > ω0 we have the overdamped case. The natural frequencies are
s1 = −α + α 2 − ω02 = −1 ,
s2 = −α − α 2 − ω02 = −9 ,
and the solution vC (t ) is given by (1.15) repeated below
vC = k1e s1t + k 2 e s2t + Vs .
We set into this equation s1 = −1 , s2 = −9 , Vs = 2
vC = k1e − t + k2e −9t + 2 .
(1.31)
To find k1 and k 2 so that the initial conditions are satisfied we create another equation
i=C
(
)
dvC 1
= − k1e − t − 9k2e − 9t .
dt
9
Next we write equations (1.31) and (1.32) at t = 0
i (0 ) = iL (0) = I 0 = 1
(1.32)
and apply the relationship
vC (0) = 3 = k1 + k2 + 2 ,
i (0 ) =
1 1
= (− k1 − 9k2 ) .
3 9
We solve the above set of equations, using substituting method, finding k1 =
Then we substitut k1 and k 2 into (1.31) and (1.32)
1
3
vC (t ) = e − t − e − 9t + 2 ,
2
2
1
1
i (t ) = − e − t + e − 9t .
2
6
The waveforms of vC (t ) and iL (t ) are plotted in Figs. 1.4 and 1.5.
8
3
1
, k2 = − .
2
2
vC(t)
3.154
3
2
1
0
0.137
0.412
t
Fig. 1.4
i(t)
1/3
0
-0.098
0.137
0.412
t
Fig. 1.5
Example 2
Consider the series RLC circuit with initial voltage across the capacitor vC (0) = 100 V . At
t = 0 the circuit is short circuited (see Fig. 6). Find i (t ) .
i
C
R
L
C = 50 μF ,
L = 10 mH ,
R = 10 Ω .
vC
t=0
Fig. 1.6
R
1
= 500 , ω0 =
= 1414 . Since α < ω0 we
2L
LC
have the underdamped case, hence, the natural frequencies are complex conjugate:
We calculate the constants α and ω0 : α =
s1 = −α + jωd ,
9
s2 = −α − jωd ,
where α = −500 and ωd = ω02 − α 2 = 1323. Using (1.17) yields
vC = ke −500t cos(1323t + θ ) .
(1.33)
To determine k and θ we find the current
i = iC = C
(
dvC
= 50 ⋅ 10− 6 − 500ke −500t cos (1323t + θ ) −
dt
− ke −500t 1323sin (1323t + θ ) =
)
= −(0.025cos (1323t + θ ) + 0.066sin (1323t + θ
))ke−500t .
We write the equations (1.33) and (1.34) at t = 0 and substitute vC (0) = 100 V , i0 (0) = 0
100 = k cosθ ,
0 = −k (0.025cosθ + 0.066 sinθ ) .
Hence, we find
tanθ = −0.379 ,
or
θ = −20.74° ,
and
k=
100
= 106.9 .
cos(- 20.74°)
Substituting k and θ into (1.34) yields
i = −106.9e −500t (0.025cos (1323t − 20.74°) + 0.066 sin (1323t − 20.7°)) .
After rearrangements we obtain
i (t ) = −7.48e −500t sin 1323t .
1.5.
State equations
Let us consider the equation (1.3) repeated below
vR + vL + vC − vs = 0
10
(1.34)
and substitute
vR = Ri
and
vL = L
diL
,
dt
where iL can be considered as a current flowing through the inductor L . After simple
manipulations we obtain
diL
R
1
1
= − iL − vC + vs .
dt
L
L
L
(1.35)
Differential equation (1.35) contains two variables iL and vC . Another equation including
these variables is
iL = iC = C
dvC
,
dt
which can be rewritten in the form
dvC 1
= iL .
dt
C
(1.36)
The equations (1.35) and (1.36) form a set of state equations describing the circuit of Fig. 1.1.
The current iL flowing through the inductor L and the voltage vC across the capacitor C are
called the state variables. The initial conditions (initial state) are i (0 ) = iL (0) = I 0 , vC (0) = V0 .
Summarizing we conclude that the second order dynamic circuit shown in Fig. 1.1 can be
described by the state equations
diL
R
1
1
= − iL − vC + vs ,
dt
L
L
L
(1.37)
dvC 1
= iL ,
dt
C
(1.38)
with the initial conditions iL (0) = I 0 , vC (0) = V0 .
The solution of the state equations is specified by vC (t ) and iL (t ) for all t ≥ 0 . At t = 0
vC (0 ) = V0 , iL (0) = I 0 can be considered as the coordinates of a point on vC − iL plane. As t
increases from 0 to ∞ the point (vC (t ), iL (t )) traces a directed curve starting at (V0 , I 0 ) . This
curve generated by the initial conditions vC (t ) , iL (t ) is called a trajectory.
To illustrate this idea we take into account the circuit considered in Example 1.1. The solution
of this circuit is specified by vC (t ) , iL (t ) plotted in Figs. 1.4 and 1.5. On the basis of these
waveforms we create the trajectory on the plane vC − iL as shown in Fig. 1.7.
11
iL
1/3
t =0.137
2.98
0
1
2
vC
3.154
-0.098
t =0.412
Fig. 1.7
1.6.
Parallel RLC circuit
Let us consider a parallel connection of a resistor, an inductor and a capacitor (see Fig.
1.8).
iR
iC
is (t)
vC
C
R
iL
L
Fig. 1.8
We apply KCL at the top node
iC + iL + iR − is = 0
and substitute into equation (1.39) the following relationships
iR = GvC = GL
iC = C
diL
,
dt
dvC
d 2i
= CL 2L .
dt
dt
12
(1.39)
Hence, we obtain the second order differential equation
d 2i
di
LC 2L + GL L + iL = is .
dt
dt
We derive both sides of this equation LC
d 2iL G diL
1
1
+
+
iL =
is
2
dt
C dt LC
LC
(1.40)
and define the parameters α and ω0 as follows
α=
G
,
2C
ω0 =
1
.
LC
The parameter α is called the damping constant and the parameter ω0 is called the angular
resonant frequency. Using α and ω0 we rewrite equation (1.40) as follows
d 2 iL
di
+ 2α L + ω02iL = ω02is .
2
dt
dt
(1.41)
The necessary initial conditions are iL (0) = I 0 and vC (0) = V0 . The second initial condition can
be expressed in terms of the derivative of iL
V0 = vC (0) = vL (0) = L
diL
(0)
dt
or
L
diL
(0) = V0 .
L
dt
From mathematical point of view equation (1.41) has the same form as equation (1.5). Hence,
we conclude that the solution of equation (1.41) is
iL = in + i f ,
where in is the solution of the homogenous equation
d 2in
di
+ 2α n + ω02in = 0
2
dt
dt
and i f is a particular solution of the original (nonhomogenous) equation. In a special case
where is is a DC current source, i.e. is (t ) = I s , it holds
if = Is .
13
The solution in of the homogenous equation depends on the relative values of α and ω0
similarly as in the case of the series RLC circuit. The form of in in the overdamped, critically
damped and underdamped cases is the same as vn in the series RLC circuit considered in
Section 1.1. Thus, in order to find in we identify the case and use the corresponding equation
derived in Section 1.1 replacing vn by in .
The approach leading to the solution iL (L ) of the parallel RLC circuit is the same as in the
case of the series RLC circuit and the procedure described in Section 1.1 can be repeated step
by step replacing vC by iL , Vs by I s V0 by I 0 as well as L by C and vice versa.
14