First-Order Circuit
First-Order Circuits
vL = L
iC = C
diL
dvC
dt
dt
A first-order circuit is characterized by a first-order
differential equation
EEE110 Electric Circuits
Anawach Sangswang
Dept. of Electrical Engineering
KMUTT
iL (t ) = iL (t1 ) +
1
t
∫ vL dσ
L t1
vc (t ) = vc (t1 ) +
1
t
∫ ic dσ
C t1
2
Source-Free RC Circuit
Source-Free RC Circuit
Rate of decay
The capacitor is initially charged
(energy is already stored v(0) V0)
d  v 
1 −t / τ
  =− e
t =0
τ
dt  V0  t =0
By KCL
iR + iC = 0
Ohms law
dv
v
+C
=0
dt
R
=−
Capacitor
Solving the above equation
Time constant
τ=RC
dv
1
=
−
∫ v RC ∫ dt
v(t ) = V0 e −t / RC = V0 e −t / τ
t=τ
V0 e − t / τ = V0 e −1 = 0.368V0
3
t
τ
2τ
3τ
4τ
5τ
1
τ
v(t)/V0
0.36788
0.13534
0.04979
0.01832
0.00674
Decays more slowly
Decays faster
4
Source-Free RC Circuit
Example 7.1
The natural response of a circuit = the behavior (in
terms of voltages and currents) of the circuit itself,
with no external sources of excitation
The time constant τ of a circuit is the time required
for the response to decay by a factor of 1/e or 36.8%
of its initial value.
v decays faster for small τ and slower for large τ
The key to working with a source-free RC circuit is
finding:
The initial voltage v(0) = V0 across the capacitor.
The time constant τ = RC.
Let vC(0) = 15V.
Find vC , vx and ix for t> 0
τ = Req C = 0.4 s
Req =
5 × 20
= 4Ω
5 + 20
v = vC = v( 0 )e− t / τ = 15e− t / 0.4 = 15e −2.5t
vx , voltage divider
ix
vx =
12
15e −2.5t = 9e−2.5t
12 + 8
ix =
vx
= 0.75e−2.5t
12
5
Example 7.2
6
Source-Free RL Circuit
A first-order RL circuit consists of an inductor L (or
its equivalent) and a resistor R (or its equivalent)
The switch is open at t = 0
Find v(t) for t ≥ 0, and the
initial energy stored in the capacitor
t < 0,
9
vC =
20 = 15V
3+9
By KVL
vL + v R = 0
L
di
+ iR = 0
dt
t > 0, Req = 9 + 1 = 10Ω
Inductors law
τ = Req C = 10 × 20 ×10−3 = 0.2 s
Solving for i(t)
v( t ) = v( 0 )e − t / τ = 15e − t / 0.2 = 15e −5t
Initial energy
1
wC ( 0 ) = CvC2 ( 0 ) = 2.25 J
2
Ohms law
i (t )
7
t
di
R
=
−
∫i (0) i
∫0 L dt
ln i (t ) − ln i (0) = −
Rt
+0
L
8
Source-Free RL Circuit
Source-Free RL Circuit
The key to working with a source-free RL
circuit is finding:
A general form representing RL
i(t ) = I 0 e − t / τ
i (t ) = I 0 e −t /τ
L
where τ =
R
The time constant τ of a circuit is the time required for the
response to decay by a factor of 1/e or 36.8% of its initial
value.
i(t) decays faster for small τ and slower for large τ
The general form is very similar to a RC source-free circuit.
where τ =
L
R
The initial voltage i(0) = I0 through the
inductor.
The time constant τ = L/R
9
Example 7.3 Find i, ix
i (0) = 10
1) Equivalent resistance
10
Example 7.3
2) Solving the differential equation
Loop1
di1
1 di1
+ 2(i1 − i2 ) = 0
+ 4i1 − 4i2 = 0
dt
2 dt
di1 2
+ i1 = 0
dt 3
5
i2 = i1
6
2(i1 − i2 ) + 1 = 0
i1 − i2 = −
6i2 − 2i1 − 3i1 = 0
5
i2 = i1
1
2
i1 = −3 A
i (t )
1
t
di
2
∫i (0) i11 = − 3 ∫0 dt
io = −i1 = 3 A
6
3
= 2=
Req = RTh = = ,τ = L
Req 1
2
io
3
3
vo
1
Loop2 6i2 − 2i1 − 3i1 = 0
Rearrange and integrate both sides
2 t
i (t )
ln i i (0) = − t 0 i (t ) = i (0)e− t /τ = 10e −2t / 3
3
The inductor voltage
i (t ) = i (0)e− t /τ = 10e −2t / 3
11
vL = L
di
2
10
= 0.5 ×10 ×  −  e −2t / 3 = − e −2t / 3
dt
3
 3
ix =
vL
5
= − e −2t / 3
2
3
12
Example 7.5
Comparison between a RL and RC circuit
Find io, vo, i
t<0
10
i (t ) =
= 2 A, vo = 3i (t ) = 6V
2+3
io = 0
t>0
A RL source-free circuit
A RC source-free circuit
L
R
v(t ) = V0 e − t /τ where τ = RC
i (t ) = I 0 e − t / τ where
τ=
RTh = 3 6 = 2Ω, τ = L / RTh = 1
i (t ) = i (0)e −t /τ = 2e −t A
di
v0 (t ) = −vL = − L = 4e −tV
dt
v
2
i0 (t ) = L = − e −t A
6
3
Singularity
Singularity (or switching)
functions are functions that
either discontinuous or have
discontinuous derivatives
The unit step function u(t) is 0
for negative values of t and 1 for
positive values of t.
 0,
u (t − to ) = 
1,
t < to
t > to
 0,
u (t ) = 
1,
13
14
Unit Step Function
t<0
t>0
Representation on a circuit
Voltage source
Current source
 0,
u (t + to ) = 
1,
t < − to
t > − to
Unit impulse function (delta function)
t <0
0,
d

δ (t ) = u(t ) = undefined t = 0
dt
 0,
t >0

15
16
Unit Impulse Function
Example 7.6
Properties
Mathematically express
the given signal & its derivative
Not physically realizable
It is a very short duration pulse of unit area and can be
0+
expressed as
∫
0−
δ (t )dt = 1
The unit area is known as the strength of the impulse
function
dv
= 10 [δ (t − 2) − δ (t − 5)]
dt
Sampling property
∫
b
a
b
f (t )δ (t − t0 )dt = ∫ f (t0 )δ (t − t0 )dt
v(t ) = 10u(t − 2) − 10u(t − 5)
a
= 10 [u(t − 2) − u(t − 5)]
b
= f (t0 )∫ δ (t − t0 )dt = f (t0 )
a
∫
b
a
f (t )δ (t − t0 )dt = f (t0 )
17
Step Response: RC Circuit
Step Response: RC Circuit
The step response of a circuit is its behavior when
the excitation is the step function, which may be a
voltage or a current source.
Rearrange and integrate both sides
∫
v (t )
V0
• Initial condition:
t dt
dv
= −∫
0 RC
v − Vs
ln ( v − Vs ) − ln (V0 − Vs ) = −
Combine and take the exponential
v − Vs
−t
= e RC = e−t /τ , τ = RC
V0 − Vs
v(0-) = v(0+) = V0
• Applying KCL,
Uncharged capacitor (V0 = 0)
dv v − Vs u (t )
C +
=0
dt
R
or
18
v(t ) = Vs (1 − e−t /τ )
i(t ) = C
t
RC
Complete response
,t <0
V0
v=
−t /τ
Vs + (V0 − Vs )e , t > 0
dv Vs −t /τ
= e
dt R
dv
v − Vs
=−
u (t )
dt
RC
• Where u(t) is the unit-step function
19
20
Step Response: RC Circuit
Step Response: RC Circuit
Integrating both sides and considering the initial
conditions, the solution of the equation is:
Complete Response =
Initial value
at t = 0
Natural response
(stored energy)
=
V0e–t/τ
+
+
vn
Complete Response
t<0
V0
v(t ) = 
−t / τ
Vs + (V0 − Vs )e
Final value
at t -> ∞
Another way of looking at it
t >0
Source-free
Response
=
( V0 - Vs)e–t/τ
+
steady-state response
(permanent)
+
Vs
Typical form
The initial capacitor voltage v(0)
The final capacitor voltage v(∞) — DC voltage across C
The time constant τ.
Note: if the switch operates at time t = t0, the form becomes
v (t) = v (∞) + [v (t0 ) − v (∞)] e−(t−t0 )/τ
Vs(1–e–t/τ)
21
22
Step Response: RL Circuit
Example 7.10: v(t)
The step response of a circuit is its behavior when
the excitation is the step function, which may be a
voltage or a current source.
Find v(t) @ t = 1s, 4s
t < 0: switch @A
v(0− ) =
transient response
(temporary)
v (t) = v (∞) + [v (0+) − v (∞)]e−t /τ
To determine the step response of an RC circuit:
Forced Response
(independent source)
vf
=
5
× 24 = 15V
5+3
• Initial current
i(0-) = i(0+) = Io
• Final inductor current
i(∞) = Vs/R
• Time constant τ = L/R
t > 0: switch @B, v(0) = v(0-) = v(0+)
τ = RThC = 4k Ω⋅ 0.5mF = 2s
From circuit, v(∞) = 30V
v(t ) = v(∞) + [ v(0) − v(∞)] e−t /τ = 30 + (15 − 30)e−t / 2 = 30 − 15e−.5tV
−0.5
At t = 1, v(1) = 30 − 15e = 20.9V
At t = 4, v(4) = 30 − 15e−2 = 27.97V
23
i = itransient + isteady −state
V
τ = L/ R
i = Ae−t /τ + s
R
V
• At t = 0,
I0 = A + s
R
V 
V

i =  I 0 − s  e− t /τ + s
R
R

I0 = 0
24
Step Response: RL Circuit
Example 7.12: i(t), t > 0
t<0
V 
V

i =  I 0 − s  e− t /τ + s
R
R

i(0− ) =
10
= 5A
2
i(0) = i(0− ) = i(0+ )
t>0
i (t ) = i (∞) + [i (0+ ) − i (∞)] e − t /τ
i(∞) =
10
= 2 A, RTh = 2 + 3 = 5Ω
2+3
τ = L / RTh = 1/15
Three steps to find out the step response of an RL
circuit:
−t /τ
−15t
The current i(t ) = i(∞) + [i(0) − i(∞)] e = 2 + 3e
Check: KVL
The initial inductor current i(0) at t = 0+
The final inductor current i(∞)
The time constant τ
Note: This technique applies only for the step responses
−10 + 5i +
25
1 di
=0
3 dt
1
−10 + 5(2 + 3e−15t ) + (3 × −15e−15t ) = 0
3
26