First-Order CircuitsPart II
Instructor: Chia-Ming TsaiElectronics Engineering
National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.
Contents• Introduction
• The Source-Free RC Circuit
• The Source-Free RL Circuit
• Singularity Functions
• Step Response of an RC Circuit
• Step Response of an RL Circuit
• First-Order Op Amp Circuits
• Applications
Singularity Functions
• To aid the understanding of transient analysis
• To serve as good approximations to the switching signal in circuits
• Singularity functions are functions that either are discontinuous or have discontinuous derivatives.
• Three most widely used types are introduced– Unit step function– Unit impulse function– Unit ramp function
Unit Step Function
0 ,1
0 ,0
t
ttu
Unit Step Function (Cont’d)
0
00 ,1
,0
tt
ttttu
0
00 ,1
,0
tt
ttttu
Unit Step Function (Cont’d)
00
00
0
,
,0
ttuVtv
ttV
tttv
00
00
0
,
,0
ttuIti
ttI
ttti
Unit Impulse Function
0 ,0
0 ,Undefined
0 ,0
t
t
t
tudt
dt
10
0
dtt
Unit Impulse Function (Cont’d)
Unit Impulse Function (Cont’d)
00
00
000
0
,Consider
tfdttttf
dttttf
dttttfdttttf
b ta
b
a
b
a
b
a
b
a
(Sampled at t0)
Unit Ramp Function
0 ,
0 ,0
tt
ttr
ttudttutrt
Unit Ramp Function (Cont’d)
00
00 ,
,0
tttt
ttttr
00
00 ,
,0
tttt
ttttr
Summary
dt
tdrtu
dt
tdut
dttutr
dtttu
t
t
Example 1
5210
510210
tutu
tututv
=
Example 1 (Cont’d)
5210 tututv 5210 tt
dt
tdv
Step Response of an RC Circuit
RC
Vv
dt
dv
t
tuRC
V
RC
v
dt
dvR
tuVv
dt
dvC
R
tuVvi
dt
dvCi
ii
Vvv
S
S
S
SRC
RC
0,For
)(or
0)(
)( and
0
gives KCL Applying
)0()0( Assume 0
=
iC
iR
Cont’d
0 ,
0 ,)(
0 ,)(
where,
ln
ln
0
0
0
0
0
0
)(
||0
teVVV
tVtv
teVVVtv
RC τeVVVv
RC
t
VV
VvRC
tVv
RC
dt
Vv
dv
tSS
tSS
tSS
S
S
ttv
VS
S
Forced Response (V0=0)
)()(
),( 1)(
0 , 1
0 ,0)(
,0 i.e.
initially, unchargedcapacitor Assume
0
tueR
V
dt
dvCti
RCτ tueVtv
teV
ttv
V
tS
tS
tS
Step Response (I)• Complete Response
= Natural Response (vn) + Forced Response (vf)
(stored energy) (independent source)
tSf
tn
fn
tS
t
tSS
eVv
eVv
vv
eVeV
eVVVv
1
where
1
0
0
0
Step Response (II)• Complete Response
= Transient Response + Steady-State Response
(temporary part) (permanent part)
where
0)( and
0
0
Sss
tSt
t
tss
tSS
Vv
eVVv
v
vv
eVVVv
Short-Cut Method• Three items required to describe the response
– The initial capacitor voltage v(0) (or v(t0) )
– The final capacitor voltage v()– The time constant = RC
0)()()()(
or
)()0()()(
0tt
t
evtvvtv
evvvtv
v(0)
v()
Example 1
s 2105.0104
104 and V 30)(
V 152435
5)0()0(
33Th
3Th
CR
Rv
vv
0 V, 1530
)3015(30
)()0()()(
5.0
5.0
te
e
evvvtv
t
t
t
Example 2
t < 0 t > 0
10)0( v t
Th
evvvtv
Rv
)()0()()(
32020||10 ,20)(
Step Response of an RL Circuit
tSS
S
S
St
Sf
tn
fn
eR
VI
R
Vti
R
VIA
IR
VAi
R
VAei
R
Vi
RLτAei
iii
0
0
0
)(
)0(
,
iL
iR
=
Forced Response (I0=0)
)()(
),( 1)(
0 , 1
0 ,0)(
,0)0( i.e.
current,inductor initial no Assume
0
tueVdt
diLtv
R
Lτ tue
R
Vti
teR
Vt
ti
iI
tS
tS
tS
Short-Cut Method• Three items required to describe the response
– The initial inductor current i(0) (or i(t0) )
– The final inductor current i()– The time constant = L/R
0)()()()(
or
)()0()()(
0tt
t
eitiiti
eiiiti
i(0)
i()
Example 1
0 A, 32)25(2
)()0()()(
s 15
1
5
31
532 and
A 232
10)(
A 52
10)0()0(
1515
Th
Th
tee
eiiiti
R
L
R
i
ii
tt
t
Find i(t).
Example 2
A )1(4)40(4
)()0()()(
s 2
1
10
5
1064 and
A 464
40)(
40 )2(
0)0( ,0 (1)
.considered are intervals timeThree
22
Th
Th
tt
t
ee
eiiiti
R
L
R
i
t
it
Find i(t).
Example 2 (Cont’d)
)4(
15
22
/)4(
Th
Th
Th
8
273.111
30
)()4()()(
22
15
322
5
727.211
30
322
20)(
3
22
4121
16
20
4)1(4)4(
4 (3)
t
t
S
S
t
e
eiiiti
R
L
R
Vi
R
V
ei
t
Example: OP AMP Circuits
tt
t
o
o
o
o
ee
evvvtv
CR
vR
vv
v
v
v
v
1010
/
6Th
Th
12)012(0
)()4()()(
1.010520000
200001
200000120000
(b).circuit in KVLapply , find To
12)( and
12)0(
080000
)0(0
20000
3
2, nodeat KCLapply ,)0( find To
(b)
(a)
Applications: Delay Circuit
t
v70 V
I II
I: II:
= (R1+R2)C = RC
R
Applications: Relay Circuit
closed
, If
2
0
S
Vv
v+
_
Applications: Ignition Circuit
• Two steps to work– S is closed to build the inductor current
– Open S to force the inductor current to pass through the air gap
S
RVi S