ESE211 Lecture 3
Phasor Diagram of an RC Circuit
V(t)=Vm sin(ωt)
R
Vi(t)
C
Vo(t)
• Current is a reference in series circuit
KVL: Vm = VR + VC
VR
ϕ
VC
Vm
Im
ESE211 Lecture 3
Phasor Diagram of an RL Circuit
V(t)=Vmsin(ωt)
VR (t)
R
Vi(t)
L
Vo(t)
KVL: Vm = VR + VL
Vm
VL
ϕ
VR
Im
ESE211 Lecture 3
Phasor Diagram of a Series RLC Circuit
VC
C
VL
L
Vi
R
KVL:
VR
Vm = VC + VL + VR
VL
Vm
VL-C
VR
Im
VC
• Voltages across capacitor and inductor compensate each other
ESE211 Lecture 4
Integrating Circuit
VR
Vi(t)
R
Vo(t)
C
Vi = VR + VC
or
2
Vi =
VR + VC
2
• Assume high frequency 1/ωC << R then VR >> VC
VR ≈ Vi
t
t
1
1
VO = ∫ Ii dt ≈
V dt
C0
RC ∫0 i
Vi
VO ∝ ∫ Vi dt
• Accurate integrating can be obtained at high
frequency that leads to a low output signal
ESE211 Lecture 4
Differentiating Circuit
VC
Vi(t)
C
Vo(t)
R
Vi = VR + VC
or
Vi =
2
VR + VC
2
• Assume low frequency 1/ωC >> R then VR << VC
VR
t
Vi
VC ≈ Vi
1
d
Vi ≈ ∫ Ii dt ⇒ Ii = C Vi
dt
C0
d
VO = VR = Ii R ≈ RC Vi
dt
d
VO ∝ Vi
dt
• An accurate result can be achived at low frequency.
ESE211 Lecture 4
Transient Processes
in Passive Circuits
RC Circuit without a Source
• The circuit response is due only to the energy stored in the
capacitor
t=0
C
R
• The capacitor C is precharged to the voltage V0
Applying KCL:
C dV/dt +V/R = 0
• This is the 1st order differential equation
ESE211 Lecture 4
Solution of the Differential Equation
dV/dt + V/RC = 0
dV/V = -dt/RC
ln V = -t/RC +a
V = A e -t/RC ,
A=e+a
• e is the base of the natural logarithms, e = 2.718…
• Continuity requires the initial condition:
At t = 0
V(0) = V0
V(t) = V0 e -t/RC
• The response governed by the elements themselves
without external force is called a natural response
ESE211 Lecture 4
The Time Constant
V(t)
V0
0.368 V0
0.05 V0
τ
time
3τ
• The time constant τ = RC is the rate at which the natural
response decays to zero
• Time τ is required to decay by a factor of 1/e=0.368
• After the time period of 3τ the transient process is considered to
be completed
ESE211 Lecture 4
Determination of the Time Constant
V0
V
V1
V/e
τ
t
1) From the solution of the equation
By definition: as the time when the signal decreases by 1/e
V(t+τ)/V(t) = 1/e
The time t can be chosen arbitrarily
2) From the slope of the line at t=0
Differentiating the solution
dV/dt = -{V0/τ } e –t/τ
V1(t) = -{V0/τ }t +V0
V1(τ) = 0
ESE211 Lecture 4
Transient Response of
an Integrating RC Circuits
VR
Vi
VC
KVL:
Vi = VR + VC
Vi
t
VO =VC
VR
t
t
ESE211 Lecture 4
Transient Response of
a Differentiating RC Circuits
VC
Vi
VR
KVL:
Vi = VC + VR
Vi
t
VO = VR
t
VC
ESE211 Lecture 5
Transfer Function
Vi(t)
Vo(t)
Vi(t) = Vimcos(ωt)
Vo(t) = Vomcos(ωt+ϕ)
Vom(ω)
Vo(t) =Re { Vo e jωt},
ϕ(ω)
Vo = Vome jϕ
Vo (ω )
T (ω ) =
Vi
Amplitude response
Phase response
|T(ω)| = Vom(ω)/Vim
∠T(ω) = ϕ(ω)
ESE211 Lecture 5
The Transfer Function of
an Integrating RC Circuit
R
Vi(t)
C
Vo(t)
1
V o (ω )
1
jωC
T (ω ) =
=
=
1
Vi
1 + jωRC
R+
jωC
1
T (ω ) =
1 + jωτ
τ = RC
ESE211 Lecture 5
Amplitude Response of
the Integrating RC Circuit
• The magnitude of the transfer function is
1
T (ω ) =
1 + ω 2τ 2
• The Integrating RC circuit is a LOW-PASS filter
0.707
0.5
0.0
0.001ω
ο
-3
-2
0.01ω
ο
-1
0.1ω
ο
0
ωο
Frequency (ω)
1
10ω
ο
100ω
ο
3
1000ω
ο
0
slope
-20 dB/dec
-3 dB
10-1
-20
-40
10-2
10-3
-3
0.001ω
ο
-2
0.01ω
ο
-1
0
1
ωο
0.1ω
10ω
ο
ο
Frequency ( ω)
2
100ω
ο
3
-60
1000ω
ο
0
100
|Vo/Vin|
2
V [dB]
|Vo/Vin|
1.0
ESE211 Lecture 5
Phase Response of
the Integrating RC Circuit
The angle of the transfer function is
Im[T (ω )]
Re[T (ω )]
∠T (ω ) = arctan
1 − jωτ
∠T (ω ) = ∠
= − arctan(ωτ )
(1 + jωτ )(1 − jωτ )
Vo
0
-45
-90
-3
-2
0.001 ωο 0.01ωο
-1
0
0.1ωο
ωο
Frequency ( ω)
1
10 ωο
2
3
100 ωο 1000 ωο
ESE211 Lecture 5
The Transfer Function of
a Differentiating RC Circuit
C
Vi(t)
R
T (ω ) =
T (ω ) =
Vo(t)
Vo (ω )
I (ω ) ⋅ R
=
Vi
I (ω ) ⋅ X RC (ω )
jωRC
R
R
=
=
X RC (ω ) R + 1
1 + jωRC
j ωC
jωτ (1 − jωτ ) ωτ (ωτ + j )
T (ω ) =
=
2 2
1+ ω τ
1 + ω 2τ 2
τ = RC
ESE211 Lecture 5
Amplitude response of the Differentiating
RC Circuit
• Amplitude response of the RC circuit is the magnitude of T(ω)
ωτ (ωτ + j )
ωτ
=
1 + ω 2τ 2
1 + ω 2τ 2
T (ω ) =
• The Differentiating RC circuit is a HIGH-PASS filter
V =0.707 V
0
in
o
in
0.5
0.0
0.001ω
-3
ο
-2
-1
0.01ω
ο
0
0.1ω
ο
ωο
1
2
10ω
ο
Frequency ( ω)
3
100ω
ο
1000ω
ο
0
slope
-20 dB/dec
or
-6 dB/oct
10-1
o
-3 dB point at ω
-20
0
0
in
|V /V |
100
-40
10-2
10-3
-3
-2
0.001ω
ο
-1
0.01ω
ο
V [dB]
|V /V |
1.0
0
1
0.1ω
ωο
10ω
ο
ο
Frequency ( ω)
2
3
100ω
ο
-60
1000ω
ο
ESE211 Lecture 5
Phase response of the Differentiating
RC Circuit
• Phase response of the RC circuit is the phase angle of T(ω)
ωτ (ωτ + j ) 
∠T (ω ) = ∠
2 2 
+
ω
τ 
1

∠T (ω ) = arctan
1
ωτ
V0
90
45
0
-3
0.001ωο
-2
0.01ωο
-1
0
0.1ωο
ωο
Frequency ( ω)
1
10ωο
2
100ωο
3
1000ωο
ESE211 Lecture 5
Frequency Response of a Series RLC Circuit
VL
VC
C
L
Vi
R
KVL:
VR
Vi = VC + VL + VR
VL
VL-C
Vi
VR
Im
VC
• Voltages across capacitor and inductor compensate each other
ESE211 Lecture 5
Amplitude Response of a Series RLC Circuit
R
=
1
+R
j ωL +
j ωC
| V 0 / V IN |=
R
1 

R +  ωL −

ωC 

2
2
1.0
|Vo/Vin|
V0=0.707 V
in
0.5
0.0
0.001ωο
-3
-2
-1
0.01ωο
0
0.1ωο
1
ωο
Frequency ( ω)
2
10ωο
3
100ωο
1000ωο
|VC/Vin|, |VL/Vin|
1.5
1.0
|VL/Vin|
|VC/Vin|
0.5
0.0
-3
0.001ωο
-2
0.01ωο
-1
0
1
0.1ωο
ωο
10ωο
Frequency ( ω)
2
100ωο
3
1000ωο
ESE211 Lecture 5
Amplitude and Phase Response of
the Ladder RC Circuit
R
R
Vin
C
C
Vo
Vo
0
-90
-180
-3
-2
0.001 ωο 0.01ωο
-1
0
0.1ωο
1
ωο
10 ωο
2
3
100 ωο 1000 ωο
Frequency ( ω)
slope
10-4
10-6
-80
-40 dB/dec
-3
0.001ω
ο
-2
0.01ω
ο
-1
0
1
ωο
0.1ω
10ω
ο
ο
Frequency ( ω)
2
100ω
ο
3
-120
1000ω
ο
0
|Vo/Vin|
-40
10-2
V [dB]
0
100
ESE211 Lecture 5
Amplitude and Phase Response
of a Wien-Bridge Circuit
R
C
−1
Vin
R
C
T (ω ) =
Vo
1

 + jωC 
R

=
−1
1

1
 + jωC  + R +
jω C

R
Z OUT
ZTOTAL
100
50
Vo
0
-50
-100
0.001ωο 0.01ωο
-3
-2
-1
0
0.1ωο
1
ωο
Frequency (ω)
2
10ωο
3
100ωο
1000ωο
0.33
0.24
5
|Vo/Vin|
V0=0.707 V
max
BandWidth
0.0
0.001ωο
-3
-2
0.01ωο
-1
0
0.1ωο
ωο
Frequency ( ω)
1
10ωο
2
100ωο
3
1000ωο