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 Im
VC
V m 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 VL
C L
Vi R VR
KVL: 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
22 VViR=+VC or VViR=+VC
• Assume high frequency 1/ωC << R then VR >> VC
VVRi≈ 11t t V =≈Idt Vdt Oi∫∫i C 00RC
Vi VV∝ dt Oi∫
• 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
22 VViR=+VC or VViR=+VC
• Assume low frequency 1/ωC >> R then VR << VC VR 1 t d Vii≈⇒Idt Iii=C V Vi VVCi≈ ∫ C 0 dt d VV==IR≈RCV ORi dt i d VOi∝ V 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
-t/RC V(t) = V0 e
• 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
–t/τ dV/dt = -{V0/τ } e
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) ϕ(ω)
jωt jϕ Vo(t) =Re { Vo e }, Vo = Vome
V (ω) T (ω) = o Vi
Amplitude response |T(ω)| = Vom(ω)/Vim
Phase response ∠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) jωC 1 T(ω) = = = V 1 1+ jωRC i 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
1.0
0.707 | in /V
o 0.5 |V
0.0 -3 -2 -1 0 1 2 3 0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) 0 100 slope -20 dB/dec -1 -3 dB -20 | 10 ] in /V [dB o 0
|V -2 -40 10 V
10-3 -3 -2 -1 0 1 2 3 -60
0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) ESE211 Lecture 5
Phase Response of
the Integrating RC Circuit
The angle of the transfer function is
Im[]T(ω) ∠T (ω) = arctan Re[]T(ω)
1− jωτ ∠T(ω) = ∠ = −arctan(ωτ ) (1+ jωτ )(1− jωτ )
0
o
V -45
-90
-3 -2 -1 0 1 2 3
0.001 ωο 0.01ω ο 0.1ω ο ωο 10 ωο 100 ωο 1000 ωο Frequency ( ω) ESE211 Lecture 5
The Transfer Function of
a Differentiating RC Circuit
C
Vi(t) R Vo(t)
V (ω) I(ω) ⋅ R T(ω) = o =
Vi I(ω) ⋅ X RC (ω)
R R jωRC T (ω) = = = X (ω) 1 1+ jωRC RC R + jωC
jωτ (1− jωτ ) ωτ (ωτ + j) T (ω) = 2 2 = 2 2 1+ω τ 1+ω τ
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 T(ω) = 2 2 = 1+ ω τ 1+ ω 2τ 2
• The Differentiating RC circuit is a HIGH-PASS filter
1.0
| V =0.707 V
in 0 in V o/ 0.5 |V
0.0 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency ( ω) 0 100 slope
-1 -20 dB/dec -20 ] | 10 -3 dB point at ω
in or 0 /V -6 dB/oct [dB o 0 V |V 10-2 -40
10-3 -3 -2 -1 0 1 2 3 -60
0.001ω 0.01ω 0.1ω ω 10 100 1000 ο ο ο ο ωο ωο ω Frequency ( ω) 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) = ∠ 1+ ω 2τ 2
1 ∠T(ω) = arctan ωτ
90 0 V 45
0
-3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency ( ω) ESE211 Lecture 5
Frequency Response of a Series RLC Circuit
VC VL
C L
Vi R VR
KVL: 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 R |V 0 /V IN |= = 2 1 1 jωL + + R R 2 + ωL − jωC ωC
1.0
V =0.707 V
| 0 in in
/V 0.5 o |V
0.0-3 -2 -1 0 1 2 3 100 1000 0.001ωο 0.01ωο 0.1ωο ωο 10ωο ωο ωο Frequency ( ω) 1.5 | in /V L 1.0 |V
|, |VC/Vin| |VL/Vin| in /V
C 0.5 |V
0.0 -3 -2 -1 0 1 2 3
0.001 0.01 0.1 ω ωο ωο ωο ο 10ωο 100ωο 1000ω Frequency (ω) ο ESE211 Lecture 5
Amplitude and Phase Response of
the Ladder RC Circuit
R R
Vin C C Vo
0
o
V -90
-180
-3 -2 -1 0 1 2 3 0.001ωο 0.01ωο 0.1ωο ωο 10ωο 100 ωο 1000 ωο Frequency ( ) ω 0 0 10
-2 -40 | 10 ]
in
/V [dB o slope 0 |V -4 -80 10 V -40 dB/dec
10-6 -3 -2 -1 0 1 2 3 -120
0.001ω 0.01ω 0.1ω ω 10ω 100ω 1000ω ο ο ο ο ο ο ο Frequency (ω) ESE211 Lecture 5
Amplitude and Phase Response of a Wien-Bridge Circuit
R C
−1 1 + jωC ZOUT R T(ω) = = −1 Vin R C Vo ZTOTAL 1 1 + jωC + R + R jωC
100
50
0 o V -50
-100 -3 -2 -1 0 1 2 3 0.001 0.01 0.1 ω 10 100ω 1000ω ωο ωο ωο ο ωο ο ο Frequency (ω )
0.33
0.24 V =0.707 V
| 0 max n i 5 BandWidth /V o
|V
0.0-3 -2 -1 0 1 2 3
100 1000 0.001ωο 0.01ωο 0.1ωο ωο 10ωο ωο ωο Frequency ( ω)