MIT Lecture 04
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Transcript of MIT Lecture 04
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7/30/2019 MIT Lecture 04
1/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Summary from previous lecture
Laplace transform
Transfer functions and impedances
L [f(t)] F(s) =
Z+0
f(t)estdt.
L [u(t)] U(s) = 1s.
L
eat
=1
s + a.
L
hf(t)
i= sF(s) f(0).
L
Zt0
f()d
=
F(s)
s.
f(t) x(t)
F(s) X(s)TF(s) =
X(s)
F(s)
Z(s) =
F(s)
X(s)
Ts(s) (s)J
b
TF(s) :=(s)
Ts(s)= 1
Js + b.
ZJ = Js; Zb = b; TF(s) =
1
ZJ + Zb
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7/30/2019 MIT Lecture 04
2/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Goals for today
Dynamical variables in electrical systems: charge,
current,
voltage. Electrical elements:
resistors,
capacitors,
inductors,
amplifiers.
Transfer Functions of electrical systems (networks)
Next lecture (Friday):
DC motor (electro-mechanical element) model
DC motor Transfer Function
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7/30/2019 MIT Lecture 04
3/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Electrical dynamical variables: charge, current, voltage
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v(t)
i(t) :=dq(t)
dt
charge qcharge flow current i(t)
voltage (aka potential) v(t)
Coulomb [Cb]Ampere [A]=[Cb]/[sec]
Volt [V]
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7/30/2019 MIT Lecture 04
4/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
v(t) = Ri(t) V(s) = RI(s) V(s)
I(s)= R ZR
Collisions between the mobile charges and the material fabric (ions,generally disordered) lead to energy dissipation
Electrical resistance
(loss). As result,energy must be expended to generate current along the resistor;i.e., the current flow requires application of potential across the
resistor
The quantity ZR=R is called the resistance (unit: Ohms, or)
The quantity GR=1/R is called the conductance (unit: Mhos or-1)
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v(t)
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i(t)
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7/30/2019 MIT Lecture 04
5/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Capacitance
Since similar charges repel, the potential v is necessary to prevent
the charges from flowing away from the electrodes (discharge)
Each change in potential v(t+t)=v(t)+v results in change of the
energy stored in the capacitor, in the form of charges movingto/away from the electrodes ( change in electric field)
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v(t)
electrode
(conductor)
electrode
(conductor)dielectric
(insulator)
+
i(t) i(t)E(t)
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7/30/2019 MIT Lecture 04
6/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Capacitance
Capacitance C:
in Laplace domain:
q(t) = Cv(t) dq(t)
dt i(t) = C
dv(t)
dt
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v(t)
electrode
(conductor)
electrode
(conductor)dielectric
(insulator)
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i(t) i(t)E(t)
I(s) = CsV(s) V(s)
I(s) ZC(s) =1
Cs
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7/30/2019 MIT Lecture 04
7/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Inductance
Current flow i around a loop results in magnetic field B pointing
normal to the loop plane. The magnetic field counteracts changes incurrent; therefore, to effect a change in current i(t+t)=i(t)+i a
potential v must be applied (i.e., energy expended)
Inductance L:
in Laplace domain:
v(t)
B(t)
i(t)
v(t) = L di(t)dt
V(s) = LsI(s) V(s)
I(s) ZL(s) = Ls
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7/30/2019 MIT Lecture 04
8/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Summary: passive electrical elements; Sources
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Voltage source:v(t) independent
of current through.
Electrical inputs: voltage source, current source
Current source:i(t) independent
of voltage across.
Ground:potential reference
v(t) = 0always
Table removed due to copyright restrictions.
Please see: Table 2.3 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
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7/30/2019 MIT Lecture 04
9/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Combining electrical elements: networks
Kirchhoff Current Law (KCL)
charge conservation
Kirchhoff Voltage Law (KVL)
energy conservation
v(t) vC(t)
V(s) VC(s)Network analysis relies on two physical principles
i1
ik
+
Pik(t) = 0
++
v1
vk
Pvk(t) = 0
PVk(s) = 0PIk(s) = 0
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Courtesy of Prof. David Trumper. Used with permission.
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7/30/2019 MIT Lecture 04
10/16Lecture 04 Wednesday, Sept. 122.004 Fall 07
Impedances in series and in parallel
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I1 I2
V2V1
V+
V+
Z1 Z2
IZ
V+
IZ
+
Impedances in parallelKCL: I= I1 + I2.
KVL: V1 + V2 V.From definition of impedances:
Z1 =V1I1
; Z2 =V2I2
.
Therefore, equivalent circuit has
1
Z=
1
Z1+
1
Z2
G = G1 + G2.
Impedances in seriesKCL: I1 = I2 I.
KVL: V = V1 + V2.From definition of impedances:
Z1 =V1I1
; Z2 =V2I2.
Therefore, equivalent circuit has
Z= Z1 + Z2
1
G=
1
G1+
1
G2.
I1 I2 V2V1
IZ1 Z2 +
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7/30/2019 MIT Lecture 04
11/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
The voltage divider
+
V2Vi
Z1
Z2+
+
I
Vi V2
Since the two impedances are in series, they combine to an equivalent impedance
Z= Z1 + Z2.
The current flowing through the combined impedance is
I=V
Z.
Therefore, the voltage drop across Z2 is
V2 = Z2I= Z2VZ V
2
Vi= Z
2
Z1 + Z2.
Z2
Z1 + Z2
Vi Z+
+
I
Equivalent circuit for computing the current I.
Block diagram & Transfer Function
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7/30/2019 MIT Lecture 04
12/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Example: the RC circuit
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Vi +
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Vi VCZ2 =1
CsVC
We recognize the voltage divider configuration, with the voltage across the ca-pacitor as output. The transfer function is obtained as
TF(s) =VC(s)
Vi(s)=
1/Cs
R + 1/Cs=
1
1 + RCs=
1
1 + s,
where RC. Further, we note the similarity to the transfer function of therotational mechanical system consisting of a motor, inertia Jand viscous friction
coefficient b that we saw in Lecture 3. [The transfer function was 1/b(1 + s),i.e. identical within a multiplicative constant, and the time constant wasdefined as J/b.] We can use the analogy to establish properties of the RCsystem without rederiving them: e.g., the response to a step input Vi = V0u(t)(step response) is
VC(t) = V0
1 et/u(t), where now = RC.
1
1 + RCs
Block diagram & Transfer Function
Z1 = R
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7/30/2019 MIT Lecture 04
13/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Interpretation of the RC step response+
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Vi +
+
Z2 =1
CsVC
Z1 = R
VC(t) = V0
1 et/
u(t), = RC.
VC
(t)[Vo
lts]
t [msec]
V0 = 1 Volt R = 2k C= 1F
Charging of a capacitor:
becomes progressively more
difficult as charges accumulate.
Capacity (steady-state) is reached
asymptotically (VCV0 as t)
C
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7/30/2019 MIT Lecture 04
14/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Example: RLC circuit with voltage source
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Ls R
1
Cs
V(s)VC(s)
VL(s) VR(s)+
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Figure 2.3
Figure 2.4
1V(s)
VC(
s)LC
s2 + +1
LC
R
Ls
Figure by MIT OpenCourseWare.
v(t) C
R L
vC(t)
i(t)
++
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Figure by MIT OpenCourseWare.
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7/30/2019 MIT Lecture 04
15/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
Example: two-loop network
Images removed due to copyright restrictions.
Please see: Fig. 2.6 and 2.7 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
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7/30/2019 MIT Lecture 04
16/16
Lecture 04 Wednesday, Sept. 122.004 Fall 07
The operational amplifier (op-amp)(a) Generally, vo = A (v2 v1),where A is the amplifier gain.
(b) When v2 is grounded, as is oftenthe case in practice, then vo = Av1.
(Inverting amplifier.)
(c) Often, A is large enough that
we can approximate A.Rather than connecting the input directly,
the opamp should then instead be used in thefeedback configuration of Fig. (c).
We have:
V1 = 0; Ia = 0
(because Vo must remain finite) therefore
I1 + I2 = 0;
Vi V1 = Vi = I1Z1;
Vo V1 = Vo = I2Z2.
Combining, we obtain
Vo(s
)Z
=Vi(s)
2(s
) .Z1(s)
Figure 2.10
+v1(t)
+V
A
-V
+v2(t)
vo(t)
+
-
v1(t)
A
vo(t)
+
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V1(s)I2(s)
Ia(s)I1(s)
Z1(s)
Z
2
(s)
Vi(s)
Vo(s)
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Figure by MIT OpenCourseWare.