Summary from previous lecture
•
Laplace transform
L [f (t)] ≡ F (s) =
Z
+∞
f (t)e−st dt.
0−
L [u(t)] ≡ U (s) =
L
1
.
s+a
·Z
¸
t
f (ξ)dξ =
0−
F (s)
.
s
Transfer functions and impedances
f (t)
x(t)
⇔
•
£
¤
L e−at =
1
.
s
h
i
L f˙(t) = sF (s) − f (0−).
F (s)
TF(s) =
X(s)
Z(s) =
2.004 Fall ’07
J
Ts (s)
X(s)
F (s)
b
TF(s) :=
F (s)
X(s)
Ω(s)
ZJ = Js;
Lecture 04 – Wednesday, Sept. 12
Ω(s)
1 .
=
Ts (s)
Js + b
Zb = b;
TF(s) =
1
ZJ + Zb
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
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
Electrical dynamical variables: charge, current, voltage
charge q
charge flow ≡ current i(t)
voltage (aka potential) v(t)
+
–
Coulomb [Cb]
Ampére [A]=[Cb]/[sec]
Volt [V]
+
+
+
+
–
2.004 Fall ’07
+
–
−
v(t)
+
+
+
+
+
+
dq(t)
i(t) :=
dt
Lecture 04 – Wednesday, Sept. 12
Electrical resistance
+
−
v(t)
+
+
+
+
+
+
+
+
+
i(t)
•
Collisions between the mobile charges and the material fabric (ions,
generally disordered) lead to energy dissipation (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
V (s)
v(t) = Ri(t) ⇒ V (s) = RI(s) ⇒
= R ≡ ZR
I(s)
•
•
The quantity ZR=R is called the resistance (unit: Ohms, or Ω)
The quantity GR=1/R is called the conductance (unit: Mhos or Ω-1)
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
Capacitance
+
+
+
i(t)
+
+
+
+
electrode
(conductor)
•
+
−
−
+
+
−
−
+
+
+
+
+
−
+
+
+
−
+
+
+
•
−
v(t)
+
E(t)
+
−
+
−
+
−
dielectric
(insulator)
−
−
−
i(t)
−
+
+
+
−
−
−
−
−
−
−
−
electrode
(conductor)
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+Δt)=v(t)+Δv results in change of the
energy stored in the capacitor, in the form of charges moving
to/away from the electrodes (↔ change in electric field)
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
Capacitance
+
−
v(t)
+
+
i(t)
+
+
+
+
−
−
+
+
−
−
+
+
+
+
+
−
+
+
+
−
+
+
+
electrode
(conductor)
+
+
E(t)
+
−
+
−
+
−
dielectric
(insulator)
−
−
−
i(t)
−
+
+
+
−
−
−
−
−
−
−
−
electrode
(conductor)
dv(t)
dq(t)
≡ i(t) = C
• Capacitance C: q(t) = Cv(t) ⇒
dt
dt
V (s)
1
≡ ZC (s) =
• in Laplace domain: I(s) = CsV (s) ⇒
I(s)
Cs
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
Inductance
v(t)
B(t)
+
−
i(t)
•
•
•
Current flow i around a loop results in magnetic field B pointing
normal to the loop plane. The magnetic field counteracts changes in
current; therefore, to effect a change in current i(t+Δt)=i(t)+Δi a
potential v must be applied (i.e., energy expended)
Inductance L:
di(t)
v(t) = L
dt
in Laplace domain:
2.004 Fall ’07
V (s)
≡ ZL (s) = Ls
V (s) = LsI(s) ⇒
I(s)
Lecture 04 – Wednesday, Sept. 12
Summary: passive electrical elements; Sources
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.
Electrical inputs: voltage source, current source
+
−
Voltage source:
v(t) independent
of current through.
2.004 Fall ’07
Current source:
i(t) independent
of voltage across.
Lecture 04 – Wednesday, Sept. 12
Ground:
potential reference
v(t) = 0
always
Combining electrical elements: networks
Ω
+
-
Ω
v(t)
vC (t)
V (s)
VC (s)
Courtesy of Prof. David Trumper. Used with permission.
Network analysis relies on two physical principles
•
Kirchhoff Current Law (KCL)
– charge conservation
i1
Kirchhoff Voltage Law (KVL)
– energy conservation
−
···
ik
P
2.004 Fall ’07
•
P
+
ik (t) = 0
Ik (s) = 0
v1
−
Lecture 04 – Wednesday, Sept. 12
···
P
P
+−
vk
+
vk (t) = 0
Vk (s) = 0
Impedances in series and in parallel
I1
Z1
I2
Z2
+ V1 −
I
Z1 =
V1
;
I1
Z2 =
V2
.
I2
Therefore, equivalent circuit has
¶
µ
1
1
1
=
+
.
Z = Z1 + Z2 ⇔
G
G1
G2
I
Z
+
2.004 Fall ’07
V
V1
−
I2
−
−
Impedances in series
KCL: I1 = I2 ≡ I.
KVL: V = V1 + V2 .
From definition of impedances:
Z2
+
I1
+ V2 −
V
+
Z1
+
V2
−
Impedances in parallel
KCL: I = I1 + I2 .
KVL: V1 + V2 ≡ V .
From definition of impedances:
Z1 =
V1
;
I1
Z2 =
V2
.
I2
Therefore, equivalent circuit has
´
1
1 ³
1
=
⇔ G = G1 + G2 .
+
Z
Z1
Z2
I
Z
+
Lecture 04 – Wednesday, Sept. 12
V
−
The voltage divider
Z1
Equivalent circuit for computing the current I.
+
+
Vi
+
−
I
Z2
V2
−
+
Vi
−
+
−
I
Z
−
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
Block diagram & Transfer Function
Therefore, the voltage drop across Z2 is
V2 = Z2 I = Z2
2.004 Fall ’07
V2
Z2
V
⇒
=
.
Z
Vi
Z1 + Z2
Lecture 04 – Wednesday, Sept. 12
Vi
Z2
Z1 + Z2
V2
Example: the RC circuit
Z1 = R
+
+
Vi
1
Z2 =
Cs
+
−
−
VC
Block diagram & Transfer Function
Vi
−
1
1 + RCs
VC
We recognize the voltage divider configuration, with the voltage across the capacitor as output. The transfer function is obtained as
TF(s) =
1/Cs
1
1
VC (s)
=
=
=
,
Vi (s)
R + 1/Cs
1 + RCs
1 + τs
where τ ≡ RC. Further, we note the similarity to the transfer function of the
rotational mechanical system consisting of a motor, inertia J and viscous friction
coefficient b that we saw in Lecture 3. [The transfer function was 1/b(1 + τ s),
i.e. identical within a multiplicative constant, and the time constant τ was
defined as J/b.] We can use the analogy to establish properties of the RC
system without re—deriving them: e.g., the response to a step input Vi = V0 u(t)
(step response) is
³
´
−t/τ
u(t),
where now τ = RC.
VC (t) = V0 1 − e
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
Interpretation of the RC step response
+
Z1 = R
+
+
+
+ +
−
+
−
VC
+
+
+
+ +
+
+
+ +
+
+
+
+
+
+ + +
Vi
1
Z2 =
Cs
+
−
+
VC (t) = V0 1 − e
−t/τ
´
+
³
u(t),
V0 = 1 Volt R = 2kΩ
τ = RC.
C
C = 1μF
−
−
−
− − −
−
− −
−
+
+
−
−
−
− − −
−
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
+
t [msec]
− −
−
VC (t) [Volts]
−
Charging of a capacitor:
becomes progressively more
difficult as charges accumulate.
Capacity (steady-state) is reached
asymptotically (VC→V0 as t→∞)
Example: RLC circuit with voltage source
+
R
v(t)
-
VR (s)
Ls
R
L
+
C
i (t)
+
-
vC(t)
V (s)
− +
VL (s)
Figure 2.4
2.004 Fall ’07
VC (s)
−
Figure 2.3
1
LC
s2 +
+
1
Cs
+
−
Figure by MIT OpenCourseWare.
V(s)
−
VC(s)
R
1
s +
L
LC
Figure by MIT OpenCourseWare.
Lecture 04 – Wednesday, Sept. 12
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.
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12
The operational amplifier (op-amp)
(a) Generally, vo = A (v2 − v1 ),
where A is the amplifier gain.
+V
+v1(t)
-
+v2(t)
+
v1(t)
vo(t)
A
+
A
vo(t)
-V
Z2(s)
Vi(s)
Z1(s)
I1(s)
(b) When v2 is grounded, as is often
the 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 op—amp should then instead be used in the
feedback configuration of Fig. (c).
We have:
V1 = 0;
V1(s)
Ia(s)
-
I2(s)
Vo(s)
Ia = 0
(because Vo must remain finite) therefore
I1 + I2 = 0;
+
Vi − V1 = Vi = I1 Z1 ;
Vo − V1 = Vo = I2 Z2 .
Figure 2.10
Figure by MIT OpenCourseWare.
Combining, we obtain
Vo (s)
Z2 (s)
.
=−
Z1 (s)
Vi (s)
2.004 Fall ’07
Lecture 04 – Wednesday, Sept. 12