Control Systems Peter Grogono December 2003
Contents 1 Syllabus
3
2 Transforms
3
2.1 The Fourier Transform . 2.2 The Laplace Transform 2.2.1 Partial Fractions 2.3 The ZTransform . . . .
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3 A Simple Control System
3 5 9 10 11
4 Characteristics of Electrical and Mechanical Systems 13
4.1 Passive Electrical Components . 4.2 Phase Control . . . . . . . . . . . 4.2.1 Lead Compensation . . . 4.2.2 Lag Compensation . . . . 4.2.3 Lag/Lead Compensation . 4.3 Operational Ampli ers . . . . . . 4.4 Mechanical Systems . . . . . . . 4.5 Quality Factor . . . . . . . . . . 4.6 SecondOrder Systems . . . . . .
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5 Stability Criteria
5.1 5.2 5.3 5.4
RouthHurwitz Nyquist . . . . Bode . . . . . . RootLocus . .
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6 Case Study: Motor Controller
6.1 6.2 6.3 6.4
Zero Input . . . Stability . . . . Compensation . Normal Form .
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34 35 35 36
7 SI Units
37
8 Useful Formulas
41
2
List of Figures 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
A periodic pulse train . . . . . . . . . . . . . . . . . . General properties of the Laplace transform . . . . . . Particular Laplace transforms . . . . . . . . . . . . . . Common ztransforms . . . . . . . . . . . . . . . . . . A simple control system . . . . . . . . . . . . . . . . . Formulas for passive components . . . . . . . . . . . . Diagram for phase modi cation circuits . . . . . . . . Kinds of compensation: K is the compensation circuit Compensation: summary . . . . . . . . . . . . . . . . Operational ampli er . . . . . . . . . . . . . . . . . . . Typical applications of operational ampli ers . . . . . Simulating an inductance . . . . . . . . . . . . . . . . Analogies between mechanical and electrical systems . Rules for Bode diagrams . . . . . . . . . . . . . . . . . Summary of root locus rules . . . . . . . . . . . . . . . Basic units . . . . . . . . . . . . . . . . . . . . . . . . Derived units . . . . . . . . . . . . . . . . . . . . . . . Laplace: rules . . . . . . . . . . . . . . . . . . . . . . . Laplace: special cases . . . . . . . . . . . . . . . . . . Bode diagrams . . . . . . . . . . . . . . . . . . . . . . Root locus construction . . . . . . . . . . . . . . . . .
5 7 8 11 12 14 15 15 18 18 19 22 24 30 32 37 38 41 42 44 45
List of Examples 1 2 3 4 5 6 7 8 9 10
Fourier transform of a pulse train . . . . Laplace transform of e−at . . . . . . . . Laplace transforms of cos ωt and sin ωt ztransform of unit step . . . . . . . . . ztransform of unit ramp . . . . . . . . . ztransform of exponential decay . . . . Ampli er with proportional feedback . . Dierentiation . . . . . . . . . . . . . . Filtering . . . . . . . . . . . . . . . . . . Simulating an inductance . . . . . . . .
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4 6 7 10 10 11 20 20 20 21
1
3
SYLLABUS
1
Syllabus
This is the syllabus for the exam that I required to take: Nodal and mesh analysis of linear, nite, passive circuits; equivalent networks. Steady state AC response of lumped constant, timeinvariant networks. Time and frequency response of linear systems; impulse response and transfer functions, Laplace transform analysis, frequency response, including steadystate sinusoidal circuits. Models, transfer functions, and system response. Root locus analysis and design. Feedback and stability; Bode diagrams. Nyquist criterion, frequency domain design. State variable representation. Simple PID control systems.
2
Transforms
2.1
The Fourier Transform
A periodic function f(t) with period T can be expressed as a sum of sine and cosine functions: f(t) =
∞ 2X a0 + (an cos ωn t + bn sin ωn t) T T n=1
where the angular frequency of each term is given by ωn =
2πn . T
The coecients an and bn are de ned by Z T/2 an = Z−T/2 T/2 bn = −T/2
f(t) cos ωn t dt, n = 0, 1, 2, 3, . . . f(t) sin ωn t dt,
n = 1, 2, 3, . . .
Using the formulas cos θ = 21 (ejθ + e−jθ ) and sin θ = 2j (ejθ − e−jθ ), we can express this sum using complex exponents as f(t) =
∞ i a0 1 Xh (an − jbn )ejωn t + (an + jbn )e−jωn t + T T n=1
where
Z T/2 f(t)e−jωn t dt
an − jbn = −T/2
Z T/2
f(t)ejωn t dt
an + jbn = −T/2
(1)
2
4
TRANSFORMS
Since ωn = 2πn/T , ω−n = −ωn and (1) can be written f(t) =
Z ∞ T/2 1 X f(t)e−jωn t dt ejωn t . T n=−∞ −T/2
We can split this into two parts: ∞ 1 X Cn ejωn t T n=−∞
f(t) =
where
Z T/2 f(t)e−jωn t dt.
Cn = −T/2
Example 1: Fourier transform of a pulse train. De ne a pulse by A, if −p/2 ≤ t ≤ p/2, and p(t) = 0, otherwise.
and assume that f(t) is this function extended periodically with period T  in other words, f is a pulse train. (Clearly, we require p < 2T .) Then Z p/2 Ae−jωn t dt
Cn = −p/2
A −jωn t p/2 e − jωn −p/2 2A ω p sin n ωn 2
= =
and f(t) =
∞ 1 X 2A ωn p jωn t sin e T ωn 2 n=0 ∞
=
AX1 ωn p jωn t sin e π n 2
n=0
since
2
2 1 = . Tωn nπ
If we let the period, T , tend to in nity, we obtain Z∞
F(ω) =
f(t)e−jωt dt
−∞
which is the Fourier transform of an arbitrary (i.e., nonperiodic) function of time, f(t). The Fourier transform has an important limitation: it is valid for a function f(t) only if
Z∞
−∞
f(t)dt
has a nite value. For example, the \step function", u(t), has no Fourier transform. The Laplace transform introduces an additional \damping factor", e−st , and consequently can be used for a greater variety of functions.
2
5
TRANSFORMS
f(t) 6 p 
p 
A
0
 t
p 2
T
2T
Figure 1: A periodic pulse train 2.2
The Laplace Transform
The Laplace transform of a function f(t) is Z∞ L (f(t)) =
f(t) e−st dt.
0
By convention, we write F(s) for L (f(t)). The Laplace transform is linear:
Z∞
(A f(t) + B g(t)) e−st dt 0 Z∞ Z∞ −st = A f(t) e dt + B g(t) e−st dt
L (A f(t) + B g(t)) =
0
0
= A L (f(t)) + B L (g(t)) .
The transform of a derivative has a simple relationship to the transform of the original function. Using the formula for integration by parts Z
uv =
Z
(2)
u dv + v du
with u(t) = f(t) and v(t) = e−st gives ∞
−st
f(t) e
0
Z∞ = −s
f(t) e 0
−st
Z∞ dt +
f 0 (t) e−st dt.
0
Assuming lim f(t) e−st = 0, (3) leads to t→∞
L f 0 (t)
= s L (f(t)) − f(0+ ).
(3)
2
6
TRANSFORMS
Dierentiating again: L f 00 (t)
= sL f 0 (t) − f 0 (0+ )
= s2 F(s) − sf(0+ ) − f 0 (0+ ).
The transform of an integral can be obtained in a similar way. Substituting Zt
f(τ) dτ
u(t) =
0 −st
v(t) = e
in (2) gives Z∞ Zt
Zt
∞ e−st f(τ) dτ 0
f(τ) dτ
= −s 0
0
L
e
−st
dt +
!
f(τ) dτ
f(t) e−st dt.
0
0
The term on the left is zero since it contains the factor t = ∞. Thus Zt
Z∞
!
=
0
R0 0
f(τ) dτ at t = 0 and the factor e−st at
1 L (f(t)) . s
Suppose that f(t) is a variable in a control system. Then typical operations on f scale it by a linear factor (e.g., ampli er), dierentiate it (e.g., inductor), or integrate it (e.g., capacitor). Consequently, the Laplace transform of f is likely to be a rational function of s or, in other words, P(s) in which P(s) and Q(s) are polynomials in s. The it will be a function of the form M(s) = Q(s) roots of P are the zeroes of M and the roots of Q are the poles of M. Theorem 2.1 Initial Value
Provided that the limits exist: lim f(t) = s→∞ lim s F(s).
t→0
Theorem 2.2 Final Value
Provided that the limits exist: lim f(t) = lim s F(s).
t→∞
s→0
Example 2: Laplace transform of e−at . Let f(t) = e−at . Then Z∞ F(s) = e−at e−st dt 0 Z∞ = e−(a+s)t dt =
0 ∞ e−(a+s)t
a + s 0
1 = 0− − a+s 1 = . s+a
2
7
TRANSFORMS
Function
Transform
Remarks
f(t)
F(s)
General notation
Af(t) + Bg(t)
AF(s) + BG(s)
Linearity
f(t − T )u(t − T )
e−sT F(s)
u(t) is the unit step function
e−at f(t)
F(s + a)
f 0 (t)
sF(s) − f(0+ )
f 00 (t) Rt 0 f(τ)dτ Rt 0 f(t − τ)g(τ)dτ
s2 F(s) − sf(0+ ) − f 0 (0+ ) F(s)/s F(s)G(s)
Figure 2: General properties of the Laplace transform 2 Example 3: Laplace transforms of cos ωt and sin ωt. Let f(t) = cos ωt. Then Z∞ cos(ωt)e−st dt F(s) = 0 Z Z 1 ∞ −st+jωt 1 ∞ −st−jωt = e dt + e dt 2 0 2 0 ∞
=
∞
1 e−st+jωt 1 e−st−jωt + 2 −s + jω 0 2 −s − jω 0
1 1 1 + 2 −s + jω −s − jω 1 −s − jω −s + jω = − + 2 s2 + ω2 s 2 + ω2 s = 2 . s + ω2
= −
Let u = sin ωt du = ω cos ωt dt v = e−st dv = −se−st dt R
R
and apply the rule for integration by parts, uv = u dv + v du: sin(ωt)e
∞
−st
0
Z∞ = −s 0
sin(ωt)e
−st
Z∞ dt + ω 0
cos(ωt)e−st dt
2
8
TRANSFORMS
Function
Transform
Remarks
δ(t)
1
Dirac's δfunction
1 s 1 sn 1 s+a 1 s(s + a) s 2 s + a2 s s2 − a2 a 2 s + a2 a 2 s − a2 1 s(s2 + a2 ) 1 s2 (s2 + a2 ) 1 (s + a)2 s (s + a)2 1 2 s + 2ζω + ω2
Unit step function
u(t) tn−1 (n − 1)! e−at 1 − e−at a
cos at cosh at sin at sinh at 1 − cos at a2 at − sin at a3 te−at e−at (1 − at) e−at sin bt b
a = ζω, b = ω 1 − ζ2 , and ζ < 1 p
Figure 3: Particular Laplace transforms which is equivalent to 0 = −sL (sin ωt) + ωL (cos ωt)
and therefore L (sin ωt) = = 2
ω L (cos ωt) s ω . s2 + ω2
2
9
TRANSFORMS
2.2.1
Partial Fractions
The last step of a solution that employs Laplace transforms usually requires the inverse transforP(s) mation of an expression of the form in which P and Q are polynomials in s. Since the inverse
Q(s) 1 transform of is straightforward, we need to express such rational functions in the form s+a A2 A3 A1 + + + ··· s + s1 s + s2 s + s3
If Q(s) has only simple roots, we use the following rule: if Q(s) = (s + s1 )(s + s2 ) · · · (s + sn )
then P(s) Q(s)
=
n X Ai s + si i=1
where
Ai =
(s + si )
P(s) . Q(s) s=−si
The situation is slightly more complicated when there are repeated roots. If Q(s) = · · · (s + si )r · · ·
then the partial fraction expansion of
P(s) will include the terms Q(s)
B1 B2 Br + + ··· + 2 s + si (s + si ) (s + si )r
in which Br = Br−1 = Br−2 =
.. . B1 =
(s + si )r
P(s) (s + si ) Q(s) s=−si
P(s) 1 d2 (s + si )r 2 2! ds Q(s) s=−s 1 d 1! ds
P(s) Q(s) s=−si r
i
dr−1 1 (r − 1)! dsr−1
r P(s) (s + si ) Q(s) s=−s
i
2
10
TRANSFORMS
2.3
The ZTransform
A sampling system takes \samples" of a continuous input at discrete points in time. Each sample can be represented by Dirac's δ function with an appropriate amplitude. Thus if e(t) is a continuous signal sampled at times 0, T, 2T, 3T, . . ., the sampled signal is given by ∞ X
e∗ (t) =
e(nT )δ(t − nT )
n=0
where T is the sampling period. The Laplace transform of e∗ (t) is ∞ X
∗
E (s) =
e(nT )e−nTs .
(4)
n=0
In order to avoid the nonalgebraic terms that would be introduced by e−nTs , we change the variables by de ning z = eTs .
With this change, we have s =
ln z T
, e−nTs = eTs
∗
−n
∗
E (s) = E
ln z
T
∞ X
=
= z−n , and (4) becomes
e(nT )z−n
n=0
= E(z)
and we say that E(z) is the ztransform of e(t). Figure 4 shows some common ztransforms. In general, E(z) = e(0) z0 + e(T ) z−1 + e(2T ) z−2 + · · · + e(nT ) z−n + · · · and we can think of the nth coecient as representing the function at time nT . Example 4: ztransform of unit step. For the unit step function, r(t) = 1 for t ≥ 0 and therefore e(nT ) = 1 for all n ≥ 0. Thus R(z) =
∞ X
z−n
n=0
=
z . z−1
2 Example 5: ztransform of unit ramp. For the unit ramp function, r(t) = t for t ≥ 0 and therefore e(nT ) = nT for n ≥ 0. R(z) =
∞ X
nTz−n
n=0
= T (z−1 + 2z−2 + · · ·) z = T . (z − 1)2
3
11
A SIMPLE CONTROL SYSTEM
Time function ztransform
Remark
δ(t)
1
Impulse
z z−1 z z−1 Tz (z − 1)2 T 2 z(z + 1) 2(z − 1)3 z z − e−aT (1 − e−aT )z (z − 1)(z − e−aT ) z sin ωT 2 z − 2z cos ωt + 1
Unit step
u(t) δT (t) t 1 2 t 2 e−at 1 − e−at
sin ωt
δT (t) =
∞ X
δ(t − nT )
n=0
Figure 4: Common ztransforms 2 Example 6: ztransform of exponential decay. For the exponential decay function, r(t) = e−at for t ≥ 0 and e(nT ) = e−aTn for n ≥ 0. ∞ X
R(z) =
e−aTn z−n .
n=0
We note that
R(z) R(z)zeaT
= = zeaT
+1 + e−aT z−1 + e−2aT z−2 + · · · +1 + e−aT z−1 + e−2aT z−2 + · · ·
and therefore R(z)zeaT − R(z) = zeaT
from which R(z) = =
zeaT zeaT − 1 z . z − e−aT
2
3
A Simple Control System
In the simple control system shown in Figure 5: R(s) = System input, or stimulus signal
3
12
A SIMPLE CONTROL SYSTEM
E(s) = Actuating signal G(s) = Forward path, or openloop, transfer function C(s) = System output, or controlled signal H(s) = Feedback path transfer function B(s) = Feedback signal G(s)H(s) = Loop transfer function
All functions are expressed as Laplace transforms of the corresponding timedomain functions. E(s)
R(s)

+

− 6
G(s)

C(s)
B(s) H(s)
Figure 5: A simple control system Inspecting Figure 5, we see that: E(s) = R(s) − B(s) C(s) = E(s) G(s) B(s) = C(s) H(s)
From these equations, C(s) = E(s) G(s) = [R(s) − B(s)] G(s) = [R(s) − C(s) H(s)] G(s)
and therefore C(s) =
R(s) G(s) 1 + G(s) H(s)
and the transfer function, M(s), is given by M(s) =
C(s) G(s) = . R(s) 1 + G(s) H(s)
Since the transfer function without feedback is simply G(s), the denominator 1 + G(s)H(s) represents the eect of feedback. Note that the sign of the feedback (negative) is assumed: B(s) is subtracted from R(s). The characteristic equation is obtained by equating the denominator of the transfer function to zero. For the system in Figure 5, the characteristic equation is R(s) = 0. In many cases  for example, when G(s) is a simple polynomial  it will be 1 + G(s)H(s) = 0.
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
13
From the form of the characteristic equation, we note that G and H have reciprocal units. In the following example, the units of G are radians/volt and the units of H must therefore be volts/radian.
4
Characteristics of Electrical and Mechanical Systems
Kirchoff ’s Laws Current Law (KCL). The algebraic sum of the currents leaving/entering a node is zero. Voltage Law (KVL). The algebraic sum of the potential dierences around a loop is equal to
the algebraic sum of the e.m.f.'s acting around the loop in the same direction.
Definition 4.1 Mesh Analysis
apply KVL to every loop.
Assume an unknown current in every loop of the circuit;
Assume zero potential at a selected node and an unknown potential dierence (voltage) between this node and each other node; apply KCL at each node.
Definition 4.2 Node Analysis
4.1
Passive Electrical Components
For any circuit component, we can write: an equation that gives the relationship of the voltage dierence across the component and
the current owing through it as a function of time;
the relationship between phasors (see below) when voltage and current vary sinusoidally; and the Laplace transform of the time function.
Resistance. For resistance, the time function (and hence the other functions) is very simple: v(t) = R i(t). Capacitance. A capacitance stores charge, giving i = C
or v =
1 C
dv dt
Zt i dτ. τ=0
Inductance. An inductance has the inverse eect: v = L
di . dt
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
Component
Time
Frequency
Laplace
v = f(i)
V(ω)/I(ω)
V(s)/I(s)
Ri
R
R
1/jωC
1/sC
jωL
sL
Resistance Capacitance Inductance
1 C
R
L
i dt di dt
14
Figure 6: Formulas for passive components Phasors are derived as follows: we use inductance as an example. Assume, in the equation above, that i = Aejωt . Then di dt d jωt Ae = L dt = LAjωejωt
v = L
= jωLi
and therefore v i
= jωL
in which jωL is a phasor that expresses the impedance of the inductance. In general, we can forget about the term ejωt in our calculations, because it cancels out in all the equations. We can ignore that amplitude, A, for the same reason. The three forms are summarized in Figure 6. It is useful to note that there are several ways of combining components to give terms of particular dimensions. For example, RC has the dimensions of time and RCω is therefore a pure number. Similarly, LC has dimension sec2 and LCω is therefore a pure number. See Section 7 for more examples. 4.2
Phase Control
Figure 7 is a general circuit for phase control. If i is the current owing through Z1 and Z2 , we have Vin = i(Z1 + Z2 ) Vout = iZ2
and so the response of the circuit is Vout Vin
=
Z2 . Z1 + Z2
4
15
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
Vin
Z1
Vout
Z2
Figure 7: Diagram for phase modi cation circuits
R
K
G
C
H
(a) Cascade compensation
R
G
K
C
H
(b) Feedback compensation Figure 8: Kinds of compensation: K is the compensation circuit The eect of compensation depends on where the compensation circuit is placed. Figure 8 shows the two most common topologies: cascade compensation is placed in series with the forward control network and feedback compensation is placed in series with the feedback network. 4.2.1
Lead Compensation s+a
The transfer function for a lead compensator is , which has a zero at s = −a and a pole s+b at s = −b. For lead compensation, a < b, and the phase shift is always positive: jω + a jω + b = arg(jω + a) − arg(jω + b) ω ω = tan−1 − tan−1 a b
∆φ = arg
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
16
which is positive because tan−1 is a monotonically increasing function. We can make the circuit of Figure 7 a lead compensator by putting an RC shunt as Z1 and a resistance as Z2 . Then R1 /Cs R1 + 1/Cs R1 = 1 + R1 Cs = R2
Z1 =
Z2
The response is Vo Vi
Z1 Z1 + Z2 R2 (1 + R1 Cs) R1 + R2 + R1 R2 Cs s + R11C
= = =
s + R11C + s+a s+b
=
1 R2 C
with a = 1/R1 C and b = 1/R1 C + 1/R2 C, so that a < b, as required. The phaselead compensator is a form of highpass lter. The compensator introduces gain at high frequencies, which may increase instability, and phase lead, which tends to be stabilizing. The pole and zero are typically placed in the high frequency region. 4.2.2
Lag Compensation s+b
, which has a zero at s = −b and a pole at The transfer function for a lag compensator is s+a s = −a. For lag compensation, a < b, and the phase shift is always negative: jω + b jω + a = arg(jω + b) − arg(jω + a) ω ω = tan−1 − tan−1 b a
∆φ = arg
which is negative because tan−1 is a monotonically increasing function. We can make the circuit of Figure 7 a lag compensator by putting a resistance as Z1 and a resistance and a capacitance in series as Z2 . Then Z1 = R1 Z2 = R2 +
1 Cs
1 Cs R2 + 1/Cs R1 + R2 + 1/Cs
Z1 + Z2 = R1 + R2 + Z2 Z1 + Z2
=
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
17
s + 1/R2 C s + 1/(R1 + R2 )C s+b s+a
= =
where a = 1/(R1 + R2 )C and b = 1/R2 C, so that a < b, as required. The phaselag compensator is a simple form of lowpass lter. It reduces the gain at high frequencies, which tends to stabilize the system, and introduces phase lag, which tends to destabilize the system. The pole and zero of the compensator are usually placed in the low frequency region. 4.2.3
Lag/Lead Compensation
The transfer function for a lag/lead compensator is
(s + a1 )(s + b2 ) with a1 < b1 and a2 < b2 . (s + b1 )(s + a2 )
With an RC parallel circuit as Z1 and an RC serial circuit as Z2 , we have R1 1 + R1 C1 s 1 = R2 + C2 s R2 + 1/C2 s = R1 /(1 + R1 C1 s) + R2 + 1/C2 s (1 + R1 C1 s)(1 + R2 C2 s) = R1 C2 s + (1 + R1 C1 s)(R2 C2 s) + 1 + R1 C1 s
Z1 = Z2 Z2 Z1 + Z2
=
s2 +
s+
1 R2 C 2
+
1 R1 C1
1 R2 C1
+
s+
1 R2 C 2
1 R1 C 1
s+
1 R1 C 1 R2 C 2
which meets the requirements for lag/lead compensation with a1 = 1/R1 C1 b2 = 1/R2 C2 a1 b2 = a2 b1 a2 + b1 = a1 + b2 + 1/R2 C1 .
Table 9 summarizes the advantages and disadvantages of the various kinds of compensation. 4.3
Operational Amplifiers
Figure 10 shows the circuit symbol for an operational ampli er (\op amp") and the equivalent circuit. In designing with op amps, we assume: The input impedance is large (Zin ≈ ∞) The output impedance is small (Zout ≈ 0)
4
18
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
Compensation Advantages Phase lag LF characteristics improved
Phase lead
stability marrgins maintained or improved bandwidth reduced (useful if HF noise is a problem) improved stability margins improved HF response may be required for stability
Disadvantages at least one slow term in transient reponse reduced bandwidth may be a disadvantage possible HF noise problems may generate large signals, out of linear range of system
Figure 9: Compensation: summary The ampli er is linear with large gain (Vout = A Vin and A 1)
Figure 11 shows op amps in two typical con gurations. Each circuit uses two impedances, Z1 and Z2 , in a feedback loop. The input signal for the left circuit (a) goes to the − input of the op amp, and the circuit has negative gain. The input signal for the right circuit (b) goes to the + input of the op amp, and the circuit has positive gain. To analyze Figure 11(a), we assume that the input impedance of the op amp is high enough to ensure that the current owing into its − input is negligible. Consequently, we can assume that the same current ows through Z1 and Z2 , and we will call it i. It follows that: v − Vin = i Z1 Vout − v = i Z2 Vout = −A v
Vin
Zout
Zin Vin 6 ?
Z − Z +
A Vin
Vout
(a) Circuit Symbol
0
(b) Equivalent circuit Figure 10: Operational ampli er
Vout
4
19
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
Z2
Vout
v Vin
Z1
Z2
v
Vout
Z − +
Vin
Z
Z − +
Z
Z1
(a) Inverting ampli er with feedback
(b) Noninverting ampli er with feedback
Figure 11: Typical applications of operational ampli ers in which v is the voltage at the junction of Z1 and Z2 (as shown in the diagram), and A is the gain of the ampli er. Eliminating i, we obtain v − Vin Z1
=
Vout − v Z2
and therefore Vout Vin
= −
Z2 . Z1 + Z2 Z1 + A
If we assume that A is large, this simpli es to Vout Vin
≈ −
Z2 . Z1
We can analyze Figure 11(b) in a similar way. Again, we assume that the currents owing into the inputs of the op amps are negligible and that the current owing through Z1 and Z2 is i. Thus v = i Z1 Vout = i(Z1 + Z2 )
and, eliminating i, v Z1
=
Vo ut Z1 + Z2
or v =
Vout Z1 . Z1 + Z2
The op amp ensures that Vout = A(v − Vin ).
4
20
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
We can eliminate v, giving Vout Vin
Z1 1 − . Z1 + Z2 A
=
For large A, we have Vout Vin
≈
Z1 . Z1 + Z2
If we replace Z1 and Z2 by the Laplace transforms of the corresponding impedances, we have, as the response function for the op amp circuit: T (s) = −
Z2 Z1
for the inverting circuit and T (s) =
Z1 Z1 + Z2
for the noninverting circuit. Example 7: Amplifier with proportional feedback. If the impedances are both resistive, Z1 = R1 and Z2 = R2 , say, then T (s) = −
R2 R1
and the circuit behaves as an (inverting) linear ampli er with gain R2 /R1 . 2 Example 8: Differentiation. If Z1 = 1/Cs is a capacitor and Z2 = R is a resistor, T (s) = −RCs
and the circuit is an inverting dierentiator. 2 Example 9: Filtering. The impedance of a resistor R in parallel with a capacitor C is
If we use parallel circuits for the impedances, we have Z1 = Z2 = T (s) = = =
R . 1 + RCs
R1 1 + R1 C1 s R2 1 + R2 C2 s Z2 − Z1 R2 1 + R1 C1 s − · 1 + R2 C2 s R1 R2 1 + R1 C1 s − · R1 1 + R2 C2 s
For a d.c. signal, s = jω = 0 and T (s) = −R2 /R1 , as we would expect. For a.c. signals, the gain is g
R2 1 + R1 C1 jω = · R1 1 + R2 C2 jω v u u R2 = t
R22 C1 C2 ω2 + R1 (1 + R22 C22 ω2 ) 1 + R22 C22 ω2
!2
+
R22 C2 ω R2 C1 ω − 1 + R22 C22 ω2 R1 (1 + R22 C22 ω2 )
!2
.
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
21
For large frequencies, the gain is determined by the capacitances: lim g = C1 /C2 .
ω→∞
2 Example 10: Simulating an inductance.
Figure 12 shows a circuit that is intended to simulate an inductance. To demonstrate its action, we compute the eective impedance between the two input connections at the left of the diagram. To obtain the impedance, we apply a voltage V to these terminals and calculate the resulting current. Let i1 be the current owing through R1 and i2 be the current owing through C and R2 . We have u = V + i1 R1 , i2 , v = V+ sC i2 . V = i 2 R2 + sC
(5) (6) (7)
The eect of the op amp is to keep the voltages at points u and v approximately equal. Assuming
u = v, we obtain from (5) and (6):
V + i1 R1 = V +
i2 sC
and therefore i2 . sR1 C
i1 =
From (7), i2 =
V sC 1 + sR2 C
and so i1 + i2 = =
V sC + V s2 R1 C2 sR1 C(1 + sR2 C) V sC 1 + sR1 C · sR1 C 1 + sR2 C
The eective impedance of the circuit is therefore V i1 + i2 1 + sR1 C = R1 · . 1 + sR2 C
Z =
Setting s = jω and separating into real and imaginary parts gives R1 (1 + ω2 R1 R2 C2 ) jωR1 C(R2 − R1 ) + 1 + ω2 R21 C2 1 + ω2 R21 C2 = R 0 + jωL 0 .
Z =
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
V
c 6 c
22
u
R1
Z − +
C
v
Z
R2
Figure 12: Simulating an inductance The justi cation for writing the imaginary part as ωL 0 is that R1 (R2 − R1 )C has the dimensions of an inductance, as shown in Section 7. The quality factor (see Section 4.5) of the simulated inductor is Q = ωL 0 /R 0 ωC(R2 − R1 ) . = 1 + ω2 R1 R2 C2
In some situations, these inequalities will hold: ω2 R1 R2 C2 1 ω2 R21 C2
1
R2 R1
(8) (9) (10)
In these circumstances, R 0 ≈ R2 L0 ≈ Q ≈
R2 ω2 R1 C 1 ωR1 C
It appears from these approximations that we can obtain large values of both L 0 and Q by using a very small value of C. Small values of C, however, may invalidate the inequalities (8){(10). In principle, we can get some quite impressive inductors. For example, suppose that we use the values R1 = 10 KΩ, R2 = 1 MΩ, and C = 1 nF. At f = 1 KHz, the values of R 0 , L 0 , and Q are shown below. The values given by the approximate formulas, shown in parentheses, are not accurate at this frequency. R 0 = 13, 893 Ω (1 MΩ) L 0 = 9.86 H (2533 H) Q = 4.46 (15.9)
At a higher frequency, f = 1 MHz, the approximations are much better: R 0 = 999, 749 Ω (1 MΩ) L 0 = 2.507 mH (2.533 mH) Q = 0.0157 (0.0159)
4
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
23
These results suggest that when the approximations are useful, the inductor is not useful, and vice versa. C'est la vie. 2 4.4
Mechanical Systems
Equations for mechanical systems enable us to calculate the motion caused by a force. A mass obeys Newton's law: force = mass × acceleration or d2 x dt2 dv = m . dt
f(t) = m
In an electrical system in which e is the potential, q is the charge, i is the current, and L is an inductance, we have the analogous equations d2 q dt2 di = L . dt
e(t) = L
A spring changes its length in approximate proportion to the applied force: f(t) = kx.
This is analogous to a capacitance: e(t) =
q . C
The analogy is also apparent in the following equations: 1 df k dt de C dt
= v = i.
Friction is nonlinear in general. However, we can approximate viscous friction as a force propor
tional to velocity:
f(t) = µv
and, in this form, friction is analogous to resistance: e(t) = Ri.
Rotating systems are similar but, instead of mass, force, and velocity, we have torque, moment of inertia, and angular velocity. Figure 13 summarizes the analogies between mechanical and electrical systems.
4
24
CHARACTERISTICS OF ELECTRICAL AND MECHANICAL SYSTEMS
Linear
Symbol
Rotating
Symbol
Electrical
Symbol
Position
x
Angle
θ
Charge
q
Mass
m
Moment of Inertia
I
Inductance
L
Velocity
v=
dx dt
Angular Velocity
ω=
Torque
T = I ddt2θ
Potential
e = L ddt2q C de dt = i
dθ dt
Current
Force
f = m ddt2x
Spring
k df dt = v
Clockspring
λ dT dt = ω
Capacitance
Friction
µv = f
Angular friction
δω = T
Resistance
2
2
i=
dq dt 2
Ri = e
Figure 13: Analogies between mechanical and electrical systems 4.5
Quality Factor
The quality factor, Q, of a resonant circuit is a measure of the height of the \spike" in the response function. The Q of an LC circuit is typically determined largely by the resistance, R, of the inductor L, and Q =
ωL . R
Quality does not apply only to electronic circuits, however, and the general de nition of Q is given in terms of the bandwidth of the \spike". Definition 4.3 Quality Factor Let S be a resonant system with resonant frequency (that is, maximum response) at ω0 . Let ω1 √and ω2 , where ω1 < ω0 < ω2 , be the \halfpower points", where the amplitude is 3 dB (or 1/ 2) down. Then the quality factor for S is Q =
ω0 . ω2 − ω1
mω
Iω
Comparable quality factors are: for a linear mechanical oscillator and for a rotating µ δ mechanical oscillator, where µ and δ are linear and angular friction as described in Figure 13, and ω is the frequency of oscillation. 2π
4.6
SecondOrder Systems
For the canonical second order system G(s) =
ω2n . s2 + 2ζωn s + ω2n
5
25
STABILITY CRITERIA
The response of this system to a unit step is −1
L
ω2n s(s2 + 2ζωn s + ω2n )
!
= 1−
1 −ζωn t e sin(βωn t + θ) β
where q
β =
1 − ζ2
θ = tan−1 (β/ζ)
The time to the rst peak is Tp = the overshoot is e−πζ/
5
√
π and the amplitude at this time is 1 + e−πζ/β . Consequently, βωn
1−ζ2 .
Stability Criteria
Definition 5.1 Stability A system is stable if, for every bounded input, the output remains bounded with increasing time.
The characteristic equation (CE) is 1 + G(s)H(s) = 0.
In most case, G(s) and H(s) are rational functions of s and we can write 1 + G(s)H(s) =
(s − z1 )(s − z2 )(s − z3 ) · · · (s − p1 )(s − p2 )(s − p3 ) · · ·
(11)
in which the factors in the numerator are (mis)named the zeroes and the factors in the denominator are (mis)named the poles of the CE. Recalling that the closedloop transfer function is C(s) R(s)
=
G(s) 1 + G(s)H(s)
we see that the zeros of (11) are the poles of (12). For a given input R(s), the output is C(s) = =
R(s)G(s) 1 + G(s)H(s) k2 kn k1 + + ··· + + Cr (s) s − p1 s − p2 s − pn
and consequently c(t) = k1 ep1 t + k2 ep2 t + · · · + kn epn t + cr (t)
(12)
5
26
STABILITY CRITERIA
Since the transfer function is a polynomial, cr (t) will be bounded for bounded inputs c(t). The other terms will be bounded if all of the pi are negative. If there are quadratic factors, some of the pi will be complex, and must have negative real parts. It follows that, for stability, the roots of the CE must have negative real parts. If the CE has roots on the imaginary axis, the system is marginally stable and, in practice, will oscillate. Consequently, the stability analysis depends on being able to nd the zeros of a polynomial. We begin with some general considerations. The equation (s − r1 )(s − r2 ) · · · (s − rn ) = 0 has roots at s = r1 , s = r2 , . . . s = rn and expands to sn + an−1 sn−1 + an−2 sn−2 + · · · + a0 = 0.
We have: an−1 = −sum of all roots an−2 = +sum of products of pairs of roots an−3 = −sum of products of triples of roots ... a0 = (−1)n × product of all roots
If all of the roots real and negative, then ri < 0 for i = 1, 2, . . . , n and it follows that aj > 0 for j = n − 1, n − 2, . . . , 0. If the roots are complex, they must occur in conjugate pairs. Since the ai are real, the imaginary parts will cancel, and we have the same result: if the roots have negative real parts, then ai > 0. In summary: 1. If any ai = 0, then there are roots not in the left halfplane. 2. If any ai < 0, then at least one root is in the right halfplane. 5.1
RouthHurwitz
The RouthHurwitz method uses the ideas above to determine whether a given polynomial has roots in the right halfplane. Given the equation an sn + an−1 sn−1 + · · · + a0 = 0
the RouthHurwitz array is set up like this: sn sn−1 sn−2 sn−3 ...
an an−2 an−4 an−6 . . . an−1 an−3 an−5 an−7 . . . b1 b2 b3 ... c1 c2 ...
5
27
STABILITY CRITERIA
where b1 b2 ... c1 c2 ...
an−2 an−5 an−3 b2 an−5 b3
a n = − an−1 an−1 1 an = − an−1 an−4
1
a n−1 b1 1 an−1 = − b1 b1
1 = − b1
an−2 an−3
If an entry does not exist, it is assumed to be zero. The terms in the third and successive rows have the form − dci , in which di is a determinant formed from the two preceding rows and k is the rst term of the immediately preceding row. The columns of di consist of the rst and (i + 1)'th terms in the two preceding rows. Definition 5.2 RouthHurwitz Criterion The number of zeroes in the right halfplane of a polynomial P is equal to the number of sign changes in the rst column of its RouthHurwitz
array.
Three cases arise in practice: 1. All entries in the rst column are nonzero: the criterion can be applied directly. 2. The rst entry of a row is zero. We can proceed by assuming that the value of the zero term is actually ε and completing the calculations. Then we let ε → 0+ and ε → 0− . However, there will invariably be sign changes in one or both cases, and consequently there are zeroes in the right halfplane. 3. There is an entire row of zeroes. In this case, the polynomial has an even polynomial as a factor. (An even polynomial in s has only even powers of s.) The coecients of this factor are the entries in the line above the line of zeroes. We can divide the original polynomial by this polynomial and proceed. The RouthHurwitz array for the quadratic as2 + bs + c is s2 a c s1 b s0 c
showing that all three coecients must be positive, which is alredy obvious from the stadnard solution. For the cubic a3 s3 + a2 s2 + a1 s + a0 , the RouthHurwitz array is a3 a1 s3 2 s a2 a0 1 s a1 − a3 a0 /a2 s0 a0
showing that the coeecients must be positive and, in addition, a1 a2 > a3 a0 , for stability.
5
28
STABILITY CRITERIA
RouthHurwitz provides a simple way of deciding whether all roots of the CE lie in the negative halfplane. Advantage
RouthHurwitz does not provide any other information about the location of the roots: it is therefore not useful for system design. Disadvantage
5.2
Nyquist
The Nyquist criterion is based on a polar map of the openloop transfer function, G(jω)H(jω). The Nyquist diagram is obtained by the mapping s 7→ 1 + G(s)H(s) where s has the following locus: s starts at −j∞ and moves up the imaginary axis to j∞. If there are poles of G(s)H(s) on
the imaginary axis, the locus follows a small semicircle, with positive real part, to go around them.
The locus is closed by an in nite semicircle de ned by s = u + jv where u > 0, r = and r → ∞.
√
u2 + v2 ,
The locus is in nite only for generality; all that matters is that it must enclose all poles and zeros that lie in the right halfplane. The Nyquist criterion makes use of a theorem of Cauchy: If F(s) is analytic within a closed contour C, then N = Z − P, where N is the number of times F(s) encircles the origin as s moves around C, Z is the number of zeroes within C, and P is the number of poles within C. The number of times that 1 + G(s)H(s) encircles the origin is clearly the same as the number of times G(s)H(s) encircles the point −1 + j0. Since the contour we have chosen encloses the right halfplane, Z is the number of zeros with positive real parts which, for stability, must be zero. Also, P is the number of poles which, since the openloop transfer function should be stable, is also zero. Definition 5.3 Nyquist Criterion As s moves along of G(s)H(s) must not encircle the point −1 + j0.
the contour described above, the locus
The importance of the Nyquist criterion is that it tells us not only whether a system is stable, but how far it is from instability. Suppose that, for some value of s, G(s)H(s) = α, where α is real and α > −1. Then the openloop gain could be increased by a factor of α1 before instability occurs; the gain margin is 1 20 log10 db α
and an acceptable value for it is 12 db corresponding to α = 1/4. Let γ be the angle between the negative real axis and the point where G(s)H(s) = 1. Then γ is the phase margin of the system and an acceptable value for it is π/3 = 60◦ .
5
29
STABILITY CRITERIA
5.3
Bode
The Bode criterion is similar to the Nyquist criterion but uses approximations to simplify the calculations needed. A Bode plot of a response function R(s) has two parts, the amplitude plot showing R(jω) and the phase plot showing arg R(jω). The horizontal axis shows log ω (or, more often, log10 ω) and the vertical axis uses decibels for R(jω) and degrees or radians for arg R(jω). Figure 14 summarizes the results for rstorder functions. The loglog scales enable typical response functions to be approximated as straight lines, as follows: Constant: K. The magnitude plot is constant at 20 log10 K and the phase plot is constant at 0◦ for K ≥ 0 or 180◦ for K < 0. Integration: 1/jω. The amplitude plot is −20 log10 ω which is a line with slope −20 db/decade passing through 0 db at ω = 1. The phase plot is constant at −π/2. Differentiation: jω. The amplitude plot is 20 log10 ω which is a line with slope 20 db/decade passing through 0 db at ω = 1. The phase plot is constant at π/2. a . The amplitude plot is Phase lag: a + jω a = 20 log q 1 20 log10 10 ω2 a + jω +1 a2
s
= −20 log10
ω2 + 1. a2
We can approximate this in two parts. When ω a, the amplitude is approximately log10 1 = 0 db. When ω a s
−20 log10
ω ω2 + 1 ≈ −20 log10 2 a a
which is a straight line with a slope of −20 db/decade crossing 0 db at ω = a. The two lines (asymptotes) join at ω = a and the approximate Bode plot consists of two lines. a The phase is given by arg a+jω = tan−1 ω a . For ω a, this is approximately zero and, for ω a, it tends to π/2. We approximate this as ω < 0.1a : 0 π ω − 0.1a 0.1a ≤ ω ≤ 10a : · 2 10a − 0.1a ω > 10a : π/2
To sketch the phase plot, we can draw the rst and the third components, and then draw a straight line between their end points. a + jω . By similar reasoning, the Bode plot is constant at 0 db for ω ≤ a and a increases at −20 db/decade for ω > a. The phase is approximated as
Phase lead:
ω < 0.1a : 0 π ω − 0.1a · 2 10a − 0.1a ω > 10a : −π/2
0.1a ≤ ω ≤ 10a : −
5
30
STABILITY CRITERIA
Function
Amplitude
K jω 1/jω 1 + jω/ωn
20 log10 K 20 log10 ω −20 log10 ω 0 20(log10 ω − log10 ωn ) 0 −20(log10 ω − log10 ωn )
1 1+jω/ωn
ω ≤ ωn ω > ωn ω ≤ ωn ω > ωn
: : : :
Point ω db 1 1 ωn
0 0 0
ωn
0
Phase rule 0 ω < 0.1ωn ω > 10ωn ω < 0.1ωn ω > 10ωn
: : : :
π/2 −π/2 0 π/2 0 −π/2
Figure 14: Rules for Bode diagrams Let G(ω) denote the gain of the system and φ(ω) the phase shift. The phase crossover frequency, ωp , is de ned by φ(ωp ) = −π
and the gain margin is
1 . G(ωp )
The gain crossover frequency, ωg , is de ned by G(ωg ) = 1
or log G(ωg ) = 0. and the phase margin is
5.4
arg G(ωg ) + π.
RootLocus
Rootlocus techniques provide a way of studying the response of a system as one real parameter is varied. In principle, any parameter can be varied; in practice, it is often the gain that is varied, because this makes it easy to sketch the loci. The rootlocus is the set of paths traced by the roots of the equation 1 + KG(s) = 0 on the complex plane with 0 ≤ K ≤ ∞. K is the gain of the system. The equation is equivalent to the two real equations KG(s) = 1
arg KG(s) = −π which are called the magnitude criterion and the phase criterion respectively. The rst equation does not tell us much (since 0 ≤ K ≤ ∞, it can be satis ed by arbitrary vlues of s).
5
31
STABILITY CRITERIA
The second is more useful: it says that KG(s) is a negative real number. The root locus can be estimated by using the following rules (Shiners, pages 146{151, Philips & Harbor, pages 211.). Figure 15 summarizes the rules for sketching the root locus: detailed notes follow. 1. There is one locus for each pole of 1 + KG(s). In other words, the number of loci is the same as the order of the characteristic equation. 2. If the coecients of the CE are real, as is usually the case, roots are either real or complex conjugate pairs. Consequently, each locus is symmetrical with respect to the real axis. +··· with α = n − m > 0. To make the number of zeros 3. In general, KG(s) has the form K bmsns+··· match the number of poles, we say that KG(s) has α zeros at infinity. We can construct asymptotes to the zeros at in nity, as shown below. Since the number of poles and zeros is then equal, we have the following rule: m
As K → 0, G(s) → ∞ and as K → ∞, G(s) → 0. Consequently, each locus starts at a pole of G(s) and ends at a zero of G(s). 4. A point on the real axis is on a locus if the sum of poles and zeroes to its right is odd. Since complex conjugate poles cancel in the imaginary direction, we need to consider only real poles and zeroes. Let x+j0 be the given point on the real axis. The angular contribution of a pole or zero > x is π while the angular contribution of a pole or zero < x is 0. Thus the eect of all the poles is nπ, where n is the number of poles or zeroes > x. The angle criterion is satis ed if n is odd. 5. For large values of s, the equation 1 + KG(s) = 0 becomes 1+
Kbm sα
= 0
and the roots satisfy sα + Kbm = 0.
The angles of these roots are θ = rπ/α for r = ±1, ±2, . . .. 6. The intersection of the root locus and the real axis can be determined by applying the RouthHurwitz criterion to the formula 1 + KG(s). In some situatiopns, we can do better than this. A breakaway point is a point at which the root locus leaves the real axis and becomes complex. At a breakaway point, the CE has multiple roots and, consequently, it shares roots d d d with its derivative. Since ds (1 + KG(s)) = K ds G(s), we need to consider only ds G(s). The rule is: Breakaway points occur where 1 + KG(s) and
d ds G(s)
= 0 have a common real root.
7. Let δi be the angle at which a locus departs from pole pi and let αj be the angle at which a locus arrives at zero zj . Then: δj =
X
αi +
i
αj =
X i
X
δi + rπ
i6=j
δi +
X i6=j
αi + rπ
6
32
CASE STUDY: MOTOR CONTROLLER
Rule Note #loci = #roots of 1 + KG = 0 1 Loci are symmetrical about the real axis 2 Loci start at poles and end at zeros 3 Loci includes all real points to the left of an 4 odd number of real critical frequencies Loci are asymptotic to zeros at in nity 5 Breakaway points occur at multiple roots 6 Angles sum to 180◦ 7 Asymptotes intersect real axis 8 Figure 15: Summary of root locus rules where r = ±1, ±2, . . .. 8. The intersection of the asymptotic lines and the real axis occurs at x + j0 where P
P
poles − zeros #poles − #zeros
x =
in which finite poles and zeros only are considered.
6
Case Study: Motor Controller
The motor controller is used as a running example throughout these notes. The purpose of the system is to control the angular position of a shaft, θ(t), with an electric motor. The motor is driven by a voltage, u(t), and its equation of motion is I
d2 θ dt2
= −µ
dθ + ku dt
where I is the moment of inertia of the motor, µ is a frictional factor, and k is the coupling constant relating voltage to torque. The equation is in Joules (kg m2 s−2 ). The units for individual variables are: I : kg m2 µ : kg m2 s−1 k : Joules/volt = coulombs = A s u : volts = kg m2 s−3 A−1
The Laplace transform of this equation is IΘs2 = −µΘs + kU
6
33
CASE STUDY: MOTOR CONTROLLER
and hence G=
Θ k = radians/volt. U s(Is + µ)
Feedback is provided by a potentiometer driven by the output shaft. This provides a voltage signal proportional to the angular position, v = fθ. The transfer function of the feedback loop is therefore H = f volts/radian and the closedloop transfer function of the system is M(s) = = =
G(s) 1 + G(s)H(s) k/s(Is + µ) 1 + fk/s(Is + µ) k radians/volt 2 Is + µs + fk
and the characteristic equation is (13)
Is2 + µs + fk = 0.
The openloop transfer function is GH =
fk . s(Is + µ)
(14)
Suppose the input to the system is a step function of V volts. Then U(s) = V/s and Θ(s) = U(s)M(s) Vk = 2 s(Is + µs + fk)
Let (15)
s2 + µs/I + fk/I = (s + A)(s + B).
Then Θ(s) =
Vk I
1 1 1 + + ABs A(A − B)(s + A) B(B − A)(s + B)
and θ(t) =
Vk I
!
1 e−At e−Bt + + . AB A(A − B) B(B − A)
We can obtain the values of A and B from (15): q 1 µ/I + (µ/I)2 − 4fk/I , 2 q 1 2 µ/I − (µ/I) − 4fk/I . 2
A = B =
6
34
CASE STUDY: MOTOR CONTROLLER
Since R(A) > 0 and R(B) > 0, the zeroes of the characteristic equation have negative real parts and the system is stable. There will be oscillation if We have θ(0) =
Vk I
µ I
<
√ 4fk , that is, if µ < 2 fkI. I
1 1 1 + + AB A(A − B) B(B − A)
and, since AB = fk/I, we have
= 0
Vk V = . IAB a
lim θ(t) =
t→∞
6.1
2
Zero Input
Consider the case when there is no input voltage. Then u = 0 and the equation of motion for the motor shaft is I
dθ d2 θ +µ dt2 dt
(16)
= 0.
The Laplace transform of (16) is I(s2 Θ(s) − sθ(0+ ) −
dθ + 0 )) + µ(sΘ(s) − θ(0+ )) = 0. d(
_ + ) = ω0 , this simpli es to If we assume θ(0+ ) = 0 and θ(0 Iω0 s(Is + µ)
Θ(s) =
To split the right side into partial fractions, assume 1 s(Is + µ)
A B + . s Is + µ
=
This equation holds for all values of s. Setting s = 0 gives A = 1/µ and setting s = −µ/I gives B = −I/µ. Since I/µ has the dimensions of time, we also de ne τ = I/µ. Then: Θ(s) = =
Iω0 Iω0 − µs µ(s + µ/I) ω0 τ ω0 τ − . s s + 1/τ
The inverse Laplace transform of Θ(s) gives the shaft position as a function of time 1 s
θ(t) = ω0 τ L−1
= ω0 τ 1 − et/τ
− L−1
and we can dierentiate it to give ω = ω0 e−t/τ .
1 s + µ/I
6
35
CASE STUDY: MOTOR CONTROLLER
6.2
Stability
RouthHurwitz
The characteristic equation (13) is Is2 + µs + fk = 0.
The coecients of this equation are a2 = I, a1 = µ, a0 = fk
and the RouthHurwitz array is
s2 I fk s1 µ s0 fk.
The system is stable if all of the constants are positive. Nyquist
The open loop transfer function (14) is G(s)H(s) =
fk s(Is + µ)
and the frequency response is G(jω)H(jω) =
fk . + µjω
−Iω2
The locus does not encircle the point −1 + j0 and the system is therefore stable. 6.3
Compensation
Applying compensation
s+a gives s+b G =
k(s + a) s(Is + µ)(s + b)
and so GH =
fk(s + a) s(Is + µ)(s + b)
The CE is GH + 1 = 0 which is equivalent to Is3 + (Ib + µ)s2 + (µb + fk)s + fka = 0.
The RouthHurwitz array is
s3 I µb + fk 2 s Ib + µ fka s1 B s0 fka
6
36
CASE STUDY: MOTOR CONTROLLER
where
1 I µb + fk B = − fka Ib + µ Ib + µ =
(Ib + µ)(µb + fk) − Ifka IB + µ
The system is stable if B > 0 or (Ib + µ)(µb + fk) > Ifka. Since this is equivalent to the condition (17)
µ(Ib2 + µb + fk) + Ifk(b − a) > 0
a sucient condition for stability is a < b. This is lead compensation (Section 4.2.1 on page 15) and generally improves stability. However, we could use lag compensation provided that (17) is satis ed. 6.4
Normal Form
k 1 k can be written in the form · 2 µ + µs + fk I s + Is+ 1 paring this to the canonical form 2 , we see that s + 2ζωn s + ω2n
The closedloop transfer function
Is2
fk I
. Com
s
ωn = ζ =
fk I µ √ 2 Ifk
The time constant of the system is τ = =
1 ζωn √ 2 Ifk µ
Substituting some plausible values I = 5 × 10−6 k = 0.1 f = 0.5 µ = 0.1
gives
√ 2 5 × 10−6 × 0.5 × 0.1 τ = 0.1 = 0.01 s.
If we put these values and b = 10 (corresponding to a time constant of 0.1 s) into (17), we obtain the approximate condition 0.1 + 25 × 10−8 (10 − a) > 0
which will be satis ed for reasonable values of a.
7
7
37
SI UNITS
SI Units
Figure 16 shows the basic SI units and Figure 17 shows units derived from the basic units. We have RF = V A−1 C V−1 = A s A−1 = s
showing that the product of resistance and capacitance is a time. Thus, in circuit analysis, RC is a time and RCω is a pure number. Similarly, HF = Wb A−1 C V−1 = V s A−1 A s V−1 = s2
showing that the product of an inductance and a capacitance is a squared time. In circuit analysis, LC has dimension T 2 and LCω2 is a pure number. Furthermore, the unit of inductance is the product of resistance and time: H = Wb A−1 = V A−1 s = Ωs
This means, for example, that a circuit with impedance jωR1 R2 C behaves like an inductance. In practice, we cannot construct a circuit from resistors and capacitors with pure imaginary conductance, but we could obtain an impedance of the form R + jωR1 R2 C, corresponding to a resistor R in series with an inductor L = R1 R2 C. Quantity Length Mass Time Current Temperature Luminosity Amount
Unit metre kilogram second ampere kelvin candela mole
Symbol m kg s A K cd mol
Figure 16: Basic units The following expressions were generated by the program dimensions.cpp. The dimension of each formula is noted in the heading. Common abbreviations are used to denote values in the expressions: C = capacitance, I = current, ` = length, L = inductance, µ = friction, R = resistance, t = time, V = voltage.
7
38
SI UNITS
Quantity
Unit
Symbol
SI
SI (basic) kg
Frequency Force
hertz
Hz
s−1
newton
N
kg m s−2
Friction Torque
kg s−1 newtonmetre
Moment of inertia Angular friction
m
s −1
1
1 −2
1 −1
kg m2 s−2
1
2 −2
kg m2
1
2
kg m2 s−1
1
2 −1
Pressure
pascal
Pa
N m−2
Work, energy, heat
joule
J
Nm
1
2 −2
Power
watt
W
J s−1
1
2 −3
Charge
coulomb
C
As
Potential
volt
V
J C−1
1
Capacitance
farad
F
C V−1
−1
Resistance
ohm
Ω
V A−1
1
Conductance
siemens
S
A V−1
−1
Magnetic ux
weber
Wb
Vs
1
Flux density
tesla
T
Wb m−2
1
Inductance
henry
H
Wb A−1
1
Figure 17: Derived units
A
1 −1 −2
1
1
2 −3 −1 −2
4
2
2 −3 −2 −2
3
2
2 −2 −1 −2
−1
2 −2 −2
7
39
SI UNITS
Dimensions: pure number. ` mµ
mµ `
1 tω
tω
V IR
IR V
Rt L
R Lω
L Rt
Lω R
t2 CL
1 CLω2
L CR2
t CR
1 CRω
CR2 L
CL t2
CLω2
CR t
CRω
1 tω2
t
t2 ω
IL V
1 ω
Rt2 L
R Lω2
L R
L2 R2 t
L2 ω R2
t2 CR
1 CRω2
CV I
CL t
CLω
CR
C2 R2 t
C2 R2 ω
V2 IR2
I
I t2 ω 2
I tω
Itω
It2 ω2
I2 R V
Vt L
V Lω
V R
CV t
CVω
Dimensions: T .
Dimensions: A.
Dimensions: LM2 T −3 A−1 (volts). V2 IR V It C
IL t V t2 ω 2 I Cω
ILω V tω
IR
I2 R2 V
Vtω Vt2 ω2
7
40
SI UNITS
Dimensions: LM2 T −3 A−2 (ohms). V2 I2 R
V I
IR2 V
R2 t L
R2 Lω
L 2 t ω
L t
Ltω2
Lω
L2 Rt2
L2 ω2 R
R
R
R tω
Rtω
Rt2 ω2
t2 C2 R
C2 Rω2
1 Ctω2
t C
t2 ω C
1 Cω
L CR
CR2 t
CR2 ω
t2 ω 2 1
8
41
USEFUL FORMULAS
8
Useful Formulas Function
Transform
Remarks
f(t)
F(s)
General notation
Af(t) + Bg(t)
AF(s) + BG(s)
Linearity
f(t − T )u(t − T )
e−sT F(s)
u(t) is the unit step function
e−at f(t)
F(s + a)
f 0 (t)
sF(s) − f(0+ )
f 00 (t) Rt 0 f(τ)dτ Rt 0 f(t − τ)g(τ)dτ
s2 F(s) − sf(0+ ) − f 0 (0+ ) F(s)/s F(s)G(s)
Figure 18: Laplace: rules Express
P(s) in the form Q(s) A1 A2 A3 + + + ··· s + s1 s + s2 s + s3
Q(s) has only simple roots: if Q(s) = (s + s1 )(s + s2 ) · · · (s + sn )
then n X Ai = s + si
P(s) Q(s)
i=1
where
Ai =
P(s) (s + si ) . Q(s) s=−si
Repeated roots: if Q(s) = · · · (s + si )r · · · then P(s) Q(s)
= ··· +
B1 B2 Br + + ··· + + ··· s + si (s + si )2 (s + si )r
where Br = Br−1 =
P(s) (s + si ) Q(s) s=−si
1 d r P(s) (s + si ) 1! ds Q(s) s=−si r
8
42
USEFUL FORMULAS
Function
Transform
Remarks
δ(t)
1
Dirac's δfunction
1 s 1 sn 1 s+a 1 s(s + a) s 2 s + a2 s 2 s − a2 a s2 + a2 a 2 s − a2 1 2 s(s + a2 ) 1 2 2 s (s + a2 ) 1 (s + a)2 s (s + a)2 1 2 s + 2ζω + ω2
Unit step function
u(t) tn−1 (n − 1)! e−at 1 − e−at a
cos at cosh at sin at sinh at 1 − cos at a2 at − sin at a3 te−at e−at (1 − at) e−at sin bt b
a = ζω, b = ω 1 − ζ2 , and ζ < 1 p
Figure 19: Laplace: special cases
Br−2 =
1 d2 2! ds2
i
.. . B1 =
r P(s) (s + si ) Q(s) s=−s
dr−1 1 (r − 1)! dsr−1
r P(s) (s + si ) Q(s) s=−s
i
E(s)
R(s)

+
− 6

G(s)

B(s) H(s)
C(s)
8
43
USEFUL FORMULAS
Openloop response Closedloop response
G(s) H(s) G(s) 1 + G(s) H(s) 1 + G(s) H(s) = 0
Characteristic equation The phaselead compensator is a form of highpass lter. The compensator introduces gain at high frequencies, which may increase instability, and phase lead, which tends to be stabilizing. The pole and zero are typically placed in the high frequency region. The phaselag compensator is a simple form of lowpass lter. It reduces the gain at high frequencies, which tends to stabilize the system, and introduces phase lag, which tends to destabilize the system. The pole and zero of the compensator are usually placed in the low frequency region. Compensation Advantages Phase lag LF characteristics improved stability marrgins maintained or improved bandwidth reduced (useful if HF noise is a problem) improved stability margins improved HF response may be required for stability
Phase lead
Disadvantages at least one slow term in transient reponse reduced bandwidth may be a disadvantage possible HF noise problems may generate large signals, out of linear range of system
Z2
Vout
v Vin
Z1
Z2
v
Vout
Z − +
Vin Z − Z +
Z
Z1
(a) T (s) = − ZZ12
(b) T (s) =
For the canonical second order system G(s) =
ω2n . s2 + 2ζωn s + ω2n
Z1 Z1 +Z2
8
44
USEFUL FORMULAS
Linear Symbol Position x Mass m Velocity v = dx dt 2 Force f = m ddt2x Spring k df dt = v Friction µv = f
Rotating Symbol Angle θ Moment of Inertia I Angular Velocity ω = dθ dt 2 Torque T = I ddt2θ Clockspring λ dT dt = ω Angular friction δω = T
Function
Amplitude
K jω 1/jω 1 + jω/ωn
20 log10 K 20 log10 ω −20 log10 ω 0 20(log10 ω − log10 ωn ) 0 −20(log10 ω − log10 ωn )
1 1+jω/ωn
ω ≤ ωn ω > ωn ω ≤ ωn ω > ωn
: : : :
Electrical Symbol Charge q Inductance L Current i = dq dt 2 Potential e = L ddt2q Capacitance C de dt = i Resistance Ri = e
Point ω db 1 1 ωn
0 0 0
ωn
0
Figure 20: Bode diagrams RouthHurwitz array for an sn + an−1 sn−1 + · · · + a0 = 0: sn sn−1 sn−2 sn−3 ...
an an−2 an−4 an−6 . . . an−1 an−3 an−5 an−7 . . . b1 b2 b3 ... c1 c2 ...
where b1 b2 ... c1 c2 ...
a n = − an−1 an−1 1 an = − an−1 an−4
1
a n−1 b1 1 an−1 = − b1 b1
1 = − b1
an−2 an−5 an−3 b2 an−5 b3
an−2 an−3
Phase rule 0 ω < 0.1ωn ω > 10ωn ω < 0.1ωn ω > 10ωn
: : : :
π/2 −π/2 0 π/2 0 −π/2