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14
Robustness Margins
Three common measures of insensitivity of a closed-loop system to variations are
• the time-delay margin,
• the gain margin, and
• the percentage variation margin.
All are easily computed, and provide a quantitative measure of the insensitivity of the closedloop system to specific types of variations. The nominal closed-loop system is shown below.
r(t)
e(t)
+
- i
−6
u(t)
C
-
- y(t)
P
We assume that both P and C are governed by linear, ordinary differential equations, and
that the nominal closed-loop system is stable.
The perturbed closed-loop system is shown below, with both a gain variation in the plant,
a time-delay in the feedback measurement, and a change in process from P to P̃ .
r(t)
e(t)
+
- i
−6
u(t)
C
- γ
-
- y(t)
P̃
delay, T ¾
f (t) = y(t − T )
Note that if T = 0, γ = 1 and P̃ = P , then the perturbed system is simply the nominal
closed-loop system.
The time-delay margin is characterized by the minimum time delay T > 0 such that the
closed-loop system becomes unstable, with γ set equal to its nominal value, γ = 1, and
P̃ = P .
³
´
The gain margin is characterized by the largest interval γ, γ̄ , containing 1, such that the
³
´
closed-loop system is stable for all γ ∈ γ, γ̄ , with T set to its nominal value, T = 0, and
P̃ = P
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The percentage variation margin is a frequency-dependent bound α(ω), such that the closedloop system is stable for all P̃ which satisfy satisfies
¯
¯
¯ P̃ (jω) − P (jω) ¯
¯
¯
¯
¯ < α(ω) for all ω.
¯
¯
P (jω)
assuming γ = 1, T = 0.
Two viewpoints can be taken regarding the interpretation of these margins:
1. We are explicitly concerned about the destabilizing effects of time-delays, gain changes
and perccentage changes in frequency responses, and these margins give exact answers
with respect to those variations.
2. While not being explicitly concerned with time-delays and/or gain changes and/or
percentage changes in frequency response functions, we are cognizant of the fact that
the model differential equation we use for the plant is not entirely accurate. In that
vein, we would like some easily computable information concerning the sensitivity (or
insensitivity) of the stability of the closed-loop system to unmodeled variations. We
use the computed margins as indications of the potential for instability brought on by
variations in the plant’s behavior.
Since linear, single-input, single-output systems commute, we can redraw the diagram, preserving both the stability characteristics, and the relationship from r to y as
r(t)
+
- i
−6
e(t)
- γ
v(t)
f (t) = y(t − T )
-
C
-
L
P
- y(t)
delay, T ¾
By convention, if P denotes the nominal plant, then P̂ (s) will denote the transfer function
of the nominal plant. Similar convention for controller and in fact, any dynamical system. A
capital letter denotes the system itself while the “hatted” capital letter denotes the system’s
transfer function.
For certain computational purposes, we denote the cascade of C and P as the system L. In
terms of transfer functions, we have L̂(s) = P̂ (s)Ĉ(s) (see section 12). Assume L̂ is of the
form
b1 sn−1 + · · · + bn−1 s + bn
L̂(s) = n
s + a1 sn−1 + · · · + an−1 s + an
This is equivalent to the differential equation relating v to y as
y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b1 v [n−1] (t) + · · · + bn−1 v [1] (t) + bn v(t)
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Under both types of perturbations, we get that
v(t) = γ [r(t) − y(t − T )]
h
i
Note that for any k, v [k] (t) = γ r[k] (t) − y [k] (t − T ) . Plugging in and rearranging gives the
closed-loop delay-differential equation
y [n] (t) + a1hy [n−1] (t) + · · · + an−1 y [1] (t) + an y(t)
i
+ γ b1 y [n−1] (t − T ) + · · · + bn−1 y [1] (t − T ) + bn y(t − T )
h
i
= γ b1 r[n−1] (t) + · · · + bn−1 r[1] (t) + bn r(t)
(68)
It is from this equation that we begin our analysis, though most of our computations will
ultimately involve the transfer function L̂(s) evaluated along the imaginary axis s = jω.
14.1
Gain Margin
The setup for the gain margin problem is shown in Figure 14, with T = 0. Here, a plant
and controller are in closed-loop along with a gain γ at the plant input. We assume
³ that
´ the
closed-loop system is stable for γ = 1, and we wish to know the largest interval γ, γ̄ such
³
´
that closed-loop stability is maintained for all γ ∈ γ, γ̄ . Since the closed-loop system is
stable for γ = 1, it is clear that γ < 1 < γ̄.
With the time-delay T = 0, the differential equation (68) becomes an ordinary differential
equation,
y [n] (t) + (ah1 + γb1 ) y [n−1] (t) + · · · + (an−1 + γbn−1i ) y [1] (t) + (an + γbn ) y(t)
= γ b1 r[n−1] (t) + · · · + bn−1 r[1] (t) + bn r(t)
Clearly, the closed-loop characteristic polynomial is a function of γ, namely
pγ (λ) = λn + (a1 + γb1 )λn−1 + (a2 + γb2 )λn−2 + · · · + (an−1 + γbn−1 )λ + (an + γbn )
It is a fact from mathematics that the roots of the characteristic polynomial are continuous
functions of the parameter γ, hence as γ varies, the roots vary continuously. Therefore, the
only way for a root to move into the right-half plane is to cross the imaginary axis, either
at 0, or at some jβ, with β > 0 (since complex roots come in conjugate pairs, if they cross
the imaginary axis, one will cross positively, and one negatively - we can focus on one or the
other, and for no specific reason, we look at the positive crossing). Hence, we are looking
for values γ and β, with close to 1, and β ≥ 0 such that λ = jβ is a root of pγ . Plugging in
gives
(jβ)n + (a1 + γb1 )(jβ)n−1 + (a2 + γb2 )(jβ)n−2 + · · · + (an−1 + γbn−1 )(jβ) + (an + γbn ) = 0
Rearranging gives
[(jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an ]
.
+ γ [b1 (jβ)n−1 + b2 (jβ)n−2 + · · · + bn−1 (jβ) + bn ] = 0
(69)
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ME 132, Spring 2005, UC Berkeley, A. Packard
This is two equations (since it relates complex quantities) in two real unknowns (β and γ).
We need to find all solutions, and then isolate the solutions whose γ coordinate is closest to
1.
First, we can look for solutions to (69) by looking for solutions (ie., real β) to
(jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an = 0.
If there are roots, then coupled with γ = 0, we have found solutions to equation (69). There
likely are others, which are described below.
We can also look for solutions to (69) with (jβ)n +a1 (jβ)n−1 +a2 (jβ)n−2 +· · ·+an−1 (jβ)+an 6=
0, so that after dividing, we manipulate equation (69) into
−1 = γ
b1 (jβ)n−1 + b2 (jβ)n−2 + · · · + bn−1 (jβ) + bn
(jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an
But, this is recognized as
¯
¯
−1 = γ L̂(s)¯
s=jβ
¯
¯
This can be solved graphically, finding values of β for which L̂(s)¯
s=jβ
then taking a negative reciprical to find the associated γ value).
is a real number (and
Summarizing, a simple method to determine all solutions to equation (69) is
1. Let βi denote all (if any) of the real solutions to
(jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an = 0.
If there are any, then the pairs (βi , 0) are solutions to equation 69.
2. Next, plot (in the complex plane, or on separate magnitude/phase plots) the value of
L̂(jβ) as β ranges from 0 to ∞. By assumption, this plot will not pass through the
−1 point in the complex plane (can you explain why this is the case?).
3. Mark all of the locations where L̂(jβk ) is a real number. The frequencies βk are called
the phase-crossover frequencies.
4. At each phase crossover point, βk , determine the associated value of γ by computing
γk :=
−1
¯
L̂(s)¯¯
s=jβk
5. Make a list of all of the γ values obtained in the calculations above (step 1 and 4).
Of all solutions less than 1, find the one closest to 1. Label this γ. Similarly, Of all
solutions greater than 1, find the one closest to 1. Label this γ̄. In the event that there
are no solutions less than 1, then γ = −∞. Similarly, if there are no solutions greater
than 1, then γ̄ = ∞.
ME 132, Spring 2005, UC Berkeley, A. Packard
120
Graphically, we can determine the quantity γ̄ easily. Find the smallest number > 1 such
that if the plot of L(jβ) were scaled by this number, it would intersect the −1 point. That
is γ̄. This is easily done by computing the closest intersection of the curve L̂(jβ) (as β
varies) with the real axis, to the right of −1, but less than 0. If the intersection occurs at
the location −α, then scaling the plot of L̂ by α1 will cause the intersection. By chosing the
closest intersection, α is “close” to 1, and hence γ̄ := α1 is the desired value. Note: It is
possible that there are no intersections of the curve L̂(jβ) of the real axis to the right of
−1 but less than 0. In that case, the closed-loop system is stable for all values of γ in the
interval [1, ∞), hence we define γ̄ := ∞.
If the plot of L̂(jβ) as β varies intersect the negative real axis to the left of −1, then γ > 0,
and it is easy to determine. Simply find the largest positive number < 1 such that if the plot
of L̂(jβ) were scaled by γ, it would intersect the −1 point. This is easily done by computing
the closest intersection of the curve L̂(jβ) (as β varies) with the real axis, to the left of −1.
If the intersection occurs at the location −α, then scaling the plot of L̂ by α1 will cause the
intersection. By chosing the closest intersection, α is “close” to 1, and hence γ := α1 is the
desired value.
In the above computations, each intersection of the curve L̂(jβ) (as β varies) with the
negative real axis is associated with two numbers:
1. The location (in the complex plane) of the intersection, of which the negative reciprical
determines the gain γ that would cause instability.
2. The value of β at which the intersection takes place, which determines the crossing
point (on the imaginary axis) where the closed-loop root migrates from the left-half
plane into the right-half plane, and hence determines the frequency of oscillations just
at the onset of instability due to the gain change.
14.2
Time-Delay Margin
The homogeneous equation, with γ = 1, but T > 0 of equation (68) is
y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t)
+ b1 y [n−1] (t − T ) + · · · + bn−1 y [1] (t − T ) + bn y(t − T ) = 0
(70)
Recall that we assume that for T = 0, the system is stable, and hence all homogeneous
solutions decay exponentially to 0 as t → ∞. It is a fact (advanced) that as T increases
the system may eventually become unstable, and if it does, then at the critical value of T
for which stability is violated, sinusoidal solutions to the homogeneous equation exist. In
other words, the system “goes unstable” by first exhibiting purely sinusoidal homogeneous
solutions. The frequency of this sinusoidal solution is not known apriori, and needs to be
determined, along with the critical value of the time-delay.
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The delay terms may cause a non-decaying homogeneous solution to exist. We pose the
question: what is the minimum T and corresponding frequency ω such that a sinusoidal
solution to (70) of the form yH (t) = ejωt exists? If such a solution exists, then by plugging it
into (70), and canceling the common ejωt term, which is never 0, we get a necessary algebraic
condition
n−1
(jω)n + a1 (jω)
+ · · · + an−1 jω + an
h
i
−jωT
b1 (jω)n−1 + · · · + bn−1 jω + bn = 0
e
Rearranging gives
−1 = e−jωT
b1 (jω)n−1 + b2 (jω)n−2 + · · · + bn−1 (jω) + bn
(jω)n + a1 (jω)n−1 + a2 (jω)n−2 + · · · + an−1 (jω) + an
But, this is recognized as
¯
−1 = e−jωT L̂(s)¯¯
s=jω
This is two equations (since it relates complex quantities) in two real unknowns (T and ω).
We want to solve for solutions with T close to 0 (recall that 0 was the nominal value of T ,
and we
that the closed-loop system was stable for that). Since ωT is real,
it follows
¯ assumed
¯
¯
¯
¯ −jωT ¯
¯
¯
that ¯e
¯ = 1. This gives conditions on the possible values of ω, namely ¯L̂(jω)¯ = 1.
Once these values of ω are known, the corresponding values of T can be determined, and
the smallest such T becomes the time-delay margin.
In order to understand the graphical solution techniques described, recall that for any complex number Ψ, and any real number θ, we have
6
ejθ Ψ = θ + 6 Ψ
Hence, at any value of ω, we have
6
e−jωT L̂(jω) = −ωT + 6 L̂(jω)
Also, 6 (−1) = ±π.
Hence, a simple method to determine solutions is:
1. Plot (in the complex plane, or on separate magnitude/phase plots) the quantity L̂(jω)
as ω varies from 0 to ∞.
¯
¯
¯
¯
2. Identify all of the frequencies ωi where ¯L̂(jωi )¯ = 1. These are called the gain-crossover
frequencies.
¯
¯
3. At each gain-crossover frequency ωi , (which necessarily has ¯¯L̂(jωi )¯¯ = 1), determine
the value of T > 0 such that e−jωi T L̂(jωi ) = −1. Recall that multiplication of any
complex number by e−jωi T simply rotates the complex number ωi T radians in the
clockwise (negative) direction. Hence, each T can be determined by calculating the
angle from L̂(jω) to −1, measured in a clockwise direction, and dividing this angle
by ω. On a phase plot, determine the net angle change in the downward direction
(negative) to get to the closest odd-multiple of π (or 180◦ ).
ME 132, Spring 2005, UC Berkeley, A. Packard
122
4. Repeat this calculation for each gain-crossover frequency, and choose the smallest of
all of the T ’s obtained.
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ME 132, Spring 2005, UC Berkeley, A. Packard
14.3
Examples
14.3.1
Generic
Consider a system with
L̂(s) =
¯
4
(0.5s + 1)3
¯
The plots of ¯¯L̂(jω)¯¯ , 6 L̂(jω), and L̂(jω) as ω varies are shown below. We will do calculations
in class.
Loop Gain
1
10
Nyquist Plot
1
0.5
0
0
10
−0.5
−1
−1.5
−1
10
−2
−2.5
−2
10
0
10
0
1
10
Loop Phase
−50
−100
−150
−200
−250
0
10
1
10
−3
−1
−0.5
0
0.5
1
1.5
2
2.5
3
3.5
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ME 132, Spring 2005, UC Berkeley, A. Packard
14.3.2
124
Missile
A missile is controlled by deflecting its fins. The transfer function of a the Yaw axis of a
tail-fin controlled missile is
P̂ (s) =
−0.5(s2 − 2500)
(s − 3)(s2 + 50s + 1000)
A PI controller, with transfer function
Ĉ(s) =
10(s + 3)
s
is used. This results in a stable closed-loop system, with closed-loop roots at
−8.1 ± j14.84, −18.8, −6.97
Plots of the L̂ are shown below, along with some time responses.
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ME 132, Spring 2005, UC Berkeley, A. Packard
2
Log Magnitude
10
1
10
0
10
0
10
Frequency (radians/sec)
1
10
1
10
10
2
220
Phase (degrees)
200
180
160
140
120
100
80
0
10
Frequency (radians/sec)
10
2
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ME 132, Spring 2005, UC Berkeley, A. Packard
Nyquist Plot
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
−5
−4
−3
−2
−1
0
1
Open−Loop Missile Step Response
0.5
0.4
G’s
0.3
0.2
0.1
0
−0.1
0
0.05
0.1
0.15
Time: seconds
0.2
0.25
0.3
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ME 132, Spring 2005, UC Berkeley, A. Packard
Missile Acceleration Response
2
Yaw Acceleration: G’s
1.5
1
0.5
0
−0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
Time: Seconds
0.7
0.8
0.9
1
0.7
0.8
0.9
1
Fin Deflection
12
10
8
6
Degrees
4
2
0
−2
−4
−6
−8
14.4
0
0.1
0.2
0.3
0.4
0.5
0.6
Time: Seconds
Problems
1. Consider the multiloop interconnection of 4 systems shown below.
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ME 132, Spring 2005, UC Berkeley, A. Packard
F ¾
r -d E
−6
+?
+d
G ¾
-d - H
6
?
?
p
y
-
z
q
(a) What is the transfer function from q to z? Denote the transfer function as G q→z
(b) What is the transfer function from q to p? Denote the transfer function as G q→p
(c) Verify that 1 + Gq→p = Gq→z .
2. Suppose C(s) = 4s and P (s) = 2 in a standard Controller/Plant feedback architecture,
with negative feedback.
(a) Is the closed-loop system stable?
(b) What is the closed-loop transfer function from r to y.
(c) What is the steady-state gain from r → y?
(d) What is the time-constant of the closed-loop system.
(e) What is the time-delay margin? Denote it by Td . At what frequency will the
self-sustaining (ie., unstable) oscillations occur?
(f) Verify your answers with Simulink, using time delays of 0,
9
T , 9.9 T .
10 d 10 d
1
T , 3T , 5T , 7T ,
10 d 10 d 10 d 10 d
3. Suppose C(s) = Ks and P (s) = β in a standard Controller/Plant feedback architecture,
with negative feedback. Assume both β and K are positive.
(a) Is the closed-loop system stable?
(b) What is the closed-loop transfer function from r to y.
(c) What is the steady-state gain from r → y?
(d) What is the time-constant of the closed-loop system.
(e) What is the time-delay margin?
(f) What is the ratio of the time-constant of the closed-loop system to the time-delay
margin? Note that this ratio is not a function of β and/or K.
4. The transfer functions of several controller/process (C/P ) pairs are listed below. Let
L(s) := P (s)C(s) denote the loop transfer function. For each pair, consider the closedloop system
r(t)
e(t)
+
- i
−6
u(t)
C
-
P
- y(t)
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ME 132, Spring 2005, UC Berkeley, A. Packard
• determine the closed-loop characteristic polynomial;
• compute the roots of the closed-loop characteristic polynomial, and verify that
the closed-loop system is stable;
• for the gain-margin problem, find all of the phase-crossover frequencies, and the
value of L at those frequencies;
• determine the gain-margin of the closed-loop system;
• for the time-delay-margin problem, find all of the gain-crossover frequencies, and
the value of hL at those frequencies;
• determine the time-delay-margin of the closed-loop system;
• determine and plot (as a function of frequency) the percentage-variation margin,
considering changes in P .. For cases (a) and (b) below, plot this margin on the
same plot for comparison purposes later.
The systems are:
(b) C(s) =
2.4s+1
,
s
0.4s+1
,
s
(c) C(s) =
10(s+3)
,
s
(a) C(s) =
P (s) =
P (s) =
P (s) =
1
s−1
1
s+1
−0.5(s2 −2500)
(s−3)(s2 +50s+1000)
5. In problem 4 above, the first two cases have identical closed-loop characteristic polynomials, and hence identical closed-loop roots. Nevertheless, they have different stability
margins. In one case, the plant P is unstable, and in the other case it is stable. Check
which case has better stability margins in each of the different measures. Make a conclusion (at least in the case) that all other things being equal, it is “harder” to reliably
control an unstable plant than it is a stable one.
6. Find Leff for determining time-delay and/or gain margins at the locations marked by
1, 2 and 3.
d
−6
- C1
-d 1 −6
- G1
A
- G2
-
2
C2 ¾
3
7. A closed-loop feedback system consisting of plant P and controller C is shown below.
r(t)
e(t)
+
- i
−6
u(t)
C
-
P
- y(t)
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ME 132, Spring 2005, UC Berkeley, A. Packard
It is known that the nominal closed-loop system is stable. In the presence of gainvariations in P and time-delay in the feedback path, the closed-loop system changes
to
r(t)
e(t)
+
- i
−6
u(t)
C
f (t) = y(t − T )
- γ
-
P
- y(t)
delay, T ¾
In this particular system, there is both an upper and lower gain margin - that is, for
no time-delay, if the gain γ is decreased from 1, the closed-loop system becomes
unstable at some (still positive) value of γ; and, if the gain γ is increased from 1, the
closed-loop system becomes unstable at some value of γ > 1. Let γl and γu denote
these two values, so 0 < γl < 1 < γu .
Min Time−Delay
For each fixed value of γ satisfying γl < γ < γu the closed-loop system is stable.
For each such fixed γ, compute the minimum time-delay that would cause instability.
Specifically, do this for several (say 8-10) γ values satisfying γl < γ < γu , and plot
below.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5
GAMMA
The data on the next two pages are the magnitude and phase of the product P̂ (jω)Ĉ(jω).
They are given in both linear and log spacing, depending on which is easier for you to
read. Use these graphs to compute the time-delay margin at many fixed values of γ
satisfying γl < γ < γu .
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ME 132, Spring 2005, UC Berkeley, A. Packard
2
4
10
3.75
3.5
3.25
3
1
10
2.75
Magnitude
Magnitude
2.5
2.25
2
1.75
0
10
1.5
1.25
1
0.75
0.5
−1
10
−1
10
0
10
1
Frequency, RAD/SEC
10
2
3
10
220
210
210
207.5
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Frequency, RAD/SEC
205
200
202.5
190
200
180
197.5
170
Phase (DEGREES)
Phase (DEGREES)
4
160
150
140
130
195
192.5
190
187.5
185
182.5
180
120
177.5
110
175
100
172.5
−1
10
0
10
1
Frequency, RAD/SEC
10
2
10
170
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Frequency, RAD/SEC
8. An unstable plant, P , with differential equation relating its input u and output y,
ÿ(t) − 16ẏ(t) + 15y(t) = u̇(t) + 5u(t)
is given.
(a) Calculate the range of the parameter KP for which the closed-loop system is
stable.
+ e er KP
−6
u-
P
y
-
(b) If KP = 30, what is the steady state error, ess , due to a unit-step reference input?
(c) A integral controller will reduce the steady-state error to 0, assuming that the
closed-loop system is stable. Using any method you like, show that the closedloop system shown below, is unstable for all values of KI .
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ME 132, Spring 2005, UC Berkeley, A. Packard
+ e e- R
r −6
- KI
u-
y
-
P
(d) Find a PI controller, that results in a stable closed-loop system.
(e) Consider the system with just Proportional-control, along with a time-delay, T ,
in the feedback path. We wish to determine the maximum allowable delay before
instability occurs. Find the equation (quadratic in ω 2 ) that ω must satisfy for
there to exist homogeneous solutions yH (t) = ejωt for some ω. The equation
should involve ω, KP , and the parameters in the plant, but not the time delay, T .
+ e er KP
−6
¾
u-
y
-
P
delay, T ¾
(f) For two cases of proportional control: KP = 20 and KP = 30; determine in each
case the time delay T that will just cause instability and the frequency of the
oscillations as instability is reached.
9. Let L be the transfer function of a system. Consider the diagram below.
z6
d
−6
q
p
6
- d?
-
L
Let N denote the transfer function from q to z. Let S denote the transfer function
from q to p. Derive N and S in terms of L.
10. In the diagram below,
d
−6
- PI
- d 1−6
R
-
-
R
KD ¾2
3
ME 132, Spring 2005, UC Berkeley, A. Packard
133
derive S, N and L at each marked location. In each case, verify (after the derivation)
that
N
1−S
L=−
=
1+N
S