Questions and Solutions for Chapter 21
Linearisation of systems from the real nonlinear world
M21.1 If a1, a2 and a3 are constants, which of the following equations is linear in
x?
(a) y = a1x + a2x2 + a3 x3
(b) y = (a1 + a2 + a3) x
(c) y = x
(d) y = a1/x
M21.2 Saturation effects
(a) are found in actuators
(b) are nonlinear effects
(c) limit the control output signal
(d) all of the above
M21.3 In most cases, a nominal linear model can only be used
(a) over a wide operating range
(b) over a specified operating range
(c) at several operating points
(d) where the system output shows large nonlinearity
M21.4 A linear approximation, at the operating point xp = 1.5, to y = 3x2 , is
(a) δy = 6 δx
(b)
δy = 9 δx
(c) δy = 6.75 δx
(d)
δy = 13.5 δx
M21. 5 The linear function δx L , at the operating point up = 3, of
x (u) = 2u3 + 4
is
(a)
δxL = 6 δu
(b)
δxL = 10 δu
(c)
δxL = 54 δu
(d)
δxL = 58 δu
M21.6 A linear approximation, at the operating point xp = 0.5, up = 1.5, to
y (x,u) = 2x2 + xu + u2 , is
(a) δy = 3.5δx + 3.5δu
(b)
δy = 4δx + 2δu
(c) δy = 2δx + 3δu
(d)
δy = 4.75δx + 3.75δu
dch21 linear v3 solutions
21.1
M21.7 If we have defined a linear model δx = 3.4 δu1 + 6 δu2 , δx represents
(a) the value of x(t) at the operating point
(b) the large variations in x(t) after the linearisation process
(c) the small variations in x(t) around the operating point
(d) the change in the operating point
M21.8 The ABCD matrices resulting from a linearisation of a nonlinear model
(a) contain the nonlinear terms
(b) contain the operating points
(c) contain the linearisation coefficients
(d) contain the small change inputs
M21.9 If we use a linearised model, the output, δy, of the ABCD system will
represent
(a) the change in output from the operating point
(b) the step response of the nonlinear system
(c) the step response of the linear system
(d) the change in output from zero.
M21.10 Given a SISO nonlinear model, and a linearisation of this model, then if
we double the size of the input step to both the nonlinear and linear system,
(a) the output from each system will always double
(b) the output from the nonlinear system will always double
(c) the output from the linear system will always double
(d) the output from the nonlinear system will always be double the output from
the linear system
Multiple choice questions
M21.1 b, M21.2 d, M21.3 b, M21.4 b, M21.5 c,
M21.6 a, M21.7 c, M21.8 c, M21.9 a, M21.10 c.
dch21 linear v3 solutions
21.2
Practical skills questions
Q21.1 Prove that the following functions are linear
(a) f(x1,x2) = 4x1 + 7x2
(b) f(x) =
where
Solution Q21.1
(a) We note that f(x1, x2) = [ 4 7]
= [ 4 7 ] xa
We have to have to prove that
(i)
f(xa + xb) = f(xa ) + f( xb )
(ii)
cf(xa) = f(cxa)
To prove (i) Let xb =
[ 4 7] (
+
. Then the LHS of (i) is
) = [4 7]
= 4(x1 + x3) + 7(x2 + x4) = (4x1 + 7 x2) + (4x3
+ 7x4) = f(xa) + f(xb) + RHS of (i)
To prove (ii)
The LHS of (ii) is
cf(xa) = c (4x1 + 7 x2) = 4cx1 + 7cx2 = [ 4 7]
The function is linear.
(b) We note that f(xa) =
.
We have to have to prove that
(i)
f(xa + xb) = f(xa ) + f( xb )
(ii)
cf(xa) = f(cxa)
Consider the LHS of (i)
f(xa + xb) =
dch21 linear v3 solutions
(
+
)=
21.3
= f(cxa) = RHS of (ii)
=
= f(xa) +f(xb) = RHS of (i)
To prove (ii):
consider LHS of (ii)
cf(xa) = c
7.2x1 + 0.5x2]T
= c[ 3.5x1 –4.2x2
= [ c(3.5x1 –4.2x2)
c(7.2x1 + 0.5x2)]T
= [ (3.5cx1 –4.2cx2)
(7.2cx1 + 0.5cx2)]T
=
= f(cxa) = RHS of (ii)
The function is linear.
Q21.2 Prove that the following functions are nonlinear.
x1
(a) f(x) = x where
2
(b) f(x,u) =
where
and
Solution Q21.2
We need to demonstrate that one of the linearity properties does not hold:
(a) We show that f(cx) ≠ cf(x)
Note: cx = c
=
cx1 x1
LHS f(cx) = cx = x
2
2
x1
RHS cf(x) = cx
2
LHS ≠ RHS , therefore the function is not linear.
(b) We show that f(cx, cu) ≠ cf(x,u)
dch21 linear v3 solutions
21.4
Note cx = c
=
and cu = c
=
LHS f(cx,cu) =
Since, the LHS does not equal the RHS the function is not linear.
Q21.3 The resistance relationship for a thermistor is given by R = Ae-b/T.
(a) If the following variables are defined
1
y = ln R and x = T
show that the resistance relationship can be transformed into the linear
relationship
y = ln A –bx.
(b)Suggest what experimental use this transformation might have in calibrating a
thermistor device.
Solution Q21.3
(a) The resistance relationship is R(T) = Aexp(-b/T)
b
ln R = ln A - T
Thus
y = lnA – bx
with
y = lnR and x = 1/T
(b) A plot of ln R versus 1/T will yield a straight-line relationship. The intercept
with the x=0 line will yield A and the slope will yield parameter b. If the data does
not yield a straight-line relationship then the model for the resistance-temperature
dependence should be re-examined.
Q21.4 A process furnace unit has a dynamical state variable model given by,
= -10 x(t) + 0.2 u(t)
y(t) = 25 50 x(t)
where u(t) represents the fuel flow in m3/min
dch21 linear v3 solutions
21.5
x(t) represents the thermocouple output in volts
y(t) represents the measured temperature in oC/volt
A steady operating condition arises when the fuel flow input is 10 m3/min.
Determine the furnace operating temperature.
Solution Q21.4
In steady conditions,
=0. Therefore
-10xss + 0.2 uss = 0
and xss = 0.02 uss. The steady output temperature is then
yss = 2550 xss = 51 uss
Given that uss = 10, the furnace operating temperature = 510 o C.
P21.1 Consider a simple nonlinear tank filling operation. The dimensions of the
tank are shown below.
Tank with circular cross-section
Dimensions, data and features:
Ro : radius at the tank base ;
Ro = 0.2m
RT : radius at the tank top;
RT = 3.2m
HT height of tank;
HT = 3m
qin(t) : inflow;
qin = 0.2m3/s
qout(t): outflow;
qout =
h(t)
Rf
Rf : outflow proportionality constant: Rf = 2.74
(a) Use the volumetric conservation principle,
dV
dt = qin(t) – qout(t)
dch21 linear v3 solutions
21.6
to prove
dh
Rfqin(t) - h(t)
dt = π (α2h2(t) + 2αRoh(t) + Ro2)Rf
where
α=
RT - Ro
HT
(b) If the level h(t) is defined to be state x(t) and a flow meter is located at the
orifice, define the nonlinear system equations
= f(x(t),u(t))
y(t) = g(x(t),u(t))
(c) For constant inflow qin(t) = qIN, prove that the operating level of the tank
satisfies
hp = Rf2qIN2
(d) If the operating inflow qIN = 0.2m3/s, derive a linear state variable model for
the tank system.
Solution P21.1
(a) Conservation of volume
d {Volume}
= inflow-outflow
dt
Define
Volume as V(h(t)
Inflow = qin(t)
h(t)
Outflow = qout(t) = R
f
Thus
dV(h(t)) ∂V dh
h(t)
=
= qin(t) – qout(t) = qin(t) - R
dt
dt
∂h
f
Therefore
dh ∂V -1
h(t)
=
[q
in(t) dt ∂h
Rf ]
=
∂V –1 Rf qin(t) - h(t)
[
]
Rf
∂h
We require an expression for V(h(t)). Consider the following graphical
representation for the sloping side of the tank:
dch21 linear v3 solutions
21.7
The radius formula is Ro + αh where α = (RT – Ro)/HT.
The Volume is given by a volume of revolution integral:
h
2
V(h) = ⌠
⌡πr (x) dx
o
with r(x) = Ro + αx
0≤x≤h
Thus
∂V
= π r2(h)
∂h
= π [Ro + h α]2
The nonlinear equation is
where α = (RT – Ro)/HT.
(b) If x(t) = h(t) , u(t) = qin(t) and y(t0 = qout(t), then the state equation is given by
y(t) =
x0.5(t)
Rf
(c) If qin(t) = qIN, Prove hp = Rf2 qIN2
Set qin(t) = qIN, set
Then (RfqIN – hp0.5 ) = 0 and hp = Rf2 qIN2
(d) For qIN = 0.2 then hp = (2.74 x 0.2)2 =0.30
(i)
Linearised state variable equation
dch21 linear v3 solutions
21.8
At steady state RfqIN – x0.5 = 0, therefore
Thus the linearised state variable equation is
The linearised output equation is
x0.5
Since y(t) = R ,
f
and
P21.2 The magnetic suspension system has a state variable model given by
= x2(t)
u2(t)
= g – a x 2(t) - bx2(t)
1
where the system data gives a = 9 units; b = 9.5 s-1 and g = 9.81 ms-2. The input
for the system was in the range 0.08 to 0.175 A.
(a) Calculate a set of linear models for the input signal changing in steps of 0.02
from 0.08 to 0.18 and examine the change in model parameters over the range.
(b) If a control design is to follow the modelling exercise, which model would
you use and why?
dch21 linear v3 solutions
21.9
Solution P21.2
(a) Steady conditions requires
a
Therefore x2ss = 0 and x1ss = (g )0.5uss
Linearisation
x(t) = xss + δx(t)
u(t) = uss + δu(t)
x1(t) = x1ss + δx1(t)
x2(t) = x2ss + δx2(t)
First equation:
= x2(t)
δ
but x2ss = 0, therefore δ
= x2ss + δx2(t)
= δx2(t)
Second equation:
= f(x1(t), x2(t), u(t))
δ
=
At steady state conditions
dch21 linear v3 solutions
21.10
Summary model
where
2g1.5
-2g
a21 = a0.5u and a22 = -b. Parameter b2 = u .
ss
ss
(b) Model parameters
(i)
a22 is constant
(ii)
2g1.5 1
20.48
a21 ranges over a0.5 u = u
ss
ss
Examples:
uss = 0.08: A =
Poles of system ( eigenvalues of A matrix ( chapter 22) are: 11.9417, -21.4417
uss = 0.175 A =
Poles of system ( eigenvalues of A matrix ( chapter 22) are: 7.0649, -16.5649
Suggest the model at low input values be used since it is more unstable.
dch21 linear v3 solutions
21.11
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