instrumentation and control

1. In the Name of Allah, the Merciful, the Beneficent.
Praise Be To The Lord
Of All Worlds

2. DYNAMIC CHARACTERISTICS
• Figure 2 shows the graph of how the temperature indicated by the
thermometer varies with time. The steady-state value is 55°C and so,
since 95% of 55 is 52.25C, the 95% response time is about 228 s.

3. DYNAMIC CHARACTERISTICS OF MEASUREMENT SYSTEMS
• Many measurement systems can be modelled as linear ordinary differential
equations with constant coefficients. These linear models have the general form:
σ𝑛=0
𝑁
𝐴𝑛
𝑑𝑛𝑋𝑜𝑢𝑡
𝑑𝑡𝑛 = σ𝑚=0
𝑀
𝐵𝑚
𝑑𝑚𝑋𝑖𝑛
𝑑𝑡𝑚
where Xout is the output variable, Xin is the input variable, and An and Bn are
constant coefficients. N defines the order of the system independent of M. Many
electromechanical systems exhibit nonlinear behaviour and cannot be accurately
modelled as linear systems. However, a nonlinear system may often exhibit linear
behaviour over a specific range of inputs and a linear model can be derived that
provides an adequate approximation over this range. This process of modelling a
nonlinear system with a linear model is called linearization.
1

4. ZERO-ORDER MEASUREMENT SYSTEMS
• When N = M = 0 in Equation 1. the model represents a zero-order
measurement system whose behaviour is described by
• where K = B0 /A0 is a constant referred to as the gain or sensitivity of the
system since it represents a scaling between the input and the output.
When the sensitivity is high, a small change in the input can result in a
significant change in the output. Note that the output of a zero-order
system follows the input exactly without time delay or distortion.

5. Example
It is a variable resistance device which produces an output voltage Vout that is directly proportional to the wiper
displacement Xin. This is a result of the voltage division rule, which gives

6. FIRST ORDER MEASUREMENT SYSTEMS
• When N = 1 and M = 0 in Equation 1, the equation models a first order
measurement system whose behaviour is described by
• As with the zero-order system, the coefficient ratio on the right-hand side
is called the sensitivity or static sensitivity. The coefficient ratio on the left
side of the second equation has a special name and meaning. It is called
the time constant and is defined as
 = A1/A0
• With these definitions, the first order system equation can be written as

7. • To characterize how a system responds to various types of inputs, we apply
standard inputs to the model, including step, impulse, and sinusoidal
functions. Consider the effect of applying a step input to our system. A step
input changes instantaneously from zero to a constant value Ain and is
stated mathematically as
• The output of the system in response to this input is called the step
response of the first order system. We can find it by solving the equation
with the initial condition Xout(0) = 0
• Applying the theory of elementary differential equations where the
solution is assumed to be of the form Cet, the characteristic equation for
the homogeneous form of differential equation of first order measurement
sys. is
 + 1 = 0
FIRST ORDER MEASUREMENT SYSTEMS

8. • Since the root of this equation is  = -1/, the homogeneous or transient solution
is
(Xout)h = Ce-t/
where C is a constant determined later by applying initial conditions. A particular
or steady state solution is
(Xout)p = KAin
The general solution is the sum of the homogeneous and particular solutions
Xout(t) = (Xout)h + (Xout)p = Ce-t/ + KAin
Applying the initial condition to this equation gives
0 = C+KAin
thus, C = -KAin
FIRST ORDER MEASUREMENT SYSTEMS

9. so the resulting step response is given by
Xout(t) =KAin.(1 - e-t/ )
As illustrated in Figure this represents an exponential rise in the output toward an
asymptotic value of KAin. The rate of the rise depends only on the time constant . The
smaller the time constant, the faster the response will be. Note that after one time
constant, the output reaches 63.2% of its final value:
Xout(t) = KAin.(1 - e-1 ) = 0.632KAin
After five time constants, the step response is
Xout(t) =KAin.(1 - e-5 ) = 0.993KAin
Since this value is more than 99% of the steady state
value KAin, we usually assume that a first order
system has reached its steady state value within five
time constants.
FIRST ORDER MEASUREMENT SYSTEMS

10. EXPERIMENTAL TESTING OF FIRST ORDER MEASUREMENT SYSTEMS
• When designing a first order measurement system, you should look at
quantities that affect  and reduce them if possible. The larger  is,
the longer the measurement system takes to respond to an input.
• To characterize and evaluate an existing first order measurement
system, we need methods to experimentally determine the time
constant  and the static sensitivity K. K may be obtained by static
calibration where a known static input is implied, and the output is
observed. A common method to determine the time constant  is to
apply a step input to the system and determine the time for the
output to reach 63.2% of its final value.

11. Continued…
• An alternative method to determine a value for  is presented by
rearranging Xout(t) =KAin.(1 - e-t/ ):

t
in
in
out
e
KA
KA
X −
−
=
− 
t
in
out
e
KA
X −
=
−
1
=
If we take the natural logarithm of both sides, we get

t
KA
X
in
out
−
=








−
1
ln
If we define the left-hand side as Z, then

t
Z −
=
and a plot of Z vs. t is a straight line with slope

1
−
=
dt
dZ

12. • Therefore, if we collect experimental data from a step input and plot
Z vs. t as illustrated in figure 5, we can determine  from the slope of
the line:
Z
t


−
=

Note that if experimental data for Z vs.
t deviates from a straight line, then
the system is not first order. In this
case, the system is either higher order
or nonlinear.
Continued…

13. SECOND ORDER MEASUREMENT SYSTEMS
• A good example of a second order
measurement system is the strip
chart recorder.
• The applied force Fext is provided by
an electromagnetic coil whose core
is attached to the pen carriage. The
spring keeps the pen carriage
centred in the zero position when
the input is zero. The pen carriage
bearings and the pen-paper
interface result in the damping
force. The carriage and the pen
constitute the mass to which the
forces in the system are applied.
Applying Newton’s second law of motion to the
free body diagram gives the second order
differential equation, which is the mathematical
model of the system.

14. To characterize the unforced response of a second order system, we
need to solve the differential equation with Fext = 0. The characteristic
equation found by assuming a solution of the form x(t) = Cet, where C
is a constant, is
This quadratic equation has two roots for :
If there were no damping in the system (i.e., b = 0), the roots would be
0
2
=
+
+ k
b
m 

m
k
m
b
m
b
−






+
−
=
2
1
2
2

m
k
m
b
m
b
−






−
−
=
2
2
2
2

m
k
j
=
1

m
k
j
−
=
2


15. • and the corresponding solution would be
• where A and B are constants determined from the initial conditions x(0)
and
• This motion represents pure undamped oscillatory motion with radian
frequency:
• This is called the natural frequency of the system since it is the frequency
at which the undamped system would “naturally” oscillate if the spring
were stretched and the mass released and allowed to move without any
external force (Fext = 0).
( ) 







+








= t
m
k
B
t
m
k
A
t
xh sin
cos
( )
0
dt
dx
m
k
n =


16. • If there is damping in the system (i.e., b  0) and the radicand in the
equation of roots for λ is zero, the roots are double real roots and the
resulting transient homogeneous solution is
• This represents an exponentially decaying motion. A system with this
behaviour is said to be critically damped since it is just on the verge of
damped oscillatory motion. The damping constant that results in
critical damping is called the critical damping constant bc. It is the
value of b that makes the radicand zero, so
( ) ( ) t
h
n
e
Bt
A
t
x 
−
+
=
n
c m
km
b 
2
2 =
=

17. • The damping ratio for a noncritically damped system is defined as
• It is a measure of the proximity to critical damping where =1
• With the definitions of natural frequency and the damping ratio, the
roots of the characteristic equations can be written as
• If there is damping in the system (i.e., b  0) and the radicand is
negative <1, the roots are complex conjugates and the resulting
transient homogeneous solution is

km
b
b
b
c 2
=
=


1
2
1 −
+
−
= 


 n
n
1
2
2 −
−
−
= 


 n
n

( ) ( ) ( )
 
t
B
t
A
e
t
x n
n
t
h
n 2
2
1
sin
1
cos 




−
+
−
= −

18. • This represents an exponentially decaying output. A system with
these characteristics is said to be underdamped since it has less than
critical damping. The frequency of oscillation is
• This is called the damped natural frequency of the system.
• If there is damping in the system (i.e., b  0) and the radicand in
Equations is positive >1, the roots are distinct real roots, and the
resulting transient homogeneous solution is
2
1 

 −
= n
d

( )
t
t
h
n
n
Be
Ae
t
x





 




 −
−
−





 −
+
−
+
=
1
1 2
2
This motion represents damped oscillation consisting of sinusoidal motion with exponentially decaying
amplitude. A system with these characteristics is said to be overdamped since its damping exceeds critical
damping

19. Examples
• Examples of transient responses for all three cases of damping
(underdamped, critically damped, and overdamped) are illustrated in
figure. The curves represent unforced motion of a second order
system with different amounts of damping, when the system is
released from rest x’(0)=0) at x(0) = 1

20. STEP RESPONSE OF SECOND ORDER SYSTEMS
• As we found when analysing a first order system, an important input
used to study the dynamic characteristics of a system is a step
function. The step response consists of two parts: a transient
homogeneous solution xh(t), which is of the form presented in the
previous section for the unforced response, plus a steady state
particular solution xp(t), which is a result of the forcing function. For a
step input given by
• The general solution for the step response is then
( )




=
0
0
0
0 t
F
t
t
Fext

( )
k
F
t
xp
0
=
( ) ( ) ( )
t
x
t
x
t
x p
h +
=
where the constants in xh(t) are determined by applying the initial conditions

21. As with the unforced case, there are three distinctly different types of
response based on the amount of damping in the system, as illustrated in
figure