Deakin University SEE710 Assignment 1: Instrument and Process Control

This document presents a comprehensive solution to Assignment 1 for SEE710 Instrumentation and Process Control, covering key concepts in the field. The assignment explores the differences between active and passive instruments, providing examples and discussing their relative merits. It delves into the advantages and disadvantages of null and deflection type measuring instruments, explaining their applications. The solution clarifies the distinction between accuracy and precision, supported by a practical example involving a thermocouple's sensitivity. Furthermore, it addresses error analysis in voltage measurement using a potentiometer and examines linear regression models and system analysis. The assignment concludes with an exploration of system impulse response and best-fit line calculations. The content offers a solid understanding of instrument types, performance characteristics, and error during measurement.

SEE710 INSTRUMENTATION AND PROCESS CONTROL
By Name
Course
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Date

QUESTION ONE
Q1. The Main Difference between Active and Passive Components
Active components or devices that need external source to their operation, for example
transistor, diode and SCR. Taking example of diode for explanation. When diode is connected in
an electric circuit and then that circuit is connected to the voltage supply then the diode will
conduct current till the voltage supplied reaches 0.7 ( for silicon) and 0.3 ( for germanium).
Passive components or devices are those which do not need any external source for them to
operate. For example, capacitor, resistor and inductor. Taking resistor as example for
explanation. Unlike diode, resistors do not need 0.7 or 0.3. When resistor is connected to voltage
supply, it will start operating automatically as anticipated without any specific voltage.
QUESTION TWO
Null Type method
This is a type of instrument which tries to maintain a deflection at zero through appropriate
application of known impact opposing the produced deflection through measured quality. The
measured quantity generates some mechanical impacts which engenders equal but having a
reverse effect in the instrument part.
Advantages Null Type method of instrument
Such instruments are highly accurate
Null type instruments are readily used for measuring
Disadvantages of using null type instrument

Because of addition dissimilar weights used this type of instrument is difficult.
Deflection type method
 These are instruments where pointer system is employed to illustrate the output of the
system. The accuracy of this instrument is dependent on the calibration linearity of the
spring.
Advantages of deflection type of instrument
 In this type of instrument it is very easy read the results of deflection.
 Because of convenient use of these instruments, they are readily used for calibrations.
Disadvantages of deflection instruments
These instruments have less accuracy.
Some types of null instruments used include the following;
i. Wheatstone bridge
This is a null type electrical instrument which is employed for measuring an unknown resistance
in electrical circuit via balancing 2 legs of bridge circuit where one of the legs contains
unknown. Wheatstone bridge is illustrated using the following diagram;

Figure 1: Showing Wheatstone bridge
ii. Potentiometer
This is an instrument where a slide wire is calibrated in emf, the calibration is done through the
use of emf standard source. This type of instrument is employed for measuring the electro
motive force of particular cell. This instrument also aids in comparing the electro motive force
for different cells. The instrument can also be employed in varying the resistor in several
application like a volume knob of radio receiver. This instrument can be illustrated using the
following diagram;
Figure 2: showing potentiometer

QUESTION THREE
Part I
Accuracy means how close the measured value is close to the expected actual value. While the
precision means how close 2 or more measurement are to each other regardless of the anticipated
actual value. In other words accuracy is the degree of closeness to the expected result while
precision is degree at which a process of instrument will repeat the same recorded value after
experiment. The diagram below can further illustrates the difference between the two;
Figure 3: Showing diagram which gives the difference between the precision and accuracy
Part ii

Given at 2500C = 4.37mV
5000C = 8.74 mV
Therefore sensitivity of measurement can be obtained using equation 1 below;
SoM = ∆ voltage
∆ temperature
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Where SoM is sensitivity of measurement, ∆ voltage is change in voltage and ∆ temperatureis
change in temperature.
SoM = 8.74−4.37
500−250
SoM = 4.37
250
SoM = 0.01748mV/0C
QUESTION FOUR

Measuring voltage should be E Ri
R 1
But actual measuring voltage = E ⌊ Ri∨¿ Rm
( R 1−Ri ) + Ri∨¿ Rm ¿ ¿ ⌋
= E.
RiRm
R 1+ Rm
( R 1−Ri ) + RiRm
Ri+ Rm
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . 2
=E. RiRm
( R 1−Ri ) ( Ri+ Rm ) +RiRm
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Measurement error = E. Ri
R 1 − E . R 1 Pm
( R 1−Ri ) ( Ri+ Rm ) + RiRm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . 4
= E.Ri ⌊ Rm
R 1 Ri + R 1 Rm−R 12−RiRm+ RiRm ⌋
= E.Ri ⌊ R 1 Ri + R 1 Rm−Ri2−R 1 Rm
R 1(R 1 Ri+ R 1 Rm−Ri2 ) ⌋
= E.Ri R 1 Ri−Ri2 ¿ ¿
R 1(R 1 Ri+ R 1 Rm−Ri2 )
= E . Ri2 ( R 1−Ri)
R 1( R 1 Ri+R 1 Rm−Ri2) (proved)

From error == E . Ri2 ( R 1−Ri)
R 1(R 1 Ri+ R 1 Rm−Ri2)
Actually Rm>> Ri
Therefore error can be rewritten as
Error = E . Ri2 ( R 1−Ri)
R 1(R 1 Rm−Ri2)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
And also RiRm>> R12, therefore
Error = = E . Ri2(R 1−Ri)
Rm Ri2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
= E ¿ ¿ ¿
For the result to be maximum, derror
dri =0
R1.2Ri – 3Ri2 =0
= 2R1- 3Ri =0
Ri = 2
3 R 1
σXi
A = 2
3
σX 1
A
Xi= 2
3 X 1 (Proved)

QUESTION FIVE
Part 1
Linear regression ( y= Hθ+W ¿
x(k+1) = Ax(k) +B.u(k)
y(k)= C.x (k) +v(k) Noise
x( j)J =K 1 where 1 ≤ k 1≤ N
y(k )K=1,2.. .. N
x(j) = A j−k 1 x ( s ) + ∑
P= K 1+1
j
A j−1 B . u(i−1); N≥ j ≥ k 1+1
x(j) = A j−k 1 x ( k 1 ) − ∑
i= j +1
k 1
A j−1 B . u(i−1)
y(1), y(2) . . . . . . . . . . . . . . . . . y( N-1), y(0)
θ=x (k 1)
Part 1
The linear regression model
y(1)= C.x (1)+V(1)
y(2) = C.x (2)+ V(2)
.

.
.
y(N) C. x (N) + V (N)
⌊
y 1
y 2
yn
⌋ = ⌈
1 x 11 x 12 … … … ..
1 x 21 x 22 … … … .
1 xn 1 xn 2… …
x 1k
x 2k
xnk
⌉ ⌊
C 0
C 1
Ck
⌋+
[v 1
v 2
vn ]
Part ii
Minimum mean square error
E⌊ ¿ = E ⌊ x−9 ¿
QUESTION SIX
Part 1
From the diagram
u(n) = input
G(z) = system function
v(n)= Noise
y(n)= Output

For N input and output ( ui, yi) and the line best fit is given by the following equation
Yi= a+Bui
Best fit can be minimized by using the following cost function equation;
S = ∑
i=1
n
¿¿
Introducing line best fit into cost function
S = ∑
i=1
n
¿¿
Obtaining a from partial derivative
∂ S
∂ a [ ∑
i=1
n
( y −a−Bui ) 2
]
Through using chain rule having exponent and equation between parentheses
0= ∑
i=1
n
−2( yi−a−Bui)
Pulling out −2 then dividing each equation by −2 we obtain the folliwng
0= ∑
i=1
n
( yi−a−Bui)
Dividing the equation into 3 different parts we obtain the following;
0= ∑
i=1
n
yi−∑
i=1
n
a−B ∑
i=1
n
u i ¿ ¿. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
From equation 1 above we realized that the summation of a to n is

∑
i=1
n
a=na . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2
When equation 2 above is substituted back and B rearranged then we obtain
0= ∑
i=1
n
( yi−an−B∑
i=1
n
u i)
To obtain a we have to add na and divide by n as shown in the equation below;
a= ∑
i=1
n
( yi−B ∑
i=1
n
u i )
n
a= y – Bui
Obtaining the value of B
∂ S
∂ b =⌊ ∑
i=1
n
¿ ¿
0= ∑
i=1
n
−2ui( yi−a−Bui)
0= ∑
i=1
n
(uiji−aui−Bu i2 )
a= y−B u
0= ∑
i=1
n
(uiji− y ui )−B∑
i=1
n
x i2−¿ x xi¿