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Chapter Five
Introduction to Analogue Computing Techniques
Topics to be covered
Introduction
Components of the analog computer
Simulation of Linear Systems
Magnitude and Time Scaling
Cont..
Objective
The objective of learning this topic is realize the differential equation
using (analog computer) operational amplifier.
5.1. Introduction
 Before digital computers became so universal, analog computers were
popular for solving problems such as differential equations.
 The basic building block of the analog computer is the operational
amplifier.
 In addition to this, analog computers consists of resistor, capacitor and
inductor.
 The ideal op-amp has the following properties:
1. The voltage between the +ve and the –ve terminals is zero. That is 𝑒+
= 𝑒−
. This property is
called the virtual ground or the virtual short.
2. The input impedance is infinite. This implies that, the currents into the +ve and the –ve input
terminals is zero.
3. The output impedance of an ideal op-amp is very low i.e., 𝑧𝑜𝑢𝑡 = (50 𝑢𝑝 𝑡𝑜 75)Ω.
5.2. Components of The Analog Computer
i. Inverting Amplifier:
𝑖1 = 𝑖2
⟹
0−𝑣𝑖
𝑅1
=
𝑣𝑜−0
𝑅2
⟹
−𝑣𝑖
𝑅1
=
𝑣𝑜
𝑅2
⟹
𝑣𝑜
𝑣𝑖
= −
𝑅2
𝑅1
• If 𝑅1 = 𝑅2, then
𝑣𝑜 = −𝑣𝑖
Cont’d…
ii. Non-inverting Amplifier:
⟹
𝑣𝑜−𝑣𝑖
𝑅2
=
𝑣𝑖
𝑅1
⟹
𝑣𝑜
𝑅2
=
𝑣𝑖
𝑅2
+
𝑣𝑖
𝑅1
⟹
𝑣𝑜
𝑅2
=
𝑅1+𝑅2
𝑅1𝑅2
𝑣𝑖
⟹
𝑣𝑜
𝑣𝑖
=
𝑅1+𝑅2
𝑅1
• If 𝑅2 = 0, then
𝑣𝑜 = 𝑣𝑖
Cont’d…
iii. Summer Amplifier:
𝑖4 = 𝑖1 + 𝑖2 + 𝑖3
⟹
𝑣𝑜
𝑅4
= −
𝑣1
𝑅1
−
𝑣2
𝑅2
−
𝑣3
𝑅3
• If 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4, then
𝑣𝑜 = −(𝑣1 + 𝑣2 + 𝑣3)
Cont’d…
𝑖4 = 𝑖1 + 𝑖2 + 𝑖3
⟹
𝑣𝑜
𝑅
= −
𝑣1
𝑅
𝑎
−
𝑣2
𝑅
𝑏
−
𝑣3
𝑅
𝑐
⟹ 𝑣𝑜 = −𝑎𝑣1 − 𝑏𝑣2 − 𝑐𝑣3
Cont’d…
iv. Integrator:
𝑖1 = 𝑖2
⟹ −
𝑣𝑖𝑛
𝑅
= C
𝑑𝑣𝑜
𝑑𝑡
⟹ −
1
𝑅𝐶
𝑣𝑖𝑛 =
𝑑𝑣𝑜
𝑑𝑡
⟹ 𝑣𝑜 = −
1
𝑅𝐶
𝑣𝑖𝑛𝑑𝑡
Cont’d…
v. Differentiator:
𝑖1 = 𝑖2
⟹ −𝐶
𝑑𝑣𝑖𝑛
𝑑𝑡
=
𝑣𝑜
𝑅
⟹ 𝑣𝑜 = −𝑅𝐶
𝑑𝑣𝑖𝑛
𝑑𝑡
5.3. Simulation of Linear Systems
 The procedures for obtaining an operational amplifier realization of the linear
systems are:
1. Find the mathematical model of the system.
2. Represent the mathematical model in the analog block diagram.
• Some of the analog components symbols;
3. From the analog block diagram, construct the circuit diagram by using op-amps, resistors
and capacitors.
4. Operate the computer and observes the output variables.
Cont’d…
Example: Realize the following dynamic equations by using analog components.
i) 𝑦 = −K𝑥 where K > 0
• We will use an inverting amplifier.
ii) 𝑦 = K𝑥
Cont’d…
• We will use two inverting amplifiers in series.
iii) 𝑦 = 𝑥1 + 𝑥2 + 𝑥3
Cont’d…
• We will use a summer and an inverting amplifier in series.
iv) 𝑥 = 𝑥
Cont’d…
• It uses an integrator in series with an inverter.
where RC = 1
v) 𝑣𝑜 = −𝑣1 − 2𝑣2 + 3𝑣3 + 4𝑣4
From this
𝑣𝑜 = −[𝑣1 + 2𝑣2 − 3𝑣3 − 4𝑣4]
Cont’d…
 The analog block diagram is:
Cont’d…
 The operational amplifier realization will be:
Cont’d…
vi)
𝑥1
𝑥2
=
2 −2
1 3
𝑥1
𝑥2
+
0
1
𝑢
𝑦 = −1 1
𝑥1
𝑥2
+ 4𝑢
From this
𝑥1 = 2𝑥1 − 2𝑥2
𝑥2 = 𝑥1 + 3𝑥2 + 𝑢
𝑦 = −𝑥1 + 𝑥2 + 4𝑢
= −[𝑥1 − 𝑥2 − 4𝑢]
Cont’d…
Cont’d…
Exercise:
1. Find the op-amp realization and analog block diagram for the following.
𝑥 =
1 −1 1
1 −1 1
2 2 1
𝑥 +
0
0
1
𝑢
𝑦 = 1 1 2 𝑥 − 4𝑢
2. Develop a state variable and output equation for the operational amplifier circuit
shown in below
5.4. Magnitude and Time Scaling
Magnitude Scaling:-
It is often the case that the solution to a system differential equation in real time
varies too fast or too slowly to be captured by CRT or plotter. Examples of such
instance are chemical reaction and evaluation of state where time scaling may be
necessary.
Almost any problem where magnitude variation is either too small or too large
needs magnitude scaling.
Consider the equation of motion
𝑀𝑥 + 𝐷𝑥 + 𝐾𝑥 = 𝑓(𝑡)
𝑥 =
1
𝑀
𝑓 𝑡 −
𝐷
𝑀
𝑥 −
𝐾
𝑀
𝑥 (1)
By using an estimation or some other means, let us put a bounds on the variables as
the following
Cont’d…
𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟)
𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐)
𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐2)
𝑓 ≤ 𝑓𝑚
• And also, let the maximum voltage on the computer be 𝑉
𝑚 = 10V. By using
this maximum voltage, the following scale factors are introduced.
𝐾𝑥 =
10
𝑥𝑚
𝑣 𝑚, 𝑣 𝑚 means 𝑣𝑜𝑙𝑡 𝑚𝑒𝑡𝑒𝑟
𝐾𝑥 =
10
𝑥𝑚
𝑣 𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐
Cont’d…
𝐾𝑥 =
10
𝑥𝑚
𝑣 𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐2
𝐾𝑓 =
10
𝑓𝑚
𝑣 𝑁𝑒𝑤𝑡𝑜𝑛
• Then, Equation (1) can be written as the following;
[𝐾𝑥𝑥]
𝐾𝑥
=
[𝐾𝑓𝑓]
𝑀𝐾𝑓
−
𝐷[𝐾𝑥𝑥]
𝑀𝐾𝑥
−
𝐾[𝐾𝑥𝑥]
𝑀𝐾𝑥
[𝐾𝑥𝑥] =
𝐾𝑥
𝑀𝐾𝑓
[𝐾𝑓𝑓] −
𝐷𝐾𝑥
𝑀𝐾𝑥
[𝐾𝑥𝑥] −
𝐾𝐾𝑥
𝑀𝐾𝑥
[𝐾𝑥𝑥]
• The main point is that the terms in the square bracket that are called machine
variables, are limited with in ±10𝑉.
Cont’d…
Figure: Analog block diagram of the above system
Cont’d…
Time Scaling:
Consider the following equation of the motion
𝑥(𝑡) + 𝑎𝑥 𝑡 + 𝑏𝑥 𝑡 = 𝑓(𝑡) (2)
• Time scaling can be introduced by using change of variables.
• Let
𝜏 = 𝛼𝑡
where
0 < 𝛼 < 1 ⇝corresponding for speeding up the system operation time.
1 < 𝛼 < ∞ ⇝corresponding for slow down the system operation time.
Then
𝑑𝜏 = 𝛼𝑑𝑡 ⟹ 𝑑𝑡 = 1
𝛼𝑑𝜏
Cont’d…
𝑥 𝑡 = 𝑥(𝜏 𝛼)
𝑑𝑥(𝑡)
𝑑𝑡
=
𝑑𝑥(𝜏 𝛼)
1
𝛼
𝑑𝜏
= 𝛼
𝑑𝑥(𝜏 𝛼)
𝑑𝜏
𝑑2𝑥(𝑡)
𝑑𝑡2 =
𝑑
𝑑𝑡
𝑑𝑥(𝑡)
𝑑𝑡
=
𝑑
1
𝛼
𝑑𝜏
𝛼
𝑑𝑥(𝜏 𝛼)
𝑑𝜏
= 𝛼2 𝑑2𝑥(𝜏 𝛼)
𝑑𝜏2
Cont’d…
𝑑3𝑥(𝑡)
𝑑𝑡3 =
𝑑
𝑑𝑡
𝑑2𝑥(𝑡)
𝑑𝑡2
=
𝑑
1
𝛼
𝑑𝜏
𝛼2
𝑑2𝑥(𝜏 𝛼)
𝑑𝜏2
= 𝛼3 𝑑3𝑥(𝜏 𝛼)
𝑑𝜏3
Then, Equation (2) will be
𝛼2
𝑑2
𝑥(𝜏 𝛼)
𝑑𝜏2
+ 𝑎𝛼
𝑑𝑥(𝜏 𝛼)
𝑑𝜏
+ 𝑏𝑥 𝜏 𝛼 = 𝑓(𝜏 𝛼)
Cont’d…
𝛼2 𝑑2𝑥(𝜏 𝛼)
𝑑𝜏2 = −𝑎𝛼
𝑑𝑥 𝜏 𝛼
𝑑𝜏
− 𝑏𝑥 𝜏 𝛼 + 𝑓(𝜏 𝛼)
𝑑2𝑥(𝜏 𝛼)
𝑑𝜏2 =
1
𝛼2 𝑓 𝜏 𝛼 −
𝑎
𝛼
𝑑𝑥 𝜏 𝛼
𝑑𝜏
−
𝑏
𝛼2 𝑥 𝜏 𝛼 (3)
• Analog block diagram of Equation (3) becomes
Cont’d…
Example: Consider the following equation of motion
𝑥 =
1
100
𝑓 𝑡 −
1
100
𝑥 −
1
100
𝑥
• If the bounds on each elements (states) and the maximum operational voltage
range of the system are given by below, find the scale factors, the machine
variables and the analog and op-amp realization of the magnitude scaled system
dynamic model.
𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 100𝑚
𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 10𝑚/𝑠
𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 2𝑚/𝑠2
𝑓 ≤ 𝑓𝑚 , 𝑓𝑚 = 100𝑁𝑒𝑤𝑡𝑜𝑛
𝑣 ≤ 𝑣𝑚 , 𝑣𝑚 = 10V
Cont’d…
Solution:
• The scaling factors are:
𝐾𝑥 =
𝑣𝑚
𝑥𝑚
=
10𝑣
100𝑚
= 0.1 𝑣
𝑚
𝐾𝑥 =
𝑣𝑚
𝑥𝑚
=
10𝑣
10𝑚/𝑠
= 1 𝑣
𝑚/𝑠𝑒𝑐
𝐾𝑥 =
𝑣𝑚
𝑥𝑚
=
10𝑣
2𝑚/𝑠2 = 5 𝑣
𝑚/𝑠2
𝐾𝑓 =
𝑣𝑚
𝑓𝑚
=
10𝑣
100𝑁
= 0.1 𝑣 𝑁
• Then, the scaled dynamic model will be
[𝐾𝑥𝑥]
𝐾𝑥
=
[𝐾𝑓𝑓]
100𝐾𝑓
−
[𝐾𝑥𝑥]
100𝐾𝑥
−
[𝐾𝑥𝑥]
100𝐾𝑥
Cont’d…
[𝐾𝑥𝑥] =
𝐾𝑥
100𝐾𝑓
[𝐾𝑓𝑓] −
𝐾𝑥
100𝐾𝑥
[𝐾𝑥𝑥] −
𝐾𝑥
100𝐾𝑥
[𝐾𝑥𝑥]
[𝐾𝑥𝑥] =
5
100 ∗ 0.1
[𝐾𝑓𝑓] −
5
100 ∗ 1
[𝐾𝑥𝑥] −
5
100 ∗ 0.1
[𝐾𝑥𝑥]
[𝐾𝑥𝑥] =
5
10
[𝐾𝑓𝑓] −
5
100
[𝐾𝑥𝑥] −
5
10
[𝐾𝑥𝑥]
𝐾𝑥𝑥 = 0.5 𝐾𝑓𝑓 − 0.05 𝐾𝑥𝑥 − 0.5 𝐾𝑥𝑥
[5𝑥] = 0.5[0.1𝑓] − 0.05[𝑥] − 0.5[0.1𝑥]
Cont’d…
Cont’d…

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control_5.pptx

  • 1. Chapter Five Introduction to Analogue Computing Techniques Topics to be covered Introduction Components of the analog computer Simulation of Linear Systems Magnitude and Time Scaling
  • 2. Cont.. Objective The objective of learning this topic is realize the differential equation using (analog computer) operational amplifier.
  • 3. 5.1. Introduction  Before digital computers became so universal, analog computers were popular for solving problems such as differential equations.  The basic building block of the analog computer is the operational amplifier.  In addition to this, analog computers consists of resistor, capacitor and inductor.  The ideal op-amp has the following properties: 1. The voltage between the +ve and the –ve terminals is zero. That is 𝑒+ = 𝑒− . This property is called the virtual ground or the virtual short. 2. The input impedance is infinite. This implies that, the currents into the +ve and the –ve input terminals is zero. 3. The output impedance of an ideal op-amp is very low i.e., 𝑧𝑜𝑢𝑡 = (50 𝑢𝑝 𝑡𝑜 75)Ω.
  • 4. 5.2. Components of The Analog Computer i. Inverting Amplifier: 𝑖1 = 𝑖2 ⟹ 0−𝑣𝑖 𝑅1 = 𝑣𝑜−0 𝑅2 ⟹ −𝑣𝑖 𝑅1 = 𝑣𝑜 𝑅2 ⟹ 𝑣𝑜 𝑣𝑖 = − 𝑅2 𝑅1 • If 𝑅1 = 𝑅2, then 𝑣𝑜 = −𝑣𝑖
  • 6. Cont’d… iii. Summer Amplifier: 𝑖4 = 𝑖1 + 𝑖2 + 𝑖3 ⟹ 𝑣𝑜 𝑅4 = − 𝑣1 𝑅1 − 𝑣2 𝑅2 − 𝑣3 𝑅3 • If 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4, then 𝑣𝑜 = −(𝑣1 + 𝑣2 + 𝑣3)
  • 7. Cont’d… 𝑖4 = 𝑖1 + 𝑖2 + 𝑖3 ⟹ 𝑣𝑜 𝑅 = − 𝑣1 𝑅 𝑎 − 𝑣2 𝑅 𝑏 − 𝑣3 𝑅 𝑐 ⟹ 𝑣𝑜 = −𝑎𝑣1 − 𝑏𝑣2 − 𝑐𝑣3
  • 8. Cont’d… iv. Integrator: 𝑖1 = 𝑖2 ⟹ − 𝑣𝑖𝑛 𝑅 = C 𝑑𝑣𝑜 𝑑𝑡 ⟹ − 1 𝑅𝐶 𝑣𝑖𝑛 = 𝑑𝑣𝑜 𝑑𝑡 ⟹ 𝑣𝑜 = − 1 𝑅𝐶 𝑣𝑖𝑛𝑑𝑡
  • 9. Cont’d… v. Differentiator: 𝑖1 = 𝑖2 ⟹ −𝐶 𝑑𝑣𝑖𝑛 𝑑𝑡 = 𝑣𝑜 𝑅 ⟹ 𝑣𝑜 = −𝑅𝐶 𝑑𝑣𝑖𝑛 𝑑𝑡
  • 10. 5.3. Simulation of Linear Systems  The procedures for obtaining an operational amplifier realization of the linear systems are: 1. Find the mathematical model of the system. 2. Represent the mathematical model in the analog block diagram. • Some of the analog components symbols; 3. From the analog block diagram, construct the circuit diagram by using op-amps, resistors and capacitors. 4. Operate the computer and observes the output variables.
  • 11. Cont’d… Example: Realize the following dynamic equations by using analog components. i) 𝑦 = −K𝑥 where K > 0 • We will use an inverting amplifier. ii) 𝑦 = K𝑥
  • 12. Cont’d… • We will use two inverting amplifiers in series. iii) 𝑦 = 𝑥1 + 𝑥2 + 𝑥3
  • 13. Cont’d… • We will use a summer and an inverting amplifier in series. iv) 𝑥 = 𝑥
  • 14. Cont’d… • It uses an integrator in series with an inverter. where RC = 1 v) 𝑣𝑜 = −𝑣1 − 2𝑣2 + 3𝑣3 + 4𝑣4 From this 𝑣𝑜 = −[𝑣1 + 2𝑣2 − 3𝑣3 − 4𝑣4]
  • 15. Cont’d…  The analog block diagram is:
  • 16. Cont’d…  The operational amplifier realization will be:
  • 17. Cont’d… vi) 𝑥1 𝑥2 = 2 −2 1 3 𝑥1 𝑥2 + 0 1 𝑢 𝑦 = −1 1 𝑥1 𝑥2 + 4𝑢 From this 𝑥1 = 2𝑥1 − 2𝑥2 𝑥2 = 𝑥1 + 3𝑥2 + 𝑢 𝑦 = −𝑥1 + 𝑥2 + 4𝑢 = −[𝑥1 − 𝑥2 − 4𝑢]
  • 20. Exercise: 1. Find the op-amp realization and analog block diagram for the following. 𝑥 = 1 −1 1 1 −1 1 2 2 1 𝑥 + 0 0 1 𝑢 𝑦 = 1 1 2 𝑥 − 4𝑢 2. Develop a state variable and output equation for the operational amplifier circuit shown in below
  • 21. 5.4. Magnitude and Time Scaling Magnitude Scaling:- It is often the case that the solution to a system differential equation in real time varies too fast or too slowly to be captured by CRT or plotter. Examples of such instance are chemical reaction and evaluation of state where time scaling may be necessary. Almost any problem where magnitude variation is either too small or too large needs magnitude scaling. Consider the equation of motion 𝑀𝑥 + 𝐷𝑥 + 𝐾𝑥 = 𝑓(𝑡) 𝑥 = 1 𝑀 𝑓 𝑡 − 𝐷 𝑀 𝑥 − 𝐾 𝑀 𝑥 (1) By using an estimation or some other means, let us put a bounds on the variables as the following
  • 22. Cont’d… 𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟) 𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐) 𝑥 ≤ 𝑥𝑚 (𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐2) 𝑓 ≤ 𝑓𝑚 • And also, let the maximum voltage on the computer be 𝑉 𝑚 = 10V. By using this maximum voltage, the following scale factors are introduced. 𝐾𝑥 = 10 𝑥𝑚 𝑣 𝑚, 𝑣 𝑚 means 𝑣𝑜𝑙𝑡 𝑚𝑒𝑡𝑒𝑟 𝐾𝑥 = 10 𝑥𝑚 𝑣 𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐
  • 23. Cont’d… 𝐾𝑥 = 10 𝑥𝑚 𝑣 𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐2 𝐾𝑓 = 10 𝑓𝑚 𝑣 𝑁𝑒𝑤𝑡𝑜𝑛 • Then, Equation (1) can be written as the following; [𝐾𝑥𝑥] 𝐾𝑥 = [𝐾𝑓𝑓] 𝑀𝐾𝑓 − 𝐷[𝐾𝑥𝑥] 𝑀𝐾𝑥 − 𝐾[𝐾𝑥𝑥] 𝑀𝐾𝑥 [𝐾𝑥𝑥] = 𝐾𝑥 𝑀𝐾𝑓 [𝐾𝑓𝑓] − 𝐷𝐾𝑥 𝑀𝐾𝑥 [𝐾𝑥𝑥] − 𝐾𝐾𝑥 𝑀𝐾𝑥 [𝐾𝑥𝑥] • The main point is that the terms in the square bracket that are called machine variables, are limited with in ±10𝑉.
  • 24. Cont’d… Figure: Analog block diagram of the above system
  • 25. Cont’d… Time Scaling: Consider the following equation of the motion 𝑥(𝑡) + 𝑎𝑥 𝑡 + 𝑏𝑥 𝑡 = 𝑓(𝑡) (2) • Time scaling can be introduced by using change of variables. • Let 𝜏 = 𝛼𝑡 where 0 < 𝛼 < 1 ⇝corresponding for speeding up the system operation time. 1 < 𝛼 < ∞ ⇝corresponding for slow down the system operation time. Then 𝑑𝜏 = 𝛼𝑑𝑡 ⟹ 𝑑𝑡 = 1 𝛼𝑑𝜏
  • 26. Cont’d… 𝑥 𝑡 = 𝑥(𝜏 𝛼) 𝑑𝑥(𝑡) 𝑑𝑡 = 𝑑𝑥(𝜏 𝛼) 1 𝛼 𝑑𝜏 = 𝛼 𝑑𝑥(𝜏 𝛼) 𝑑𝜏 𝑑2𝑥(𝑡) 𝑑𝑡2 = 𝑑 𝑑𝑡 𝑑𝑥(𝑡) 𝑑𝑡 = 𝑑 1 𝛼 𝑑𝜏 𝛼 𝑑𝑥(𝜏 𝛼) 𝑑𝜏 = 𝛼2 𝑑2𝑥(𝜏 𝛼) 𝑑𝜏2
  • 27. Cont’d… 𝑑3𝑥(𝑡) 𝑑𝑡3 = 𝑑 𝑑𝑡 𝑑2𝑥(𝑡) 𝑑𝑡2 = 𝑑 1 𝛼 𝑑𝜏 𝛼2 𝑑2𝑥(𝜏 𝛼) 𝑑𝜏2 = 𝛼3 𝑑3𝑥(𝜏 𝛼) 𝑑𝜏3 Then, Equation (2) will be 𝛼2 𝑑2 𝑥(𝜏 𝛼) 𝑑𝜏2 + 𝑎𝛼 𝑑𝑥(𝜏 𝛼) 𝑑𝜏 + 𝑏𝑥 𝜏 𝛼 = 𝑓(𝜏 𝛼)
  • 28. Cont’d… 𝛼2 𝑑2𝑥(𝜏 𝛼) 𝑑𝜏2 = −𝑎𝛼 𝑑𝑥 𝜏 𝛼 𝑑𝜏 − 𝑏𝑥 𝜏 𝛼 + 𝑓(𝜏 𝛼) 𝑑2𝑥(𝜏 𝛼) 𝑑𝜏2 = 1 𝛼2 𝑓 𝜏 𝛼 − 𝑎 𝛼 𝑑𝑥 𝜏 𝛼 𝑑𝜏 − 𝑏 𝛼2 𝑥 𝜏 𝛼 (3) • Analog block diagram of Equation (3) becomes
  • 29. Cont’d… Example: Consider the following equation of motion 𝑥 = 1 100 𝑓 𝑡 − 1 100 𝑥 − 1 100 𝑥 • If the bounds on each elements (states) and the maximum operational voltage range of the system are given by below, find the scale factors, the machine variables and the analog and op-amp realization of the magnitude scaled system dynamic model. 𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 100𝑚 𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 10𝑚/𝑠 𝑥 ≤ 𝑥𝑚 , 𝑥𝑚 = 2𝑚/𝑠2 𝑓 ≤ 𝑓𝑚 , 𝑓𝑚 = 100𝑁𝑒𝑤𝑡𝑜𝑛 𝑣 ≤ 𝑣𝑚 , 𝑣𝑚 = 10V
  • 30. Cont’d… Solution: • The scaling factors are: 𝐾𝑥 = 𝑣𝑚 𝑥𝑚 = 10𝑣 100𝑚 = 0.1 𝑣 𝑚 𝐾𝑥 = 𝑣𝑚 𝑥𝑚 = 10𝑣 10𝑚/𝑠 = 1 𝑣 𝑚/𝑠𝑒𝑐 𝐾𝑥 = 𝑣𝑚 𝑥𝑚 = 10𝑣 2𝑚/𝑠2 = 5 𝑣 𝑚/𝑠2 𝐾𝑓 = 𝑣𝑚 𝑓𝑚 = 10𝑣 100𝑁 = 0.1 𝑣 𝑁 • Then, the scaled dynamic model will be [𝐾𝑥𝑥] 𝐾𝑥 = [𝐾𝑓𝑓] 100𝐾𝑓 − [𝐾𝑥𝑥] 100𝐾𝑥 − [𝐾𝑥𝑥] 100𝐾𝑥
  • 31. Cont’d… [𝐾𝑥𝑥] = 𝐾𝑥 100𝐾𝑓 [𝐾𝑓𝑓] − 𝐾𝑥 100𝐾𝑥 [𝐾𝑥𝑥] − 𝐾𝑥 100𝐾𝑥 [𝐾𝑥𝑥] [𝐾𝑥𝑥] = 5 100 ∗ 0.1 [𝐾𝑓𝑓] − 5 100 ∗ 1 [𝐾𝑥𝑥] − 5 100 ∗ 0.1 [𝐾𝑥𝑥] [𝐾𝑥𝑥] = 5 10 [𝐾𝑓𝑓] − 5 100 [𝐾𝑥𝑥] − 5 10 [𝐾𝑥𝑥] 𝐾𝑥𝑥 = 0.5 𝐾𝑓𝑓 − 0.05 𝐾𝑥𝑥 − 0.5 𝐾𝑥𝑥 [5𝑥] = 0.5[0.1𝑓] − 0.05[𝑥] − 0.5[0.1𝑥]