04 Transfer Function.pptx

1. AE 313
AE Systems & Control
04 SYSTEM MODELLING
Dr. Syed Saad Azhar Ali
75-110
syed.ali@kfupm.edu.sa

2. Catching up…
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• Trying to learn about control systems
– An interconnection of components in a configuration that helps us achieve our desired output
– The interconnecting configuration can be open loop or closed loop
• In this interconnection, there is
– The process/system/plant to be controlled
– The actuator
– The sensor
– Feedback
– Comparator
– Controller
• Once the overall scheme (control and controlled variables, reference value, specifications,
objectives) is finalized, the most important step is to know the system
• Need physical laws and respective mathematical representation - > differential equations
• solving the differential equation will take time and effort
– Time domain is easier to comprehend for us but not easy to solve
– Convert to Frequency Domain -> Laplace Transform

3. Laplace Transforms

4. Laplace Transform Properties

5. Mechanical System
Spring-Mass Damper
• Displacement of the mass M (y)
• The mass is suspended with
spring (spring constant k)
• moves inside the walls with
friction (coefficient b)
• r(t) is the applied force that
– Moves -> Ma
– Takes care of spring -> ky
– Overcomes the friction -> bv
Ma bv ky
𝒓 𝒕 = 𝑴𝒂 + 𝒃𝒗 + 𝒌𝒚
𝒓 𝒕 = 𝑴
𝒅𝟐𝒚
𝒅𝒕𝟐
+ 𝒃
𝒅𝒚
𝒅𝒕
+ 𝒌𝒚 This is the differential equation
for the spring-mass damper
𝒓 𝒕 = 𝑴𝒂 + 𝒃𝒗 + 𝒌𝒚
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6. Mechanical System
Spring-Mass Damper
Ma bv ky
𝒓 𝒕 = 𝑴
𝒅𝟐𝒚
𝒅𝒕𝟐 + 𝒃
𝒅𝒚
𝒅𝒕
+ 𝒌𝒚
The differential equation for the
spring-mass damper
Taking Laplace Transform
𝓛 𝒓 𝒕 = 𝓛 𝑴
𝒅𝟐𝒚
𝒅𝒕𝟐 + 𝒃
𝒅𝒚
𝒅𝒕
+ 𝒌𝒚
𝓛 𝒓 𝒕 = 𝓛 𝑴
𝒅𝟐𝒚
𝒅𝒕𝟐
+ 𝓛 𝒃
𝒅𝒚
𝒅𝒕
+ 𝓛 𝒌𝒚
𝓛 𝒓 𝒕 = 𝑴𝓛
𝒅𝟐𝒚
𝒅𝒕𝟐 + 𝒃𝓛
𝒅𝒚
𝒅𝒕
+ 𝒌𝓛 𝒚
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7. Mechanical System
Spring-Mass Damper
Ma bv ky
Using the Laplace of derivative of a function
𝓛 𝒓 𝒕 = 𝑴𝓛
𝒅𝟐
𝒚
𝒅𝒕𝟐
+ 𝒃𝓛
𝒅𝒚
𝒅𝒕
+ 𝒌𝓛 𝒚
𝓛
𝓛−𝟏
⇌
𝑹 𝒔 = 𝑴𝒔𝟐
𝒀 𝒔 − 𝒔𝟏
𝒚 𝟎 − 𝒚′ 𝟎 +𝒃𝒔𝒀 𝒔 − 𝒚 𝟎 +𝒌𝒀 𝒔
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8. Mechanical System
Spring-Mass Damper
Ma bv ky
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𝑹 𝒔 = 𝑴𝒔𝟐𝒀 𝒔 − 𝒔𝒚 𝟎 − 𝒚′ 𝟎 + 𝒃𝒔𝒀 𝒔 − 𝒚 𝟎 + 𝒌𝒀 𝒔
• Now assuming the system is initially at
rest or relaxed
• i.e. no displacement, velocity or
acceleration
• This means ZERO INITIAL CONDITIONS
• Zero displacement 𝒚 𝟎 = 𝟎
• Zero velocity 𝒚′ 𝟎 = 𝟎
= 𝟎
= 𝟎
= 𝟎
𝑹 𝒔 = 𝑴𝒔𝟐
𝒀 𝒔 + 𝒃𝒔𝒀 𝒔 + 𝒌𝒀 𝒔

9. Mechanical System
Spring-Mass Damper
Ma bv ky
8/23/2023 Dr Syed Saad Azhar Ali 10
𝑹 𝒔 = 𝑴𝒔𝟐𝒀 𝒔 + 𝒃𝒔𝒀 𝒔 + 𝒌𝒀 𝒔
𝑹 𝒔 = (𝑴𝒔𝟐
+ 𝒃𝒔 + 𝒌)𝒀 𝒔
𝒀(𝒔)
𝑹(𝒔)
=
𝟏
𝑴𝒔𝟐 + 𝒃𝒔 + 𝒌
This expression is called the
Transfer Function

10. TRANSFER FUNCTION
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•
𝒀(𝒔)
𝑹(𝒔)
=
𝟏
𝑴𝒔𝟐+𝒃𝒔+𝒌
• So what is a transfer function
• Output over input (ratio of output and input)
• Is this a transfer function - >
𝒚(𝒕)
𝒓(𝒕)
• Output over input in frequency domain
𝒀(𝒔)
𝑹(𝒔)
,
𝒀(𝒁)
𝑹(𝒁)
,
𝒀(𝝕)
𝑹(𝝕)
• Anything missing…
» Lets review …
NO…!!

11. TRANSFER FUNCTION
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•
𝒀(𝒔)
𝑹(𝒔)
=
𝟏
𝑴𝒔𝟐+𝒃𝒔+𝒌
• So what is a transfer function
• Output over input in frequency domain
𝒀(𝒔)
𝑹(𝒔)
,
𝒀(𝒁)
𝑹(𝒁)
,
𝒀(𝝕)
𝑹(𝝕)
• With zero initial conditions

12. Electrical System
Parallel RLC Circuit
• The total current is divided in all
the parallel branches
• r(t) is divided in
– resistance -> iR(t) = v(t)/R
– inductance -> iL(t) =
1
𝐿
v(t)𝑑𝑡
– Capacitance-> iC(t) = C
𝑑v(t)
𝑑𝑡
This is the called the integro-
differential equation
𝒓 𝒕 = 𝒊𝑹 + 𝒊𝑳 + 𝒊𝑪
𝒅𝒓(𝒕)
𝒅𝒕
= C
𝒅𝟐
v(t)
𝑑𝑡2 +
1
𝑅
𝒅v(t)
𝒅𝒕
+
1
𝐿
v(t)
𝒓 𝒕 =
v(t)
𝑹
+
1
𝐿
v(t)𝑑𝑡 + C
𝑑v(t)
𝑑𝑡
Differentiating the equation
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13. Electrical System
Parallel RLC Circuit 𝒅𝒓(𝒕)
𝒅𝒕
= C
𝒅𝟐v(t)
𝑑𝑡2 +
1
𝑅
𝒅v(t)
𝒅𝒕
+
1
𝐿
v(t)
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determine the transfer function
Don’t forget to set the initial
conditions ZERO
𝑽(𝒔)
𝑹(𝒔)
=
𝒔
𝑪𝒔𝟐 +
1
𝑅
𝒔 +
1
𝐿

14. 15
Project 1 (Competition)
• Group Project
• Competition (I will try to get some prize as well)
• Design the glider
• Get theoretical results
• Competition day - we will get real results
• 11 March 2023 – KFUPM Beach
• 5%
8/23/2023 Dr Syed Saad Azhar Ali