control engineering

1. Mathematical Modeling of
Physical Systems
Dep't: Mechanical Engineering
course Code: Meng-5272
Wallaga University
Nekemte, Ethiopia
April14, 2021

2. Chapter-2
Mathematical Modeling of Physical
Systems
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3. 2.1.Objective
o Understanding the concept of mathematical modeling.
o Showing the types of mathematical modeling.
o Describing the variables, elementary laws and interconnection laws.
o Obtaining system model of parallel and series combinations.
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4. 2.2. Modeling of mechanical systems
2.2.1. Concept of Mathematical Modeling
• It describes the pattern of physical systems.
• It reveals the relation of parameters, performances and dynamic behavior of
a systems.
• The derivation of a model is up on the fact of the dynamic systems.
• The ability to analyzing and performing depend on mathematical shows.
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5. 2.2.2. Types of mathematical modeling
The mathematical modeling reveal the pattern of
the system using:
1. Static modeling : the motion pattern time
independent
2. Dynamic modeling: the motion pattern is time
dependance.
a. External model: show the relation of input and
output. i.e. Differential equations and transfer
functions
b. Internal models: relation of input, output and
the internal variables.
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6. 2.2.3. Element Law
1. Mass:
From newton second law force acting on
a. mass equal to time rate a change of momentum.
Figure-2.1 Mass
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7. Continued…
• From the two variables the second-order derivative will be:
2. Friction:
• Mass depicted on Sliding under oil film has a laminar flow subjected to
viscosity friction. “B” the unit of N-S/m and ∆𝑉 = 𝑣2 − 𝑣1
Figure-2.2 Friction
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8. Continued…
• Considering dashpot at initial velocity zero (v1=0);
• The change will be:
Figure-2.3 Dashpot
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9. Continued…
• Second order derivative:
Figure-2.4 Second Order
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10. Continued…
3. Stiffness:
• A mechanical element undergoes the deformation relation that exists
between elongation and force.
Figure-2.5 Stiffness
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11. Continued…
• From the Hooks law:
• The relation between the parameters:
Figure-2.6 Hooks Law
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12. 2.2.4. Interconnection Law
1. D’Alembert’s Law
➢ It is the restatement of Newton second law at constant mass for a governing
law rate of momentum.
➢ The motion is constrained to fixed direction from i value.
➢ When the force will be rewritten on equilibrium position().
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13. Continued…
2. The law of reaction force
✓ The reaction force will be found after equal magnitude opposite direction of
the exerted force and stiffness force.
Figure-2.7 Reaction force
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14. 2.2.5. Obtaining The System Model
1. Free Body Diagram
Example-2.1
• Draw the free-body diagram and apply D’Alembert’s law to write a
modeling equation for the system. The mass is assumed to move horizontally
on frictionless bearings, and the spring and dashpot are linear. fa(t) applied
force, X(t) displacement and V(t) is an intermediate variables not exist in the
final differential equation.
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15. Continued…
a. Translational system
b. Free body diagram and Element laws
Figure-2.8 Translational and Free body diagram
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16. Continued…
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Solution:
Replacing “ ”
Example -2.2
• Draw the FBD for the two mass system shown and use D’Alembert’s law to
write the two differential equations .
a. Translational system

17. Continued…
b. Free body diagram and Element laws
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18. Continued…
2. Parallel Combination
❖ If two springs and dashpot are attached to the same body in both directions.
a. When the two springs have the same unstretched length
• The coefficient of single spring Keq is written as:
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19. Continued…
❖ Hence,
b. When the K1 and K2 are replaced by single spring.
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20. Continued…
3. Series Combination
o If the elements joined one of each while not connected to common junction
on both side.
Example-2.3
Draw the FBD of mass “m” and for massless junction “A”. Show the equation
as X1 while the “A” is depend on the “m” X1 and X2 proportional.
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21. Continued…
In terms:
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22. Continued…
Then, the stiffness and friction,
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23. 2.3. Equation of Electrical Network
• Due to the electrical circuit at a high frequency considered as a lumped
element,
• Calculated by simple ordinary differential equation combining the elements
and laws solved.
• To be known as a lumped element the inductance or capacitance
• It let the energy be stored at high frequency rather than transmitting along
the wire length.
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24. Continued…
2.3.1 Resistor
• An element of algebraic relation
of voltage and current across the
terminal.
• Where the voltage drop is due to
the proportion of the resistance
magnitude.
The unit is ohm(∩).
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25. Continued…
2.3.2. Capacitor
❑ An element of algebraic relation
to voltage and the charge.
❑ Where the charge is the integral of
current and flow from +ve to _ve.
2.3.3. Inductor
• An element of algebraic relation
of voltage and derivative of flux
linkage.
• The linear inductor’s voltage is
given by faradays law with a unit
of Henries(H).
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26. 2.3.4. Inter connection Law
a. Kirchhoff's voltage Law
❖ The sum algebraic of voltage
across the elements which make
the loop equal to zero.
❖ The summation of the voltage
along the loop.
❖ Counter clock wise
❖ Reverse direction
Figure-2.9
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27. Continued…
b. Kirchhoff’s Current Law
• Since not possible to accumulate
every net charge at any node, the
current will be the algebraic sum
at any node equal to zero.
• The summation will be:
• For the three element:
i1+i2+i3=0
Figure-2.10
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28. 2.3.5 Analogue Relationship
• The differential equation of physical system will be written by stating and
applying basic governing laws to specific devices.
• The differential equation for different physical system may have the same
form.
• Whereby ANALOG is the same of corresponding variables and parameters
in different system represent the differential equations.
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29. Continued…
Table 2.1: Analog variables
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30. Continued…
Figure-2.11-Analog relation
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31. Continued…
• Thereby transfer function will be
written:
a.
b.
c.
d.
Example -2.4
Apply the Kirchhoff’s law and ohm’s
law to write the modeling equation.
Where Ui(t) and Uo(t) are input and
out put voltage; R1, R2 and C are
constant.
Figure-2.12
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32. Continued…
• Using nodal method analysis
• Check: By eliminating i1(t), i2(t),
and i(t)
Example-2.5
• Using nodal method analysis
Figure-2.13
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33. Continued…
• Eliminating i1(t), i2(t) and i3(t)
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.

34. 2.4 Transfer Function
2.4.1. Definition
❖ Every block diagram has the transfer function in specific,
❖ Where the transfer function is the ratio of Laplace transform.
❖ It is the ratio of output to input Laplace transform at zero condition.
❖ Where the Fo(s) and Fi(s) Laplace transform output and input.
❖ Whereby the Transfer function is the algebraic function of “s”
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35. Continued…
• The general nth- order output and input.
• Assuming t=+0, the initial condition will be Zero.
• Using the algebraic equation of “s”, the remaining term:
• The system transform function is rational transform of “s”.
• The Y(s) out put is the product of rational transform of “s” to X(s) input.
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36. Continued…
• Transfer function will be:
2.4.2. Properties
1. Independent of input system; the characteristics not modified by input
signal(even if at working time).
2. All initial conditions are assumed to be Zero or at rest.
3. Describe only the time-invariant linear systems, the parameters not change;
little change.
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37. Continued…
4. It’s unit related to the ratio of input to output system’s type; not essential.
Example-2.6.
• Write the below differential equation to Transfer function:
1. 2.
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38. Continued…
2.4.3. Rational Polynomial form of Transfer function
• When the frequency of input is high no output; thus the expression:
• Where :The Z1, Z2, …are know zores and they are the roots of Y(S).
:The P1, P1…are known Poles and they are the roots of X(s).
:K’=bm/an
:(S-zi) zores Factor, and (S-Pi) poles factor
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39. Continued…
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• Zi and Pi may be the real or complex and the complex number will be written
by: Zi=𝜎1 + 𝑗𝜔1
• Hence, the two conjugate roots find in the conjugate pairs:

40. Continued…
➢ Hence, the conjugate pairs will be found from second-order derivatives.
Example-2.7
➢ A system with transfer function:
Where the zeros and poles are;
S1=-2, S2=-3, and S3=-1+j S4=-1-j
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41. 2.4.4.Transfer function of elements in Series Connection
✓ Using the mathematical operator of TF system the below series connection
will be simplified | Multiplication of each system|.
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42. 2.4.5.Transfer function of elements in Parallel Connection
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✓ Using the mathematical operator of TF system the below series connection
will be simplified | Summation of each system|.

43. Continued…
Example-2.8
➢ Evaluate the given Block diagram Y(s)/U(s) and Z(s)/U(s) to TF then show
in the Rational function of “s”.
➢ Starting from parallel connection
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44. Continued…
➢ Then the block diagram looks like:
▪ For series connection
*
▪ Y(s)/U(s) and Z(s)/U(s) |Unit loop|
+ =
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45. Continued…
o Hence,
o The final parallel combination to single block.
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46. 2.4.6.Transfer function for typical Link
• Some typical transfer function in Mechanical Engineering their physical
structure and principles are different.
• Link is kind of mechanism doesn’t represent a components made up of units
or parts.
• It used in complex analysis and researching systems.
General form:
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47. Continued…
Where,
a. Proportional Link
• Taking Laplace transforms re-arranging the TF will be:
It is applicable
• In speed gear system input/output.
• In Lever system input and output.
• In output voltage and rotated angle of a potentiometer.
• In output and input electric amplifier.
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48. Continued…
b. Integral link
• The differential Equation:
• TF will be:
• Example 2-9
From the schematic diagram of the hydro-cylinder, the cross-section “A”, and
the input flow is q(t), the output is the velocity V(t) of the cylinder piston. Find
TF.
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49. Continued…
• The velocity of piston
Figure-2.14
• The TF is find using the Laplace transform
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50. Continued…
c. Inertia Link
• The differential equation:
• Laplace form will be:
• By re arranging the equation the TF:
• Where “T” time constant and “K” constant of element.
Example 2.10
For the intermediate i(t), C and R, and Ui(t), Uo(t) find the TF.
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51. Continued…
Solution
• Using Kirchhoff’s law the passive circuit
Differential equation will be:
Figure-2.15
• By Cancelling current i(t) by rearranging:
• Where the Laplace transform of above :
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52. Continued…
d. Differential link
About three differential links exist:
1. Ideal differential link
2. First order differential link
3. Second order differential link
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53. Continued…
Where, T is the constant time, is damping ratio.
Example 2-11
✓ For constant C and R ; i(t)intermediate function find the TF.
✓ Using Kirchhoff’s and ohm’s law
the differential equation will be:
✓ By Cancelling current i(t) by rearranging: Figure 2-16
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54. Continued…
➢ Where the TF electrical Link:
Note: If the value of Ts=RC is very small, it will be neglected and the link will
be an ideal differential link.
e. Oscillation Link
• The differential Equation
• The Laplace Transform
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55. Continued…
• The Laplace and TF differential:
2.5. Function Block Diagram
2.5.1. Block
• It is a statement of its operation.
• It indicates the happens to the input information after transmission.
• The illustration of Simple amplifier a shown as a block.
Figure:2-17
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56. Continued…
➢ In the block diagram summing junction and point will be shown as circle
with arrows into and out symbol.
➢ The arrow identified by plus and minus sign, showing +ve and –ve signal.
➢ Where by signal out of summing point is the algebraic sum of signal into it.
Figure2.18
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57. Continued…
a. Pick/tie point
▪ It is used while more than one signal flow direction is necessary.
▪ It is node or junction where no summing point (+ve and -ve) is done.
▪ Provide extra path through the signal flow with out affecting its original.
Table:2.2
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58. Continued…
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.

59. Continued…
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2.5.1.Terminology
1. The forward path not repeated signal flow from input to output.
2. The product od expression G(s)
link will be:
3. Feed back the signal flow move
from output to input.

60. Continued…
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4. The product of every link with
expression H(s):
5. Open loop is ration of primary feedback to error:
6. Closed loop will be:

61. Continued…
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• When the H(s)=1 unit feed back system:
• The relation of open loop and closed loop with selection of unit –ve feed
back:
;
• Where G(s) and ∅(𝑠) is the open loop and closed loop TF respectively.

62. 2.5.2. Simplification of Function Body Diagram
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a. Moving Tie points:
• Moving the points forward or backwards.
• When the system G is front and back of the tie point; R=B
• Front position: R=B, where by Back position: R*G*1/G=B => R=B

63. Continued…
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Example:2.12
• Show the feedback loop system of series and parallel combinations.
Solution
• Taking the inner feedback path to the output point “A” and the inner
feedback transfer function.

64. Continued…
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b. Moving summing point
✓ With the references of block diagram
:On LHS
:On RHS.

65. Continued…
Example: 2-13
❖ By removing the right and keeping the left hand side summing junction
remodel the block diagram.
a.
b.
Figure-2.19
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66. Continued…
✓ The transfer function:
c. General Feed Case
➢ For a given error E(s) an feed back
Signal B(s).
So,
➢ From the Laplace transform
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67. Continued…
❖ The final closed-loop TF system
And,
❖ Whereby the open loop TF is independent of loop open.
❑ The steps for simplifying the block diagram.
1. Combine all series
2. Parallel combination
3. Close all inner loops and
4. Move summing junction and Tie point to left or right.
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68. Continued…
Example:2-14
✓ Simplify the block diagram to get TF.
Solution
✓ Reduce the projected
// connection.
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69. Continued…
✓ Solve the inner closed loop
Then, TF:
✓ Apply to feed back loopback substitution:
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70. Continued…
d. Multi inputs
➢ At several input the achievement will be:
1. Set all but one input to zero.
2. Determine single input to output of TF
3. Repeat step two for other
4. Add all TF for all output value.
Example: 2-15
➢ Determine the output C two input R and D.
➢ The closed loop TF for R set to zero.
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71. Continued…
Then,
or Figure-2.20
Again, by setting D to zero:
or
Total summation:
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72. 2.6. Signal Flow diagram
➢ It is an alternative pictorial representation to block diagram.
➢ All signal/variables shown by dot called nodes like x1,x2,x3…
➢ The related variables joined lines called directed branch.
➢ Each node has associated transmittance which link xi to xj with zero
transmittance from xj to xi .
➢ Input variable is source node (x1, X5.)..
➢ Output variable is sink node(x4).
➢ Pass from source to sink node without passing through any node((x1-X2).
➢ Branch forming loop is closed loop transmittance(b , c).
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73. Continued…
Note:
1. Signal at the node is the sum of the transmittance to the node.
2. Transmittance simple related to TF
3. The transmittance connected input/output node both are unity
4. The same rule of block diagram apply; transmittance maybe –ve.
Figure-2.21
➢ Where X(s) and Y(s) are source and sink node, G(s) TF.
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74. Continued…
Figure 2.22
Note: The –ve sign on signal flow shows the negative feedback
Example: 2-16
➢ Change the given block diagram to signal flow diagram.
Figure 2.23 BDG
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75. Continued…
Solution
Figure-2.24 SFD
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76. Continued…
Exercise 2.1
▪ Change FBD to SFD
Figure-2.25
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77. 2.7.Mason Gain Formula
▪ The input –output relation derived from simplified complex diagram.
▪ The original signal diagram used to do so using Mason gain formula.
▪ Mason gain state, the net transmittance “P”, from source input to sink output.
▪ Where:
1. K-number of open path from source to sink.
2. Pk- the transmittance Kth open path.
3. ∆= 1-(sum of all loop transmittance)+(sum of products of loop
transmittances of non-touching loops taken in pairs ) –(sum of similar
product taken three at a time) + etc.
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78. Continued…
4. ∆𝑘- the value of ∆ calculated for the path of the graph not touching the open
path.
Note: generally
1. All the forward path transmittance
2. All loop transmittance
3. All the non touching loop transmittance
4. Whether all the loop and transmittance touching and not.
Example: 2-17
▪ Get the transfer function of block diagram
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79. Continued…
Solution
1. Forward paths=3 and forward path
transmittances are P1=G1G2G3G4G5,
P2=G1G6G4G5, and P3=G1G2G7.
2. Feed back loops =4 and feed back
loop transmittances L1=-G4H1, L2=-G2G3G4G5H2, L3=-G6G4G5H2,
L4=-G2G7H2.
3. Non-touch feedback loops are L1 and L4
∆=1-( L1 +L2+ L3+ L4) + L1 L4(non-touch).
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80. Continued…
4. All feed loops touch the forward path P1,hence, ∆1=1
5. All feed back loops touch the forward path P2 , hence, ∆2=1
6. The feed back loops L1 doesn’t touch the forward path P3 , hence, ∆3=1-L1
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81. Regulation and
Control
Chapter-2
Math
Modeling
End
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