+ All Categories
Home > Documents > Simulink Coursework

Simulink Coursework

Date post: 21-May-2017
Category:
Upload: yaraalwazir
View: 234 times
Download: 4 times
Share this document with a friend
16
Simulink coursework CGC022 Yara al Wazir Question 1 G ( s) = 1 10 s 2 +1 This function is a second order function. In order to tackle this question, the first step is to find the natural frequency The natural frequency of the function can be found using the following equation G 2 ( s ) = 2 s 2 +2 ζωs+ω 2 Where ω represents the natural frequency, ζis the damping ratio, and K is the system gain. Letter Description Value K System gain 1 ζ Damping ratio 0 ω (natural) frequency 10 Therefore, by substituting these numbers into the equation and rearranging it, the natural frequency is found to be: G 2 ( s ) = 2 s 2 +2 ζωs+ω 2 ω= 1 10 rads 1 Thus, the system was adjusted accordingly, as seen in Figure 1
Transcript
Page 1: Simulink Coursework

Simulink coursework

CGC022

Yara al Wazir

Question 1

G (s )= 110 s2+1

This function is a second order function. In order to tackle this question, the first step is to find the natural frequency

The natural frequency of the function can be found using the following equation

G2 (s )= Kω2

s2+2ζωs+ω2

Where ω represents the natural frequency, ζ is the damping ratio, and K is the system gain.

Letter Description ValueK System gain 1

ζ Damping ratio 0ω (natural) frequency 10

Therefore, by substituting these numbers into the equation and rearranging it, the natural frequency is found to be:

G2 (s )= Kω2

s2+2ζωs+ω2

ω=√ 110 rad s−1Thus, the system was adjusted accordingly, as seen in Figure 1

Staff/Research Student, 02/12/13,
Check what w actually denotes
Page 2: Simulink Coursework

Figure 1 - setup of the system

The natural frequency of the system is seen in Figure 2

i)

Page 3: Simulink Coursework

Figure 2 - Natural frequency of the system, w=sqrt(0.1)

ii)

Page 4: Simulink Coursework

Figure 3 - Input frequency 10% higher than the natural frequency, w=1.1 x sqrt(0.1)

iii)

Page 5: Simulink Coursework

Figure 4 - Input frequency 10% lower than natural frequency, w=0.9 x sqrt(0.1)

Page 6: Simulink Coursework

iv)

Figure 5 - Input frequency 10 times greater than the natural frequency, w=10 x sqrt(0.1)

v)

Page 7: Simulink Coursework

Figure 6 - Input frequency 10 times smaller than the natural frequency, w=0.1 x sqrt(0.1)

Page 8: Simulink Coursework

Question 2

G (s )= 4(2 s+1)(5 s+1)

This is a second order transfer function that can essentially be split into two individual first order transfer functions, which will be used to build the open loop process.

G (s )= 4(2 s+1)

∙ 1(5 s+1)

The initial model can be seen in

The scope shows the model; the yellow line indicates the process model, and the pink line indicates the FOPTD model. The target is to fit the pink line to the yellow line in a superimposable fashion. This is done by altering ‘Transfer Fcn3’.

Trial and error was used to fit this model. The numerator is known (‘4’, as it never changes from the original equation), and the first denominator coefficient was altered between 1-8 until the closest fit was found at ‘6.5’; this can be seen in

Page 9: Simulink Coursework

Therefore, the equation of the model is

Gmodel (s )= 3e−2 s

6.5 s+1

b)

The FOPTD model identified can be used to design a PID controller to simulate the closed-loop response of the system

The PID model was built as seen in

However, before progressing, the parameters representing proportional gain, integral time, and derivative time must be set.

In order to work KC out, the following equation was use (As per the Ziegler and Nichols model)

KC=1.2α

( τK p

)

Where α=2 as the function is a second-order function

τ I=2α=2 ∙2=4

τ D=0.5α=0.5 ∙2=1

KC=1.22 ( 6.54 )=0.975

Staff/Research Student, 02/12/13,
Update screenshot to make tau = 6.5 not 6.3
Page 10: Simulink Coursework

Also, from the model simulated in question 2a, it was determined that τ=6.5, α=2 and that Kp= 4

Thus, using equation __, the parameters were identified

Parameter Value SourceKp 4 Model equation

τ 6.5 Model equationα 2 Model equation

Kc 0.975 Equation __τ I 4 Equation ___τ D 1 Equation ___

Inserting these parameters into the model, the following output is achieved in

c)The model was adjusted to account for the real life simulation of the process. To achieve this, the block model was altered as seen in Figure 7, inserting the coefficients ξ=0.3 and ξ=0.5 in ‘Transfer Fcn1’ and ‘Transfer Fcn5’ respectively.

Staff/Research Student, 02/12/13,
Screenshot updated
Staff/Research Student, 02/12/13,
Shouldn’t the purple and yellow lines be superimposable?
Page 11: Simulink Coursework

Figure 7- Block model simulation for true process

The product of this simulation can be seen in Figure 8

Page 12: Simulink Coursework

Figure 8 scopelarger

It is clear that the FOPTD model isn’t an accurate representation of the real life situation. The real life situation is the yellow line; the pink line does not have the initial ‘dip’ that the yellow line has. Thus, the simulation needs further modelling to represent real life sittuations. This can be seen clearly in Figure 9

Page 13: Simulink Coursework

Figure 9 diplarger

To accurately simulate the real process, the PID controllwer was introduced to the system, in addition to the coefficients. This is seen in Figure 10

Staff/Research Student, 02/12/13,
Coefficients? What are the squiggly lines officially called?
Staff/Research Student, 02/12/13,
Update screenshot to adhere to regulations regarding screenshot width
Page 14: Simulink Coursework

Figure 10 q2c2initialbuild

The aggregated results of this block model simulation appears in Figure 11. The red line represents when ξ=0.5. The cyan blue line represents when ξ=0.3. The yellow line simulates a standard step function. The purple line represents the standard step function when ξ=0.

It can be seen that the results display a similar pattern when magnified, but at different amplitudes.

Page 15: Simulink Coursework

Figure 11q2c2alltogether

Q2d

The stability of the system is inherently linked to the value of the gain. The point at which the system becomes unstable is when the response of the system changes with an increase in value of the input The threshold for instability was found to be when ξ=0.25, as seen in Figure 12. Simulation of where ξ=0.2 and ξ=0.3 can be seen in Figure 13 and Figure 14 respectively

Figure 12 simulation of instability where ξ=0.25

Page 16: Simulink Coursework

Figure 13 ξ=0.2

Figure 14 ξ=0.3


Recommended