MATLAB

 Hello i have homework that need to solve using MATLAB 

Assignment6: Simulink

Solve this equation using the Simulink.

– (optional) 𝐹 =
3𝑥3+|𝑥|

𝑥1/2

– (optional) 𝑌 = ∫ sin(𝑡) + 5 cos(𝑡) 𝑑𝑡 where y(0)=0,

– 𝑌 = ∫ sin(𝑡) + 5 cos(𝑡) 𝑑𝑡 where y(0)=1,

– (optional) 𝑦 = 10 ∗ sin(2𝑡) + 𝑛𝑜𝑖𝑠𝑒

–
𝑑𝑦

𝑑𝑡
= 𝑡4 + 10𝑦 + 𝑝𝑢𝑙𝑠𝑒𝑠

– �̈� = −30�̇� − 4𝑦2 + 10 ∗ sin(5𝑡) + 5

Assignment5: transfer function for solving ODE

Derive the differential equation for the following mass spring damper mechanism, Find the transfer

function; find its step input. Find the max overshoot, settling time, rise time.

𝑚 = 10, 𝑘2 = 3 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡 𝑖𝑠 𝑐 = 1

Determine the transfer function of the following diagram using MatLab, where

𝐺1 = 𝐺2 = 𝐺3 = 𝐺5 = 𝐺5 = 𝐺6 =
1

2𝑠 + 10

And 𝐻1 = 𝐻2 =
1

𝑠+0.1

Find the block reduction using matlab commands, then ind the step response of the system.

Assignment5: transfer function for electrical system:

– Find the transfer function for the following RC circuit. 𝑅 = 1 𝑘 𝑜ℎ𝑚 and 𝐶 = 220 𝜇𝑓

Find the step input,

Find the transfer function for the RC circuit with the step input response as shown in the following plot.

If the 𝑅 = 1000, What is the value of the C .

(optional question):

Find the differential equation of the following RLC circuit.

Find the transfer function of 𝑉𝑐(𝑠)

Find the step input with 𝑅 = 1 𝑘𝑜ℎ𝑚 𝐿 = 1 𝑚𝐻 𝐶 = 1 𝜇𝑓

Assignment 9:

Disturbance signals represent unwanted inputs which affect the control-system’s output, and result in an

increase of the system error. It is the job of the control-system engineer to properly design the control

system to partially eliminate the affects of disturbances on the output and system error.

Overview for disturbance input problem:
The construction of the tunnel under the English Channel from France to the Great Britain began in December

1987. The first connection of the boring tunnels from each country was achieved in November 1990. The tunnel

is 23.5 miles long and bored 200 feet below sea level. Costing $14 billion, it was completed in 1992 making it

possible for a train to travel from London to Paris in three hours.

The machine operated from both ends of the channel, bored towards the middle. To link up accurately in the

middle of the channel, a laser guidance system kept the machines precisely aligned. A model of the boring

machine control is shown in the figure, where Y(s) is the actual angle of direction of travel of the boring

machine and R(s) is the desired angle. The effect of load on the machine is represented by the disturbance,

Td(s).

Figure: A block diagram model of a boring machine control system

Transfer function from R(s) to Y(s)

Assume Td(s) = 0 then the block diagram will be

Due to unit step – r(s) for K=20

% Response to a Unit Step Input R(s)=1/s for K=10

numg=[1];deng=[1 1 0];plant=tf(numg,deng);

K =10;

num=[11 K]; den=[0 1];

contr=tf(num,den);

sys_s=series(contr,plant)

sys=feedback(sys_s,[1]);

stepinfo(sys)

Repeat this for K=20, 50 and 100.

Gc(s)
Controller

K + 11s

G(s)
Plant

1

s(s + 1)

–

+ R(s)
Desired

Angle

Y(s)

Angle +

+
Td(s)

Gc(s)
Controller

K + 11s
G(s)
Plant

1
s(s + 1)

–
+ R(s)
Desired
Angle
Y(s)
Angle

Transfer function from Td(s) to Y(s)

Assume R(s) = 0 then the block diagram will be

Due to unit disturbance – Td(s) for K=20

% Response to a Unit Step Input R(s)=1/s for K=10
numg=[1];deng=[1 1 0];plant=tf(numg,deng);
K =10;
num=[11 K]; den=[0 1];
contr=tf(num,den);

sys=feedback(plant,contr);

stepinfo(sys)

Repeat this for K=20, 50 and 100.

Exercise 3:
a) For each case find the percentage overshoot(%O.S.), rise time, settling time, steady state of y(t)

Unit Step Input Unit Step Disturbance

Overshoot Rise Time

Setting

time

Steady

state

Overshoot Rise Time

Setting
time
Steady
state

K=10

K=20

K=50

K=100

b) Compare the results of the two cases and investigate the effect of changing the controller gain on the
influence of the disturbance on the system output

G(s)
Plant

1
s(s + 1)

–

+ Td(s) Y(s)
Angle

Gc(s)
Controller

K + 11s

Assignment8:

Ex1: Control systems open loop and closed loop:

For the following RC circuit find its transfer function. R=10 000 ohm, C= 500*10-6 f .

–

Find the open loop time constant.

– Design a control system to drive this circuit such that the control algorithm is

𝑣𝑖 = 𝑒 ∗ 𝑘, where 𝑒

=

𝑣𝑖𝑑 − 𝑣𝑜 and 𝑣𝑖𝑑 is the set point.

– Find k that makes the closed loop time constant is 2 sec.

– Find k that makes the system pole at -10 ;

– Draw the root locus plot for the system.

– Find the final value for the closed loop system with step input.

– Use Matlab to show the step response for both the open loop and closed loop systems.

– Use Simulink to simulate the closed loop system and the open loop system.

𝑇𝑠 = 𝜋𝜏 =
4

𝜁𝜔𝑛

𝑃. 𝑂. = 100𝑒−𝜁𝜋/√1−𝜁

2

Overview Second Order Systems:

Consider the following Mass-Spring system shown in the Figure 2. Where K is the spring

constant, B is the friction coefficient,

x(t)

is the displacement and F(t) is the applied force:

Figure 2. Mass-Spring syste

m

The differential equation for the above Mass-Spring system can be derived as follows

𝑀
𝑑2𝑥(𝑡)

𝑑𝑡2
+ 𝐵

𝜕𝑥(𝑡)

𝜕𝑡
+ 𝐾𝑥(𝑡) = 𝐹(𝑡)

Applying the Laplace transformation we get

(𝑀𝑠2 + 𝐵𝑠 + 𝐾) ∗ 𝑋(𝑠) = 𝐹(𝑠)

provided that, all the initial conditions are zeros. Then the transfer function representation of the

system is given by

𝑇𝐹 =
𝑂𝑢𝑡𝑝𝑢𝑡

𝐼𝑛𝑝𝑢𝑡
=

𝑋(𝑠)

𝐹(𝑠)
=

1

(𝑀𝑠2 + 𝐵𝑠 + 𝐾)

The above system is known as a second order system.

The generalized notation for a second order system described above can be written as

2

2 2
( ) ( )

2

n

n n

Y s R s
s s



 
=

+ +

With the step input applied to the system, we obtain

2

2 2
( )

( 2 )

n
n n

Y s
s s s


 
=
+ +

for which the transient output, as obtained from the Laplace transform table (Table 2.3,

Textbook), is

2 1

2

1
( ) 1 sin( 1 cos ( ))

1

nt

n
y t e t


 





− −
= − − +

−

K

M

B

F(t)

x(t)

where 0 < ζ < 1. The transient response of the system changes for different values of damping

ratio, ζ. Standard performance measures for a second order feedback system are defined in terms

of step response of a system. Where, the response of the second order system is shown below.

The performance measures could be described as follows:

Rise Time: The time for a system to respond to a step input and attains a response equal to a

percentage of the magnitude of the input. The 0-100% rise time, Tr, measures the time to 100%

of the magnitude of the input. Alternatively, Tr1, measures the time from 10% to 90% of the

response to the step input.

Peak Time: The time for a system to respond to a step input and rise to peak response.

Overshoot: The amount by which the system output response proceeds beyond the desired

response. It is calculated as

P.O.= 100%t
p

M

f

f





−


where MPt is the peak value of the time response, and fv is the final value of the response. And it

is approximately equal to

𝑃. 𝑂. = 100𝑒−𝜁𝜋/√1−𝜁
2

Settling Time: The time required for the system’s output to settle within a certain percentage of

the input amplitude (which is usually taken as 2%). Then, settling time, Ts, is calculated as

4
s

n

T


=

Exercise 2: Effect of damping ratio ζ on performance measures. For a single-loop second order
feedback system given below

Find the step response of the system for values of ωn = 1 and ζ = 0.1, 0.4, 0.7, 1.0 and 2.0. Plot

all the results in the same figure window and fill the following table. Use “stepinfo”

ζ Rise time Peak Time % Overshoot Settling time Steady state value

0.1

0.4

0.7

1.0

2.0

R(s)

E(s)

Y(s)

+

–

Exercise 3: Design of a Second order feedback system based on performances.

For the motor system given below, we need to design feedback such that the overshoot is limited

and there is less oscillatory nature in the response based on the specifications provided in the

table. Assume no disturbance (D(s)=0).

Table: Specifications for the Transient

Response

Performance Measure

Desired Value

Percent overshoot Less than 8%

Settling time Less than 400ms

Use MATLAB, to find the system performance for different values of Ka and find which value of

the gain Ka satisfies the design condition specified. Use the following table. (hint use “stepinfo”)

Ka 20 30 50 60 80

Percent

Overshoot

Settling

time

Ka 5
1

𝑠(𝑠 + 20)

R(s)
–
+
+

–
Y(s)

Amplifier

Motor

Constant Load
D(s)

(optional) Ex4: second order system: for the following mass spring damper system. k=1, c=0.1, m=10.

The system open loop transfer function is

𝑇 =
𝑥(𝑠)

𝑓𝑎(𝑠)
=

1

𝑚𝑠2 + 𝑐𝑠 + 𝑘

– Rewrite the system to be in the form
𝐺𝑎𝑖𝑛

𝑠2+2𝜁𝑤𝑛𝑠+𝑤𝑛
2

– Find the settling time using the following relation 𝑇𝑠 = 𝜋𝜏 =
4

𝜁𝜔𝑛
;

– Find the percentage overshoot using the following relation 𝑃. 𝑂. = 100𝑒−𝜁𝜋/√1−𝜁
2
.

– Find the steady state value with step input using the final value theorem.
– Find the open loop overshoot, rise time, settling time. Using “stepinfo”

– Design a control system to drive this circuit such that the control algorithm is 𝑓𝑎 = 𝑒 ∗ 𝑘, where 𝑒 =

𝑥𝑑 − 𝑥 and 𝑥𝑑 is the set point. And find its transfer function.

– Find k that makes the settling time less that the half of its value in the open loop.

– Find k that makes the percentage overshoot less than half that in the open loop.

– Draw the root locus plot for the system.

k

m

c

fa(t)

x(t)