Mathematical Modelling Electro-Mechanical System in Matlab using Simulink and Transfer Function equations

1. Group:-6A 5
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Lab:03 Transfer Function of Mechanical
Systems .
Task: For the Mechanical System Shown in Figure#1.Find the Transfer Function.
. .
Here in Figure#1
M1= M3=1kg , M2=1.5kg. K1= K2= K3=1N/m . D=0.2Ns/m.
And also find Step response and Impulse response using MATLAB.

2. Group:-6A 5
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Solution For Task
o Mathematical Equations:- Table#1
For Mass1:-
F = M1 ̈ + k1 x1+ k2 (x1- x2)
= M1 ̈ + k1 x1+ k2x1-k2x2
= ̈ (M1) + x1(k1+ k2)- x2(k2)
For Mass3:-
0 = M3 ̈ +D ̇ + k3 (x3- x2) + k4 x3
= M3 ̈ +D ̇ + k3 x3- k3 x2+ k4 x3
= ̈ (M3) +D ̇ + x3(k3+ k4)- x2(k3)
For Mass2:-
0 = M2 ̈ + k2 (x2- x1) + k3 (x2- x3)
= M2 ̈ + k2 x2- k2 x1+ k3 x2- k3 x3
= ̈ (M2) + x2(k2+ k3)- x1(k2)- x3(k3)
o Transfer Function Equations:- Table#2
For Mass1:-
F = X1(s)[ M1s2
+ k1+ k2]- X2(s)[ k2]
For Mass3:-
0 = -X2(s)[ k3]+ X3(s)[ M3s2
+Ds + k3+ k4]
For Mass2:-
0 = -X1(s)[ k2]+ X2(s)[ M2s2
+ k2+ k3]
- X3(s)[ k3]
 Putting values of Mases ,Springs and Damper given in Eq(1), Eq(2), Eq(3).
F = X1(s)[ s2
+ 2]- X2(s)+0 --------------------------Eq(1)
0 = -X1(s)+ X2(s)[1.5s2
+2] - X3(s) --------------------------Eq(2)
0 = 0-X2(s)+ X3(s)[s2
+0.2s+2] --------------------------Eq(3)
o In Matrix Form in MATLAB
[ s^2 + 2, -1, 0]
[ -1, (3*s^2)/2 + 2, -1]
[ 0, -1, s^2 + s/5 + 2]

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o Code of Task In MATLAB
syms s F X x1 x2 x3 f
X=[x1; x2; x3]
F=[f; 0; 0]
z=[s^2+2 -1 0; -1 1.5*s^2+2 -1; 0 -1 s^2+0.2*s+2]
ZIN=inv(z)
X=ZIN*F
x1=X(1)/f
x2=X(2)/f
x3=X(3)/f
[n1,d1]=numden(x1)
[n2,d2]=numden(x2)
[n3,d3]=numden(x3)
num1=sym2poly(n1);
num2=sym2poly(n2);
num3=sym2poly(n3);
den1=sym2poly(d1);
den2=sym2poly(d2);
den3=sym2poly(d3);
g1=tf(num1,den1);
g2=tf(num2,den2);
g3=tf(num3,den3);
figure
subplot(2,1,1)
step(g1)
title('Step(x1)')
subplot(2,1,2)
impulse(g1)
title('Impulse(x1)')
figure
subplot(2,1,1)
step(g2)
title('Step(x2)')
subplot(2,1,2)
impulse(g2)
title('Impulse(x2)')

4. Group:-6A 5
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figure
subplot(2,1,1)
step(g3)
title('Step(x3)')
subplot(2,1,2)
impulse(g3)
title('Impulse(x3)')
o Code Result In Command Window
X =
x1
x2
x3
F =
f
0
0
z =
[ s^2 + 2, -1, 0]
[ -1, (3*s^2)/2 + 2, -1]
[ 0, -1, s^2 + s/5 + 2]
ZIN =
[ (15*s^4 + 3*s^3 + 50*s^2 + 4*s + 30)/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s
+ 40), (2*(5*s^2 + s + 10))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40),
10/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)]
[ (2*(5*s^2 + s + 10))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40), (2*(s^2 +
2)*(5*s^2 + s + 10))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40),
(10*(s^2 + 2))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)]
[10/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40), (10*(s^2 +
2))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40), (5*(3*s^4 + 10*s^2 +
6))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)]
X =
(f*(15*s^4 + 3*s^3 + 50*s^2 + 4*s + 30))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 +
6*s + 40) (2*f*(5*s^2 + s + 10))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)
(10*f)/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)

5. Group:-6A 5
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 Transfer Function Equations x1,x2,x3
x1 =
(15*s^4 + 3*s^3 + 50*s^2 + 4*s + 30)/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s +
40)
x2 =
(2*(5*s^2 + s + 10))/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)
x3 =
10/(15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40)
 Numenators and Denominators of Tarnsfer Equation x1,x2,x3
n1 =
15*s^4 + 3*s^3 + 50*s^2 + 4*s + 30
d1 =
15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40
n2 =
10*s^2 + 2*s + 20
d2 =
15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40
n3 =
10
d3 =
15*s^6 + 3*s^5 + 80*s^4 + 10*s^3 + 120*s^2 + 6*s + 40

6. Group:-6A 5
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o Step Response and Impulse Response of Transfer Function Equations
Figure#1.1
Figure#1.2

7. Group:-6A 5
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Figure#1.3
Conclusion:-
There are two types of responces:
1) Impulse Response: It is the reaction of any dynamic system in response to
some external change in very short time (approximately zero).
2) Step Response: The response or change of a system while giving it
continous input (constant input).
 When we applied impulse to our system, it gives us impulse response at
that time which is observed in figures (1.1, 1.2 and 1.3).
 When step input is given to the system, we got step response of our
system which is also observed in figures (1.1, 1.2 and 1.3).
 As force is directly applied on M1 so the step response of system 1 is
maximum. i-e nearly around 1.