Transfer Function State Space
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Transcript of Transfer Function State Space
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TRANSFER FUNCTIONSTATE SPACE MODELAND SIMULATION IN MATLAB INVERTED PENDULUM
Submitted To: Sir Sadaqat Ali
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Submitted By :
Somia Hassan2K12 ELE 212
Irfana Wali2k12 ELE 232Waqas Umair2k11 ELE 272Rana Shahid 2k12 ELE 244
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INVERTED PENDULUM SYSTEMS
Inverted Pendulum Dynamics:
•The mechanical system has Two Degrees of freedom (DOF), the linear motion of the cart in the X-axis, the rotational motion of the pendulum in the X-Y plane.
• Let x be the distance in m from the Y-axis,
•And q be the angle in rad w.r.t vertical.
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M – Mass of cart in kgm – Mass of Pendulum in kgJ – Moment of Inertia of pendulum in kg-m2L – Length of Pendulum in mb – Cart friction coefficient in Ns/mg – Acceleration due to gravity in m/s2
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FREE BODY DIAGRAM OF THE CART
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The horizontal forces are considered in the analysis as they only give information about the dynamics since the cart has only linear motion.
Max = F + N – B eq(1)
N=md²(x+LsinӨ)=mẍ+mӪLcosӨ-m(Ө’)²LsinӨ eq(2) dt²
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FREE BODY DIAGRAM OF PENDULUM
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Vertical Forces:
P+mg=m d² (LcosӨ) eq(3) dt²
P=mLӪsinӨ+mL(Ө’)²cosӨ-mg
•Moment due to the reaction forces P and N are resolved into X and Y directions
•Vcmt is the velocity of centre of mass
•V is the velocity of point Oin the X direction
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Summing the moments across the center we get:
-NLcosӨ - PLsinӨ =j Ӫ eq(4)
Substitution of eq(2) and eq(3) in eq(4) yields
-mLẍcosӨ -(mL² + J )Ӫ = mgLsinӨ eq(5)
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After substituting eq(2) in eq (1) we get:
we assume a very small deviation Ө from the vertical.
Ө ≈0sin Ө=-Ө
cos Ө=-1 Ө’²=0
F=(M+m)ẍ+bẋ-mL(Ө’)²sinӨ+mLӪcosӨ .........eq(6)
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CONTINUE.....
Putting in eq(5)
Putting in eq(6)
(M+m)ẍ+bẋ-mLӪ=u ........ eq(8)
mLẍ=(J+mL²) Ӫ-mgLӨ ........ eq(7)
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LINEAR MATHEMATICAL MODEL•Defined as a set of mathematical equations
•More accurate the model more complex the equations will be.
•To obtain a linear model the Taylor series expansion can be used
•The system has two equilibrium points:
• one is the stable i.e. the pendant position
•and the other one is the unstable equilibrium point i.e. the inverted position
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TRANSFER FUNCTION
To obtain the transfer,we must first take the Laplace transform of the system equations assuming zero initial conditions.
mLx(s)s² =(J+mL²)Ө(s) s²-mgLӨ(s) ......(9)
From eq(9)
(M+m) x(s)s² +bx(s)-mLӨ(s)s²=u(s).......(10)
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X(s)= [J+mL² - g ] Ө(s) ......eq(11) mL s²
Then substitute the above into the eq(10)
(M+m) )[J+mL² - g ]s²Ө(s)+b[J+mL² - g ]sӨ(s)-mLs²Ө(s)=u(s) mL s² mL s²
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Rearranging, the transfer function is then the following
S⁴+b(J+mL²) s³-(M+m)mgL s²–bmgL s q q q
mLs² q
Where q=[(M+m)(J+mL²)-(mL²)]
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mLs² q
S³+b(J+mL²) s²-(M+m)mgLs –bmgL q q q
Ө(s) =U(s
P(pend)=Rad .....(12) N
(J+mL²) s² -gmL qP(cart)=x(s) =
u(s)S⁴+b(J+mL²) s³-(M+m)mgL s²–bmgL s q q q
M .....(13)N
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The state space is of the formẋ = Ax+ Bu......... Eq(14)
The state space for the Inverted Pendulum system is obtained as
..................Eq(15)
Where σ’=MmL²+J(M+m)
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The output equation is given by
..................Eq(16)
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INVERTED PENDULUM SYSTEM PARAMETERS
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After substitution of parameters
The transfer functions in (12) and (13) are substituted by the values in Table we obtain
X (s) = 2.59s²-3.92 U(s) s⁴+0.26s³-36.296s²-7.26s
Ө(s) = 7.407s U(s) s³+0.26s²-36.296s-7.26
q=[(M+m)(J+mL²)-(mL²)]
q=5.4×10-3
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