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![• In case of a spring , we require force to deform the
spring.
• Here the force is proportional to the displacement.
• Net displacement on application of force f(t) at and X1
and X2
F(t) = K [ X1(t) - X2(t)]](https://image.slidesharecdn.com/transferfunctionandmathematicalmodeling-160723080558/85/Transfer-function-and-mathematical-modeling-16-320.jpg)
![• In case of a spring , we require force to deform the
spring.
• Here the force is proportional to the displacement.
• Net displacement on application of force f(t) at and X1
and X2
F(t) = K [ X1(t) - X2(t)]](https://image.slidesharecdn.com/transferfunctionandmathematicalmodeling-160723080558/75/Transfer-function-and-mathematical-modeling-16-2048.jpg)
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Transfer function and mathematical modeling
- 1. Transfer Function andMathematical Modeling
- 2. Transfer Function Poles And Zeros of a Transfer Function Properties of Transfer Function Advantages and Disadvantages of T.F.
- 3. Transfer functiongives us the relationship between the input and the output and hence describes the system. In control systems, the input is represented as r(t) (instead of x(t)) and the output is represented as c(t) (instead of y(t)).
- 4. Once theconcept of Transfer function is understood, Let us move a little further. Let the Transfer function of a system be represented as G (s). i.e. G (s) = Or more generally G(s) = Where C(s) :- Laplace transform of output ; R(s) :- Laplace transform of input. Both C(s) and R(s) are polynomials in s. i.e. G(s) = Vo (s) Vi (s) C(s) R(s)
- 5. = Poles:- poles of a transfer function are the values of the Laplace transform variable, s, that cause the transfer function to become infinite. When s=a1, s=a2, s=a3…s=an, the transfer function G(s) becomes infinite. Hence a1,a2,a3…an are the poles of the transfer function. If we equate the denominator of the transfer function to zero, we obtain the poles of the system. K(s-b1) (s-b2) (s-b3)…(s-bm) (s-a1) (s-a2) (s-a3)…(s-an)
- 6. Zeros :-zeros of a transfer function are the values of the Laplace transform variable, s, that cause the transfer function to become zero. When s=b1, s=b2, s=b3…s=bm, the transfer function G(s) becomes equal to zero. Hence b1,b2,b3…bm are the zeros of the transfer function. If we equate the numerator of the transfer function to zero, we obtain the zeros of the system.
- 7. 1. The transferfunction of a system is the Laplace transform of its impulse response for zero initial conditions. 2. The transfer function can be determined from system input-output pair by taking ratio Laplace of output to Laplace of input. 3. the transfer function is independent of the inputs to the system. 4. The system poles/zeros can be found out from transfer function. 5. The transfer function is defined only for linear time invariant systems. It is not defined for non-linear systems.
- 8. Advantages :- 1.It is a mathematical model that gives the gain of the given block/system. 2. Integral and differential equations are converted to simple algebraic equations. 3. Once the transfer function is known, any output for any given input, can be known. 4. System differential equation can be obtained by replacement of variable ‘s’ by ‘d/dt’ 5. The value of transfer function is dependent on the parameters of the system and independent of the input applied.
- 9. Disadvantages :- 1.Transfer function is valid only for Linear Time Invariant systems. 2. It does not take into account the initial conditions. Initial conditions loose their significance. 3. It does not give any idea about how the present output is progressing.
- 10. Translation motion Rotational motion Translation-Rotation counterparts Analogy system Force-Voltage analogy Force-Current Analogy Advantages Example
- 11. As stated earliertranslation motion refer to a type of motion in which a body or an object moves along a linear axis rather than a rotation axis. Translation motion involves moving left or right , forward or back , up and down. The following three basic element viz. 1). Mass 2). Spring 3). Damper
- 12. • A modelof the mass element assumes that the mass in concentrated at the body. • The Displacement of the mass always take place in the direction of the applied force.
- 13. We know, force= Mass * Acceleration F = M . a
- 14. If amobile phone on the table needs to be pushed from one place two other, we needs to apply force. The force that we apply will have to overcome this friction. Friction exists between a moving body and a fixed support or also between moving surface. while friction opposes motion, it is not always unesirable
- 15. The fig. showsfriction in the despot
- 16. • In caseof a spring , we require force to deform the spring. • Here the force is proportional to the displacement. • Net displacement on application of force f(t) at and X1 and X2 F(t) = K [ X1(t) - X2(t)]
- 17. In such system,force get replaced by Torque(T), displacement by angular displacement ( ), velocity by angular velocity( ) and acceleration by angular acceleration( ). • The following three element viz. 1). Inertia J 2). Damper 3). Spring
- 18. In rotationmotion, we have a concept of inertia. T = 2). Damper • As stated earlier, it’s behavior is similar to that in translation motion. T(t) = B .
- 19. • Like thedamper, the spring is also similar to the one studied in translation motion. T = K (t)
- 20. Sr. No. TranslationMotion Rotational Motion 1 Mass (M) Inertia (J) 2 Damper (B) Damper (B) 3 Spring (K) Spring (K) 4 Force (F) Force (T) 5 Displacement (X) Angular Displacement 6 Velocity = v Angular Velocity =
- 21. There aretwo main electrical analogous system : 1). Force-voltage analogy 2). Force-current analogy Now that we have Discussed mechanical system as well electrical system, it is worth nothing that exists a analogy- similarity in their equations. According to Newton’s law, the applied force will be used up to cause displacement in the spring, acceleration to the mass against the friction force. + + K x (t)
- 22. we get, F(t) =M S2 X(s) + s B(s) X(s) + K x(s) • This equation is called equilibrium equation of the mechanical system.
- 23. • Here forceis analogous to voltage. v(t) = Ri +L + . dt put, i= v(t) = R + L + q
- 24. Translation Electrical Rotational ForceF Voltage - V Torque - T Mass M Inductance - L Inertia J Dumper B Resistance - R Damper -B Spring K Elasticity – D = Spring K Displacement X Charge -q Displacement Velocity -V Current - i = Velocity - ω ω
- 25. • Standard equationfor a Translational system is + + K x (t) The following analogies F V M L B R K X Q
- 26. Standard equationfor a Rotational system is T = J + B . + K The following analogies • T V J L B R K Q
- 27. Here forceis analogous to current. I = + ∫ V. dt C put, v = I = . + + C i(s)=1/R.S. .S
- 28. Translation Electrical Rotational ForceF Voltage - i Torque - T Mass M Capacitance - C Inertia J Spring K Resistance of inductance - Spring K Damper B Conductance = Damper - B Displacement X Flux linkege - Displacement - Velocity Voltage V = Velocity
- 29. Standard equationfor a Translation system is + + K x (t) The following analogies • F I • M C B K X
- 30. Standard equationfor a Rotational system is T = J + B . + K The following analogies • T I J C B • K Q
- 31. • Equation ofthe system can be converted into another. • Trial design in one system involving changinge of the values M , B , K may be costlier than changing in R , L , C .
- 32.