Transfer Function PPT PDF

MECHATRONICS (UME-410) Transfer Function Dr. Anant Kr Singh Professor Department of Mechanical Engineering Thapar Institute of Engineering and Technology Patiala-147004 Transfer Function Ratio of Laplace transform of the output to the Laplac...

MECHATRONICS (UME-410) Transfer Function Dr. Anant Kr Singh Professor Department of Mechanical Engineering Thapar Institute of Engineering and Technology Patiala-147004 Transfer Function Ratio of Laplace transform of the output to the Laplace transform of the inputs, considering initial conditions to zero. Input Output 𝑢(𝑡) Process 𝑦(𝑡) g(t) ℒ𝑢(𝑡) = 𝑈(𝑠) ℒ𝑦(𝑡) = 𝑌(𝑠) 𝑌(𝑠) Transfer function (𝐺(𝑠)) = 𝑈(𝑠) Input Output 𝑈(𝑠) Process 𝑌(s) 𝐺(𝑠) 2 Transfer Function…  Example 1: Kirchhoff's voltage law 𝑉 𝑡 − 𝑉𝑅(𝑡) − 𝑉𝐿(𝑡) − 𝑉𝐶(𝑡) = 0 (1) Voltage across Resistor 𝑉𝑅 = 𝑅𝑖(𝑡) Laplace Transformation 𝑉𝑅(𝑠) = 𝑅𝐼(𝑠) (2) 𝑑𝑖(𝑡) Voltage across Inductor 𝑉𝐿 = L 𝑑𝑡 Dr. Anant Kumar Singh , Professor, Mechanical Engineering Department, TIET, 3 Patiala Transfer Function… Laplace Transformation 𝑉𝐿(𝑠) = 𝐿𝑠𝐼(𝑠) (3) Voltage across Capacitor 1 𝑉𝑐= ‫𝑡𝑑 𝑡 𝑖 ׬‬ 𝐶 1 Laplace Transformation 𝑉𝐶 (𝑠) = 𝐼(𝑠) (4) 𝐶𝑠 1 𝑉 𝑠 − 𝑅𝐼(𝑠) − 𝐿𝑠𝐼(𝑠) − 𝐼(𝑠)= 0 𝐶𝑠 1 𝑉 𝑠 = (𝑅 + 𝐿𝑠 + )𝐼(𝑠) (5) 𝐶𝑠 𝑉(𝑠) 𝐼 𝑠 = (6) 1 (𝑅 + 𝐿𝑠 + 𝐶𝑠) Dr. Anant Kumar Singh , Professor, Mechanical Engineering Department, TIET, Patiala 4 Transfer Function… Transfer Function… Transfer Function taking voltage across resistance as output 𝑉 𝑅 (𝑠) 𝑅 = 1 (7) 𝑉(𝑠) 𝑅+𝐿𝑠+ 𝐶𝑠 Transfer Function taking voltage across inductance as output 𝑉𝐿(𝑠) 𝐿𝑠 = (8) 𝑉(𝑠) 𝑅+𝐿𝑠+ 1 𝐶𝑠 Transfer Function taking voltage across capacitance as output 𝑉𝐶(𝑠) 1 = (9) 𝑉(𝑠) 𝑅𝐶𝑠+𝐿𝐶𝑠2+1 Dr. Anant Kumar Singh, Professor , Mechanical Engineering Department, TIET, Patiala 6 Transfer Function…  Example 2: KVL 1 𝑅1 𝑉𝑖 𝑠 = 𝐶𝑠 𝐼 𝑠 + 𝑅 𝐼(𝑠) 1 2 𝑅1 + 𝐶𝑠 or (𝑅1+𝑅2) + 𝑅1𝑅2𝐶𝑠 𝑉𝑖 𝑠 = 𝐼 𝑠 (𝑅1𝐶𝑠 + 1) Voltage across resistor 𝑉0 𝑠 = 𝑅2𝐼 𝑠 Transfer Function 𝑉0(𝑠) 𝑅2(𝑅1𝐶𝑠+1) = 𝑉𝑖 𝑠 (𝑅1+𝑅2)+𝑅1𝑅2𝐶𝑠 Dr. Anant Kumar Singh , Professor, Mechanical Engineering Department, TIET, Patiala 7 Transfer Function… Transfer Function 𝑉0(𝑠) 𝑅2(𝑅1𝐶𝑠+1) 𝑅2 = α= 𝑉𝑖 𝑠 (𝑅1+𝑅2)+𝑅1𝑅2𝐶𝑠 (𝑅1 +𝑅2 ) 𝑉0(𝑠) 𝑅2 (𝑅1𝐶𝑠+1) = 𝑉𝑖 𝑠 (𝑅1+𝑅2) 1+ 𝑅 1 𝑅 2 𝐶𝑠 𝑇1 = 𝑅1𝐶 (𝑅 1 +𝑅 2) 𝑉0(𝑠) 𝝰(1+𝑠𝑇1) 𝑅1𝑅2 = 𝑇2 = 𝐶 𝑉𝑖 𝑠 (1+𝑠𝑇2) (𝑅1+𝑅2) Input α(1 + 𝑠𝑇1) Output 𝑉𝑖(𝑠) (1 + 𝑠𝑇2) 𝑉0(𝑠) Dr. Anant Kumar Singh , Professor, Mechanical Engineering Department, TIET, Patiala 8 Transfer Function… 9 Transfer Function… Laplace Transformation 𝐹 𝑠 = 𝑀𝑠2𝑌(𝑠) +𝐵𝑠𝑌(𝑠) + 𝐾𝑌 𝑠 Transfer Function of the System 𝑌(𝑠) 1 =𝐺 𝑠 = 𝐹(𝑠) 𝑀𝑠2 + 𝐵𝑠 + 𝐾 𝐹 𝑠 𝑌 𝑠 𝐺 𝑠 1 Transfer Function… Consider the following ODE where 𝑥(𝑡) is input of the system and 𝑦 𝑡 is the output, considering initial condition to zero 𝑑2𝑦 𝑑𝑦 𝑑𝑥 𝐴 +𝐵 −𝐶 =0 𝑑𝑡 2 𝑑𝑡 𝑑𝑡 Taking Laplace Transformation 𝐴 𝑠2𝑌(𝑠) − 𝑠𝑦 0 − 𝑦 ′ 0 + 𝐵[𝑠𝑌(𝑠) − 𝑦 0 ] - 𝐶[𝑠𝑋(𝑠) − 𝑥 0 ] = 0 𝑦 0 =0 𝑦′ 0 = 0 and 𝑥 0 =0 𝐴𝑠2𝑌(𝑠) + 𝐵𝑠𝑌(𝑠) − C𝑠𝑋(𝑠) = 0 𝐴𝑠2 + 𝐵𝑠 𝑌(𝑠) = 𝐶𝑠𝑋(𝑠) 11 Transfer Function… Transfer Function 𝐺(𝑠) = 𝑌(𝑠) = 𝐶𝑠 𝑋(𝑠) 𝐴𝑠2 + 𝐵𝑠  Problem Find the functions of time 𝑦 𝑡 , for a given differential equation: 𝑑 2 𝑦(𝑡) + 3 𝑑𝑦(𝑡) + 2𝑦(𝑡) = 5𝑢(𝑡). Were 𝑢(𝑡) is the unit step function. 𝑑𝑡 2 𝑑𝑡 The initial conditions are y 0 = −1, 𝑦 ′ 0 = 2 12 Transfer Function… Transfer Function are used to Determine the response of the system for any given input To determine the stability of the system Differential equation of time domain is first transformed to algebraic equation of frequency domain. After Solving algebraic equation in frequency domain, result is finally transformed to time domain. Laplace transformation is a method of solving the differential equation  Example 𝑠 𝑠 is a complex variable (complex frequency )  ℒ cos(ω𝑡) = 𝑠2 + ω2 𝑠 = 𝜎 + 𝑗ω 1  ℒ𝑒 −𝑎𝑡 = 𝑠+𝑎 13 Laplace Transformation of Integrals 𝑁(𝑠) 𝑠−𝑍1 𝑠−𝑍2 𝑠−𝑍3 …….(𝑠−𝑍𝑛) 𝑇. 𝐹 = 𝐺 𝑠 = = 𝐷(𝑠) 𝑠−𝑃1 𝑠−𝑃2 𝑠−𝑃3 …….(𝑠−𝑃𝑛)  Poles The poles of a transfer function are defined as the roots of the polynomial of the denominator of the transfer function. 10(𝑠+2) T. F = (𝑠+1)(𝑠+3) 10(𝑠 + 2)/[(𝑠 + 1)(𝑠 + 3)] has a pole at 𝑠 = −1 and a pole at 𝑠 = −3 Complex poles always appear in complex-conjugate pairs The transient response of system is determined by the location of poles 14 Laplace Transformation of Integrals  Zeros The zeros of a transfer function are defined as the roots of the polynomial of the numerator of the transfer function. 10(𝑠 + 2)/[(𝑠 + 1)(𝑠 + 3)] has a zero at 𝑠 = −2 Complex zeros always appear in complex-conjugate pairs Poles of the system are represented by ‘x’ and zeros of the system are represented by ‘o’ Order of the system is equal to the number of the poles 14 Laplace Transformation of Integrals Stability A system is stable if bounded inputs produce bounded outputs i.e. 𝑓(𝑡) → 0 as 𝑡 → ∞. The complex s-plane is divided into two regions: the stable region, which is the left half of the plane, and the unstable region, which is the right half of the 𝑠-plane. All the poles have negative real parts. 16 Laplace Transformation of Integrals Stability  A system is unstable, if there is unbounded outputs for bounded inputs i.e. 𝑓(𝑡) → ∞ as 𝑡 → ∞. The poles have a positive real part, i.e. at least one pole in the right hand plane. Unstable Marginally stable A system is marginally stable if for bounded inputs 𝑓 𝑡 does not decay to 0 or go to ∞ as 𝑡 → ∞. One or more distinct poles on the imaginary axis, and any remaining poles have negative real parts. 17 Laplace Transformation of Integrals Summary Location of poles in S-plane affect the stability of system. Poles are roots of denominator of T. F. and Zeros are roots of numerator of T.F. Stability 1) Stable: A system is stable if and only if real part of all poles are –ve means all poles lie left half of S-plane. 2) Unstable: A system is unstable if real part of atleast one pole is +ve means atleast one pole lies in right half. 3) Marginally stable: A system is marginally stable if atleast one pole is purely imaginary (has no real part) and no pole has +ve real part (means must have –ve real part i.e left half). 18 Laplace Transformation of Integrals Stability 𝑠+2 𝑠+2 𝐺 𝑠 = = 𝑠2+7𝑠+12 (𝑠+4)(𝑠+3) Zeros; 𝑠 + 2 = 0; 𝑠 = −2 Poles; 𝑠 + 4 𝑠 + 3 = 0; 𝑠 = −4, 𝑠 = −3 ×× -4 -3 -2 Stable System 19 Laplace Transformation of Integrals Stability 𝐺 𝑠 = 𝑠 𝑠2+9 Poles; 𝑠2 + 9 = 0; 𝑠 = ±3𝑗 3j -3j Marginally Stable System 20 Laplace Transformation of Integrals Stability 1 𝐺 𝑠 = 𝑠−4 Poles; 𝑠 − 4 = 0; 𝑠 = 4 4 Unstable System 21 Laplace Transformation of Integrals Stability 3𝑆 𝐺(𝑆) = 𝑆 2 + 4𝑆 + 8 Zeros; 𝑠=0 Poles; 𝑠 = −2+2𝛚, 𝑠 = −2 −2𝛚 X -2 X Stable System 22  Thank You Dr. Anant Kumar Singh , Professor, Mechanical Engineering Department, TIET, Patiala 20