TIME RESPONSE OF SECOND ORDER SYSTEM USING MATLAB
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El documento detalla un sistema de control de segundo orden, definiendo su función de transferencia y analizando su respuesta temporal ante una entrada de escalón unitario. Se exploran diferentes casos según el coeficiente de amortiguamiento, incluyendo condiciones de subamortiguamiento, sobreamortiguamiento y amortiguamiento crítico. Además, se incluye un análisis del código MATLAB utilizado para realizar simulaciones de estas respuestas.
TIME RESPONSE OF SECOND ORDER SYSTEM USING MATLAB
- 1. Time response of Second order system Using MATLAB-SIMULINK Subject: Control Systems Name& Roll No: P.SANJAY KUMAR 18981A0241 1. Block diagramof secondorder system: The order of the control system is determined by the power of‘s’ in the denominator of its function. If the power of s in the denominator of the transfer function of a control system is 2 then the system is said to be a second order system. The general equation of the transfer function of a second order system is given as. The block diagram of a second order system :- 𝑪( 𝒔) 𝑹(𝒔) = 𝝎 𝟐 𝒏 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔+ 𝝎 𝟐 𝒏 Where 𝜔 𝑛 = natural frequency 𝜺 = damping ratio 𝝉 = time constant Then make the denominator equal to zero
- 2. 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔 + 𝝎 𝟐 𝒏 = 𝟎 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔 + 𝝎 𝟐 𝒏 + 𝜺 𝟐 𝝎 𝟐 𝒏 − 𝜺 𝟐 𝝎 𝟐 𝒏 = 𝟎 (𝑠 + 𝜀𝜔 𝑛)2 − 𝜺 𝟐 𝝎 𝟐 𝒏 + 𝝎 𝟐 𝒏 (𝑠 + 𝜀𝜔 𝑛)2 + (𝝎 𝒏√ 𝟏− 𝜺 𝟐) 𝟐 = 𝟎 Where s1 and s2 are the solutions of the equsations S1 = −𝜺𝝎 𝒏 + 𝒋𝝎 𝒏√𝟏− 𝜺 𝟐 S2 = −𝜺𝝎 𝒏 − 𝒋𝝎 𝒏√𝟏− 𝜺 𝟐 These two roots of the equation of s represents the poles of the transfer function of that system.the real parts of the systemrepresentsthe damping and imaginary partb representsdampedfrequency of the system. 2. Time response for unit step input : The transfer function can be written as a C(s) = R(s) * 𝝎 𝟐 𝒏 𝒔 𝟐+𝟐𝜺𝝎 𝒏 𝒔+𝝎 𝟐 𝒏 Here R(s) = 1 𝑠 𝑪( 𝒔) 𝑹(𝒔) = 𝝎 𝟐 𝒏 𝒔(𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔+ 𝝎 𝟐 𝒏) By using the partial fractions 𝝎 𝟐 𝒏 𝒔( 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔+ 𝝎 𝟐 𝒏) = 𝑨 𝑺 + 𝑩𝒔 + 𝑪 ( 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔 + 𝝎 𝟐 𝒏) 𝝎 𝟐 𝒏 = 𝑨( 𝒔 𝟐 + 𝟐𝜺𝝎 𝒏 𝒔 + 𝝎 𝟐 𝒏) + (𝑩𝑺 + 𝑪)( 𝑺) 𝝎 𝟐 𝒏 = 𝑨 𝒔 𝟐 + 𝑩𝑺 𝟐 + 𝑨𝟐𝜺𝝎 𝒏 𝒔 + 𝑪𝑺 + 𝑨𝝎 𝟐 𝒏 𝑩𝒚 𝒄𝒐𝒎𝒑𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕𝒔 𝝎 𝟐 𝒏 = 𝑨𝝎 𝟐 𝒏 𝑨 = 𝟏 𝑨 𝒔 𝟐 + 𝑩𝑺 𝟐 = 𝟎 A + B = 0 B = -1 𝑨𝟐𝜺𝝎 𝒏 𝒔 + 𝑪𝑺 = 𝟎 𝑪 = −𝟐𝜺𝝎 𝒏
- 3. 𝑪( 𝒔) = 𝟏 𝑺 − 𝒔 + 𝟐𝜺𝝎 𝒏 ( 𝒔 𝟐+𝟐𝜺𝝎 𝒏 𝒔+𝝎 𝟐 𝒏) 𝑪( 𝒔) = 𝟏 𝑺 − 𝒔 + 𝜺𝝎 𝒏+𝜺𝝎 𝒏 ( 𝒔 𝟐+𝟐𝜺𝝎 𝒏 𝒔+𝝎 𝟐 𝒏+𝜺 𝟐 𝝎 𝟐 𝒏−𝜺 𝟐 𝝎 𝟐 𝒏) 𝑪( 𝒔) = 𝟏 𝑺 − 𝒔 + 𝜺𝝎 𝒏 ( 𝒔 +𝜺𝝎 𝒏) 𝟐−𝝎 𝟐 𝒏(𝟏−𝜺 𝟐) − 𝜺𝝎 𝒏 ( 𝒔 +𝜺𝝎 𝒏) 𝟐−𝝎 𝟐 𝒏(𝟏−𝜺 𝟐) By taking the cases Case 1:- 𝜺 = 𝟏 ( 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒅𝒂𝒎𝒑𝒊𝒏𝒈 ) 𝑪( 𝒔) = 𝟏 𝒔 − 𝟏 𝒔 + 𝝎 𝒏 + 𝝎 𝒏 (𝒔 + 𝝎 𝒏) 𝟐 By taking the inverse Laplace transformations 𝐶( 𝑡) = 1 − 𝑒 𝜔 𝑛 𝑡 − 𝝎 𝒏 𝑒 𝜔 𝑛 𝑡 ∗ 𝑡 𝐶( 𝑡) = 1 − 𝑒 𝜔 𝑛 𝑡 (1 + 𝑡𝝎 𝒏) At t = 0 >> c (t) = 1 – 𝑒0(1 + 0) = 1 − 1 = 0 At t = ∞ ≫ 𝑐( 𝑡) = 1 − 𝑒−∞(1 + ∞) = 1 − 0 = 1 Case 2 :- 𝜀 < 1 ( 𝑢𝑛𝑑𝑒𝑟 𝑑𝑎𝑚𝑝𝑖𝑛𝑔) 𝑐( 𝑡) = 1 − 𝑒−𝜔 𝑛 𝑡 cos( 𝝎 𝒏 (√ 𝟏 − 𝜺 𝟐) 𝒕)+ 𝜺 √𝟏 − 𝜺 𝟐) 𝐬𝐢𝐧( 𝝎 𝒏 (√ 𝟏 − 𝜺 𝟐) 𝒕) 𝑐( 𝑡) = 1 − 𝑒−𝜔 𝑛 𝑡 𝐜𝐨𝐬𝐡( 𝝎 𝒅 𝒕) + 𝜺 √𝟏 − 𝜺 𝟐) 𝐬𝐢𝐧( 𝝎 𝒅 𝒕) Case 3:- 𝜀 > 1 ( 𝑜𝑣𝑒𝑟 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 ) 𝑐( 𝑡) = 1 − 𝑒−𝜔 𝑛 𝑡 cos( 𝝎 𝒏 (√ 𝟏 − 𝜺 𝟐) 𝒕)+ 𝜺 √𝟏 − 𝜺 𝟐) 𝐬𝐢𝐧( 𝝎 𝒏 (√ 𝟏 − 𝜺 𝟐) 𝒕) 𝑐( 𝑡) = 1 − 𝑒−𝜔 𝑛 𝑡 𝐜𝐨𝐬𝐡( 𝝎 𝒅 𝒕) + 𝜺 √𝟏 − 𝜺 𝟐) 𝐬𝐢𝐧( 𝝎 𝒅 𝒕)
- 4. 3. Time response plot: mat lab code 4. Response onmat lab :-