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Definition of First Order Systems • the General Form of Transfer Function of Firs

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Definition of First Order Systems • the General Form of Transfer Function of Firs Response of first order systems Outline: Definition of first order systems The general form of transfer function of first order systems Response of first order systems to some common forcing functions (predict and understand how it responds to an input) Time behavior of a system is important. When you design a system, the time behavior may well be the most important aspect of its behavior (How quickly a system responds is important) What is a first order system? It is a system whose dynamic behavior is described by a first order differential equation. Synonyms for first order systems are first order lag and single exponential stage. Transfer function The transfer function is defined as the ratio of the output and the input in the Laplace domain. It describes the dynamic characteristics of the system. 표푢푡푝푢푡 퐺(푠) = 푛푝푢푡 General rules to develop a transfer function 1. Make unsteady state balance (mass, heat or momentum) 2. Make steady state balance 1 3. Subtract the steady state equation from the unsteady state equation (why?) 4. Transform the resulting equation into the Laplace domain 5. Rearrange the equation to get the ratio of the (out/in) in one side and the other parameters in the other side (the resulting is the transfer function) Example on first order systems A mercury thermometer: consider the mercury thermometer shown in the figure Cross view of thermometer Basic assumptions: 1. All the resistance to heat transfer resides in the film surrounding the bulb (conduction resistance is neglected). 2. All the thermal capacity is in the mercury. 3. The mercury assumes a uniform temperature throughout. 4. The glass wall containing the mercury does not expand or contract during the transient response. 5. Constant properties. To develop the transfer function of the thermometer we will follow the steps mentioned earlier; 1. Unsteady state heat balance Input – Output = Rate of change 2 푑푦 ℎ퐴(푥 − 푦) − 0 = 푚푐 푝 푑푡 Temperature profile in the thermometer Note: output from the thermometer = 0.0 (mercury is expanded when heated) 2. Steady state balance ℎ퐴(푥푠 − 푦푠) = 0 3. Unsteady state balance – steady state balance 푑(푦 − 푦푠) ℎ퐴[(푥 − 푥 ) − (푦 − 푦 )] = 푚푐 푠 푠 푝 푑푡 푑(푦−푦 ) 푑푦 Note: 푠 = 푑푡 푑푡 Write the above equation in terms of the deviation variables 푑푌 ℎ퐴[푋 − 푌] = 푚푐 푝 푑푡 푚푐푝 푑푌 [푋 − 푌] = ℎ퐴 푑푡 4. Laplace transform 3 푚푐푝 [푋(푠) − 푌(푠)] = 푠푌(푠) ℎ퐴 5. Output/Input 푌(푠) 1 = 푚푐 푋(푠) 푝 푠 + 1 ℎ퐴 푚푐 The parameter 푝 has the units of time and is called the time ℎ퐴 constant of the system and is denoted by 휏 푌(푠) 1 = 푋(푠) 휏푠 + 1 Note: The time constant is a measure to how fast be the system response. The smaller is the time constant, the more responsive is the system. 휏 also called “dead time” or “dynamic lag” Standard form of first order transfer functions 푌(푠) 퐾 = 푋(푠) 휏푠 + 1 The important characteristics of the standard form are as follows: The denominator must be of the form 휏푠 + 1 The coefficient of the s term in the denominator is the system time constant 휏 The numerator is the steady-state gain K. 푌(푠) 2 Example 1: A first order system has a transfer function = 1 . Identify 푋(푠) 푠+ 3 the time constant and the steady state gain. Properties of transfer functions: Superposition is applicable 푌(푠) 퐺(푠) = 푋(푠) If 푋(푠) = 푎1푋1(푠) + 푎2푋2(푠) 4 푌(푠) = 퐺(푠) 푋(푠) 푌(푠) = 푎1퐺(푠)푋1(푠) + 푎2퐺(푠)푋2(푠) = 푎1푌1(푠) + 푎2퐺(푠)푌2(푠) i.e. The output of multi-input is the sum of the output to the individual inputs. Response of first order systems to some common forcing functions 1. Step response (푋(푡) = 퐴 푢(푡); 푌(푡) =? ? ) Required: Sketch of the input and the corresponding output General equation of the output The ultimate and maximum value of the output and their corresponding time. To get the response to any input follow the following steps (scheme) ℒ푇 푇퐹 ℒ−1 [퐼푛푝푢푡] 푋(푡) → 푋(푠) → 푌(푠) → 푌(푡)[푂푢푡푝푢푡] 푌(푠) 퐾 = 푋(푠) 휏푠 + 1 퐴 푋(푠) = 푠 퐴 퐾 퐶1 퐶2 푌(푠) = = + 푠 휏푠 + 1 푠 휏푠 + 1 Solving by partial fractions 퐾퐴 퐾퐴 푌(푠) = − 푠 1 푠 + 휏 푡 − 푌(푡) = 퐾퐴[1 − 푒 휏] The above equation is the general form of first order system response to step change. 5 Input X(t) Output Y(t) 푡 푋(푡) = 퐴 푢(푡) − 푌(푡) = 퐾퐴[1 − 푒 휏] Ultimate value = KA Maximum value = KA Characteristics of step response A. The value of the output reaches 63.2% of its ultimate value after 푡 = 휏 B. If the initial rate of change is maintained the response will be completed after 푡 = 휏 C. The response is completed after 푡 = 5휏 D. The speed of the response of a first-order system is determined by the time constant for the system. as t increases, it takes longer for the system to respond to the step disturbance. 6 Effect of the time constant on the step response of a first order system 2. Impulse response (푋(푡) = 퐴 훿(푡); 푌(푡) =? ? ) 푌(푠) 퐾 = 푋(푠) 휏푠 + 1 푋(푠) = 퐴 퐾퐴 퐾퐴/휏 푌(푠) = = 휏푠 + 1 1 푠 + 휏 푡 퐾퐴 − 푌(푡) = 푒 휏 휏 Input X(t) Output Y(t) 푡 푋(푡) = 퐴 훿(푡) 퐾퐴 − 푌(푡) = 푒 휏 휏 The response rises immediately and then decays exponentially. Ultimate value = 0 퐾퐴 Maximum value = at t = 0 휏 3. Pulse response 퐻 0 < 푡 < 푇 푋(푡) = { 0 푇 < 푡 Write the function X(t) in terms of the unit step function 푋(푡) = 퐻 푢(푡) − 퐻푢(푡 − 푇) 퐻 퐻 푋(푠) = − 푒−푇푠 푠 푠 퐻 퐻 퐾 푌(푠) = ( − 푒−푇푠) 푠 푠 휏푠 + 1 퐻 퐾 퐻 퐾 푌(푠) = − 푒−푇푠 푠 휏푠 + 1 푠 휏푠 + 1 7 푡 (푡−푇) − − 푌(푡) = 퐾퐻 [1 − 푒 휏] 푢(푡) − 퐾퐻 [1 − 푒 휏 ] 푢(푡 − 푇) Input X(t) Output Y(t) 푡 푋(푡) = 퐴 훿(푡) 퐾퐴 − 푌(푡) = 푒 휏 휏 Ultimate value = 0 푇 − Maximum value = 퐾퐻 [1 − 푒 휏 ] at t =T 4. Ramp response (푋(푡) = 퐴푡 푢(푡); 푌(푡) =? ? ) where A is the slope of the ramp function 푌(푠) 퐾 = 푋(푠) 휏푠 + 1 퐴 푋(푠) = 푠2 퐾퐴 퐾퐴 휏 퐶1 퐶2 퐶3 푌(푠) = 2 = = 2 + + 푠 (휏푠 + 1) 2 1 푠 푠 1 푠 (푠 + 휏) (푠 + 휏) Solving by partial fractions 퐶1 = 퐾퐴 퐶2 = −퐾퐴휏 퐶3 = 퐾퐴휏 퐾퐴 퐾퐴휏 퐾퐴휏 푌(푠) = − + 푠2 푠 1 (푠 + 휏) 8 푡 − 푌(푡) = 퐾퐴푡 − 퐾퐴휏 + 퐾퐴휏 푒 휏 푡 − 푌(푡) = 퐾퐴푡 − 퐾퐴휏 (1 − 푒 휏) Input X(t) Output Y(t) 푡 푋(푡) = 퐴푡 푢(푡) − 푌(푡) = 퐾퐴푡 − 퐾퐴휏 (1 − 푒 휏) Ultimate value = ∞ Maximum value = ∞ Physical examples of first order systems The objective of this part is to develop the transfer function of some first order systems and to confirm that the time constant depends on the system parameters. 1. Liquid level A tank of uniform cross sectional area A, the inlet flow is q and the outlet is 푞표. The liquid level in the tank is h and the tank has a linear flow resistance h at the outlet (e.g. valve). (q = ) o R 9 Liquid level system 퐻(푠) 푄 (푠) Required: the transfer function and 표 (note: this system is called 푄(푠) 푄(푠) single input multi-output SIMO) Basic assumptions: a) Constant density b) Linear resistance c) Constant cross sectional area i. Unsteady state mass balance 푑(휌푉) 푑(휌퐴ℎ) 휌푞 − 휌푞 = = 표 푑푡 푑푡 푑ℎ 푞 − 푞 = 퐴 표 푑푡 ii. Steady state mass balance 푞푠 − 푞표푠 = 0 iii. Subtract the steady state equation from the unsteady state one 푑퐻 푄 − 푄 = 퐴 표 푑푡 Where: 푄 = 푞 − 푞푠 푄표 = 푞표 − 푞표푠 퐻 = ℎ − ℎ푠 The above parameters are called deviation variables. iv. Taking the transform of the resulting equation 푄(푠) − 푄표(푠) = 퐴푠퐻(푠) H Q = o R 10 H(s) 푄(푠) − = 퐴푠퐻(푠) R H(s) 푅 = 푄(푠) 퐴푅푠 + 1 H(s) 푅 = 푄(푠) 휏푠 + 1 휏 = 퐴푅 Note: tanks having small cross sectional area are more responsive. Qo(s) 1 = 푄(푠) 휏푠 + 1 N.B: the steady state gain depends on the input and the output in the last transfer function it is dimensionless because the input and the output have the same units. The steady state gain is a conversion factor that relates the input and the output at steady state. 2. Liquid level with constant flow outlet Liquid level with constant flow outlet In this example the resistance R is replace by a constant flow pump (푞표(푡) = 푞표,푠) H(s) Required: the transfer function 푄(푠) Basic assumptions: d) Constant density e) Constant flow outlet f) Constant cross sectional 11 (푞표(푡) = 푞표,푠) area i. Unsteady state mass balance 푑ℎ 푞 − 푞 = 퐴 표 푑푡 ii. Steady state mass balance 푞푠 − 푞표푠 = 0 iii. Subtract the steady state equation from the unsteady state one 푑퐻 푄 = 퐴 푑푡 Where: 푞표 − 푞표푠 = 0 iv. Taking the transform of the resulting equation 푄(푠) = 퐴푠퐻(푠) H(s) 1 = 푄(푠) 퐴푠 푄(푠) H(s) = 퐴푠 Note: Inverse of the above equation yields: 1 푡 H(t) = ∫ 푄(푡)푑푡 퐴 0 Ex: Find the response of this system to a unit step change in input 푄(푡) = 1 푢(푡) 1 푄(푠) = 푠 1 퐻(푠) = 퐴푠2 1 퐻(푡) = 푡 퐴 12 The response is a ramp function that grows with time without limit.
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