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Lab # 4 Time Response Analysis Islamic University of Gaza Feedback Control Systems Lab Faculty of Engineering Eng. Tareq Abu Aisha Computer Engineering Dep. Lab # 4 Lab # 4 Time Response Analysis What is the Time Response …? It is an equation or a plot that describes the behavior of a system and contains much information about it with respect to time response specification as overshooting, settling time, peak time, rise time and steady state error. Time response is formed by the transient response and the steady state response. Time response = Transient response + Steady state response Transient time response (Natural response) describes the behavior of the system in its first short time until arrives the steady state value and this response will be our study focus. If the input is step function then the output or the response is called step time response and if the input is ramp, the response is called ramp time response … etc. Delay Time (Td): is the time required for the response to reach 50% of the final value. Rise Time (Tr): is the time required for the response to rise from 0 to 90% of the final value. Settling Time (Ts): is the time required for the response to reach and stay within a specified tolerance band ( 2% or 5%) of its final value. Peak Time (Tp): is the time required for the underdamped step response to reach the peak of time response (Yp) or the peak overshoot. Percent Overshoot (OS%): is the normalized difference between the response peak value and the steady value This characteristic is not found in a first order system and found in higher one for the underdamped step response. It is defined as: Ymax − Yss Percent Overshoot (PO%) = ∗ 100% Yss Steady State Error (ess): indicates the error between the actual output and desired output as ‘t’ tends to infinity, and is defined as: ( ) = ( ) − ( ) = ( ) ∞ ∞ G(s) E(s) = U(s) − Y(s),and Y(s) = ∗ U(s) 1+G(s) G(s) U(s) ⎯⎯ E(s) = U(s) − ∗ U(s) = 1+G(s) 1+G(s) s ∗ U(s) lim sE(s) = lim 1 + G(s) DC Gain: The DC gain is the ratio of the steady state step response to the magnitude of a step input. For example if your input is step function with amplitude = 1 and found the step response output = 5 then the DC gain = 5/1 = 5. In other words it is the value of the transfer function when s=0. Step time response: We know that the system can be represented by a transfer function which has poles (values make the denominator equal to zero), depending on these poles the step response divided into four cases: 1. Underdamped response: In this case the response has an overshooting with a small oscillation which results from complex poles in the transfer function of the system. 2. Critically response: In this case the response has no overshooting and reaches the steady state value (final value) in the fastest time. In other words it is the fastest response without overshooting and is resulted from the existence of real & repeated poles in the transfer function of the system. 3. Overdamped response: In this case no overshooting will appear and reach the final value in a time larger than critically case. This response is resulted from the existence of real & distinct poles in the transfer function of the system. 4. Undamped response: In this case a large oscillation will appear at the output and will not reach a final value and this because of the existence of imaginary poles in the transfer function of the system and the system in this case is called "Marginally stable". Characteristic of various systems: First Order system: The first order system can take the general form: b G(s) = = (s+a) ( . +1) The DC gain Kdc= is found by setting s = 0 at the transfer function. ü Time constant t: is the time to reach 63% of the steady state value for a step input or to decrease to 37% of the initial value and t= is found. It is special for the first order system only. ü Rise Time (Tr):T = . ü Settling Time (Ts):T = . The first order system has no overshooting but can be stable or not depending on the location of its pole. The first order system has a single pole at -a. If the pole is on the negative real axis (LHP), then the system is stable. If the pole is on the positive real axis (RHP), then the system is not stable. The zeros of a first order system are the values of s which makes the numerator of the transfer function equal to zero. Second Order system: The general form of second order system is: ü Natural frequency ωn: is the frequency of oscillation of the system without damping. ü Damping Ratio: Note that the system has a pair of complex conjugate poles at: ω: damped frequency of oscillation. ü DC Gain: is = ü Percent Overshoot: ü Settling Time: . ü Rise Time: = MATLAB Work · stepinfo(sys): this command is used to Compute step response characteristics. For the following transfer functions we will find the settling time, rise time, overshoot and steady state error: 1. G(s) = clear all clc X = tf([2],[1 3]) step(X) %%%%%%%%% stepinfo(X) Results: By MATLAB Calculations: RiseTime: 0.7328 RiseTime = 2.2/3 = 0.73 SettlingTime: 1.3041 SettlingTime = 4/3 = 1.33 Overshoot: 0 Overshoot: 0 Steady state error = 1 + 3 lim = lim = 0.6 G(s) + 1 + 5 2. G(s) = ( ) clear all clc X = tf([12],[1 2 9]) step(X) %%%%%%%%% stepinfo(X) Results: By MATLAB Calculations: RiseTime: 0.4629 RiseTime = 0.3951 SettlingTime: 3.7000 SettlingTime = 4/1 = 4 Overshoot: 32.6877 Overshoot: 32.93 Steady state error = 1 + 2 + 9 9 lim = lim = = 0.4286 (1 + ( )) + 2 + 21 21 3. G(s) = clear all clc X = tf([12],[1 6 9]) step(X) %%%%%%%%% stepinfo(X) Results: By MATLAB Calculations: RiseTime: 1.1196 RiseTime = 1.167 SettlingTime: 1.9447 SettlingTime = 4/3 = 1.33 Overshoot: 0 Overshoot: 0 Steady state error = 1 s + 6s + 9 9 lim = lim = = 0.4286 (G(s) + 1) s + 6s + 21 21 4. G(s) = ( ) clear all clc X = tf([15],[1 7 12]) step(X) %%%%%%%%% stepinfo(X) Results: By MATLAB Calculations: RiseTime: 0.9849 RiseTime = 1.02 SettlingTime: 1.7183 SettlingTime = 4/3.5 = 1.143 Overshoot: 0 Overshoot: 0 Steady state error = 1 + 7 + 12 12 lim = lim = = 0.4444 (1 + ( )) + 7 + 27 27 5. G(s) = clear all clc X = tf(9],[1 0 3]) step(X) %%%%%%%%% stepinfo(X) Results: By MATLAB Calculations: RiseTime: NaN RiseTime: NaN SettlingTime: NaN SettlingTime: NaN Overshoot: NaN Overshoot: NaN Steady state error = No steady state error because the system is marginally stable. Exercises (1): for the following system: Let ωn = 1, Obtain the poles of the system for ξ = 0, 0.1, 0.7, 1, 5 and guess the behavior of step response for every value with explanation. Plot the step response for every ξ above on the same figure and obtain the behavior of it from the graph. Let ωn = 1, ξ = 0.8. Plot the step response and obtain the OS%, Ts, Tr, S.S.E & DC gain by : 1. From the graph. 2. Mathematical by equations. Exercise (2): consider the transfer function below describes an elevator system: For this system we want some specifications to be suitable to users: ü Percent overshooting less than 10% (do not shock the users at beginning). ü Settling time less than 4 sec (don't make users feel boring). Design the system by finding ξ and ωn, and then give the step response of the complete transfer function of the system to make sure from the plot. .
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