"Question 1A second order system is under-damped with a damping ratio of 0.4 and a natural frequencyof 10 Hz.a) Transfer functionb) Time response when it is subjected to a unit step inputc) Percentage overshootd) Rise time, peak time, settling timeSolution Given the natural frequency as: ? = 10?? and the damping ratio ? = 0.4, then: ? ? = 2? ? = 20? ??? /? = 62.83 ??? /?? ? Consequently, the Damping frequency is: 2 2 v ? = ? v1 - ? = 20? × 1 - 0.4 = 57.586 ??? /?? ? a) The transfer function for a closed-loop second-order system is given by: 2 ( ) ? ? ? ? ? ( ) ? ? = =? 2 2 ( ) 1 + ? ? ? + 2? ? ? + ? ? ? Substituting the values of? ??? ? : ? C(s) 3947.84 ( ) ? ? = = ? 2 R(s) s + 50.265s+ 3947.84b) The step-input output response when the system is subjected to a unit-step input isrepresented by the general function: 1 -? ? ? ? ( ) ? ? = 1 - ( * ? * sin(? ? + Ø))? 2 v1 - ? 2 v 1-? -1 -1 ° WhereØ = ??? = ??? ? = 66.42 = 1.159 ???? Hence -? ? ? ? (? ) = ? - [? . ???? * ? * ??? (?? . ???? + ? . ??? )] EEE 2411Time Domain Analysis (Assignment 1)This can be obtained from the Transfer function in part (a) above by the Laplace Transformsas follows: 1 / The unit-step input is given by a function: R(s) = s 3947.84 1 Therefore the Transfer function:? (? )=× 2 ? +50.265? +3947.84 ? 3947.84 ( ) Which can be represented by:? ? = 2 2 [ ] ? (s +? ? ) + ?? ? Using partial fractions the function becomes:A B ? (? ) =+2 2 s (s + ??n) + ?d 2 2 ?n ?n For A = ?? ? = 0 = = 12 2 2 2 2 2 [(s + ??n) + ?d ] ? ?n +?n (1- ? ) 2 ?n 2 2 For B =[(S + ?? ) + ? ] = 0 ? = -? ? + ? ?? ? ? ?S 2 ?n ? = -??n + j?d 2 -? (?? + j? ) ? ? ? The conjugate being = (?? - j? )( ?? + j? )? ? ? ? Giving B as – (S+2?? )therefore B = -(S + 8)? The output equation C(s) becomes: 1 s+2??nC(s) =- 2 2 S (S + ??n) + ?d The Laplace inverse of the output equation:??n -?? t -??nt ? C (t) = 1 - [e * Cos ? t +e Sin ? t ]? ? ?d -?? t ? e 2 C (t)= 1 - [v1 - ? * Cos? t + ? Sin? t ]? ? 2 v1 - ? Substituting values of ? ??? ? : ? -?? .???? ? C (t) =? - [? . ???? * ????? . ??? ? + ? . ???? ?? . ???? ]? . 2 | P a g e???? ??EEE 2411Time Domain Analysis (Assignment 1)c) Percentage Overshoot-?? -0.4? 2 v1-? 2 v1-0.4 %?? = 100 × ? = 100 × ? = ?? . ?? %d) Rise time (? ) , Peak time (? ) and Settling time (? )? ? ? ? - ? ? - 1.159 ???? ???? , ? = == ? . ??????? ? ? 20? × 0.84 ? v ? ? ???? ???? , ? : = = ? . ??????? ? 2 ? ? ? v1 - ? ? 4 ???????? ???? ? : = = ? . ??? ???? ? ? ? Question 216 Given an under-damped system with transfer function, ? (? ) = * ? (? ) is subjected1 2 ? +6? +16 to a unit step input. Using MATLAB, investigate and discuss the following effects and plotthe response obtained.i. The effect of increasing the damping (?), on the Rise time (? ), Peak time (? ),? ? Settling time (? ) and maximum % overshoot, while keeping the natural frequency? 2 (? ), constant. Set ? (? ) = ? + ?? + ? , with b= 6,5,4,3 Discuss the design? requirements and the implications of these parameters.DESIGN REQUIREMENTSIf ? < 0, the poles are located on the right half of the s-plane, hence the system isunstable. If? < 1then the system has two real poles. Generally, the unit-step response isdetermined to a large extent by a dominant pole in this case. The system is said to be over- damped. If? = 1 then the system has a double pole at -? ? . This is a critically damped? system. Ideally, the value of zeta ? < 0.4 whereby the system shoots and is said to be under- damped and ? > 0.8 makes the system “sluggish” thus the required value of ? to have a goodresponse time should lie between the two values above. 3 | P a g e EEE 2411Time Domain Analysis (Assignment 1)The MATLAB simulation for the system given by the Transfer Function: ?? ? ? (? )= * ? (? ); subjecting it to a unit step valuewill be: ? ? ? +?? +?? ? Fig 1.0: Output response of a Unit Step Input function with b = 6The required design parameters for the different values of zeta can therefore be calculatedjust as in question 1 above as shown in the table below: ? = v16 = 4 ??? /?? ? = ?? ? ???? , ? ? ? ? %O.S? ? ? ? 6 0.75 1.187 0.9142 1.333 2.8385 0.625 1.006 0.7193 1.600 8.0804 0.5 0.9069 0.6046 2.000 16.303 0.375 0.8472 0.5273 2.667 28.06Table 1.0: Values obtained when varying b4 | P a g e EEE 2411Time Domain Analysis (Assignment 1)The resultant graphs with b = 5, 4 and 3 respectively are: Fig 1.1: Output response of a Unit Step Input function with b = 5Fig 1.2: Output response of a Unit Step Input function with b = 45 | P a g e EEE 2411Time Domain Analysis (Assignment 1)Fig 1.3: Output response of a Unit Step Input function with b = 3DISCUSSION 1 Fig 1.4: Comparison of the Output responses for various values of b6 | P a g e EEE 2411Time Domain Analysis (Assignment 1)As ? is constant (=4), step responses for various damping ratios shows that: the? overshoot increases as zeta decreases. This is however at the expense of rise time? . While the response becomes more sluggish, it settles quicker with increase in zeta.? Damping is the characteristic ability of a system to oppose the oscillatory nature of itstransient response. Smaller values of the damping coefficient produces transientresponses with more oscillations before settling begins. ii. The effect of including a zero at s= -12,-6,-3, 0 and +3 on the unit step response.Set N(s) = 16, 16(s+12), 16(s+6). The effect of the parameters? , ? , ? , ? , %? ? ? overshoot, ? and zeros on the output step response.? Fig 2.0: Comparison of the Output responses with ZEROS7 | P a g e EEE 2411Time Domain Analysis (Assignment 1)DISCUSSION 2The addition of zeros in the left half-plane speeds up transients, making rises and fallssharper. Smaller values of zeros makes this effect more prominent. As a result, an additionalzero in the left half-plane makes the system faster and more oscillatory.A zero towards the origin, time to first peak decreases monotonically whilepercentage overshoot increases monotonically. It takes longer for the system to settle to thefinal value of the response.Zeros on the right-half plane retards the system and produces an under-shoot. Thepercent undershoot decreases as zero moves along toward infinity. The system oscillates forlonger. REFERENCES 1. http://seit.unsw.adfa.edu.au/staff/sites/valu/TEACHING/CT1/effects.pdf accessed on Junend 22 , 20152. Lecture Notes on Control engineering, EEE24113. Hasan S. Saeed, “Automatic control system”, Katsons Books.4. N.S. Nice, “Control system engineering”, Third Edition. Wiley, USA, 2000. 5. K. Ogata, “Modern Control Engineering”, 3rd ed. Upper saddle River, NJ: Prentice Hall1997. 8 | P a g e "