![=>
𝑅(𝑠)
1+𝐿(𝑠)
= 𝑅(𝑠) − 𝐶(𝑠)
=> 𝑅(𝑠) = 𝑅(𝑠) − 𝑅(𝑠)𝐿(𝑠) − 𝐶(𝑠) − 𝐶(𝑠)𝐿(𝑠)
=>
𝐶(𝑠)
𝑅(𝑠)
=
𝐿(𝑠)
1+𝐿(𝑠)
but 𝐿(𝑠) =
𝜔𝑛
2
𝑠(𝑠+2𝜁𝜔𝑛)
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
𝑠(𝑠+2𝜁𝜔𝑛)
1+
𝜔𝑛
2
𝑠(𝑠+2𝜁𝜔𝑛)
simplifying this equation gives:
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
𝑆2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2
2. Obtain analytically, the time response of the system c(t) for a unit step input r(t) when ﻉ <1.
ANSWER
For a unit step function, r(t) when >ﻉ1 as the input of the system,
𝐶(𝑠)
𝑅(𝑠)
can be written as;
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
(𝑠+𝜁𝜔𝑛+𝑗𝜔𝑑)(𝑠+𝜁𝜔𝑛−𝑗𝜔𝑑)
Where 𝜔𝑑 = 𝜔𝑛√1 − 𝜁2 is called the damped natural frequency
𝐶(𝑠) = 𝑅(𝑠).
𝜔𝑛
2
𝑆2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2 for a unit-step input, 𝑅(𝑠) =
1
𝑠
=> 𝐶(𝑠) =
𝜔𝑛
2
(𝑆2+2𝜁𝜔𝑛+𝜔𝑛
2)𝑠
Taking the partial fraction, we have;
𝐶(𝑠) =
1
𝑠
−
𝑠+𝜁𝜔𝑛
(𝑠+𝜁𝜔𝑛)2+𝜔𝑑
2 −
𝜁𝜔𝑛
(𝑠+𝜁𝜔𝑛)2+𝜔𝑑
2
Hence 𝑐(𝑡) = 𝐿−1[𝐶(𝑠)]
𝑐(𝑡) = 1 − 𝑒−𝜁𝜔𝑛𝑡
(𝑐𝑜𝑠𝜔𝑑𝑡 +
𝜁
√1−𝜁2
𝑠𝑖𝑛𝜔𝑑𝑡)
𝑐(𝑡) = 1 −
𝑒−𝜁𝜔𝑛𝑡
√1−𝜁2
𝑠𝑖𝑛(𝜔𝑑𝑡 + 𝜙) where 𝜙 = tan−1 √1−𝜁2
𝜁
3) For 𝑤 𝑛 = 10 𝑟 𝑎 𝑑 𝑠 /𝑠 𝑒 𝑐 , 𝑎 𝑛 𝑑 =ﻉ 0.4, calculate (i) the damped natural frequency, (ii) the peak response,
(iii) the time to peak, (iv) the rise time, (v) the settling time for a 5% tolerance, and (vi) the maximum overshoot.
ANSWER
𝑤𝑛 = 10 𝑟𝑎𝑑𝑠/𝑠𝑒𝑐, 𝑎𝑛𝑑 4.0 = ع](https://image.slidesharecdn.com/feedbackreport-240614124224-c1cc4c4b/85/FEEDBACK-REPORT-for-control-system-docx-9-320.jpg)
FEEDBACK REPORT for control system .docx
- 1. UNIVERSITY OF BUEA FACULTY OF ENGINEERING AND TECHNOLOGY COMPUTER ENGINEERING FEEDBACK CONTROL SYSTEM COURSE INSTRUCTOR: DR BAZIL NAME MATRICULE SPECIALTY CHE KASSINA KUM FE21A158 SE EEF460 EXPERIMENT 1 AND 2
- 2. EXPERIMENT 1 a) Differential Equation: The RLC network can be described using Kirchhoff's voltage law (KVL). Applying KVL to the loop containing V2(t), the resistor R, the inductor L, and the capacitor C, we get: V1(t)−VR(t)−VL(t)−V2(t)=0 Where: VR(t) is the voltage drop across the resistor, which is R⋅i(t). VL(t) is the voltage drop across the inductor, which is L⋅ 𝑑𝐼(𝑡) 𝑑𝑡 V1(t) and V2(t) are the voltages across the capacitor and the inductor, respectively. Since) V1(t)=VC(t) andV2(t)=VL(t), we can rewrite the equation as: V1(t)−R⋅i(t)−L. 𝑑𝑖(𝑡) 𝑑𝑡 - V2(t)=0 V1(t) = i(t)R + 𝐿 𝑑𝑖 𝑑𝑡 + ∫ 𝑖 𝑐 𝑖(𝑡)𝑑𝑡 V2(t) = ∫ 𝑖 𝑐 𝑖(𝑡)𝑑𝑡 b)This is the block diagram
- 3. c)Taking the Laplace transform of V1 and V2 we have V1(S) = I(s)R + LI(s)S + 𝐼(𝑠) 𝐶𝑆 V2(S) = 𝐼(𝑠) 𝐶𝑆 T.F = V2(S)/V1(S) T.F = 𝐼(𝑠) 𝐶𝑆 I(s)R + LI(s)S + 𝐼(𝑠) 𝐶𝑆 T.F = 1 𝑅𝐶𝑆+𝐿𝐶𝑆2+1 d) W(S) = 𝐺(𝑆) 1+𝐿(𝑆) L(S) = G(S)H(S) but for a unity feedback, H(S) = 1 L(S) = G(S) W(S) = 𝐿(𝑆) 1+𝐿(𝑆) W(S) = 1 𝑅𝐶𝑆+𝐿𝐶𝑆2+1 1+ 1 𝑅𝐶𝑆+𝐿𝐶𝑆2+1 W(S) = 1 𝐿𝐶𝑆2+𝑅𝐶𝑆+1 e) 𝑆2 + 𝑅 𝐿 𝑆 + 1 𝐿𝐶 = 0 S=(-R+-√𝑅2 + 4 𝐿 𝐶 ⁄ )2C R=2√ 𝐿 𝐶 EXERCISE1 f) Model the differential equation of the system using Simulink blocks To model the differential equation of the given electrical system using Simulink blocks, we first express the system's behavior in terms of differential equations. The given electrical system comprises a resistor (R), an inductor (L), and a capacitor (C) connected in series. The differential equation governing this system is derived from Kirchhoff's voltage law (KVL): Vin(t)−VR(t)−VL(t)−VC(t)=0
- 4. g) This MATLAB code calculates the transfer function G(s) using the given values of R and L, then plots the step response of the system. h) The plot show how the voltage across the component changes over time in response to the step input. The shape of the response curve depend on the characteristics of the components (R, L, C) and the input voltage. I analyze parameters like rise time, settling time, overshoot, and steady- state value from the plot to understand the system's dynamic behavior.
- 5. i) the settling time it is deduce from the graph the settling time for a 5% tolerance, 𝑡𝑠 = 1.2 EXERCISE 2 m as the mass of the object.
- 6. x as the displacement of the mass from the equilibrium position. f(t) as the input force. The equation of motion can be written as: mdt2d2x=−kx−cdtdx+f(t) Where: k is the spring constant. c is the damping coefficient. Taking the Laplace transform of the equation yields: ms2X(s)=−kX(s)−csX(s)+F(s) Where the Laplace transforms of x(t) and f(t) respectively. Rearranging terms to isolate X(s), we get: X(s)(ms2+cs+k)=F(s) So the transfer function G(s) of the system is: (s)=F(s)X(s)= 1 𝑚𝑠2+𝑐𝑠+𝑘 This represents the relationship between the input force f(t) and the output displacement (t) in the frequency domain. a) This code generate a plot showing the step response of the electrical system and analyze its behavior over time. m=2 kg (mass of the object) k=16 N/m (spring constant) b=4 N·s/m (damping coefficient) f ranges from 0 to 8 N (input force) Using these parameters, we can calculate the transfer function of the system: G(s)= 1 𝑚𝑠2+𝑐𝑠+𝑘1 = 1 2𝑠2+4𝑠+161 Now, we can analyze the system's response to different input forces f using MATLAB. We'll plot the system's response for f ranging from 0 to 8 N.
- 7. b) Analyzing the plot provide various aspects of the system's behavior, such as rise time (time taken to reach a specified percentage of the final value), settling time (time taken to reach and stay within a specified tolerance of the final value), overshoot (maximum deviation from the final value before settling), and steady-state value. These parameters help evaluate the system's dynamic performance. c)
- 8. D) the settling time the settling time for a 5% tolerance, 𝑡𝑠 = 2. EXPERIMENT 2 1. Show that the closed-loop transfer function 𝐶(𝑠) 𝑅(𝑠) = 𝜔𝑛 2 𝑆2+2𝜁𝜔𝑛𝑠+𝜔𝑛 2 for unity gain feedback. ANSWER 𝐸(𝑠) = 𝑅(𝑠) − 𝐶(𝑠) but 𝐸(𝑠) = 𝑅(𝑠) 1+𝐿(𝑠)
- 9. => 𝑅(𝑠) 1+𝐿(𝑠) = 𝑅(𝑠) − 𝐶(𝑠) => 𝑅(𝑠) = 𝑅(𝑠) − 𝑅(𝑠)𝐿(𝑠) − 𝐶(𝑠) − 𝐶(𝑠)𝐿(𝑠) => 𝐶(𝑠) 𝑅(𝑠) = 𝐿(𝑠) 1+𝐿(𝑠) but 𝐿(𝑠) = 𝜔𝑛 2 𝑠(𝑠+2𝜁𝜔𝑛) 𝐶(𝑠) 𝑅(𝑠) = 𝜔𝑛 2 𝑠(𝑠+2𝜁𝜔𝑛) 1+ 𝜔𝑛 2 𝑠(𝑠+2𝜁𝜔𝑛) simplifying this equation gives: 𝐶(𝑠) 𝑅(𝑠) = 𝜔𝑛 2 𝑆2+2𝜁𝜔𝑛𝑠+𝜔𝑛 2 2. Obtain analytically, the time response of the system c(t) for a unit step input r(t) when ﻉ <1. ANSWER For a unit step function, r(t) when >ﻉ1 as the input of the system, 𝐶(𝑠) 𝑅(𝑠) can be written as; 𝐶(𝑠) 𝑅(𝑠) = 𝜔𝑛 2 (𝑠+𝜁𝜔𝑛+𝑗𝜔𝑑)(𝑠+𝜁𝜔𝑛−𝑗𝜔𝑑) Where 𝜔𝑑 = 𝜔𝑛√1 − 𝜁2 is called the damped natural frequency 𝐶(𝑠) = 𝑅(𝑠). 𝜔𝑛 2 𝑆2+2𝜁𝜔𝑛𝑠+𝜔𝑛 2 for a unit-step input, 𝑅(𝑠) = 1 𝑠 => 𝐶(𝑠) = 𝜔𝑛 2 (𝑆2+2𝜁𝜔𝑛+𝜔𝑛 2)𝑠 Taking the partial fraction, we have; 𝐶(𝑠) = 1 𝑠 − 𝑠+𝜁𝜔𝑛 (𝑠+𝜁𝜔𝑛)2+𝜔𝑑 2 − 𝜁𝜔𝑛 (𝑠+𝜁𝜔𝑛)2+𝜔𝑑 2 Hence 𝑐(𝑡) = 𝐿−1[𝐶(𝑠)] 𝑐(𝑡) = 1 − 𝑒−𝜁𝜔𝑛𝑡 (𝑐𝑜𝑠𝜔𝑑𝑡 + 𝜁 √1−𝜁2 𝑠𝑖𝑛𝜔𝑑𝑡) 𝑐(𝑡) = 1 − 𝑒−𝜁𝜔𝑛𝑡 √1−𝜁2 𝑠𝑖𝑛(𝜔𝑑𝑡 + 𝜙) where 𝜙 = tan−1 √1−𝜁2 𝜁 3) For 𝑤 𝑛 = 10 𝑟 𝑎 𝑑 𝑠 /𝑠 𝑒 𝑐 , 𝑎 𝑛 𝑑 =ﻉ 0.4, calculate (i) the damped natural frequency, (ii) the peak response, (iii) the time to peak, (iv) the rise time, (v) the settling time for a 5% tolerance, and (vi) the maximum overshoot. ANSWER 𝑤𝑛 = 10 𝑟𝑎𝑑𝑠/𝑠𝑒𝑐, 𝑎𝑛𝑑 4.0 = ع
- 10. ii) the damped natural frequency 𝜔𝑑 = 𝜔𝑛√1 − 𝜁2 => 𝜔𝑑 = 10√1 − 0 ⋅ 42 𝜔𝑑 = 9.17 iii) the peak response 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1 + 𝑒 −𝜁( 𝜋 √1−𝜁2 ) 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1 + 𝑒 −0.4( 𝜋 √1−0.42 ) Which give 𝑐(𝑡)𝑃𝑒𝑎𝑘 ≈ 1.25 iv) the time to peak 𝑡𝑝 = 𝜋 𝜔𝑛√1−𝜁2 => 𝑡𝑝 = 𝜋 10√1−0.42 which gives 𝑡𝑝 ≈ 0.29 v) the rise time 𝑡𝑟 = 𝜋−𝜃 𝜔𝑑 but 𝑐𝑜𝑠 𝜃 = 𝜁 => 𝜃 = cos−1 𝜁 => 𝑡𝑟 = 𝜋−cos−1 𝜁 𝜔𝑑 𝑡𝑟 = 𝜋−cos−1 0.4 9.17 which gives 𝑡𝑟 ≈ 0.219 vi) the settling time for a 5% tolerance, 𝑡𝑠 = 3 𝜁𝜔𝑛 (5% tolerance) => 𝑡𝑠 = 3 0.4(10) which gives => 𝑡𝑠 = 0.75 vii) the maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% . Now, 𝑐(∞) = 𝑐(𝑡)𝑡→∞ ≈ 1 √1−𝜁2 = 𝑐(𝑡)𝑓𝑖𝑛𝑎𝑙 => 𝑐(∞) ≈ 1 √1−0.42 ≈ 1.091 and 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.254 Therefore, 𝑀𝑝 = 1.254−1.091 1.091 × 100% which gives 𝑀𝑝 = 14.940% 4) Plot using MATLAB (two separate plots): (i) the Pole-Zero map and (ii) the time response for 𝑤𝑛 = 10 𝑟𝑎𝑑𝑠/𝑠𝑒𝑐, 𝑎𝑛𝑑 =ع 0.4. ANSWER Below is a figure that show the MATLAB script code and the screenshot result for the pole zero map
- 11. The damping ratio (zeta) affects the pole positions along the real axis and their imaginary parts, impacting the system's damping characteristics. With zeta = 0.4, the poles are in the left-half plane, indicating underdamped behavior. The closer the poles are to the imaginary axis, the more oscillatory the system's response. No numerator coefficients result in no zeros, meaning no specific frequencies where the response is cancelled out. The positions of poles depend on zeta and undamped natural frequency (omega_n), indicating the system's damping behavior and stability. ii) The figure below shows the time response and the code snippet for the system. The graph shows the system's response to a unit step input, providing insights into its transient behavi damping characteristics, and stability. The system appears stable, with damped oscillations that eventually reach a steady-state value, indicating underdamped behavior but overall stability.
- 12. 5) From the time response plot, determine (i) the damped natural frequency, (ii) the peak response, (iii) the time to peak, (iv) the rise time, (v) the settling time for a 5% tolerance, and (vi) the maximum overshoot. Compare these results with those obtained in (3) above and conclude. ANSWERS i) the damped natural frequency, 𝜔𝑑 The damped natural frequency 𝜔𝑑, is related to the period of oscillations By the formula 𝜔𝑑 = 2𝜋 𝑇 , where T is the period of oscillation. From the graph, we see that 𝑇 = 0.9 Therefore, 𝜔𝑑 = 2𝜋 0.9 ≈ 6.98 ii) the peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.25 iii) the time to peak, 𝑡𝑝 = 0.33 iv) the rise time, 𝑡𝑟 ≈ 0.26 v) the settling time for a 5% tolerance, => 𝑡𝑠 = 0.40 vi) the maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% From the graph, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.25 and 𝑐(∞) = 1.00 Therefore, 𝑀𝑝 = 1.25−1 1 × 100% giving 𝑀𝑝 = 25% When comparing these values with the theoretical ones obtained earlier, we observe a close match, indicating that the theoretical analysis is in good agreement with the actual behavior of the system. 6) For the same 𝑤𝑛 and 0 ≤ ﻉ ≤ 1.2 (𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 𝑏𝑦 0.1), plot: (i) the pole-zero map for the system and (ii) the time response for the system.
- 13. i)The pole-zero maps for the system. 𝜔𝑛 = 10 and 0 ≤ 𝜁 ≤ 1.2 (increment by 0.1) i) ii) The time response for the system. Each curve in the time response plot corresponds to a different damping ratio. 7) Observe carefully the different curves of time response plotted on the same graph as ﻉ varies and obtain all the indices of time response for each curve as requested in (5) above ANSWERS a) 𝜁 = 0 i) The damped natural frequency. The damped natural frequency 𝜔𝑑, is related to the period of oscillations. By the formula 𝜔𝑑 = 2𝜋 𝑇 , where T is the period of oscillation. From the graph, we see that 𝑇 = 0.6
- 14. Therefore, 𝜔𝑑 = 2𝜋 0.63 ≈ 9.973 ii) the peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 2.00 iii) the time to peak, 𝑡𝑝 = 0.314 iv) the rise time, 𝑡𝑟 ≈ 0.157 v) the settling time for a 5% tolerance, => 𝑡𝑠 = ∞ vi) the maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% From the graph, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 2.00 and 𝑐(∞) = 𝑁𝑜 𝑠𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 Therefore, 𝑀𝑝 ℎ𝑎𝑠 𝑜𝑛 𝑣𝑎𝑙𝑢𝑒 b) 𝜁 = 0.1 i) The damped natural frequency. The damped natural frequency 𝜔𝑑, is related to the period of oscillations. By the formula 𝜔𝑑 = 2𝜋 𝑇 , where T is the period of oscillation. From the graph, we see tha 𝑇 = 0.60 Therefore, 𝜔𝑑 = 2𝜋 0.60 ≈ 10.472 ii) the peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.725 iii) the time to peak, 𝑡𝑝 = 0.31 iv) the rise time, 𝑡𝑟 ≈ 0.160 v) the settling time for a 5% tolerance, => 𝑡𝑠 = 4.9 vi) the maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% From the graph, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.725 and 𝑐(∞) = 1.0 Therefore, 𝑀𝑝 = 1.725−1.0 1.0 = 0.725 c) 𝜁 = 0.2 The damped natural frequency. The damped natural frequency 𝜔𝑑, is related to the period of oscillations. By the formula 𝜔𝑑 = 2𝜋 𝑇 , where T is the period of oscillation. From the graph, we s that 𝑇 = 0.60 Therefore, 𝜔𝑑 = 2𝜋 0.60 ≈ 10.472 the peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.525
- 15. the time to peak, 𝑡𝑝 = 0.32 the rise time, 𝑡𝑟 ≈ 0.169 the settling time for a 5% tolerance, 𝑡𝑠 = 2.4 the maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% From the graph, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.525 and 𝑐(∞) = 1.0 Therefore, 𝑀𝑝 = 1.525−1.0 1.0 = 0.525 d) 𝜁 = 0.3 Damped natural frequency 𝑇 = 0.675 => 𝜔𝑑 = 2𝜋 0.675 ≈ 9.308 Peak response 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.35 Time to peak, 𝑡𝑝 = 0.33 Rise time, 𝑡𝑝 = 0.201 Settling time for a 5% tolerance, 𝑡𝑠 = 1.7 Maximum overshoot, 𝑀𝑝 = 𝑐(𝑡)𝑃𝑒𝑎𝑘−𝑐(∞) 𝑐(∞) × 100% From the graph, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.35 and 𝑐(∞) = 1.0 Therefore, 𝑀𝑝 = 1.35−1.0 1.0 = 0.35 e) 𝜁 = 0.4 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.9 ≈ 6.98 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.25 Time to peak, 𝑡𝑝 = 0.33 Rise time, 𝑡𝑟 = 0.26 Settling time for a 5% tolerance, 𝑡𝑠 = 0.40 Maximum overshoot, 𝑀𝑝 = 1.25−1.0 1.0 × 100% = 25% f) 𝜁 = 0.5 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.9 ≈ 6.981 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.116 Time to peak, 𝑡𝑝 = 0.34 Rise time, 𝑡𝑟 = 0.300 Settling time for a 5% tolerance, 𝑡𝑠 = 0.9 Maximum overshoot, 𝑀𝑝 = 1.116−1.0 1.0 × 100% = 11.6%
- 16. g) 𝜁 = 0.6 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.85 ≈ 7.392 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.10 Time to peak, 𝑡𝑝 = 0.35 Rise time, 𝑡𝑟 = 0.299 Settling time for a 5% tolerance, 𝑡𝑠 = 1.0 Maximum overshoot, 𝑀𝑝 = 1.10−1.0 1.0 × 100% = 10% h) 𝜁 = 0.7 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.60 ≈ 8.976 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.05 Time to peak, 𝑡𝑝 = 0.45 Rise time, 𝑡𝑟 = 0.261 Settling time for a 5% tolerance, 𝑡𝑠 = 0.90 Maximum overshoot, 𝑀𝑝 = 1.05−1.0 1.0 × 100% = 5% i) 𝜁 = 0.8 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.75 ≈ 7.953 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.025 Time to peak, 𝑡𝑝 = 0.48 Rise time, 𝑡𝑟 = 0.314 Settling time for a 5% tolerance, 𝑡𝑠 = 0.70 Maximum overshoot, 𝑀𝑝 = 1.025−1.0 1.0 × 100% = 2.5% j) 𝜁 = 0.9 Damped natural frequency, 𝜔𝑑 = 2𝜋 0.8 ≈ 7.854 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.005 Time to peak, 𝑡𝑝 = 0.65 Rise time, 𝑡𝑟 = 0.343 Settling time for a 5% tolerance, 𝑡𝑠 = 0.65 Maximum overshoot, 𝑀𝑝 = 1.005−1.0 1.0 × 100% = 0.5% k) 𝜁 = 1.0 Damped natural frequency, 𝑁𝑜 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛𝑠 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.0 Time to peak, 𝑡𝑝 = 1.01 Rise time, 𝑁𝑜 𝑣𝑎𝑙𝑢𝑒 Settling time for a 5% tolerance, 1.0 Maximum overshoot, 𝑁𝑜𝑛𝑒 l) 𝜁 = 1.1 Damped natural frequency, 𝑁𝑜 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛
- 17. Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 1.0 Time to peak, 𝑡𝑝 = 1.0 Rise time, 𝑁𝑜 𝑣𝑎𝑙𝑢𝑒 Settling time for a 5% tolerance, 0.9 Maximum overshoot, 𝑁𝑜 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛𝑠 m) 𝜁 = 1.2 Damped natural frequency, 𝑁𝑜 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 Peak response, 𝑐(𝑡)𝑃𝑒𝑎𝑘 = 0.9 Time to peak, 𝑡𝑝 = 1.0 Rise time, 𝑁𝑜 𝑣𝑎𝑙𝑢𝑒 Settling time for a 5% tolerance, 0.9 Maximum overshoot, 𝑁𝑜𝑛𝑒 8) Plot Time to peak Tp= f( )ﻉ and conclude. Note: The values of Tp are those obtained in (7) for different values of ANSWER This code will generate a plot showing how Tp changes with different damping ratios (ζ). Adjust the natural frequency (ωn) as needed. 9) What conclusion can you draw by observing the poles when =ﻉ 0, ζ < 1, ζ = 1, ζ >1? ANSWER By observing the poles of a system with varying damping ratios (ζ), I can draw the following conclusions:
- 18. When ζ=0: Undamped system, poles are purely imaginary. 0<ζ<1: Underdamped system, poles are complex conjugate with a real part less than 1. When ζ=1: Critically damped system, poles are real and repeated. ζ>1: Overdamped system, poles are real and distinct, both negative. 10 0ζ<0: Potentially unstable behavior. 10<ζ<1: Underdamped response with oscillations. ζ=1: Critically damped response without oscillations. 1ζ>1: Overdamped response without oscillations.