Digital Control Systems (DCS) Lecture-1-2 Lead Compensation Dr. Imtiaz Hussain Associate Professor Mehran University of Engineering & Technology Jamshoro, Pakistan email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/ 7th Semester 14ES Note: I do not claim any originality in these lectures. The contents of this presentation are mostly taken from the book of Ogatta, Norman Nise, Bishop and B C. Kuo and various other internet sources.

Introduction Necessities of compensation A feedback control system that provides an optimum performance without any necessary adjustment is rare. Compensator: A compensator is an additional component or circuit that is inserted into a control system to equalize or compensate for a deficient performance. Necessities of compensation A system may be unsatisfactory in： Stability. Speed of response. Steady-state error.

Compensation via Root Locus Performance measures in the time domain: Peak time; Overshoot; Settling time for a step input; Steady-state error for test inputs

Commonly Used Compensators Among the many kinds of compensators, widely employed compensators are the Lead compensators Lag compensators Lag–Lead compensators

Lead Compensation 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛼 𝑇𝑠+1 𝛼𝑇𝑠+1 , (0<𝛼<1) Generally Lead compensators are represented by following transfer function or 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝛼 𝑇𝑠+1 𝛼𝑇𝑠+1 , (0<𝛼<1) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 , (0<𝛼<1)

Lead Compensation 𝐺 𝑐 𝑠 =3 𝑠+1 𝑠+10 , (𝛼=0.1)

Electronic Lead Compensator Following figure shows an electronic lead compensator using operational amplifiers. 𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 2 𝑅 4 𝑅 1 𝑅 3 𝑅 1 𝐶 1 𝑠+1 𝑅 2 𝐶 2 𝑠+1

Electronic Lead Compensator 𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 2 𝑅 4 𝑅 1 𝑅 3 𝑅 1 𝐶 1 𝑠+1 𝑅 2 𝐶 2 𝑠+1 This can be represented as Where, Then, Notice that 𝐸 𝑜 (𝑠) 𝐸 𝑖 (𝑠) = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑠+ 1 𝑅 1 𝐶 1 𝑠+ 1 𝑅 2 𝐶 2 𝐾 𝑐 = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑇= 𝑅 1 𝐶 1 𝑎𝑇= 𝑅 2 𝐶 2 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 , (0<𝛼<1) 𝑅 1 𝐶 1 > 𝑅 2 𝐶 2

Lead Compensation Techniques Based on the Root-Locus Approach. The procedures for designing a lead compensator by the root-locus method may be stated as follows: Step-1: Analyze the given system via root locus.

Step-2 From the performance specifications, determine the desired location for the dominant closed-loop poles.

Step-3 From the root-locus plot of the uncompensated system (original system), ascertain whether or not the gain adjustment alone can yield the desired closed loop poles. If not, calculate the angle deficiency. This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.

Step-4 Assume the Lead Compensator to be: Where α and T are determined from the angle deficiency. Kc is determined from the requirement of the open-loop gain.

Step-5 If static error constants are not specified, determine the location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle. If no other requirements are imposed on the system, try to make the value of α as large as possible. A larger value of α generally results in a larger value of Kv, which is desirable. Larger value of α will produce a larger value of Kv and in most cases, the larger the Kv is, the better the system performance.

Step-6 Determine the value of Kc of the lead compensator from the magnitude condition.

Example-1 Consider the position control system shown in following figure. It is desired to design an Electronic lead compensator Gc(s) so that the dominant closed poles have the damping ratio 0.5 and undamped natural frequency 3 rad/sec.

Step-1 (Example-1) Draw the root Locus plot of the given system. The closed loop transfer function of the given system is: The closed loop poles are

Step-1 (Example-1) Determine the characteristics of given system using root loci. The damping ratio of the closed-loop poles is 0.158. The undamped natural frequency of the closed-loop poles is 3.1623 rad/sec. Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable.

Step-2 (Example-1) From the performance specifications, determine the desired location for the dominant closed-loop poles. Desired performance Specifications are: It is desired to have damping ratio 0.5 and undamped natural frequency 3 rad/sec.

Desired Closed Loop Pole Step-2 (Example-1) Alternatively desired location of closed loop poles can also be determined graphically Desired ωn= 3 rad/sec Desired damping ratio= 0.5 Desired Closed Loop Pole

Desired Closed Loop Pole Step-3 (Exampl-1) From the root-locus plot of the uncompensated system ascertain whether or not the gain adjustment alone can yield the desired closed loop poles. Desired Closed Loop Pole

Desired Closed Loop Pole Step-3 (Exampl-1) If not, calculate the angle deficiency. To calculate the angle of deficiency apply Angle Condition at desired closed loop pole. -1 Desired Closed Loop Pole -2 120o 100.8o

Step-3 (Exampl-1) Alternatively angle of deficiency can be calculated as. Where are desired closed loop poles

Step-4 (Exampl-1) This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles. Note that the solution to such a problem is not unique. There are infinitely many solutions.

Step-5 (Exampl-1) Solution-1 If we choose the zero of the lead compensator at s = -1 so that it will cancel the plant pole at s =-1, then the compensator pole must be located at s =-3.

Step-5 (Example-1) 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1 𝑠+3 Solution-1 The pole and zero of compensator are determined as 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1 𝑠+3 The Value of 𝛼 can be determined as 1 𝑇 =1 yields 𝑇=1 1 𝛼𝑇 =3 yields 𝛼=0.333

Step-6 (Example-1) 𝐺 𝑐 𝑠 =0.9 𝑠+1 𝑠+3 Solution-1 The Value of Kc can be determined using magnitude condition. 𝐾 𝑐 (𝑠+1) 𝑠+3 10 𝑠(𝑠+1) 𝑠=−1.5+𝑗2.5981 =1 𝐾 𝑐 10 𝑠(𝑠+3) 𝑠=−1.5+𝑗2.5981 =1 𝐾 𝑐 = 𝑠(𝑠+3) 10 𝑠=−1.5+𝑗2.5981 =0.9 𝐺 𝑐 𝑠 =0.9 𝑠+1 𝑠+3

Example-1 Determine the circuit values Solution-1 Determine the circuit values We must choose standard values of components Let us choose 𝑅 1 =30𝐾Ω 𝐺 𝑐 𝑠 = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑠+ 1 𝑅 1 𝐶 1 𝑠+ 1 𝑅 2 𝐶 2 =0.9 𝑠+1 𝑠+3 1 𝑅 1 𝐶 1 =1 ⇒ 𝑅 1 𝐶 1 =1 𝐶 1 =33𝜇𝐹

Example-1 Determine the circuit values Solution-1 Determine the circuit values We must choose standard values of components Let us choose 𝑅 2 =27𝐾Ω 𝐺 𝑐 𝑠 = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑠+ 1 𝑅 1 𝐶 1 𝑠+ 1 𝑅 2 𝐶 2 =0.9 𝑠+1 𝑠+3 1 𝑅 2 𝐶 2 =3 ⇒ 𝑅 2 𝐶 2 =0.33 𝐶 2 =1.2𝜇𝐹

Example-1 Determine the circuit values Solution-1 Determine the circuit values We must choose standard values of components If we choose 𝑅 3 =10𝐾Ω 𝐺 𝑐 𝑠 = 𝑅 4 𝐶 1 𝑅 3 𝐶 2 𝑠+ 1 𝑅 1 𝐶 1 𝑠+ 1 𝑅 2 𝐶 2 =0.9 𝑠+1 𝑠+3 𝑅 4 𝐶 1 𝑅 3 𝐶 2 =0.9 𝑅 4 ×33𝜇 𝑅 3 ×1.2𝜇 =0.9 ⇒ 𝑅 4 =0.0327 𝑅 3 𝑅 4 =327Ω (not a standard value) 𝑅 4 =330Ω

Final Design Check Solution-1 The open loop transfer function of the designed system then becomes The closed loop transfer function of compensated system becomes. 𝐺 𝑐 𝑠 𝐺(𝑠)= 9 𝑠(𝑠+3) 𝐶(𝑠) 𝑅(𝑠) = 9 𝑠 2 +3𝑠+9

Final Design Check Solution-1 𝐺(𝑠)= 10 𝑠(𝑠+1) 𝐺 𝑐 𝑠 𝐺(𝑠)= 9 𝑠(𝑠+3)

Final Design Check Solution-1

Final Design Check Solution-1 The static velocity error constant for original system is obtained as follows. The steady state error is then calculated as 𝐾 𝑣 = lim 𝑠→0 𝑠𝐺(𝑠) 𝐾 𝑣 = lim 𝑠→0 𝑠 10 𝑠(𝑠+1) =10 𝑒𝑠𝑠= 1 𝐾 𝑣 = 1 10 =0.1

Final Design Check Solution-1 The static velocity error constant for the compensated system can be calculated as The steady state error is then calculated as 𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠) 𝐾 𝑣 = lim 𝑠→0 𝑠 9 𝑠(𝑠+3) =3 𝑒𝑠𝑠= 1 𝐾 𝑣 = 1 3 =0.333

Final Design Check Solution-1

Step-5 (Exampl-1) Solution-2 Solution-2 -1 -2 90o 49.2o -3

Step-5 (Exampl-1) 𝐺 𝑐 𝑠 =1.03 𝑠+1.5 𝑠+3.6 Solution-2 Solution-2 -1 -2 -3 𝐺 𝑐 𝑠 =1.03 𝑠+1.5 𝑠+3.6

Step-5 (Exampl-1) Solution-2 Solution-2 Uncompensated System 𝑅(𝑠) 1.03 𝑠+1.5 𝑠+3.6 10 𝑠(𝑠+1) 𝑅(𝑠) 𝐶(𝑠) Compensated System

Root Locus

Step Response

Steady State Error

Step-5 (Example-1) Solution-3 If no other requirements are imposed on the system, try to make the value of α as large as possible. A larger value of α generally results in a larger value of Kv, which is desirable. Procedure to obtain a largest possible value for α. First, draw a horizontal line passing through point P, the desired location for one of the dominant closed-loop poles. This is shown as line PA in following figure. Draw also a line connecting point P and the origin O. P A -1 -2 -3 O

Step-5 (Example-1) Solution-3 Bisect the angle between the lines PA and PO, as shown in following figure. O P A -2 -1 -3 -2 -1

Step-5 (Example-1) Solution-3 Draw two lines PC and PD that make angles ± 𝜃 𝑑 2 with the the bisector PB. The intersections of PC and PD with the negative real axis give the necessary locations for the pole and zero of the lead network. O P A C D -2 -1 -3 -2 -1 B

𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458 Step-5 (Example-1) Solution-3 The lead compensator has zero at s=–1.9432 and pole at s=–4.6458. Thus, Gc(s) can be given as -1 -2 B -3 O P A C D 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458

𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458 Step-5 (Example-1) Solution-3 For this compensator value of 𝛼 is Also 𝐺 𝑐 𝑠 = 𝐾 𝑐 𝑠+ 1 𝑇 𝑠+ 1 𝛼𝑇 = 𝐾 𝑐 𝑠+1.9432 𝑠+4.6458 1 𝑇 =1.9432 yields 𝑇=0.514 1 𝛼𝑇 =4.6458 yields 𝛼=0.418

Step-6 (Example-1) Solution-3 Determine the value of Kc of the lead compensator from the magnitude condition. 𝐺 𝑠 𝐺 𝑐 𝑠 𝐻(𝑠)= 10 𝐾 𝑐 (𝑠+1.9432) 𝑠(𝑠+1)(𝑠+4.6458) 10 𝐾 𝑐 (𝑠+1.9432) 𝑠(𝑠+1)(𝑠+4.6458) 𝑠=−1.5+𝑗2.5981 =1

Step-6 (Example-1) The Kc is calculated as Solution-3 The Kc is calculated as Hence, the lead compensator Gc(s) just designed is given by 𝐾 𝑐 =1.2287 𝐺 𝑐 𝑠 =1.2287 𝑠+1.9432 𝑠+4.6458

Final Design Check Solution-3 Compensated System Uncompensated System Desired Closed Loop Pole Uncompensated System Desired Closed Loop Pole Compensated System

Final Design Check Solution-3

Final Design Check Solution-3 It is worthwhile to check the static velocity error constant Kv for the system just designed. Steady state error is 𝐾 𝑣 = lim 𝑠→0 𝑠 𝐺 𝑐 𝑠 𝐺(𝑠) 𝐾 𝑣 = lim 𝑠→0 𝑠 1.2287 𝑠+1.9432 𝑠+4.6458 10 𝑠(𝑠+1) =5.139 𝑒𝑠𝑠= 1 𝐾 𝑣 = 1 5.139 =0.194

Final Design Check Solution-3

Final Design Check

Exampl-2 Design a lead compensator for following system. The damping ratio of closed loop poles is 0.5 and natural undamped frequency 2 rad/sec. It is desired to modify the closed loop poles so that natural undamped frequency becomes 4 rad/sec without changing the damping ratio.

Electrical Lead Compensator 𝑉𝑜(𝑠) 𝑉𝑖(𝑠) 𝑉𝑜(𝑠) 𝑉𝑖(𝑠) 𝑐 = 𝑅 2 𝑅 1 + 𝑅 2 𝑅 1 𝐶𝑠+1 𝑅 1 𝑅 2 𝑅 1 + 𝑅 2 𝐶𝑠+1 𝑎𝑇= 𝑅 1 𝑅 2 𝐶 𝑅 1 + 𝑅 2 𝑇= 𝑅 1 C 𝑎= 𝑅 2 𝑅 1 + 𝑅 2 𝐾 𝑐 =1

Example-3 Consider the model of space vehicle control system depicted in following figure. Design an Electrical lead compensator such that the damping ratio and natural undamped frequency of dominant closed loop poles are 0.5 and 2 rad/sec.

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