Control System

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 1 of 11

YTU FACULTY OF ELECTRICAL & ELECTRONICS ENGINEERING DEPARTMENT OF CONTROL & AUTOMATION ENGINEERING

KOM3711 CONTROL SYSTEMS, Sections 1&2, HOMEWORK-1&2

Important Notes: - When you plot, use the line width: 2 and different line styles (solid, dash, etc.) and colors for a good visibility (refer to Sln. 5). - Label each curve you plot and add figure captions under each figure like “Figure P1-2 Input f(t) and output x(t) versus time”. - You can use pen and paper for hand computations. You may print out this file and add the hand-written explanations and

computations. Matlab and Simulink figures should be also included. Do not use black background in the plots obtained from Simulink. Please note that your report should be well-organized, neat and legible.

- Submit your print outs with you name and signature in a sheet protector (transparent envelop). - No copying, no cheating, no plagiarism! They are all actual, intended or attempted deception and dishonest actions. - Individual work is expected! Otherwise you may face for undesirable consequences!

Problem 1. The translational system given has the following mass, friction and spring related values. M=2kg, fv=4N-s/m and K=8N/m,

(a) Write the equation of motion (as a differential equation). (b) Represent the system in the state-space. (c) Write the transfer function of 𝐺 𝑠 = 𝑋(𝑠)/𝐹(𝑠) (d) What would be the output 𝑥 𝑡 in meters in steady-state when the

applied input is 𝑓 𝑡 = 2 N (e) What would be the maximum value of the output? (f) Plot the input and output on the same plane in Matlab.

Solution 1.

Name, Surname: Student number: Signature: Date: 3rd December, 2018 Due Date: 19th December, 2018

Marking: Score: Marking: Score: Problem 1: 20 Problem 6: 20 Problem 2: 20 Problem 7: 20 Problem 3: 20 Problem 8: 20 Problem 4: 20 Problem 9: 20 Problem 5: 20 Problem 10: 20

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 2 of 11

Problem 2. You are requested to design an automotive suspension or shock absorber system. In order to simplify the problem to one dimensional multiple mass-spring-damper system, a quarter vehicle model is used. The system parameters and free-body diagram of such a system is shown below.

M1: Automobile body mass : 2500 kg M2: Wheel and suspension mass : 320 kg K1: Spring constant of suspension system : 80,000 N/m K2: Spring constant of wheel and tire : 500,000 N/m B: Damping constant of shock absorber : 350 N.s/m

(a) Obtain the transfer function of

𝑇' 𝑠 = GH , I ,

and 𝑇J 𝑠 = GH , (G* ,

I ,

in terms of the parameters of mass, damper and elastance (M, B and K). (b) Express the 𝑇' 𝑠 and 𝑇J 𝑠 with numerical values. (c) Plot the 𝑥' 𝑡 and 𝑥' 𝑡 − 𝑥J 𝑡 outputs of this passive suspension system for the input torque of

𝑓 𝑡 = 2,000 N PS. You may refer to the site below for a similar example, http://ctms.engin.umich.edu/CTMS/index.php?example=Suspension&section=SimulinkModeling

Solution 2.

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 3 of 11

Problem 3. The figure below on the left shows an armature controlled dc servomotor driving a load through a gear train, which is commonly used in a closed-loop control system. The schematic diagram below on the right represents the armature circuit rotating simply due to the voltage 𝑒M(𝑡) applied and the fixed magnetic field B by a permanent magnet. The armature voltage as an electrical parameter 𝑒M(𝑡) is considered to be the input to the system. The resistance and inductance of the armature circuit are 𝑅M and 𝐿M, respectively. 𝑣Q(𝑡) is the back emf and directly proportional to the rotational speed of the armature as 𝑣Q 𝑡 = 𝐾Q𝜔T(𝑡), where 𝐾Q is a constant of proportionality called the back emf constant. The torque developed by the motor is proportional to the armature current, 𝑇T 𝑡 = 𝐾U𝐼M(𝑡), where 𝐾U is the constant of proportionality and called the motor torque constant. When motor drives a load, the equivalent inertia and viscous damping at the armature are 𝐽T and 𝐷T, respectively. These entities include the corresponding armature and load parameters.

(a) Obtain the transfer function of 𝐺' 𝑠 = YZ , [\(,)

and 𝐺J 𝑠 = Y] , [\(,)

in terms of electrical and mechanical

parameters like 𝐾U, 𝐾Q, 𝐽T, 𝐷T, 𝑅M, 𝐿M. (b) Obtain 𝐺' 𝑠 and 𝐺J 𝑠 with the assumption of 𝐿M ≈ 0, which is usual for dc machines since 𝐿M ≪ 𝑅M. (c) A dc motor develops 60 Nm of torque at a speed of 500 rad/s when 12 volts are applied. It stalls out at

this voltage with 120 Nm of torque. If the inertia and damping of the armature are 7 kg-m2 and 3 Nm- s/rad, respectively, find the transfer functions, 𝐺' 𝑠 and 𝐺J 𝑠 , of this motor with the assumption of 𝐿M ≈ 0, if it drives a load with 108 kg-m2 inertia and 9 Nm-s/rad damping through a gear train as shown below on the left.

(d) Find also the transfer function from the input voltage to the output speed, i.e., 𝐺` 𝑠 = %](,) [\(,)

.

(e) Plot the step responses of GJ s and G` s for t= 0 to 20 sec. Remember that the input voltage is 12 V. (f) Determine the number of rotations within the first 18 secs from the θd vs. t plot. (g) What is the steady-state value of the speed in rad/sec and also in rpm? Confirm it from the ωd vs. t plot.

Solution 3.

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 4 of 11

Problem 4. Considering the four different responses recorded at different occasions to a step input of r(t)=2u(t) on the right-hand side,

(a) Propose a transfer function that produced the underdamped response.

(b) Write a general expression for the response of G1.

(c) What would be the order of the system G3? Why? Propose a transfer function.

(d) Which system would be critically damped? Why? Write its transfer function using the ωn of G1.

(e) Which system would be overdamped? Why?

Solution 4.

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 5 of 11

Problem 5. For each of the second order transfer function given below, where 𝐺 𝑠 = h(i) j(,)

(a) Find the locations of the poles, natural frequency and damping ratio values. (b) Determine the forms (undamped, etc.) and write the general expressions of the unit step responses. (c) Find the steady-state, css, and peak values, cmax, of the responses (depending on the forms of the

responses, you may use the formulas at the bottom of the page). (d) Revise each transfer function to produce unit step responses approximately two times slower in terms of

the settling time while maintaining the same forms and steady-state values. (e) Plot each response pairs (the original and its slower version) on the same plane as seen in the example

below.

(i) 𝐺' 𝑠 = 'J

(,;J)(,;k) , (ii) 𝐺J 𝑠 =

Jl ,*;m,;nk

, (iii) 𝐺` 𝑠 = nJJ

,o;J.,*;'mm,;':..

(iv) 𝐺. 𝑠 = n.

,*;'J,;`m , (v) 𝐺n 𝑠 =

J.(,;J:) ,;. *(,;J.)

PS. Check if the 2nd order approximations of the 3rd order systems are valid to use the formulas given. Please fill in the table below with your results you found with your computations.

(a) (b) (c) (d)

Poles, Zeros 𝝎𝒏 𝜻 Form General Expression 𝒄𝒔𝒔 𝒄𝒎𝒂𝒙 Proposed Tr. Fn.

(i)

(ii)

(iii)

(iv)

(v)

Solution 5. i) 𝐺' 𝑠 =

'J (,;J)(,;k)

= 'J ,*;':,;'m

(a) Poles: −2,−6, 𝜔y = 16 = 4 rad/s, 𝜁 = 10/(2 ∙ 4) = 1.25 (b) Since 𝜁 > 1 the form of the response will be overdamped.

𝑐 𝑡 = 𝐾' + 𝐾J𝑒(JU + 𝐾`𝑒(kU

(c) 𝑐,, = 'J 'm = 0.75. Since there is no overshoot, 𝑐TM„ = 𝑐,, = 0.75

(d) As could be seen from the formulas of both the settling and peak times, you should half the natural frequency while keeping the damping ratio unchanged to get slower responses with the same forms and

steady-state values. Therefore, the new natural frequency 𝜔yy = 'm J = 2 rad/s and the same old

damping ratio 𝜁 = 1.25 will be used in the the proposed transfer function as follows,

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 6 of 11

𝐺'' 𝑠 = 0.75 ∙ 𝜔yyJ

𝑠J + 2 ∙ 1.25 ∙ 𝜔yy𝑠 + 𝜔yyJ =

3 𝑠J + 5𝑠 + 4

(e) The original step response and its slower version are plotted as requested, can be seen below. You may use the simple Matlab command of

>> step(G1, G11), grid

Figure P5-1 The original output c1(t): solid line and its slower version c2(t): dashed line versus time.

ii) 𝐺J 𝑠 = Jl

,*;m,;nk

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 7 of 11

Problem 6. For the system given in the figure on the right-hand side, where the input is four-unit step i.e., 𝑟 𝑡 = 4𝑢(𝑡).

(a) Find the value of gain K that yields max. 20% overshoot, (b) Find the peak time for this gain, (c) Find the settling time for this gain, (d) Find the steady-state and peak values of the response for

this gain. (e) Find the value of gain K that yields max. 30% overshoot, (f) Find the peak time for this gain, (g) Find the settling time for this gain (h) Find the steady-state and peak values of the response for this gain (i) Determine the value of gain K to get the fastest response without oscillation. (j) Find the settling time for this new gain and compare it with the previous two settling times. (k) Simulate this control system in Simulink and plot the responses with 20%, 30% and fastest but no

oscillation on the same plane. Indicate settling times of the responses on the plot and list them from the smallest to the largest such as, 𝑇,„ < 𝑇,‰ < 𝑇,Š.

Solution 6.

𝐾 𝑠J + 𝑠 + 2

+ R(s) C(s)

−

1 − 2 𝐾

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 8 of 11

R(s) C(s)

−

+ +

−

Problem 7. For the feedback control system given in the figure on the right,

(a) Find the closed-loop transfer function, 𝐶(𝑠)/𝑅(𝑠).

(b) Determine the system’s stability range for gain K using Routh-Hurwitz criterion.

(c) What is the type of the system? (You need to convert the system to a simple unity feedback.)

(d) Sketch the root-locus in Matlab (you need to use the open-loop transfer function with 𝐾 = 1) and validate the stability region you found in (b).

(e) Plot the step response up to 10 seconds for the input of 1.5𝑢(𝑡) and the gain value that makes the system marginally stable.

(f) Find the steady-state error for an input step of 1.5𝑢(𝑡) for 𝐾 = 22 and plot this response together with the input up to 10 seconds in order to display the steady-state error you computed.

(g) Find the steady-state error for an input ramp of 1.5𝑡𝑢(𝑡) for 𝐾 = 22 and plot this response together with the input up to 10 seconds in order to indicate the steady-state error you computed.

Solution 7.

K

1 𝑠(𝑠 + 2)(𝑠 + 5)

𝑠

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 9 of 11

R(s) C(s) – + 𝐺i(𝑠)

PD Controller Plant

Gp(s)

Problem 8. For the PD controlled positioning system, where G•(s) = 𝐾(s + 2) and G–(s) = s+3 s(s+1)

(a) Plot the root-locus of the system using the Matlab command of “rlocus” for the positive values of gain. (b) Using the root-locus you plotted, write the range of gain for stability of the system 
 (c) What would be the value of the gain

to have a closed-loop pole at −2.5? Find it by hand and show it on the root-locus you plotted.

(d) Where would be the second pole for the same value of gain? Find it by hand and show it on the root-locus you plotted.

(e) Suppose that the plant has an additional unstable pole at +4 and hence looks as follows,

𝐺"(𝑠) = 𝑠 + 3

𝑠(𝑠 + 1)(𝑠 − 4)

Plot the root-locus of this new system, 𝐺i(𝑠)𝐺"(𝑠), for the positive values of gain in Matlab. (f) Using this root locus, write the stability range for gain of this new system. (g) What would be the frequency of sustained oscillation in Hz? (h) Plot this response for the first 4-5 periods and show that the response confirms what you found in (g).

Solution 8.

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 10 of 11

R(s) C(s) − + 𝐺i(𝑠) 𝐺"(𝑠)

Problem 9. For a similar PD controlled positioning system given in the previous problem, where

G•(s) = 𝐾(𝑠 + 4) and G–(s) = '

,(,;J)(,*;':,;.')

sketch the root-locus of the system for 𝐾 = 1 by means of the computations requested below. A tyipical cascaded control system is presented in the figure below.

(a) When sketching the root locus, if necessary, make use of the asymptotes finding σa and qa that are the intersecting point and angles with the real axis, respectively with the following formula,

𝜎M = ˜™5™š› –œ4›( ˜™5™š› ž›Ÿœ

#˜™5™š› –œ4›(#˜™5™š› –ž›Ÿœ and 𝜃M =

J¢;' $ #˜™5™š› –œ4›(#˜™5™š› –ž›Ÿœ

, where 𝑘 = 0, ±1, ±2, …

(b) If the root locus intersects the jω-axis, find the values of poles at crossing points, the value of gain K at the crossings points and write the range of gain K making the system stable.

(c) If there are complex poles find the angle of departure. (d) Find the breakaway and break-in points if they exist. (e) Plot the root-locus of the system for the positive values of gain in MATLAB. (f) Repeat the same steps for a PID controlled positioning system that has the following transfer functions.

G•(s) = ¨ ,;. ,;:.n

, and G–(s) =

' (,;J)(,*;':,;.')

Solution 9.

𝑇" = $

%& '()* , 𝑇, ≅

. )%&

, %OS= 100. 𝑒()$ '()* , 𝜁 = (45 %78/':: $*;45*(%78/'::)

, wish you all the success, Ş.N. Engin page 11 of 11

Problem 10. The root-locus of a control system, which involves a pair of complex conjugate poles and a simple zero is presented below. Specify the statements with respect to this plot as True or False, then give brief explanations for each.

i. The system is stable after a finite positive value of the gain. ii. The system is stable up to a finite positive value of the gain. iii. The system is stable for all positive values of the gain. iv. For large values of gain the system is overdamped and for small values of gain it is underdamped

providing the system is stable. v. For small values of the gain the system is undamped and for large values of gain it is underdamped. vi. The system has pure oscillations for the gain of around 6 with a frequency of 0.842 Hz. vii. The system is unstable for the values of the gain smaller than 6. viii. The system is underdamped for the gain value of 21.3. ix. The system is underdamped and making 10% max. overshoot for the gain value of 14. x. The system is marginally stable at a finite value of the gain.

Solution 10. i. True (or False), the system is stable for K…

ii. False (or True), because…

iii. …