f(t) m x(t) fd(t) LINEAR CONTROL C (Ns/m) k (N/m) x(t) is the output variable and is measured by a displacement sensor. The output of a sensor is always an electrical output mainly voltage. In the Figure, f(t) is the actuating force that excites the system in order to produce the desired output x(t). On the other hand, fd(t) is the disturbance that works against actuating force and prevents the system to reach the desired output.

Actuators: V1 Gact (Force, Moment, etc.) Actuators are used to drive the engineering systems. DC motors, servo motors, hydraulic and pneumatic system elements are the basic type of actuating elements. The inputs of the actuators are also voltage. One can write a transfer function in order to model the behaviour of an actuator. (Voltage) Gact V1 (Force, Moment, etc.)

Sensors: Sensors are the basic components of a control system that are used to measure the output/outputs of the system in order to obtain a feedback signal. The output of a sensor is mainly a voltage signal. One can write a transfer function in order to relate the input and output of the sensor. The input of a sensor is a physical event, temperature, pressure, displacement, velocity, acceleration, magnetic flux, rate of the fluid flow etc.) (Voltage) Displacement, pressure, temperature V2 Gsens

+ Gact OPEN-LOOP CONTROL -Fd Gsens V2 V1 F(t) x(t) 0.1 Volt/m 100 N/Volt 0.1 Volt/m

Let’s take Fd=0

Objective: The desired steady-state value of x(t) is 0. 05 m Objective: The desired steady-state value of x(t) is 0.05 m. In this case, the sensor output voltage is calculated as The input voltage which is sufficient to obtain the output voltage, V2 is calculated as m=200 kg c=300 Ns/m k=5000 N/m The steady-state behaviour doesn’t include time variations and the system considered as static, then

V2(t) V2ss=0.005 Volt The Figure shows that, desired x(t) value is achieved if the external disturbance Fd is zero. If Fd is not zero!

Let’s take Fd=80 N.

Time (second) V2(t) Fd=0 Fd=80 N

+ - Gcont Gact -Fd Gsens V2 V1 F(t) x(t) E Vref Can we obtain desired x(t) even if Fd is not zero. This is the main concern of a control process. This can be achieved via the impementation of a closed-loop control. Vref -Fd Gsens V2 + Gact V1 F(t) x(t) 100 N/Volt 0.1 Volt/m Gcont - E (Error) Feedback

The form of the transfer function of the controller is the most important parameter in a control application. Gcont can be Proportional, Derivative, Integral or a suitable combination of these three choices.

For proportional control, Gcont=Kp Kp is the proportional constont, Ki is the integral constant and Kd is the derivative constant. For proportional control, Gcont=Kp Vref=0.005 Volt, Fd=80 N Let’s take Kp=1000

Open-loop Closed-loop, Kp=1000. Desired V2(t) Time (second)

For Kp=10000

ess ess V2(t) Time (second) Kp Closed-loop, Kp=10000 Desired Open-loop

By MATLAB Simulink

For proportional-derivative (PD) control, Gcont=Kp+Kds Kp=10000, Kd=100

PD Control by Simulink Reference Voltage:0.005 Volt Disturbance:-80 N

PD Control by Simulink Desired Time (second) V2(t)

PID Control by Simulink Gc Controller

Effect of the PID parameters on the output P Control: Kp=1000 PD Control: Kp=1000, Kd=1000 PI Control: Kp=1000, KI=1000 PID Control: Kp=1000, KI=1000, Kd=1000 Desired Output

Effect of the PID parameters on the output

STABILITY OF LINEAR CONTROL SYSTEMS Among the many forms of performance specifications used in design, the most important requirement is that the system be stable. An unstable system is generally considered to be useless. The stability of a control system is directly related to the location of the roots of the charactersitic equation D(s). s-plane j Stable region Unstable region  Stable region Unstable region

Example 1: Consider an open-loop control system 100 V1(s) V2(s) D(s)

s-plane D(s)=s2+9s+15 by MATLAB s1=-2.2087 >>a=[1 9 15]; -6.7913 by MATLAB s1=-2.2087 s2= -6.7913 >>a=[1 9 15]; >> roots (a) The two roots of D(s) are in the left hand side of the s-plane. It can be easily said that the open-loop system is stable. Time (s) V2(t) V2ss=6.66 Response of the system to a unit step input can be obtained by MATLAB as >>ns=[100]; >>ds=[1 9 15], >>step(ns, ds)

Example 2: V1(s) V2(s) 100 D(s) >>a=[1 3 75]; >> roots (a) s1=-1.5+8.53i s2=-1.5-8.53i

Response of the system to a unit step input can be obtained by MATLAB as >>ds=[1 3 75], >>step(ns, ds) s-plane -1.5+8.53 i V2ss=1.33 -1.5-8.53 i

100 V1(s) V2(s) Example 3: D(s) >>a=[1 0 15]; >> roots (a) s1=+3.8730i s2= -3.8730i The real part of the roots are zero. No roots on the right-half s-plane and system is said to be marginally stable.

Response of the system to a unit step input can be obtained by MATLAB as >>ds=[1 0 15], >>step(ns, ds) 3.873 i -3.873 i s-plane V2(t) Time (s)

Example 4: V1(s) V2(s) 100 D(s) The real part of the roots are positive. Two roots are on the right-half s-plane and system is said to be unstable. >>a=[1 -2 16]; >> roots (a) s1=1+3.8730i s2= 1-3.8730i

Response of the system to a unit step input can be obtained by MATLAB as >>ds=[1 -2 16], >>step(ns, ds) 1+3.873 i 1-3.873 i s-plane V2(t) Unstable Time (s)

K + V2(s) R(s) - Example 4: System Consider a closed-loop control system V2(s) K R(s) + - Controller Mason Formula:

K R(s) + - V2(s)

K + V2(s) R(s) - By using Mason formula D(s) Forward Path Loop By using Mason formula D(s) Is the closed-loop control system stable? For what values of K, the system is said to be stable?

Routh-Hurwitz criterion: This criterion is an algebraic method that provides information about the stability of a linear time invariant system that has a characteristic equation with constant coefficients. The criterion tests whether any of the roots of characteristic equation lie in the right-half s-plane. The number of roots that lie on the j axis and in the right half plane is also indicated. s3 1 K+9 s2 5 3K+5 s1 0 s0 3K+5 0 All terms must have the same sign. The number of positive roots is equal to the change in the sign.

K>-1.66 System is stable for all positive K values. K+9 >0 , K>-9 3K+5 > 0, K> -1.66 0.4K+8>0, 0.4K>-8, K>-8/0.4, K>-20 K>-1.66 System is stable for all positive K values. s1= -1.1333 + 3.3941i s2= -1.1333 - 3.3941i s3= -2.7334 For K=10; Response of the system to a unit step input can be obtained by MATLAB as >>ns=[10 30]; >>ds=[1 5 19 35], >>step(ns, ds) s-plane -1.1333+3.3941i -2.7334 -1.1333-3.3941i

Response to a unit step input, V2(t) Time (s) Stable

Example 1: K R(s) + - V2(s) D(s)

K>0, 24K>0, 80K>0  K>0 s5 1 50000 24K s4 600 K 80K s3 0 s2 0 s1 s0

K=3x107 K=2.1386x107 K=5x105 K=2.34x105 K=1x105 Unstable Stable Unstable

K Example 2: + V2(s) R(s) - s3 1 200 s2 30 K s1 0 s0 K<6000 Kcritical=6000

By MATLAB, >>K=2000; >>ns=[K]; >>ds=[1 30 200 K]; >>step(ns,ds) V2(t) Time (sec) K=2000 Stable

By MATLAB, >>K=6000; Marginally stable >>ns=[K]; >>ds=[1 30 200 K]; >>step(ns,ds) Marginally stable K=6000 V2(t) D(s)=s3+30s2+200s+6000 >>ds=[1 30 200 6000]; >>roots(ds) p1=-30.0000 p2= 0.0000 +14.1421i p3=0.0000 -14.1421i Time (sec)

By MATLAB, >>K=8000; >>ns=[K]; >>ds=[1 30 200 K]; >>step(ns,ds) K=6000 V2(t) Time (sec) Unstable

ZIEGLER-NICHOLS DESIGN The coefficients of the controller can be chosen according to Kcritical and critical which are calculated from Routh tabulation. Ziegler and Nichols recommended below relationships while determining the controller coefficients. For P control  For PI control  For PIDcontrol 

Example 2: + V2(s) R(s) Gc(s) -

APPLICATION OF THE MASON FORMULA K K1 + - R(s) Y(s)

Example: K K1 + - R(s) Y(s) N(s) s

With respect to R(s) With respect to N(s)

STEADY STATE ERRORS Y(s) R(s) E(s) + K - The steady-state error value can be found for different inputs, R(s).

Step Input: R(t) R(t)=Au(t) A t(s) (Step input error coefficient) The steady-state error for a step input with magnitude A.

Ramp Input: R(t) R(t)=At  A=Tan() t(s) Ramp input error coefficient

Error coefficients and steady-state errors are valid for stable systems. Unity feedback systems are considered. Example: Find the step input and ramp input error coefficients and steady-state errors for the closed loop system whose forward path is given as There is no error for step input, A=1.

System has steady-state error for ramp input, A=1.

ROOT-LOCUS Method Consider a control system, K + R(s) Y(s) E(s) -

Root-Locus Plot: s+1=0, s=-1 Zeros: Np(s)=0 According to stability, the roots of the denominator of H(s) should lie on the left side of s-plane. Root Locus method gives the locations of the roots of the denominator with respect to varying controller gains, K. Root-Locus Plot: The root-locus plot of a closed loop system can be obtained easily by using the forward path transfer function of a closed loop control system. Example: Find the root locations of a closd loop cntrol system whose process transfer function is Zeros: Np(s)=0 s+1=0, s=-1 m = (number of zeros)=1

Plotting the root-locus : Poles, Dp(s)=0, n=Number of poles = 4. p1=0 p2=-5 p3=-2+2i p4=-2-2i n-m=4-1=3 Plotting the root-locus : Determine the zeros and poles of the Gp(s) Determine the angles of asymptotes, Determine the intersection of asypmtotes, Determine the breakaway points. For the given system, zeros and poles are calculated as p1=0 p2=-5 p3=-2+2i p4=-2-2i z1=-1 n-m=4-1=3

Angles of asymptotes: Poles of Gp(s) Zeros of Gp(s) >>ns=[1 1]; Root-locus by MATLAB Poles of Gp(s) Zeros of Gp(s) >>ns=[1 1]; >>ds=[1 9 28 40 0] >>rlocus(pay,payda)

Intersection of asymptotes: Root-locus by MATLAB K 180o 60o Root-Locus curve is symmetrical with respect to the real axis. K=0 K K=0 K=0 Kc 300o K=0 Critical K value can be found by Routh tabulation. -2.66 K

Breakaway points: A Breakaway point is the intersection point of the root-locus curve on the real axis. These points can be found by,

s1=-7.1447 s2=-2.36 s3= -0.4955 + 0.9687i s4= -0.4955 - 0.9687i There are no breakaway points because the root-locus curve doesn’t intersect the real axis.

Example: Plot the Root-Locus graph of the Closed-loop control system whose process transfer function is given as K + R(s) Y(s) -

Example: Plot the Root-Locus graph of the Closed-loop control system whose process transfer function is given as K + R(s) Y(s) - Process Transfer Function Solution: Zeros and Poles of Gp(s) are found, s+3=0  s=-3, Number of zeros m=1 s1=0, s2=-5, s3=-6, s4=-1+1i, s5=-1-1i, Number of poles=5 Angles of the Asymptotes

Angles of the Asymptotes Intersection point of the Asymptotes The Breakaway point is calculated as,

Np(s)=s+3 Dp(s)=s5+13s4+ 54s3+ 82s2+ 60s Breakaway point

>>rlocus(ns,ds) Root Locus by MATLAB, >>ns=[1 3]; >>ds=[1 13 54 82 60 0]; >>rlocus(ns,ds) Asymptote -2.5 45O 135O Intersection point K  Breakaway Point -5.53 Real Imaginary K=0

BODE PLOTS The bode plot gives information about the frequency response of a system. Consider a system whose input is a harmonic function F(s) Y(s) A A Cos(t) A Sin(t)  +A -A Y(s)=G(s) F(s) The response y(t) can be written as,

Example: Find the response of the system (G(s)) , to a harmonic excitation given as f(t)=3*Cos (5*t). Take n=1 rad/sec, =0.3. Solution: A = 3, s=5 i, The response of the system to a harmonic input, A*cos(t), can be found for different excitation frequencies, . For harmonic response replace s by i. >> w=0:0.001:30; >> s=i*w; >>wn=1; >>ksi=0; >>gs=wn^2./(s.^2+2*ksi*wn+wn^2); >>plot(w,abs(gs))

Frequency Response of The System, n=1 rad/sec. |G(i)| =0 =0.1 =0.5 =0.3 =0.707 =1 Frequency Response of The System, n=1 rad/sec. n Rezonance Rezonance Rezonance Frequency

BODE PLOT by MATLAB >>wn=1; >>ksi=0.3; >>ns=[wn^2]; >>ds=[1 2*ksi*wn wn^2]; >>bode(ns,ds)

BODE PLOT Y(s) R(s) + K - 1+G(s)=0  G(s)=-1 = -1 G(i) G(s) plane Gain Margin Gm Phase Margin m G(i) Log10 -180 Gm m Log10|G(i)|

Example: Find the Gain and Phase margins of the system whose transfer function is given below, >>ds=[1 3400 1200000 0]; >>bode(ns,ds) Gm=32.2 dB m=76.91

EXAMPLES E1: Draw the approximate Root-Locus graph of a control system given below. - + R Y K Solution: Find Zeros ans Poles of the Gp(s) m=1 n=3 Angles of the Asymptotes:

Real part of the roots goes to -4.5 as K goes to infinity. Intersection of the Asymptotes with real axis -6 270 -2.5 -3i 3i -4.5 K=0 Real Imaginary K There is no breakaway point. The system remains stable for All positive K values. Real part of the roots goes to -4.5 as K goes to infinity.

E2. The forward path transfer function of a closed-loop control system is given as. Find the Kt value which makes the ramp input error coefficient Kr=2. Find the steady state errors for unit step and unit ramp input for this Kt value. Kt=1.5 System has no steady-state error for unit step input. System has steady-state error for unit ramp input.

E3. The root-locus graph of a closed-loop control system whose denominator is D(s)=s3+2s2+(6+K)s+(3K+1) is given below. Find the coefficients (Kp, Kı, Kd) of a PID controller which will be employed in this control system. K -3 0.5  s= 4.3211 i s= -4.3211 i Solution: First of all, find Kcritical by Routh tabulation.

K<11 Kcritical=11 PID D(s)=s3+2s2+(6+K)s+(3K+1) s3 1 6+K s2 2 3K+1

Kp=6.6, KI= 9.078, Kd=1.2