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Chapter 10 – The Design of Feedback Control Systems 
PID 
Compensation Networks
Different Types of Feedback Control 
On-Off Control 
This is the simplest form of control.
Proportional Control 
A proportional controller attempts to perform better than the On-off type by 
applying power in proportion to the difference in temperature between the 
measured and the set-point. As the gain is increased the system responds faster to 
changes in set-point but becomes progressively underdamped and eventually 
unstable. The final temperature lies below the set-point for this system because 
some difference is required to keep the heater supplying power.
Proportional, Derivative Control 
The stability and overshoot problems that arise when a proportional 
controller is used at high gain can be mitigated by adding a term proportional 
to the time-derivative of the error signal. The value of the damping can be 
adjusted to achieve a critically damped response.
Proportional+Integral+Derivative Control 
Although PD control deals neatly with the overshoot and ringing 
problems associated with proportional control it does not cure the 
problem with the steady-state error. Fortunately it is possible to eliminate 
this while using relatively low gain by adding an integral term to the 
control function which becomes
The Characteristics of P, I, and D controllers 
A proportional controller (Kp) will have the effect of reducing the 
rise time and will reduce, but never eliminate, the steady-state 
error. 
An integral control (Ki) will have the effect of eliminating the 
steady-state error, but it may make the transient response worse. 
A derivative control (Kd) will have the effect of increasing the 
stability of the system, reducing the overshoot, and improving the 
transient response.
Proportional Control 
By only employing proportional control, a steady state error 
occurs. 
Proportional and Integral Control 
The response becomes more oscillatory and needs longer to 
settle, the error disappears. 
Proportional, Integral and Derivative Control 
All design specifications can be reached.
The Characteristics of P, I, and D controllers 
CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR 
Kp Decrease Increase Small Change Decrease 
Ki Decrease Increase Increase Eliminate 
Kd Small Change Decrease Decrease Small Change
Tips for Designing a PID Controller 
1. Obtain an open-loop response and determine what needs to be improved 
2. Add a proportional control to improve the rise time 
3. Add a derivative control to improve the overshoot 
4. Add an integral control to eliminate the steady-state error 
5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall 
response. 
Lastly, please keep in mind that you do not need to implement all three controllers 
(proportional, derivative, and integral) into a single system, if not necessary. For 
example, if a PI controller gives a good enough response (like the above 
example), then you don't need to implement derivative controller to the system. 
Keep the controller as simple as possible.
Open-Loop Control - Example 
num=1; 
den=[1 10 20]; 
step(num,den) 
G(s) 
1 
s2 + 10s + 20
Proportional Control - Example 
The proportional controller (Kp) reduces the rise time, increases 
the overshoot, and reduces the steady-state error. 
MATLAB Example 
Kp=300; 
num=[Kp]; 
den=[1 10 20+Kp]; 
t=0:0.01:2; 
step(num,den,t) 
Time (sec.) 
Amplitude 
Step Response 
1 
0.9 
0.8 
0.7 
0.6 
0.5 
0.4 
0.3 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
1.4 
1.2 
1 
0.8 
0.6 
0.4 
0.2 
0 
From: U(1) 
To: Y(1) 
T(s) 
Kp 
s2 + 10×s + (20 + Kp) 
Time (sec.) 
Amplitude 
Step Response 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
0.2 
0.1 
0 
From: U(1) 
To: Y(1) 
K=300 K=100
Proportional - Derivative - Example 
The derivative controller (Kd) reduces both the overshoot and the 
settling time. 
MATLAB Example T(s) 
Time (sec.) 
Amplitude 
Step Response 
1 
0.9 
0.8 
0.7 
0.6 
0.5 
0.4 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
1.4 
1.2 
1 
0.8 
0.6 
0.4 
0.2 
0 
From: U(1) 
To: Y(1) Kp=300; 
Kd=10; 
num=[Kd Kp]; 
den=[1 10+Kd 20+Kp]; 
t=0:0.01:2; 
step(num,den,t) 
Kd×s + Kp 
s2 + (10 + Kd)×s + (20 + Kp) 
Time (sec.) 
Amplitude 
Step Response 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
0.3 
0.2 
0.1 
0 
From: U(1) 
To: Y(1) 
Kd=10 
Kd=20
Proportional - Integral - Example 
The integral controller (Ki) decreases the rise time, increases both 
the overshoot and the settling time, and eliminates the steady-state 
error 
MATLAB Example 
Time (sec.) 
Amplitude 
Step Response 
1.4 
1.2 
1 
0.8 
0.6 
0.4 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
1.4 
1.2 
1 
0.8 
0.6 
0.4 
0.2 
0 
From: U(1) 
To: Y(1) Kp=30; 
Ki=70; 
num=[Kp Ki]; 
den=[1 10 20+Kp Ki]; 
t=0:0.01:2; 
step(num,den,t) 
T(s) 
Kp×s + Ki 
s3 + 10×s2 + (20 + Kp)×s + Ki 
From: U(1) 
Time (sec.) 
Amplitude 
Step Response 
0.2 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
To: Y(1) 
Ki=70 
Ki=100
RLTOOL 
Syntax 
rltool 
rltool(sys) 
rltool(sys,comp)
RLTOOL
RLTOOL
RLTOOL
RLTOOL
Example - Practice 
Consider the following configuration:
Example - Practice 
The design a system for the following specifications: 
· Zero steady state error 
· Settling time within 5 seconds 
· Rise time within 2 seconds 
· Only some overshoot permitted
Lead or Phase-Lead Compensator Using Root Locus 
A first-order lead compensator can be designed using the root locus. A lead compensator 
in root locus form is given by 
Gc(s) 
(s + z) 
(s + p) 
where the magnitude of z is less than the magnitude of p. A phase-lead compensator 
tends to shift the root locus toward the left half plane. This results in an improvement in 
the system's stability and an increase in the response speed. 
When a lead compensator is added to a system, the value of this intersection will be a 
larger negative number than it was before. The net number of zeros and poles will be the 
same (one zero and one pole are added), but the added pole is a larger negative number 
than the added zero. Thus, the result of a lead compensator is that the asymptotes' 
intersection is moved further into the left half plane, and the entire root locus will be 
shifted to the left. This can increase the region of stability as well as the response speed.
Lead or Phase-Lead Compensator Using Root Locus 
In Matlab a phase lead compensator in root locus form is implemented by using the 
transfer function in the form 
numlead=kc*[1 z]; 
denlead=[1 p]; 
and using the conv() function to implement it with the numerator and denominator 
of the plant 
newnum=conv(num,numlead); 
newden=conv(den,denlead);
Lead or Phase-Lead Compensator Using Frequency Response 
A first-order phase-lead compensator can be designed using the frequency response. A lead 
compensator in frequency response form is given by 
Gc(s) 
( 1 + a×t×s) 
a×( 1 + t×s) 
p 
1 
t 
z 
1 
at 
wm z×p sin(fm) a - 1 
a + 1 
In frequency response design, the phase-lead compensator adds positive phase to the system 
over the frequency range. A bode plot of a phase-lead compensator looks like the following
Lead or Phase-Lead Compensator Using Frequency Response 
Additional positive phase increases the phase margin and thus increases the stability of 
the system. This type of compensator is designed by determining alfa from the amount of 
phase needed to satisfy the phase margin requirements, and determining tal to place the 
added phase at the new gain-crossover frequency. 
Another effect of the lead compensator can be seen in the magnitude plot. The lead 
compensator increases the gain of the system at high frequencies (the amount of this gain 
is equal to alfa. This can increase the crossover frequency, which will help to decrease the 
rise time and settling time of the system.
Lead or Phase-Lead Compensator Using Frequency Response 
In Matlab, a phase lead compensator in frequency response form is 
implemented by using the transfer function in the form 
numlead=[aT 1]; 
denlead=[T 1]; 
and using the conv() function to multiply it by the numerator and 
denominator of the plant 
newnum=conv(num,numlead); 
newden=conv(den,denlead);
Lag or Phase-Lag Compensator Using Root Locus 
A first-order lag compensator can be designed using the root locus. A lag compensator in root 
locus form is given by 
Gc(s) 
(s + z) 
(s + p) 
where the magnitude of z is greater than the magnitude of p. A phase-lag compensator tends to 
shift the root locus to the right, which is undesirable. For this reason, the pole and zero of a lag 
compensator must be placed close together (usually near the origin) so they do not appreciably 
change the transient response or stability characteristics of the system. 
When a lag compensator is added to a system, the value of this intersection will be a smaller 
negative number than it was before. The net number of zeros and poles will be the same (one 
zero and one pole are added), but the added pole is a smaller negative number than the added 
zero. Thus, the result of a lag compensator is that the asymptotes' intersection is moved closer 
to the right half plane, and the entire root locus will be shifted to the right.
Lag or Phase-Lag Compensator Using Root Locus 
It was previously stated that that lag controller should only minimally change the 
transient response because of its negative effect. If the phase-lag compensator is 
not supposed to change the transient response noticeably, what is it good for? The 
answer is that a phase-lag compensator can improve the system's steady-state 
response. It works in the following manner. At high frequencies, the lag controller 
will have unity gain. At low frequencies, the gain will be z0/p0 which is greater 
than 1. This factor z/p will multiply the position, velocity, or acceleration constant 
(Kp, Kv, or Ka), and the steady-state error will thus decrease by the factor z0/p0. 
In Matlab, a phase lead compensator in root locus form is implemented by using 
the transfer function in the form 
numlag=[1 z]; 
denlag=[1 p]; 
and using the conv() function to implement it with the numerator and 
denominator of the plant 
newnum=conv(num,numlag); 
newden=conv(den,denlag);
Lag or Phase-Lag Compensator using Frequency Response 
A first-order phase-lag compensator can be designed using the frequency response. A 
lag compensator in frequency response form is given by 
Gc(s) 
( 1 + a×t×s) 
a×( 1 + t×s) 
The phase-lag compensator looks similar to a phase-lead compensator, except that a is 
now less than 1. The main difference is that the lag compensator adds negative phase to 
the system over the specified frequency range, while a lead compensator adds positive 
phase over the specified frequency. A bode plot of a phase-lag compensator looks like 
the following
Lag or Phase-Lag Compensator using Frequency Response 
In Matlab, a phase-lag compensator in frequency response form is 
implemented by using the transfer function in the form 
numlead=[a*T 1]; 
denlead=a*[T 1]; 
and using the conv() function to implement it with the numerator and 
denominator of the plant 
newnum=conv(num,numlead); 
newden=conv(den,denlead);
Lead-lag Compensator using either Root Locus or Frequency 
Response 
A lead-lag compensator combines the effects of a lead compensator with those of a lag 
compensator. The result is a system with improved transient response, stability and 
steady-state error. To implement a lead-lag compensator, first design the lead 
compensator to achieve the desired transient response and stability, and then add on a lag 
compensator to improve the steady-state response
Exercise - Dominant Pole-Zero Approximations and Compensations 
The influence of a particular pole (or pair of complex poles) on the response is mainly determined 
by two factors: the real part of the pole and the relative magnitude of the residue at the pole. The 
real part determines the rate at which the transient term due to the pole decays; the larger the real 
part, the faster the decay. The relative magnitude of the residue determines the percentage of the 
total response due to a particular pole. 
Investigate (using Simulink) the impact of a closed-loop negative real pole on the overshoot of a 
system having complex poles. 
T(s) 
pr wn 
2 
× 
(s + pr) s2 + ( 2×z×wn×s) wn 
2 
+ éë 
ùû 
× 
Make pr to vary (2, 3, 5) times the real part of the complex pole for different values of z (0.3, 0.5, 
0.7). 
Investigate (using Simulink) the impact of a closed-loop negative real zero on the overshoot of a 
system having complex poles. 
T(s) 
(s + zr) 
ùû 
s2 + ( 2 
2×z×wn×s) + wéën 
Make zr to vary (2, 3, 5) times the real part of the complex pole for different values of z (0.3, 0.5, 0.7).
Exercise - Lead and Lag Compensation 
Investigate (using Matlab and Simulink) the effect of lead and lag compensations on the two 
systems indicated below. Summarize your observations. Plot the root-locus, bode diagram 
and output for a step input before and after the compensations. 
Remember 
lead compensation: z<p (place zero below the desired root location or to the left of the first two 
real poles) 
lag compensation: z>p (locate the pole and zero near the origin of the s-plane) 
Lead Compensation (use z=1.33, p=20 and K =15).
Lag Compensation (use z=0.09 , and p=0.015, K=1/6 ) 
Summarize your findings
Problem 10.36 
Determine a compensator so that the percent overshoot is less than 20% and Kv 
(velocity constant) is greater than 8.
Lead-lag controller
Lead-lag controller

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Lead-lag controller

  • 1. Chapter 10 – The Design of Feedback Control Systems PID Compensation Networks
  • 2. Different Types of Feedback Control On-Off Control This is the simplest form of control.
  • 3. Proportional Control A proportional controller attempts to perform better than the On-off type by applying power in proportion to the difference in temperature between the measured and the set-point. As the gain is increased the system responds faster to changes in set-point but becomes progressively underdamped and eventually unstable. The final temperature lies below the set-point for this system because some difference is required to keep the heater supplying power.
  • 4. Proportional, Derivative Control The stability and overshoot problems that arise when a proportional controller is used at high gain can be mitigated by adding a term proportional to the time-derivative of the error signal. The value of the damping can be adjusted to achieve a critically damped response.
  • 5. Proportional+Integral+Derivative Control Although PD control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the problem with the steady-state error. Fortunately it is possible to eliminate this while using relatively low gain by adding an integral term to the control function which becomes
  • 6. The Characteristics of P, I, and D controllers A proportional controller (Kp) will have the effect of reducing the rise time and will reduce, but never eliminate, the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving the transient response.
  • 7. Proportional Control By only employing proportional control, a steady state error occurs. Proportional and Integral Control The response becomes more oscillatory and needs longer to settle, the error disappears. Proportional, Integral and Derivative Control All design specifications can be reached.
  • 8. The Characteristics of P, I, and D controllers CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR Kp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate Kd Small Change Decrease Decrease Small Change
  • 9. Tips for Designing a PID Controller 1. Obtain an open-loop response and determine what needs to be improved 2. Add a proportional control to improve the rise time 3. Add a derivative control to improve the overshoot 4. Add an integral control to eliminate the steady-state error 5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to implement derivative controller to the system. Keep the controller as simple as possible.
  • 10. Open-Loop Control - Example num=1; den=[1 10 20]; step(num,den) G(s) 1 s2 + 10s + 20
  • 11. Proportional Control - Example The proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady-state error. MATLAB Example Kp=300; num=[Kp]; den=[1 10 20+Kp]; t=0:0.01:2; step(num,den,t) Time (sec.) Amplitude Step Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.4 1.2 1 0.8 0.6 0.4 0.2 0 From: U(1) To: Y(1) T(s) Kp s2 + 10×s + (20 + Kp) Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.2 0.1 0 From: U(1) To: Y(1) K=300 K=100
  • 12. Proportional - Derivative - Example The derivative controller (Kd) reduces both the overshoot and the settling time. MATLAB Example T(s) Time (sec.) Amplitude Step Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.4 1.2 1 0.8 0.6 0.4 0.2 0 From: U(1) To: Y(1) Kp=300; Kd=10; num=[Kd Kp]; den=[1 10+Kd 20+Kp]; t=0:0.01:2; step(num,den,t) Kd×s + Kp s2 + (10 + Kd)×s + (20 + Kp) Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.3 0.2 0.1 0 From: U(1) To: Y(1) Kd=10 Kd=20
  • 13. Proportional - Integral - Example The integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state error MATLAB Example Time (sec.) Amplitude Step Response 1.4 1.2 1 0.8 0.6 0.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.4 1.2 1 0.8 0.6 0.4 0.2 0 From: U(1) To: Y(1) Kp=30; Ki=70; num=[Kp Ki]; den=[1 10 20+Kp Ki]; t=0:0.01:2; step(num,den,t) T(s) Kp×s + Ki s3 + 10×s2 + (20 + Kp)×s + Ki From: U(1) Time (sec.) Amplitude Step Response 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 To: Y(1) Ki=70 Ki=100
  • 14. RLTOOL Syntax rltool rltool(sys) rltool(sys,comp)
  • 19. Example - Practice Consider the following configuration:
  • 20. Example - Practice The design a system for the following specifications: · Zero steady state error · Settling time within 5 seconds · Rise time within 2 seconds · Only some overshoot permitted
  • 21. Lead or Phase-Lead Compensator Using Root Locus A first-order lead compensator can be designed using the root locus. A lead compensator in root locus form is given by Gc(s) (s + z) (s + p) where the magnitude of z is less than the magnitude of p. A phase-lead compensator tends to shift the root locus toward the left half plane. This results in an improvement in the system's stability and an increase in the response speed. When a lead compensator is added to a system, the value of this intersection will be a larger negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a larger negative number than the added zero. Thus, the result of a lead compensator is that the asymptotes' intersection is moved further into the left half plane, and the entire root locus will be shifted to the left. This can increase the region of stability as well as the response speed.
  • 22. Lead or Phase-Lead Compensator Using Root Locus In Matlab a phase lead compensator in root locus form is implemented by using the transfer function in the form numlead=kc*[1 z]; denlead=[1 p]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead);
  • 23. Lead or Phase-Lead Compensator Using Frequency Response A first-order phase-lead compensator can be designed using the frequency response. A lead compensator in frequency response form is given by Gc(s) ( 1 + a×t×s) a×( 1 + t×s) p 1 t z 1 at wm z×p sin(fm) a - 1 a + 1 In frequency response design, the phase-lead compensator adds positive phase to the system over the frequency range. A bode plot of a phase-lead compensator looks like the following
  • 24. Lead or Phase-Lead Compensator Using Frequency Response Additional positive phase increases the phase margin and thus increases the stability of the system. This type of compensator is designed by determining alfa from the amount of phase needed to satisfy the phase margin requirements, and determining tal to place the added phase at the new gain-crossover frequency. Another effect of the lead compensator can be seen in the magnitude plot. The lead compensator increases the gain of the system at high frequencies (the amount of this gain is equal to alfa. This can increase the crossover frequency, which will help to decrease the rise time and settling time of the system.
  • 25. Lead or Phase-Lead Compensator Using Frequency Response In Matlab, a phase lead compensator in frequency response form is implemented by using the transfer function in the form numlead=[aT 1]; denlead=[T 1]; and using the conv() function to multiply it by the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead);
  • 26. Lag or Phase-Lag Compensator Using Root Locus A first-order lag compensator can be designed using the root locus. A lag compensator in root locus form is given by Gc(s) (s + z) (s + p) where the magnitude of z is greater than the magnitude of p. A phase-lag compensator tends to shift the root locus to the right, which is undesirable. For this reason, the pole and zero of a lag compensator must be placed close together (usually near the origin) so they do not appreciably change the transient response or stability characteristics of the system. When a lag compensator is added to a system, the value of this intersection will be a smaller negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a smaller negative number than the added zero. Thus, the result of a lag compensator is that the asymptotes' intersection is moved closer to the right half plane, and the entire root locus will be shifted to the right.
  • 27. Lag or Phase-Lag Compensator Using Root Locus It was previously stated that that lag controller should only minimally change the transient response because of its negative effect. If the phase-lag compensator is not supposed to change the transient response noticeably, what is it good for? The answer is that a phase-lag compensator can improve the system's steady-state response. It works in the following manner. At high frequencies, the lag controller will have unity gain. At low frequencies, the gain will be z0/p0 which is greater than 1. This factor z/p will multiply the position, velocity, or acceleration constant (Kp, Kv, or Ka), and the steady-state error will thus decrease by the factor z0/p0. In Matlab, a phase lead compensator in root locus form is implemented by using the transfer function in the form numlag=[1 z]; denlag=[1 p]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlag); newden=conv(den,denlag);
  • 28. Lag or Phase-Lag Compensator using Frequency Response A first-order phase-lag compensator can be designed using the frequency response. A lag compensator in frequency response form is given by Gc(s) ( 1 + a×t×s) a×( 1 + t×s) The phase-lag compensator looks similar to a phase-lead compensator, except that a is now less than 1. The main difference is that the lag compensator adds negative phase to the system over the specified frequency range, while a lead compensator adds positive phase over the specified frequency. A bode plot of a phase-lag compensator looks like the following
  • 29. Lag or Phase-Lag Compensator using Frequency Response In Matlab, a phase-lag compensator in frequency response form is implemented by using the transfer function in the form numlead=[a*T 1]; denlead=a*[T 1]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead);
  • 30. Lead-lag Compensator using either Root Locus or Frequency Response A lead-lag compensator combines the effects of a lead compensator with those of a lag compensator. The result is a system with improved transient response, stability and steady-state error. To implement a lead-lag compensator, first design the lead compensator to achieve the desired transient response and stability, and then add on a lag compensator to improve the steady-state response
  • 31. Exercise - Dominant Pole-Zero Approximations and Compensations The influence of a particular pole (or pair of complex poles) on the response is mainly determined by two factors: the real part of the pole and the relative magnitude of the residue at the pole. The real part determines the rate at which the transient term due to the pole decays; the larger the real part, the faster the decay. The relative magnitude of the residue determines the percentage of the total response due to a particular pole. Investigate (using Simulink) the impact of a closed-loop negative real pole on the overshoot of a system having complex poles. T(s) pr wn 2 × (s + pr) s2 + ( 2×z×wn×s) wn 2 + éë ùû × Make pr to vary (2, 3, 5) times the real part of the complex pole for different values of z (0.3, 0.5, 0.7). Investigate (using Simulink) the impact of a closed-loop negative real zero on the overshoot of a system having complex poles. T(s) (s + zr) ùû s2 + ( 2 2×z×wn×s) + wéën Make zr to vary (2, 3, 5) times the real part of the complex pole for different values of z (0.3, 0.5, 0.7).
  • 32. Exercise - Lead and Lag Compensation Investigate (using Matlab and Simulink) the effect of lead and lag compensations on the two systems indicated below. Summarize your observations. Plot the root-locus, bode diagram and output for a step input before and after the compensations. Remember lead compensation: z<p (place zero below the desired root location or to the left of the first two real poles) lag compensation: z>p (locate the pole and zero near the origin of the s-plane) Lead Compensation (use z=1.33, p=20 and K =15).
  • 33. Lag Compensation (use z=0.09 , and p=0.015, K=1/6 ) Summarize your findings
  • 34. Problem 10.36 Determine a compensator so that the percent overshoot is less than 20% and Kv (velocity constant) is greater than 8.