Design Via Root locus
Design K such that the settling time is 2.5 sec and damping ratio is 0.6 in closed loop
-1.6±j2128
This closed loop pole should lie on the root locus
But , the closed loop pole lie on root locus along 0.6 damping ratio line is -0.714 +0.926i
The settling time is 5.5 sec. The equivalent K value cannot be obtained
One way to solve our problem is to replace the existing system with a system whose root locus intersects the desired design point, B. Unfortunately, this replacement is expensive and counterproductive.
Rather than change the existing system, we augment, or compensate, the system with additional poles and zeros, so that the compensated system has a root locus that goes through the desired pole location for some value of gain.
Both methods change the open-loop poles and zeros, thereby creating a new root locus that goes through the
desired closed-loop pole location.
Compensators that use pure integration for improving steady-state error or pure differentiation for improving transient response are defined as ideal compensators
Ideal compensators must be implemented with active networks, which, in the case of electric networks, require the use of active amplifiers and possible additional power sources
An advantage of ideal integral compensators is that steady-state error is reduced to zero.
Compensators implemented with passive elements such as resistors and capacitors are not ideal compensators.
Advantages of passive networks are that they are less expensive and do not require additional power sources for their operation.
Their disadvantage is that the steady-state error is not driven to zero
Systems that feed the integral of the error to the plant are called integral control systems.
Thus, we use the name PI controller interchangeably with ideal integral compensator, and we use the name lag compensator when the cascade compensator does not employ pure integration.
Improving Steady-State Error
via Cascade Compensation
Lag Compensation
Ideal integral compensation-with its pole on the origin-active network
passive networks- the pole and zero are moved to the left, close to the origin, as shown
This placement of the pole and zero although it does not increase the system type, does yield an improvement in the static error constant over an uncompensated system.
In order to reduce the steady state error, the gain is increased.
The compensated steady state gain Kpc = p times the uncompensated gain kpu
The factor p = Zc/Pc
For the lag compensator pole should be nearer to origin compared to zero.
Also both compensator pole and zero should be close to origin to retain the transient response specifications.
Compensate the system to improve the steady-state error by a factor of 10 if the system is operating with a damping ratio of 0.174.
From the root locus, the corresponding gain value for damping ratio of 0.174 is K =162.
The error constant Kpu = 8.1. The steady state error is = 1/1+kpu = 0.109.
To improve the steady state error requires to reduce the error by 10 times = 0.0109.
The equivalent Kpc = 91.59
Construct a line at an angle of 10° (or less) with δ line from sd. Intersection with real axis is the compensator zero(zc)
The compensator pole (pc) = zc/β
The open loop transfer function of the compensated system is
Improving Transient Response via Cascade Compensation- Lead Compensator
The objective is to design a response that has a desirable percent overshoot and a shorter settling time than the
uncompensated system.
For this, ideal derivative compensation, a pure differentiator (a proportional-plus-derivative (PD) controller) is added to the forward path of the feedback control system.
Adding differentiation is the addition of a zero to the forward-path transfer function (s+Zc). This type of compensation requires an active network for its realization.
The second technique does not use pure differentiation. Instead, it approximates differentiation with a passive network by adding a zero and a more distant pole (a lead compensator). to the forward-path transfer function
The advantages of a passive lead network over an active PD controller are that
Lead Compensator
Select a desired dominant, second-order pole on the s-plane from the given specifications
The sum of the angles from the uncompensated system's poles and zeros to the design point can be found
The difference between 180° and the sum of the angles must be the angular contribution required of the compensator.
Design a lead compensator for the given system that will
reduce the settling time ts by a factor of 2 while maintaining
the overshoot as 30%.
2. Translate the transient response specifications into a pair of
complex dominant root
Given overshoot = 0.30.
Determine the equivalent damping ratio δ =0.36
find θ= cos-1(δ)
Draw a δ line with an angle of cos-1(δ) from the negative real axis. The point of intersection of δ line on the root locus, is the complex dominant root (sd).
Sd =-1.01+j2.6
3. Calculate settling time ts= 4/ δωn =4/1.01 =3.96 sec
required settling time = tsn =ts/2 =3.96/2 =1.98sec
Determine the desired dominant root with the revised settling time tsn :
Real part δωn = 4/1.98 = 2.02
desired dominant root -2.02+j6
Place the compensating zero on the real axis in the region below the desired dominant root sd. If the uncompensated system has an open loop pole on the real axis in the region below sd, then the compensating zero should lie to the left of the pole.
Let the compensator zero (zc) be at -2.5
Find the angle contribution at sd by all the poles and zeros (θc )
The angle should be contributed by the compensator pole to get -180° is θd = -180 – (θc)
= -180 – (-150) = 30°
6. Determine the compensator pole (pc) using the formula
Pc = -12.9
The open loop transfer function of the compensated system is
where the gain (Knew) at that point sd is determined by
K new= 540