RHP pole

From the Nyquist stability criterion, let N k, G(iuj)) be the net number of clockwise encirclements of a point (k, 0) of the Nyquist contour. Assume that all plants in the family tt, expressed in equation (9.132) have the same number ( ) of right-hand plane (RHP) poles. 

Robust stability ean therefore be stated as If all plants G(.v) in the family tt have the same number of RHP poles and that a partieular eontroller C(.v) stabilizes the nominal plant Gmfv), then the system is robustly stable with the eontroller C(.v) if and only if the eomplementary sensitivity funetion T s) for the nominal plant Gmfv) satisfies the following bound... 

Processes with RHP poles in their openloop transfer functions are openloop unstable. The irreversible, exothermic chemical reactor is the classical chemical engineering example. These systems can sometimes be stabilized by using... 

In addition, for an unstable plant with a real RHP-pole at 5=p we approximately need ... 

As shown in Section 5.2, the system has two real poles at s = p and two real zeros at i = z. Thus the system has both a right-half plane (RHP) pole and an RHP zero as discussed in the textbooks [25, 26], this imposes fundamental restrictions on feedback control performance. In particular, as discussed by Skogestad and Postlethwaite [26, Section 5.9, p. 185], system bandwidth (as measured by the critical frequency, coc, [26, Section 2.4.5, p. 36]) should be approximately bounded by... 

Having determined that a flow input is required and that the system is inherently hard to control due to the RHP pole and zero, the next step is to design a controller. A state-space approach is used here and so a state-space model must first be derived from the bond graph of Fig. 5.2c. This system represents a differential-algebraic equation (DAE) which can, however, be rewritten as a state-space equation. In particular, system state-space equations can be derived from Fig. 5.2c as follows. Defining the angular momenta of the two pendula as h and /12, respectively, the torque eo (which drives the controlled pendulum) is, from the left-hand side of Fig. 5.2e,... 

The fundamental control limitations that can exist in a plant include time delays, RHP poles and RHP zeros [9]. Time-delays are usually caused by measurements and material transport, and are in general not affected by feedback effects. Feedback will, however, move the poles of a system, and can hence affect stability by moving poles across the imaginary axis. 

Because H(s) and Gql ) have the same denominator, they have the same number of RHP poles. 

E. If a pole has a negative real part, it is in the left-hand plane (LHP). Conversely, if a pole has a positive real part, it is in the right-hand plane (RHP) and the time-domain solution is definitely unstable. 

This is the idea behind the plotting of the closed-loop poles—in other words, construction of root locus plots. Of course, we need mathematical or computational tools when we have more complex systems. An important observation from Example 7.5 is that with simple first and second order systems with no open-loop zeros in the RHP, the closed-loop system is always stable. 

The important point is that the phase lag of the dead time function increases without bound with respect to frequency. This is what is called a nonminimum phase system, as opposed to the first and second transfer functions which are minimum phase systems. Formally, a minimum phase system is one which has no dead time and has neither poles nor zeros in the RHP. (See Review Problems.)... 

The effect of adding a lag or a pole is to pull the root locus plot toward the unstable region. The two curves that s ait ats=— Jand.s=—1 become complex conjugates and curve off into the RHP. Therefore this third-order system is closed-loop unstable if is greater than = 20. This was the same result that we obtained in Example 10.5,... 

Nevertheless linear techniques are very useful in looking at stability near some operating level. Mathematically, if the system is openloop unstable, it must have an openloop transfer function that has at least one pole in the RHP. 

Note that this is Jiot a hrsl-order lag because of the negative sign in the denominator. The system has an openloop pole in the RHP at s = -I- 1/tj,. The unit step response of this system is an exponential that goes olT to inhnity as time increases. 

The root locus curves are shown in Fig. 11.10c. The loci start at the poles of the openloop transfer function . s = — 1 and 5 = — Since the loci must end at the zeros of the openloop transfer function (.t — + ) the curves swing over into the RHP. Therefore the system is closedloop unstable for gains greater than 2. 

If the process is openloop unstMe, Gm<.) will have one or more poles in the RHP, so F(,y = 1 + Gm( , B( will also have one or more poles in the RHP. We can find out how many poles there are by solving for the roots of the openloop characteristic equation or by using the Routh stability criterion on the openloop characteristic equation (the denominator of Gj, ). Once the number of poles P is known, the number of zeros can be found from Eq. (13.6). 

This system is openloop stable (with three poles that are all in the left half of the s plane and none in the RHP), so P = 0. 

The Nyquist stabihty criterion can be used for openloop unstable processes, but we have to use the complete, rigorous version with P (the number of poles of the closedloop characteristic equation in the RHP) no longer equal to zero. 

First of all, we know immediately that the openloop system transfer function has one pole (at s = -I- 1/ip) in the RHP, Therefore the closedloop... 

Ensure that the crossover frequency is well below any troublesome poles or zeros — like the RHP zero in continuous conduction mode (boost and buck-boost — with voltage mode or current mode control), and the subharmonic instability pole in continuous conduction mode (buck, boost, and buck-boost — with current mode control). The latter pole is discussed later. 

However everyone does seem to agree that current mode control alters the poles of the system (as compared to voltage mode control), but the zeros are unchanged. So the boost and the buck-boost still have the same RHP zero, as we discussed earlier. And care is still needed to ensure that the RHP zero is at a much higher frequency than the chosen crossover frequency. 

It is well known from classical root-locus analysis that as the feedback gain increases towards infinity, the closed-loop poles migrate to the positions of open-loop zeros. Thus, the presence of RHP-zeros implies high-gain instability. 

Representing the roots of the characteristic equation in the complex domain offers a simple way to perform a stability analysis. The system is stable if and only if all the poles are located in the open left-half-plane (LHP). If there is at least one pole in the right-half-plane (RHP), the system is unstable. The representation is similar wiA the well-known root-locus plot used to evaluate the stability of a closed-loop system. 

If we pick a contour that goes completely around the right half of the s plane and plot 1 + Gm(s)Gc s) Eq. (11.1) tells us that the number of encirclements of the origin in this (1 + G Gq) plane will be equal to the difference between the zeros and poles of 1 + GmGq that lie in the RHP. Figure 11.2 shows a case where there are two zeros in the RHP and no poles. There are two encirclements of the origin in the (1 + GmGc) plane. 

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