The most important problem in linear control systems concerns stability. It is the most important system specification among the three requirements enter into the design of a control system: transient response, stability and steady-state error. That is, under what conditions will a system become unstable? If it is unstable, how should we stabilize the system?
The total response of a system is the sum of the forced and natural responses:
Using this concept, we present the following definition of stability, instability and marginal stability:
- A linear time-invariant system is stable if the natural response approaches to zero as time approaches infinity.
- A linear time-invariant system is unstable if the natural responses grows without bound as time approaches infinity.
- A linear time-invariant system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates as time approaches infinity.
Thus the definition of stability implies that only the forced response remains as the natural response approaches zero. An alternate definition of stability is regards the total response and implies the first definition based upon the natural response:
- A system is stable if every bounded input yields a bounded output. We call this statement a bounded-input, bounded-output (BIBO) definition of stability.
We now realize that if the input is bounded but the output is unbounded, the system is unstable. If the input is unbounded we will see an unbounded total response and we cannot draw any about conclusion about stability.
Physically, an unstable system whose natural response grows without bound can cause damage to the system, to adjacent property, or to human life. Many times systems are designed with limited stops to prevent total runaway.
Negative feedback tends to improve stability. From the study of the system poles we must recall that poles in the left half-plane (lhp) yield either pure exponential decay or damped sinusoidal natural responses. These natural responses decay to zero when time approaches infinity. Thus, if the closed-loop system poles are in the left half of the plane and hence have a negative real part, the system is stable. That is:
- Stable systems have closed-loop transfer functions with poles only in the left half-plane.
Poles in the right half-plane (rhp) yield either pure exponentially increasing or exponentially increasing sinusoidal natural responses, which approche infinity when time approaches infinity. Thus:
- Unstable systems have closed-loop transfer functions with at least one pole in the right half-plane and /or poles of multiplicity greater than 1 on the imaginary axis.
Finally, the system that has imaginary axis poles of multiplicity 1 yields pure sinusoidal oscillations as a natural response. These responses neither increase nor decrease in amplitude. Thus,
- Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity one and poles in the left half-plane.
Figure 6.1a shows a unit step response of a stable system, while Figure 6.1b shows an unstable system.
Routh’s Stability Criterion
Routh’s Stability criterion tell us whether or not there are unstable roots in a polynomial equation without actually solving for them. This stability criterion applies to polynomials with only a finite number of terms. When the criterion is applied to a control system, information about absolute stability can be obtained directly from the coefficients of the characteristic equation.
The method requires two steps: 1) Generate a data table called a Routh Table and 2) Interpret the to tell how many closed-loop system poles are in the left half-plane, the right half-plane, and on the jw-axis. The power of the method lies in design rather than analysis. For example, if you have an unknown parameter in the denominator of a transfer function, it is difficult to determine via a calculator the range of this parameter to yield stability. We shall see that The Routh-Hurwitz criterion can yield a closed-form expression for the range of the unknown parameter.
- Generating a basic Routh Table. Considering the equivalent closed-loop transfer function in Figure 6.3. Since we are interested in the system poles, we focus our attention on the denominator. We first create the Routh Table shown in Table 6.1:
Begin by labeling the rows with powers of s from the highest power of the denominator of the closed-loop transfer function to s^0. Next start with the coefficient of the highest power of s in the denominator and list, horizontally in the first row every other coefficient.
In the second row, list horizontally, starting with the next higher power of s, every coefficient that was skipped in the first row. The remain entries are filled as follows:
Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row. The left-hand column of the determinant is always the first column of the previous two rows, and the right-hand column is the elements of the column above and to the right.
Figures 6.4 shows an example of building the Routh Table :
The complete array of coefficients is triangular. Note that in developing the array an entire row maybe divided or multiplied by a positive number in order to simplify the subsequent numerical calculation without altering the stability conclusion.
Consider the following characteristic equation, example 5-13:
The first two rows can be obtained directly from the given polynomial. The second is divided by two but we arrive to the same conclusion:
- Interpreting the basic Routh Table. Simply stated, the Routh-Hurwitz criterion declares that the number of roots of the polynomial that are in the right-half plane is equal to the number of sign changes in the first column.
If the closed-loop transfer function has all poles in the left-hand plane the system is stable. Thus, the system is stable if there are no sign changes in the first column of the Routh Table.
The last case, Example 5-13, is one of an unstable system. In that example the number of sign changes in the first column is equal to two. This means that there are two roots with positive real parts. Table 6.3 is also an unstable system. There, the first change occurs from 1 in the s^2 row to -72 in the s^1 row. The second occurs from -72 in the s^1 row to 103 in the s^0 row. Thus, the system has two poles in the right-half plane.
Routh’s stability criterion is of limited usefulness when applying to control system analysis because it does not suggest how to improve relative stability or how to stabilize a unstable system. It is possible, however, to determine the effects of changing one or two parameters of the system by examining the values that causes instability. In the following we consider the problem of determining the stability range of a parameter value. Consider the system of the Figure 5-38. Let us determine the range of K for stability.
The characteristic equation is:
And the Routh Table:
For stability, K must be positive and all coefficient in the first column must be positive. Therefore:
When K=14/9 the system becomes oscillatory and, mathematically, the oscillation is sustained at constant amplitude.
Routh-Hurwitz Criterion Special Cases
Two special cases can occur: (1) The Routh table sometimes will have a zero only in the first column of a row, (2) The Routh table sometimes will have an entire row that consists of zeros
- Zero only in the first column. If the first elemento of a row is zero, division by zero will be required to form the next row. To avoid this phenomenon, an epsilon ε is assigned to replace the zero in the first column. The value is then allowed to approach zero from either the positive or the negative side, after which the signs of the entries in the first column can be determined. To see the application of this, let us look the follow example: determine the stability of the closed-loop transfer function T(s):
The solution is shown in table 6.4:
We must begin by assembling the Routh table down to the row where a zero appears only in the first column (the s^3 row). Next, replace the zero by a small number ε complete the table. To begin the interpretation we must first assume a sign, positive or negative for the quantity ε. Table 6.5 shows the first column of table 6.4 along with the resulting signs for choices of ε positive and ε negative.
If is chosen ε positive Table 6.5 shows a sign change from the s^3 row to the s^2 row, and there will be another sign change from the s^2 row to the s^1 row. Hence the system is unstable and has two poles in the right half-plane. Alternatively, we could chose ε negative. Table 6.5 then shows a sign change from the s^4 row to the s^3 row. Another sign change would occur from the s^3 row to the s^2 row. Our result would be exactly the same as that for a positive choice for ε. Thus, the system is unstable.
- Entire Row is zero. We now look at the second special case. Sometimes while making a Routh table, we can find that an entire row consists of zeros because there is a even polynomial that is a factor of the original polynomial. This case must be handled differently from the previous case. Next example shows how to construct and interpret the Routh table wne an entire row of zeros is present.
Determine the number of right half-plane poles in the closed-loop transfer function T(s):
Start by forming the Routh table for the denominator. We get Table 6.7:
At the second we multiply by 1/7 for convenience. We stop at the third row since the entire row consist of zeros and use the following procedure. First we return to the row immediately above the row of zeros and form an auxiliary polynomial using the entries in that row as coefficients. The polynomial will start with the power of s in the label column and continue by skipping every other power of s. Thus, the polynomial formed for this example is:
Next we differentiate the polynomial with respect to s and we obtain:
Finally we use the coefficients of this last equation to replace the row of zeros. Again, for convenience the third row is multiplied by ¼ after replacing the zeros. The remainder of the table is formed in a straightforward manner by following the standard form shown in Table 6.2
We get Table 6.7. It shows that all entries in the first column are positive. Hence, there are no right half-plane poles and the system is stable.
- Control Systems Engineering, Nise pp 301-320
- Sistemas de Control Automatico Benjamin C Kuo pp
- Modern_Control_Engineering, Ogata 4t pp 288,
Written by: Larry Francis Obando – Technical Specialist – Educational Content Writer.
Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, Caracas.
Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, Valle de Sartenejas.
Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.
Contact: Caracas, Quito, Guayaquil, Jaén, Villafranca de Ordizia- Telf. +34633129287
If what you need is to solve urgently a problem of a «Mass-Spring-Damper System » (find the output X (t), graphs in Matlab of the 2nd Order system and relevant parameters, etc.), or a «System of Electromechanical Control «… to deliver to your teacher in two or three days, or with greater urgency … or simply need an advisor to solve the problem and study for the next exam … send me the problem…I Will Write The Solution To any Control System Problem…
, …I will give you the answer in digital and I give you a video-conference to explain the solution … it also includes simulation in Matlab. In the link above, you will find the description of the service and its cost.