routh root locus

8/21/2019 Routh Root Locus

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Routh-Hurwtz Criterion & Root-Locus CriteriaStability of Feedback Control Systems



Let si be the poles of rational G.Then, G is:

• stable if Re(si)<0 for all i

• marginally stable if

1. Re(si)<=0 for all i, and

2. simple root for Re(si)=0

• unstable if it is neither stablenor marginally stable

Stability



This is for LTI systems with a polynomialdenominator (without sin, cos, exponential etc.)

It determines if all the roots of a polynomial lie inthe open LHP (left half-plane), or equivalently, havenegative real parts.

It also determines the number of roots of apolynomial in the open RHP (right half-plane).

It does NOT explicitly compute the roots.

Routh-Hurwitz criterion



Write the polynomial as thecharacteristic equation

Step 1



Write the coefficients in 2rows

• 1st row starts with an

• 2nd row start with an-1

• Other coefficientsalternate between rows

• Both rows should be thesame length• Continue until no

coefficients are left• Add zero as last

coefficient if necessary

Step 2



Complete the 3rd row• Call the new entries b1 , …, bk

• The 3rd row will be the same length as the 1st two

• The denominator is the 1st entry from the previous row• The numerator is the determinant of the entries from the

previous 2 rows:• The first column• The next column following the coefficient• If a coefficient doesn’t exist, substitute 0

Step 3



• Treat each following row in the same way as the 3rd row.

• There should be n+1 rows total, including the first row.

Step 4



• Now examine the 1st column

• Note: Any row can be multiplied by any positiveconstant without changing the result



1. The necessary and sufficient condition for all the rootsof the characteristic equation to have negative realparts (stable system) is that all the elements of the firstcolumn of the Routh array (a0 , a1 , b1 , c1 , etc.) bepositive and nonzero.

2. If some of the elements in the first column arenegative, the number of roots with a positive real part(in the right half plane) is equal to the number of signchanges in the first column.

3. If one pair of roots is on the imaginary axis,equidistant from the origin, and all other roots are in

the left half plane, all the elements of the nth row willvanish and none of the elements of the preceding rowwill vanish. The location of the pair of imaginary rootscan be found by solving the equation Cs2 + D = 0,where the coefficients C and D are the elements of thearray in the (n-1)th row as read from left to right,respectively.

Theorems of the Routh Test



• Consider the quadratic polynomial function:

where all the ai are positive. The array of coefficients becomes

where the coefficient a1 is the result of multiplying a1 by a2 andsubtracting a0(0) then dividing the result by a2. In the case of asecond order polynomial, we see that Routh’s stability criterionreduces to the condition that all ai be positive.

Example 2



• Consider the generic cubic polynomial

• where all the ai are positive. The Routh array is

• so the condition that all roots have negative real partsis

Example 3



• A root locus plot is a figure that shows how the rootsof the closed-loop characteristic equation vary as thegain of the feedback controller is varied from zero toinfinity. The abscissa is the real part of the closed-

loop roots; the ordinate is the imaginary part. Sincewe are plotting closed-loop roots, the time constantsand damping coefficients that we will pick off theseroot locus plots are all closed-loop time constants andclosed-loop damping coefficients.

•

Graphical procedure for finding the roots of 1 + G = 0,as one of the parameters of G varies continuously.

Root-Locus Method



Consider the closed loop transfer function

• How do the poles of the closed-loop system changeas a function of the gain K?

The closed-loop transfer function is:

The characteristic equation is:

The closed loop poles are



Example of Root-Locus Plots



Let us start with the simplest of all processes, a first-

order lag. We will choose a proportional controller.The system and controller transfer functions are

The closed-loop characteristic equation is

Sample Problem



Solving for the closed-loop root gives



For a 1st order system, the closed-loop root is always real, so the

system can never be underdamped or oscillatory. The closed-loop damping coefficient of this system is always greater thanone. The larger the value of controller gain, the smaller theclosed-loop time constant because the root moves farther awayfrom the origin (the time constant being the reciprocal of thedistance from the root to the origin).

The 1st order system can never be closed-loop unstable becausethe root always lies in the LHP. No real system is only 1st order.There are always small lags in the process, in the control valve

or in the instrumentation, that make all real systems of higherorder than 1st.

routh root locus

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