ROUTH HURWITZ ANALYSIS

Transcription

1 ROUTH HURWITZ ANALYSIS The Routh Hurwitz analyi tell you how many root are located in the a) let-hand plane, ) right-hand plane, and c) on the jω-axi. The technique i illutrated here with an example. The Routh Hurwitz analyi involve creating an array o value. There are two pecial cae which may e involved, 1) a row o the array i all zero (thi happen in our example) and ) a row egin with zero ut i not all zero (thi doe not occur in our example ut i covered on the lat page). THE PROBLEM Given the ytem equation ind the numer o root located in the let-hand plane, right-hand plane, and on the jω-axi. FIND THE ZEROS Since the equation can e actored y, there are two zero in the polynomial, i.e. two root at the origin ( ) STARTING THE ROUTH HURWITZ ARRAY We can drop the actor and conider only the remaining equation The order o thi polynomial i n6, o we will e contruction an array having n+1 or 7 row. Taking the coeicient, we contruct the irt two row o the array in thi order tomzap@eden.com page 1 o 5

2 THE NEXT ROWS OF THE ROUTH HURWITZ ARRAY Suequently, each new row i derived rom the two row immediately aove it, which I will call the working row. The irt column o the matrix in our expreion i alway the irt column o the two working row o the Routh Hurwitz array. The remaining column i the column o the working row jut to the right o the poition o the unknown. The denominator o the expreion i the irt numer o the lower o the two working row So now we have SPECIAL CASE #1: WHEN A ROW IS ALL ZEROS For the ourth row 1 16 c c c c c c and we have 1 16 We take element o the lat non-zero row and orm an auxiliary equation: ( n K) ( n K) ( n K) where n i the order o the polynomial (6 in thi cae), and K i the row numer o the row jut aove the, rom which thee value are taken ( in thi cae). Note the pattern o utracting nothing rom the irt exponent, utracting rom the econd, and utracting 4 rom the third, etc. So the auxiliary equation ecome tomzap@eden.com page o 5

3 Now we take the derivative with repect to d d ( ) We replace row 4 with thee new coeicient THE FIFTH ROW For the ith row d d d d d 16 d So we have tomzap@eden.com page o 5

4 THE SIXTH ROW For the ixth row e e e e e e So we have THE SEVENTH ROW For the eventh row So we have We now have the required n+1 row. tomzap@eden.com page 4 o 5

5 THE NUMBER OF ROOTS IN THE RHP Count the numer o ign change in the irt column ( in thi cae). Thi i the numer o root in the right-hand plane. THE NUMBER OF ROOTS ON THE jw-axis Solving the auxiliary equation, we ind 4 root to that equation, o there are 4 root on the jω-axi (o ar). THE NUMBER OF ROOTS IN THE LHP Since n6 and we have accounted or 4 o thee root, the remaining root are in the let-hand plane. THE FINAL RESULT Returning to the eginning o the prolem, we had actored out an rom the original polynomial. Thee are zero o the polynomial and are thereore on the jω-axi. And ince the polynomial i in the denominator o the traner unction, thee are additional root on the jω-axi and are added to the other 4 root we have ound. So the inal reult i 6 root on the jω-axi, root in the righthand plane and root in the let-hand plane. SPECIAL CASE #: WHAT IF A ROW BEGINS WITH? Thi pecial cae did not occur in our example, o it i covered here. I a row hould egin with a zero and the row i not all zero, we replace thi leading zero with an ε. Thi ε i conidered to e a mall poitive numer and expreion containing ε that appear in column 1 o uequent row are determined to e poitive or negative aed on thi characteritic. tomzap@eden.com page 5 o 5