GATE 2018 EE Q.34: solution and answer

GATE 2018 · EEQ.34 · 2 marks · MCQ

Q.34 The number of roots of the polynomial, $s^{7} + s^{6} + 7 s^{5} + 14 s^{4} + 31 s^{3} + 73 s^{2} + 25 s + 200$ , in the open left half of the complex plane is

(A)
3
(B)
4
(C)
5
(D)
6

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Answer and solution

ANSWER

The polynomial has 3 roots in the open left half of the s-plane: option (A).

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

Routh array, first two rows (from the coefficients): $s^{7} : 1, 7, 31, 25$ and $s^{6} : 1, 14, 73, 200$ .
Row $s^{5}$ : $\frac{1 \cdot 7 - 1 \cdot 14}{1} = - 7, \frac{1 \cdot 31 - 1 \cdot 73}{1} = - 42, \frac{1 \cdot 25 - 1 \cdot 200}{1} = - 175$ . Dividing a whole row by a positive number does not change any sign, so divide by $7$ : $- 1, - 6, - 25$ .
Row $s^{4}$ (from the $s^{6}$ and $s^{5}$ rows): $\frac{( - 1 ) ( 14 ) - ( 1 ) ( - 6 )}{- 1} = 8, \frac{( - 1 ) ( 73 ) - ( 1 ) ( - 25 )}{- 1} = 48, \frac{( - 1 ) ( 200 ) - ( 1 ) ( 0 )}{- 1} = 200$ . Divide by $8$ : $1, 6, 25$ .
Row $s^{3}$ : $\frac{( 1 ) ( - 6 ) - ( - 1 ) ( 6 )}{1} = 0$ and $\frac{( 1 ) ( - 25 ) - ( - 1 ) ( 25 )}{1} = 0$ . The whole row is zero, so form the auxiliary polynomial from the $s^{4}$ row: $A (s) = s^{4} + 6 s^{2} + 25$ .
Replace the zero row with $d A / d s$ : $\frac{d A}{d s} = 4 s^{3} + 12 s$ , which gives $s^{3} : 1, 3$ after dividing by $4$ . Then $s^{2} : \frac{1 \cdot 6 - 1 \cdot 3}{1} = 3, 25$ ; $s^{1} : \frac{3 \cdot 3 - 1 \cdot 25}{3} = - \frac{16}{3}$ ; $s^{0} : 25$ .
First column (rows $s^{7}$ to $s^{0}$ , after the positive scalings): $1, 1, - 1, 1, 1, 3, - \frac{16}{3}, 25$ . It changes sign four times: $1 \to - 1$ , $- 1 \to 1$ , $3 \to - \frac{16}{3}$ and $- \frac{16}{3} \to 25$ . So there are 4 roots in the right half-plane.
Check for roots on the $j ω$ axis: a zero row means the auxiliary polynomial's roots are also roots of $P (s)$ . From $A (s) = 0$ : $s^{2} = \frac{- 6 \pm 36 - 100}{2} = - 3 \pm j 4$ . Since $(1 + j 2)^{2} = - 3 + j 4$ and $(1 - j 2)^{2} = - 3 - j 4$ , the roots are $s = \pm (1 + j 2), \pm (1 - j 2)$ , i.e. $1 \pm j 2$ and $- 1 \pm j 2$ . None lies on the imaginary axis. Two are in the RHP and two in the LHP, which matches the two sign changes from the $s^{4}$ row downward ( $1, 1, 3, - \frac{16}{3}, 25$ ).
Cross-check by division: $P (s) = (s^{4} + 6 s^{2} + 25) (s^{3} + s^{2} + s + 8)$ ; expanding gives $s^{7} + s^{6} + 7 s^{5} + 14 s^{4} + 31 s^{3} + 73 s^{2} + 25 s + 200$ . For the cubic factor, the Routh column is $1, 1, \frac{1 \cdot 1 - 1 \cdot 8}{1} = - 7, 8$ : two sign changes, so the cubic has 2 RHP roots and 1 LHP root.
Count: RHP roots $= 2 + 2 = 4$ , $j ω$ -axis roots $= 0$ , so LHP roots $= 7 - 4 - 0 = 3$ (1 from the cubic and 2 from $A (s)$ , namely $- 1 \pm j 2$ ). The answer is (A) 3.

KEY IDEA

Routh–Hurwitz with an all-zero row: the zero row at $s^{3}$ means an even auxiliary polynomial $A (s) = s^{4} + 6 s^{2} + 25$ divides $P (s)$ . Replace the zero row with $d A / d s$ and continue. The sign changes in the first column count the RHP roots (4). Then check whether any roots of $A (s)$ lie on the $j ω$ axis (none here; they are $\pm 1 \pm j 2$ ). LHP count = total − RHP − $j ω$ = 7 − 4 − 0 = 3.

WATCH OUT FOR

Seeing a zero row and assuming it always means roots on the $j ω$ axis. That makes a student subtract four imaginary-axis roots, or count $A (s)$ 's roots as marginal, and so pick a wrong option. An even polynomial's roots are symmetric about the origin: they can be real pairs $\pm a$ , imaginary pairs $\pm j ω$ , or complex quadruples $\pm σ \pm j ω$ . Here $s^{2} = - 3 \pm j 4$ gives the quadruple $\pm 1 \pm j 2$ , so none are on the axis. A second trap: people read "4 sign changes" as 4 LHP roots (option B). The sign changes count right-half-plane roots.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 15 Routh–Hurwitz questions, with notes

GATE 2018 EE Q.34: solution and answer

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