Question Q#7: Using Routh Hurwitz Criterion, discuss the stability of the system when characteristic polynomial is: S8+ 10s? + 50s + 500 = 0 (a) No sign change in the first column of Routh table, so system is stable (b) One sign change in the first column of Routh table, so system is unstable (c) No sign change in the first column of Routh table, so system is marginally stable (d) One sign change in the first column of Routh table, so system is marginally stable

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Transcribed Image Text: Q#7: Using Routh Hurwitz Criterion, discuss the stability of the system when characteristic polynomial is: S8+ 10s? + 50s + 500 = 0 (a) No sign change in the first column of Routh table, so system is stable (b) One sign change in the first column of Routh table, so system is unstable (c) No sign change in the first column of Routh table, so system is marginally stable (d) One sign change in the first column of Routh table, so system is marginally stable

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Given polynomials^{3}+10 s^{2}+50 s+500=0This criterion is based on arranging the coefficients of characteristic equation into an array called "Routh array".q(s)=a_{0} s^{n}+a_{1} s^{n-1}+a_{2} s^{n-2}+\cdots+a_{n}=0 .Routh array: \quad a_{0}=1 ; a_{1}=10 ; a_{2}=50 ; a_{3}=500s^{n} \quad a_{0} \quad a_{2} \quad a_{4} \quad a_{6} \ldotss^{n-1} \quad a_{1} \quad a_{3} \quad a_{5} \quad a_{7} \ldots where;\begin{array}{l}s^{n-2} \quad b_{1} b_{2} b_{3} \ldots \ldots \quad b_{1}=\frac{a_{1} a_{2}-a_{0} a_{3}}{a_{1}} \\s^{n-3} c_{1} \quad c_{2} \quad c_{3} \ldots b_{2}=\frac{a_{1} a_{4}-a_{0} a_{5}}{a_{1}} \\c_{1}=\frac{b_{1} a_{3}-a_{1} b_{2}}{b_{1}} \\s^{0} \quad a_{n} \\c_{2}=\frac{b_{1} a_{5}-a_{1} b_{3}}{b_{1}} \\5^{3} 150 \\s^{2} \quad 10 \quad 500 \\s^{\prime} \quad 0 \quad 0 \\b_{1}=\frac{10 \times 50-500}{10} \\b_{1}=0 \\\end{array}No sign change in the first column of Routh table, we got zero, so the system is marginally stable.\rightarrow Answer is option (C) ...