Mathematics, 22.04.2020 01:02 sheram2010

For the characteristic polynomialp(s) =s5+ 2s4+ 24s3+ 48s2−25s−50(a) Use the Routh-Hurwitz Criterion to determine the number of roots ofp(s) in the right-half plane, in the left-half plane, and on thejω-axis.(b) Use Matlab to determine the roots ofp(s), and verify your results in part 2a.

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Answer from: idontknow11223344

1 root in the right half-plane1 conjugate pair on the imaginary axis2 roots in the left half-plane

Step-by-step explanation:

Without using the Routh-Hurwitz criterion at all, you know there is one positive real root. Descartes' rule of signs tells you the number of positive real roots is equal to the number of sign changes in the coefficients (perhaps less a multiple of 2). There is one sign change in + + + + - - , so there is one positive real root.

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(a) The Routh array starts as two rows of the polynomial's coefficients, alternate coefficients on each row. For this odd-degree polynomial, the number of coefficients is even, so no zero-padding is necessary at the right end of the second row. That is, we start with ...

$\begin{array}{cccc}s^5&1&24&-25\\s^4&2&48&-50\end{array}$

The next row is formed from combinations of coefficients in the two rows above. The computation is similar to that of a determinant. By matching the numbers to those in the array, you can see the pattern of the computation.

The next row values are ...

$\begin{array}{ccc}s^3&\dfrac{(2)(24)-(1)(48)}{2}&\dfrac{(2)(-25)-(1)(-50)}{2}\end{array}$

Simplifying, we find this row to be ...

$\begin{array}{ccc}s^3&0&0\end{array}$

The zero row is a special case that requires we proceed as follows. The row above (identified with s⁴) represents an "auxiliary polynomial":

2s^4 +48s^2 -50

To continue the process, we replace the zero row by the coefficients of the derivative of this auxiliary polynomial. Proceeding as before, the array now becomes ...

$\begin{array}{cccc}s^5&1&24&-25\\s^4&2&48&-50\\s^3&8&96\\s^2&24&-50\\s^1&112\frac{2}{3}&0\\s^0&-50\end{array}$

The number of sign changes in the first column (1) tells the number of roots in the right half-plane. The auxiliary polynomial will give us the remaining two pairs of roots:

$2s^4+48s^2-50=0\\\\2(s^2+25)(s^2-1)=0\\\\s=\pm 5i,\ s=\pm 1$

So, we have determined there to be ...

1 root in the right half-plane2 roots on the jω axis2 roots in the left half-plane

(b) The original polynomial can be factored as ...

p(s) = (s +2)(s² +25)(s +1)(s -1)

p(s) = (s +2)(s +1)(s -5i)(s +5i)(s -1)

This verifies our result from part (a).

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Additional comments

Any row can be multiplied by a convenient factor to simplify the arithmetic. Here, it would be convenient to divide the second row by 2 and the third row by 8.

A zero element (not row) in the first column is replaced by "epsilon" (a small positive number) and the rest of the arithmetic is continued as normal. That row is not counted (it is ignored) when counting sign changes in the first column.

Answer from: Quest

It would be the second one because it passes the vertical line test
Which graph represents a function?

Answer from: Quest

answer: sine

step-by-step explanation:

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