GATE EE 2016 Set 1 — Question 43

We are given the polynomial equation $P(s) = s^3 + 5.5 s^2 + 8.5 s + 3 = 0$.
We need to find the number of roots with real parts strictly less than $-1$, i.e., $\text{Re}(s) < -1$.To do this, we can shift the origin of the s-plane to $s = -1$. Let's define a new complex variable $z = s + 1$, which implies $s = z - 1$. The condition $\text{Re}(s) < -1$ is equivalent to $\text{Re}(z) < 0$, which means we need to find the number of roots of the transformed polynomial in $z$ that lie in the left-half of the z-plane.Substitute $s = z - 1$ into the polynomial $P(s)$:
$P(z-1) = (z-1)^3 + 5.5(z-1)^2 + 8.5(z-1) + 3 = 0$.Expand the terms:
$(z^3 - 3z^2 + 3z - 1) + 5.5(z^2 - 2z + 1) + 8.5(z - 1) + 3 = 0$
$z^3 - 3z^2 + 3z - 1 + 5.5z^2 - 11z + 5.5 + 8.5z - 8.5 + 3 = 0$Combine like terms to get the new polynomial $Q(z)$:
$Q(z) = z^3 + (-3 + 5.5)z^2 + (3 - 11 + 8.5)z + (-1 + 5.5 - 8.5 + 3) = 0$
$Q(z) = z^3 + 2.5z^2 + 0.5z - 1 = 0$Now, we apply the Routh-Hurwitz stability criterion to $Q(z)$ to find the number of its roots in the right-half of the z-plane (which corresponds to roots of $P(s)$ with $\text{Re}(s) > -1$).The Routh array for $Q(z)$ is:$b_1 = \frac{(2.5)(0.5) - (1)(-1)}{2.5} = \frac{1.25 + 1}{2.5} = \frac{2.25}{2.5} = 0.9$
$c_1 = \frac{(b_1)(-1) - (2.5)(0)}{b_1} = -1$The first column of the Routh array is $[1, 2.5, 0.9, -1]$.
There is one sign change in the first column (from $0.9$ to $-1$). This indicates that there is one root of $Q(z)$ in the right-half of the z-plane.
This means there is one root of the original polynomial $P(s)$ with $\text{Re}(s) > -1$.Let's check for roots on the line $s = -1$. Substitute $s=-1$ into $P(s)$:
$P(-1) = (-1)^3 + 5.5(-1)^2 + 8.5(-1) + 3 = -1 + 5.5 - 8.5 + 3 = 8.5 - 9.5 = -1 \neq 0$.
Since $P(-1) \neq 0$, there are no roots exactly on the line $\text{Re}(s) = -1$.The polynomial $P(s)$ is of degree 3, so it has a total of 3 roots.

• Number of roots with $\text{Re}(s) > -1$: 1
• Number of roots with $\text{Re}(s) = -1$: 0
• Number of roots with $\text{Re}(s) < -1$: Total roots - (roots with $\text{Re}(s) > -1$) - (roots with $\text{Re}(s) = -1$) = $3 - 1 - 0 = 2$.

Therefore, there are 2 roots with real parts strictly less than $-1$.