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Chapter 7: Problem 14

Calculate the poles of the rational function $$ G(s)=\frac{s+5}{s^{2}+3 s+2} $$

Short Answer

Expert verified
Answer: The poles of the given rational function are \(s_1 = -1\) and \(s_2 = -2\).

Step by step solution

01

Factor the denominator

In order to find the roots of the denominator, we need to factor the polynomial \(s^2 +3s + 2\). We will look for two binomials whose product is equal to the given polynomial. This requires finding two numbers that multiply to 2 and add to 3, which are 1 and 2. So we can factor the denominator as follows: $$ s^2 +3s + 2 = (s+1)(s+2) $$ Note that the numerator doesn't influence the poles, so it doesn't need to be factored.
02

Find the roots of the factored denominator.

Next, we'll find the values of \(s\) where the factored denominator is equal to 0, i.e., \((s+1)(s+2)=0\). To do this, set each factor in the polynomial equal to 0 and solve for s. * For the first factor: \(s+1=0\), we find \(s=-1\). * For the second factor: \(s+2=0\), we find \(s=-2\).
03

State the poles of the rational function

Since the roots of the denominator polynomial are -1 and -2, the poles of the given rational function \(G(s)\) are: $$ s_1 = -1 \quad \text{and} \quad s_2 = -2 $$ The poles represent the values of \(s\) that make the rational function infinite. These are crucial in control systems and signal processing applications.

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Most popular questions from this chapter

For each fraction state the degrees of the numerator and denominator, and hence determine which are proper and which are improper: (a) \(\frac{x+1}{x}\) (b) \(\frac{x^{2}}{x^{3}-x}\) (c) \(\frac{(x-1)(x-2)(x-3)}{x-5}\)

Verify that \(x=2\) and \(x=3\) are both solutions of \(x^{2}-5 x+6=0\)

Solve the inequality \(|3 x+2| \leq 4\)

Solve the following quadratic equations by an appropriate method. (a) \(x^{2}+16 x+64=0\) (b) \(x^{2}-6 x+3=0\) (c) \(2 x^{2}-6 x-3=0\) (d) \(x^{2}-4 x+1=0\) (e) \(x^{2}-22 x+121=0\) (f) \(x^{2}-8=0\)

Verify that the given value is a solution of the given equation. $$ \frac{1}{3} x+\frac{4}{3}=2, x=2 $$
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