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## Homework Statement:

- We are working with the closed loop transfer function: $$ P(s) = \frac{K(s) G(s)}{1+K(s)G(s)} $$, where $$ G(s) = \frac{3s^2 - 2s + 3}{(s - 1)^2} $$ and ## K(s) = K ## where K is a real constant. Find the range of values of K to ensure closed loop stability

## Relevant Equations:

- Closed loop transfer function

Hi,

I was just working through a closed loop stability problem in Control Theory and I don't really understand how the answer has arrived at the solution so quickly.

$$ P(s) = \frac{K(s) G(s)}{1+K(s)G(s)} $$, where $$ G(s) = \frac{3s^2 - 2s + 3}{(s - 1)^2} $$ and ## K(s) = K ## where K is a real constant. Find the range of values of K to ensure closed loop stability

For stability, we want the poles to be in the left hand side (< 0)

I have tried to calculate the poles and just enforce the condition that they are < 0 (in the left half plane). Also note that this question is 3 marks so I don't really see how one can be expected to do lots and lots of algebra.

Nonetheless, if we consider ## 1 + KG(s) = 0 ## to find the open loop poles, we get the equation:

$$ 1 + K \frac{3s^2 - 2s + 3}{(s - 1)^2} = 0 $$

which leads to:

$$ s^2 (3K + 1) + s( -2 - 2K) + (1 + 3K) = 0 $$

if we use the quadratic equation, we get:

$$ s = \frac{(2 + 2K) \pm \sqrt {(2+2K)^2 - 4(3K+1)^2}}{2(3K+1)} $$

For now, I just choose to look at the first part of the term ## \frac{2 + 2K}{2(1 + 3K)} ##

if we want this to be < 0, then either the numerator or denominator have to be negative, but not both:

Case 1: numerator is negative

$$ 2 + 2K < 0 \rightarrow 1 + K < 0 \rightarrow K < -1 $$

and

$$ 2(1 + 3K) > 0 \rightarrow K > \frac{-1}{3} $$

thus, there are no values of K for this solution which meet both constraints

Case 2: denominator is negative

$$ 2(1 + 3K) < 0 \rightarrow K < \frac{-1}{3} $$

and $$ 2 + 2K > 0 \rightarrow K > -1 $$

which means that the following range ensures closed loop stability: ## -1 < K < \frac{-1}{3} ## (after I need do check that those values work for the discriminant of the pole in the quadratic equation)

However, this is different than the quoted solution. They have the closed loop transfer function the same as me.

1) How did they get the answer so quickly? Is there some additional theory that allows us to do this?

2) Where did I go wrong such that my range is the opposite of theirs?

Any help is greatly appreciated. Thanks.

I was just working through a closed loop stability problem in Control Theory and I don't really understand how the answer has arrived at the solution so quickly.

**Problem:**We are working with the closed loop transfer function:$$ P(s) = \frac{K(s) G(s)}{1+K(s)G(s)} $$, where $$ G(s) = \frac{3s^2 - 2s + 3}{(s - 1)^2} $$ and ## K(s) = K ## where K is a real constant. Find the range of values of K to ensure closed loop stability

**My attempt:**For stability, we want the poles to be in the left hand side (< 0)

I have tried to calculate the poles and just enforce the condition that they are < 0 (in the left half plane). Also note that this question is 3 marks so I don't really see how one can be expected to do lots and lots of algebra.

Nonetheless, if we consider ## 1 + KG(s) = 0 ## to find the open loop poles, we get the equation:

$$ 1 + K \frac{3s^2 - 2s + 3}{(s - 1)^2} = 0 $$

which leads to:

$$ s^2 (3K + 1) + s( -2 - 2K) + (1 + 3K) = 0 $$

*At this point the solution just says: "*## 1 + 3K > 0 \rightarrow K > \frac{-1}{3} ##*or*## 1 + K < 0 \rightarrow K < -1 ##" and then combines those ranges for the answerif we use the quadratic equation, we get:

$$ s = \frac{(2 + 2K) \pm \sqrt {(2+2K)^2 - 4(3K+1)^2}}{2(3K+1)} $$

For now, I just choose to look at the first part of the term ## \frac{2 + 2K}{2(1 + 3K)} ##

if we want this to be < 0, then either the numerator or denominator have to be negative, but not both:

Case 1: numerator is negative

$$ 2 + 2K < 0 \rightarrow 1 + K < 0 \rightarrow K < -1 $$

and

$$ 2(1 + 3K) > 0 \rightarrow K > \frac{-1}{3} $$

thus, there are no values of K for this solution which meet both constraints

Case 2: denominator is negative

$$ 2(1 + 3K) < 0 \rightarrow K < \frac{-1}{3} $$

and $$ 2 + 2K > 0 \rightarrow K > -1 $$

which means that the following range ensures closed loop stability: ## -1 < K < \frac{-1}{3} ## (after I need do check that those values work for the discriminant of the pole in the quadratic equation)

However, this is different than the quoted solution. They have the closed loop transfer function the same as me.

**Questions:**1) How did they get the answer so quickly? Is there some additional theory that allows us to do this?

2) Where did I go wrong such that my range is the opposite of theirs?

Any help is greatly appreciated. Thanks.

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