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Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) over the range of K=-10.5 to 0 as plotted below confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

As Dr. Cheever summarized, break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this should results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) over the range of K=-10.5 to 0 as plotted below confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

As Dr. Cheever summarized, break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Note: Root locus was plotted using Octave with the rlocus command as part of the control toolbox

pkg load control %if not already loaded
sys=tf([1 -1],[1 3 2]);
rlocus(sys,.1,-10.5,0);

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) further confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

Break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this should results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) over the range of K=-10.5 to 0 as plotted below confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

As Dr. Cheever summarized, break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this should results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) further confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

Break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this should result in the break out close to -1.4 as shown in the root locus above. Combining that with the other characteristics of the root locus as summarized on Dr. Cheever's page can be used to prove the solution does not exist (as demonstrated by the root locus itself).

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) further confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

Break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this should results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.