Determine values of K>0 such that the poles are in left-hand plane?

[annas425]

Given the transfer function G(s) = 1 / (s(s+1)(s^2 + 4s + 13)), how would I determine the range of the values of K>0 such that the closed-loop poles are in the left-hand plane?

Picture of block diagram with transfer function:

Not sure if this is right at all, but I know that a system is stable when K>0, so if the poles were to be in the left-hand plane (i.e., be stable), would it just be for all K>0? I am assuming it's more involved than just that, so I would really appreciate some help please! :)

Thank you in advance!

 

[maajdl]

Start by solving question 3).

Note also that in question 2 you are given the information that for too large K, the system become unstable.
Therefore you hypothesis that K>0 for stability must be wrong. (there was no reason for that)

 

[annas425]

Start by solving question 3).

Note also that in question 2 you are given the information that for too large K, the system become unstable.
Therefore you hypothesis that K>0 for stability must be wrong. (there was no reason for that)

Well, doesn't the prompt just say that K must be greater than 0?? I am confused...

 

[rude man]

To give you a feel for how k has to range as a function of ω, form an expression for the phase shift of G(s) as a function of ω. This is not hard. The phase shift will be a transcendental equation in ω. The phase shift is independent of k. Then use Excel to run ω over a range of values starting at zero with constant small increments, and ask it to compute the phase shift for each ω. When the phase shift gets to π, put that value of ω into G(s) and solve for the range of k for which |G(s)| < 1. That range of k gives a stable closed loop.

There may be more than one value of ω for which the phase shift is π, so don't quit when you hit pay dirt the first time.

Routh-Hurwitz is seldom used in 'real life' because usually the transfer function is known as gain and phase plots but not in closed form.

I never learned how to do Root-Locus, don't regret it, so can't help you there either. Sorry about both.

 

[rezeew]

To determine the range of values of K>0 for which the closed-loop poles are in the left-hand plane, we need to analyze the stability of the system. The stability of a system is determined by the location of its poles in the complex plane. In general, for a system to be stable, all of its poles must be in the left-hand plane.

In this case, we have a transfer function G(s) = 1 / (s(s+1)(s^2 + 4s + 13)). The denominator of this transfer function can be factored as (s+1)(s^2 + 4s + 13). This means that the system has three poles, one at s=-1 and two at the roots of the quadratic equation s^2 + 4s + 13 = 0.

Using the quadratic formula, we can find the roots of this equation to be s = -2 ± 3i. This means that the two poles at the roots of the quadratic equation are complex conjugates, and they have a negative real part (since -2 < 0). Therefore, these two poles are already in the left-hand plane.

Now, we need to determine the range of values of K>0 for which the pole at s=-1 is also in the left-hand plane. This can be done by analyzing the Routh-Hurwitz stability criterion, which states that for a system to be stable, all the coefficients of the characteristic polynomial (in this case, s^3 + 5s^2 + 13s + K) must be positive.

Using the Routh-Hurwitz criterion, we can set up the following table:

s^3 | 1 13
s^2 | 5 K
s^1 | (65-K)/5
s^0 | K

For the pole at s=-1 to be in the left-hand plane, the first two rows of this table must have positive coefficients. This means that K>0 and (65-K)/5>0. Solving for K, we get K<65.

Therefore, the range of values of K>0 for which the closed-loop poles are in the left-hand plane is 0<K<65. Any value of K within this range will result in a stable system with all poles in the left-hand plane.