Conditions of Stability for second order system

[Chunkysalsa]

Homework Statement

Consider the transfer function $$H(s)=\cfrac{1}{a_{2}s^{2}+a_{1}s+a_{0}}$$

where real-valued coefficients $a_{2},a_{1}, a_{0}$ are arbitrary except that $a_{2}$ is nonzero. Verify that the system is stable iff the coefficients $a_{2},a_{1}, a_{0}$ have the same sign.

Homework Equations

Professor gave us a hint to observe the quadratic formula

The Attempt at a Solution

I really don't know how to start it. I've tried a prove by cases but got bogged down fairly quickly. I can post what I have if this is the correct approach but I have doubts that it is. I just kept gaining more and more cases. A couple of hints towards the right direction would be most helpful

EDIT: I should note that I understand that for the system to be stable all the poles (2 in this case) of the transfer function should be in the LHP. That is that the the Re(p)<0 where p is the pole.

 

[jake929]

Since you know the condition for stability real part of the root is less than 0 then you need to determine what is required of your constants to ensure that condition is met. Use your professors advice the quadratic equation: (-B±√B^2-4AC)/2A. There are really only two cases 1) the roots contain an imaginary part 2) the roots contain all real parts. If B^2-4AC is negative then there is an imaginary part. Show for each of the two cases what is required to get the real part to be negative.

 

[Chunkysalsa]

Alright this is essentially what I did. I'm a bit iffy on a bit of the parts so I'll post it later today if my prof isn't able to help. I just thought they're might be a better approach then examining all possible cases.

Am I correct in saying that I'll also need to prove that if the system is stable then the coefficients are of the same sign since the statement says if and only if? The converse is easily proven but I'm just wondering if that's explicitly needed.

 

[jake929]

Alright this is essentially what I did. I'm a bit iffy on a bit of the parts so I'll post it later today if my prof isn't able to help. I just thought they're might be a better approach then examining all possible cases.

Am I correct in saying that I'll also need to prove that if the system is stable then the coefficients are of the same sign since the statement says if and only if? The converse is easily proven but I'm just wondering if that's explicitly needed.

If b^2-4ac<0 then the roots contain imaginary parts, the real part is -B/2A, what is required for the real part to be negative? Then what is required of c given what you have already determined for A and B?

Second case is no imaginary part or B^2-4AC>0, B^2 is always greater than 0 so what is required of AC in order to make B^2-4AC greater than 0? don't forget the ± just before the radical. Now determine what is required of -b given what you know about A and C.

From your post it looks like what you are doing is trying out what happens for each combination of signs and trying to make a conclusion. This may work too but then you have to prove that two out of the 8 possible combinations actually work. My preference would be to set boundaries on the constants based on whether there is an imaginary part of the root or not.

 

[DeeNos]

I am not sure how to relate that to the coefficients of the transfer function though.

One approach to solving this problem is to use the Routh-Hurwitz stability criterion. This criterion states that for a system to be stable, all the coefficients of its characteristic polynomial (in this case, the denominator of the transfer function) must have the same sign.

In this case, the characteristic polynomial is a_{2}s^{2}+a_{1}s+a_{0}. Using the quadratic formula, we can find the roots of this polynomial:

s = (-a_{1} +/- sqrt(a_{1}^{2}-4a_{2}a_{0})) / 2a_{2}

For the system to be stable, both of these roots must have negative real parts. This means that the quantity under the square root, a_{1}^{2}-4a_{2}a_{0}, must be positive.

Since a_{2} is nonzero, we can divide both sides of the equation by a_{2}. This gives us:

s = (-a_{1}/a_{2} +/- sqrt((a_{1}/a_{2})^{2}-4(a_{0}/a_{2})))

Now, we can see that the sign of the coefficient of s^{2} (which is a_{2}) is the same as the sign of the coefficient of the square root, since we divided both sides by a_{2}. Similarly, the sign of the coefficient of s (which is a_{1}) is the same as the sign of the coefficient inside the square root.

Therefore, for the system to be stable, all three coefficients (a_{2}, a_{1}, and a_{0}) must have the same sign. If they have different signs, then the quantity under the square root will be negative, and the system will have at least one pole with a positive real part, making it unstable.

In conclusion, we can verify that the system is stable if and only if all three coefficients have the same sign. This is consistent with the Routh-Hurwitz stability criterion and shows the importance of having the same sign for all coefficients for a stable system.