NCEES 130

The question shows a feedback loop, input is R(s), output is Y(s), and there is a "black box" containing K/((s+1)(s+2)). It asks you to pick a range of values for K that make the system stable.

I'm really confused about what NCEES lists in its solution: C/R=G/(1+G)=N(s)/D(s), with D(s)=(s+1)(s+2)+K, so K>-2. What are C and R? Do N and D stand for numerator and denominator?

The best I can come up with is that if G=(black box eqn), then for a closed loop (per EERM p.63-11), the denominator is G+1=1+K/((s+1)(s+2))= K/(s^2+3s+2+K). How do they get from here to D(s)=s^2+3s+2+K?

I understand that once you get to that point, K has to be >-2 to prevent the denomintor from decreasing in size and approaching zero.

Thanks in advance for any guidance. This problem is worded so generically that I can't find any specific info to help me online.

grownupsara said:

The question shows a feedback loop, input is R(s), output is Y(s), and there is a "black box" containing K/((s+1)(s+2)). It asks you to pick a range of values for K that make the system stable.
I'm really confused about what NCEES lists in its solution: C/R=G/(1+G)=N(s)/D(s), with D(s)=(s+1)(s+2)+K, so K>-2. What are C and R? Do N and D stand for numerator and denominator?

The best I can come up with is that if G=(black box eqn), then for a closed loop (per EERM p.63-11), the denominator is G+1=1+K/((s+1)(s+2))= K/(s^2+3s+2+K). How do they get from here to D(s)=s^2+3s+2+K?

I understand that once you get to that point, K has to be >-2 to prevent the denomintor from decreasing in size and approaching zero.

Thanks in advance for any guidance. This problem is worded so generically that I can't find any specific info to help me online.

Click to expand...

Basically you are right on everything, up to a point.

I don't have the book but C is the output of the network, R is the inut and C/R is the symbol for the closed loop transfer function, which is one way to figure stability. N and D mean numerator and Denominator. When you have a feedback loop with KG (plant) in the forward path and H in the feedback path the closed loop TF is KG/(1+KGH). With unity feedback H =1. Stick in that function in the box for KG, then multiply both top and bottom by (s+1)(2+2) and you'll end up with the s^2+3s+2+K in the denominator.

I think there is an easier way to do this with the characteristic equation and a simple Routh Array.

As far as stability - I may be wrong, but I think you have to actually set up a very simple Routh-Hurwicz array based on that characteristic equation. I am sure Jim will probably answer with the definitive answer before I get home, but if not I'll try to give you at least one fairly simple way to solve this.

I can't help too much because I don't have the NCEES book or EERM here at home, but solving simple closed loop stability problems is usually accomplished by using the Routh-Hurwitz stability criterion. I know this is covered in the EERM so you can try looking it up there.

I don't want to say too much because I am recalling this from memory, but I believe you have to write out the transfer function of the system, and then use the coefficients of the denominator of the transfer function to formulate a table. Wikipedia has an entry for this under Routh-Hurwitz stability criterion but they make it sound way more complicated than it actually is.

In the equation you wrote, I'm not sure what C and R are, but G/(1+G) is the transfer function and yes, N(s) and D(s) are the numerator and denominator of the transfer function respectively. I believe the trick is to manipulate the transfer function so it is in the proper form--that is the denominator has to be of the form C+a1s + a2s2 +a3s3 + ... + ansn (which I believe is referred to as the characteristic polynomial). I think this algebraic manipulation is the step you are missing.

So, your transfer function T(s) is G(s)/(1+G(s)) = (K/((s+1)(s+2)))/(1+K/(s+1)(s+2)). If you multiply the numerator and denominator by (s+1)(s+2), you should get T(s) = K/((s+1)(s+2)+K). Then the denominator of T(s) is s2+3s+2+K, which is what they got. In this case, you can see by inspection that K has to be greater than -2 to keep the denominator from approaching zero, but in other cases it won't be that easy to see and you have to use the tabular method of Routh-Hurwitz to solve for the stable values of K. I can't try to explain Roth-Hurwitz without my references.

Hopefully someone else can explain the process better; otherwise I'll try to remember to bring my EERM home tomorrow and have a try at it.

And please if anyone sees errors in this post, correct them! I'm just recalling this from memory.

hee hee, it's always fun when we're answering the same question at the same time.

I'll leave the explanation of Routh to benbo because I know he's better at controls stuff than me. To me it seemed complicated at first but it's really not that bad (I didn't think EERM did a very good job of explaining it), and I think it probably has a good chance of showing up on the morning session of the exam.

benbo said:

I am sure Jim will probably answer with the definitive answer before I get home, but if not I'll try to give you at least one fairly simple way to solve this.

Click to expand...

Don't look at me on this one. You guys have a much better grip on the controls stuff than I do. I can get to the right answer on the simple controls problems but I have no idea how to explain it to somebody else.

Jim

Turns out the best way to calculate the characteristic equation was the way my compatriots explained it. I think this is actually a little hard for the AM.

So assuming you now know how to get to the s^2+3s+2+K you look at the coefficients of the polynomial, and put them inot the Routh Array. I think this might even be in the EERM.

I initially made a mistake here, so to avoid confusion, to see how to set up the array see Chicago's post below.

The only thing you need for stability in this one is for the signs of the coefficients in the first column to NOT change.

Good luck.

:reading:

whut the....

so add Transfer functions BJTs, Diodes, pretty much all electronics....oh an i HATE the way they word the economic problems...

I'm wondering if i should just skip the morning...lol

jdd18vm said:

:reading:
whut the....

so add Transfer functions BJTs, Diodes, pretty much all electronics....oh an i HATE the way they word the economic problems...

I'm wondering if i should just skip the morning...lol

Click to expand...

THere will be some real low lying fruit in the AM - for you especially in power if that is your discipline.

And although it isn't necessarily true for every exam, the economics problems on my test were REALLY trivial.

benbo said:

THere will be some real low lying fruit in the AM - for you especially in power if that is your discipline.
And although it isn't necessarily true for every exam, the economics problems on my test were REALLY trivial.

Click to expand...

Love low flying fruit

thanks

Wow, thanks for all the replies everyone! I just got into work for the morning, but I read through all your responses, and I think I understand now how to approach this type of problem now. I'm going to sit down and go through everything in more detail over lunch today, so I'll let you know if I have any more questions.

This is the last problem from the AM part of the NCEES exam that I was having trouble with, so it will be good for my peace of mind to work this one out.

grownupsara said:

I'm really confused about what NCEES lists in its solution: C/R=G/(1+G)=N(s)/D(s), with D(s)=(s+1)(s+2)+K, so K>-2. What are C and R? Do N and D stand for numerator and denominator?

Click to expand...

grownupsara,

The weird symbology they use is actually referred to in Fig. 63-10, p. 63-13 in EERM. Personally, I would just use Figure 63.5(3) on p. 63-6 with H(s)=1 instead.

As far as the alternate Routh array is concerned, one correction needs to be noted from the earlier post.

The Routh array should be setup as follows (it will still give you the same answer, but technically this is the way the coefficients should be correctly setup):

_______________

| 1 2+K

| 3 0

| 3(2+K)/3 0

Solve 3(2+K)/3 greater than zero.

The 3's in the num. & den. cancel out.

That leaves you with (2+K)=0

K>-2

Oops. Good catch chicago. I'm going to modify my initial post so nobody get's messed up by it.

I should have looked it up instead of relying on my memory.

benbo, I thought the Routh-Hurwitz criterion for this problem was:

s^2| 1 K+2

s^1| 3 0

s^0| [-(1*0-3*(K+2))]/3 0

1, 3 are > 0 so, K+2 should be >0 for the system to be stable --> K>-2

I am confused now; when you say 1 & K+2 should be in the same column and needs to be in the same sign. Is it a short-cut to determine stability for quadratic equations.

I referred this topic in Schaum's Basic Electrical Engineering. It is a very good reference. It also addresses the special cases when zeros occur in the first column.

Thanks,

Ilan.

benbo said:

Turns out the best way to calculate the characteristic equation was the way my compatriots explained it. I think this is actually a little hard for the AM.So assuming you now know how to get to the s^2+3s+2+K you look at the coefficients of the polynomial.

In this case (and I can't imagine one more complicated on the AM) you are only concerned with the coefficient of the s^2 term (which is one) and the s^0 (the units) term (this coefficient is 2+K). THese go in the first column, so the first column of your Routh array becomes

1

2+k

The only thing you need for stability in this one is for the signs of the coefficients in the first column to NOT change.

Since the sign of 1 is postiive,

the sign of 2+K has to be positive.

so

2+K> 0

K> -2

Good luck.

Click to expand...

Ilan said:

benbo, I thought the Routh-Hurwitz criterion for this problem was:
s^2| 1 K+2

s^1| 3 0

s^0| [-(1*0-3*(K+2))]/3 0

1, 3 are > 0 so, K+2 should be >0 for the system to be stable --> K>-2

I am confused now; when you say 1 & K+2 should be in the same column and needs to be in the same sign. Is it a short-cut to determine stability for quadratic equations.

I referred this topic in Schaum's Basic Electrical Engineering. It is a very good reference. It also addresses the special cases when zeros occur in the first column.

Thanks,

Ilan.

Click to expand...

You are right to be confused because I was wrong. Glad someone caught it.

I guess I am too slow in typing, chicago caught it too.

Really useful info, everyone. I worked through the problem using both the simple "reasoning it out" method and also using Routh-Hurwitz (using EERM p.63-13), and got both of them to work. So, I think my game plan for the exam will be to see if I can use visual inspection of the equation to figure out how to keep denominator >0. If it's too complicated for that, I'll use Routh-Hurwitz. Thanks again!

grownupsara said:

The question shows a feedback loop, input is R(s), output is Y(s), and there is a "black box" containing K/((s+1)(s+2)). It asks you to pick a range of values for K that make the system stable.
I'm really confused about what NCEES lists in its solution: C/R=G/(1+G)=N(s)/D(s), with D(s)=(s+1)(s+2)+K, so K>-2. What are C and R? Do N and D stand for numerator and denominator?

The best I can come up with is that if G=(black box eqn), then for a closed loop (per EERM p.63-11), the denominator is G+1=1+K/((s+1)(s+2))= K/(s^2+3s+2+K). How do they get from here to D(s)=s^2+3s+2+K?

I understand that once you get to that point, K has to be >-2 to prevent the denomintor from decreasing in size and approaching zero.

Thanks in advance for any guidance. This problem is worded so generically that I can't find any specific info to help me online.

Click to expand...

From the equation s^2 + 3s +(2+K), you would find the range of K by using the Routh Array. This is done be looking at the coefficients in the denominator of the closed loop transfer function. How you would set up this array in the following manner:

1st Row: 1 (2+K) <-- Looking at the coefficients of s^2 and s^0

2nd Row: 3 0 <-- Looking at the coefficient of s (the 0 is there since it is only a second degree polynomial in the denominator

3rd row: a0 <--- a0 is computed as follows: a0 = 3(2+K) - (1)(0) = 3(2+K)

if you look at the first column in order for stability to happen all of the coefficients calculated in the first column has to be of one sign (i.e. no sign changes), thus in order for this to happen a0 > 0. Hence, 3(2+K) > 0 solving for K gives K > -2.

Good luck. hope that this helps. arty-smiley-048: