Question 1 - Control (cant see where I went wrong)

[fouadalnoor]

Hey guys, I have attached Q1 of my tutorial sheet on here. If you look at part B it says that I should show how the system is damped and that the "damped frequency of osciallation" is 4rad/s

But in my case, I got the transfer function to be:

1/(s^2 + s(6+k) -4k + 25), which give me two complex poles: -3±4j

Now (unless my transfer function is wrong) this system is an undamped system since 0<ζ<1
and Wn = 5.

What am I doing wrong?

Hope you can help!

Fouad.

 

[Ratchit]

fouadalnoor,

Both the transfer function and the roots are correct.

Ratch

 

[fouadalnoor]

fouadalnoor,

Both the transfer function and the roots are correct.

Ratch

ahh so I guess the lecturer was wrong? (since it explicity says in the question to prove that the "damped" system has a Wn = 4?

Thanks!

 

[MrAl]

ahh so I guess the lecturer was wrong? (since it explicity says in the question to prove that the "damped" system has a Wn = 4?

Thanks!

Hi,

Well, there's nothing wrong with saying that the system is a 'damped oscillation'. If it wasnt damped, it would be an oscillator or totally unstable. The question is only whether or not it is underdamped, overdamped, or critically damped, but all three of these are still considered 'damped'. In other words, a damped system has response like this:
f(t)=e^at *(A*sin(wt)+B*cos(wt)), {a<0}

or similar. If you look at that kind of response you'll see that it oscillates but gradually 'damps' out after some time t rather than continually oscillating because of the "a" exponent being negative. Thus it is called a "damped oscillation".

You found roots with real part -3 and imaginary part +/- 4, so the damped frequency of oscillation is 4, so w=4 in the equation for f(t) above.
The exponential factor is -3, so a above would be equal to -3.

So you did it right, you just didnt interpret the results correctly. Overdamped is damped so much there's no oscillation, underdamped is damped less such that there is still an oscillation for some time period less than infinity, and critically damped is right in the middle somewhere.

Oh yeah, that's not exactly the transfer function is it? What happened to the numerator?


The system is always damped when 'a' is less than zero.

 

[fouadalnoor]

Hmm from my calculation for part b) the numerator should be 1...or else I will get a different frequency of oscillation and different poles.

Also, if you look at part c) I get the following:

When there is an unit impulse for the input I get:

K = 4, y(t) = (1/12.64)e^(1.32t) + (1/-12.64)e^(-11.32t)

and I get an unstable system (as expected from the poles) being an exponential function going up to infinity

k = 6.25, y(t) = (4/41) + (1/-10.25)e^(-10.25t)

we now get another exponential function, starting at 0 and then going up to 4/41 as t increases where it is stable.

Is this correct?

 

[MrAl]

Hi,

Im sorry i must have been zoomed when i looked at that block diagram so i didnt see the 1/(s+a).

But, your first post said you were looking at part B which says K=0, so i did K=0, and came up with the transfer function:
H(s)=1/(s^2+6*s+25)

which has roots at -3 +/- 4 which means it is a damped sinusoidal as noted previously. I'll check out other values of K next and we can compare notes.
In the mean time, see what you get for K=0.

Ok, here are the results for other K:

K=4, two real roots, -9 and -1, which says it's a damped exponential, no sine or cosine.

K=6.25, roots 0 and -12.25, which says it's a constant plus (or minus) an exponential, no sine or cosine. The constant term integrates to t times a constant
and the other term integrates to a damped exponential with no sine or cosine, so it tends to infinity.

So in cases K=0 and K=4 the system is stable, but in case K=6.25 it is unstable. In fact, one of the roots lines on the imaginary axis at K=6.25, and anything over 6.25 is unstable.

For a damped sinusoidal response, we have K such that:
-30.1245<K<2.1245
approximately, and we have infinite response for:
K>=6.25

 

[fouadalnoor]

Thanks for your response,

Can you please explain to me part C)?

I got the values for the poles at k=4 and k=6.25

But I don't quite get how to calculate the response for an 'impulse' etc

Do I have to find y(s) when y(s)=R(s)G(s)
(unit impulse R(s) = e^as, a=0, R(s) = 1)
When k=4 and k=6.25

Sorry, I'm a little confused about the terminology..

 

[MrAl]

Hello,

Lets see if this helps...

Say we have a transfer function like this:
1/((s+a)*(s+b))

As you can see we have roots -a and -b.

The Impulse response for this is:
h(t)=(e^(-a*t)-e^(-b*t))/(b-a)

and you can see we can get this knowing only the two roots. If the roots are complex it might take a little more work to get h(t) into a more likable form which might require some simplification.

For this:
1/((s-i*b+a)*(s+i*b+a))

We have roots:
s=-i*b-a
s=i*b-a
which is a negative real part plus and minus the imaginary part,

and the impulse response is:
((i*e^(-(i*b+a)*t))-(i*e^(-(a-i*b)*t)))/(2*b)

which simplifies using Euler identities to:
e^(-a*t)*sin(b*t)/b

so you can visualize the impulse response from this, which is a damped sinusoid with exponential envelope.
Note that with the real part of the root being negative it's a damped sinusoid, but if the real part came out positive then it would be an increasing exponential with ever increasing slope too.

For a quicker visual, you can imagine plotting the poles on the s plane, then think about what kind of exponential you have and what kind of sinusoidal you have as to the frequency. That gives a quicker idea what kind of response to expect. This works because along the real axis we have the exponential exponent, and along the jw axis we have the frequency; for exponent getting closer and closer to zero we have slower and slower decay (or slower rising), while for the imaginary part getting closer and closer to zero we have decreasing frequency, and for exponent to the left of zero we have decay and for exponent to the right of zero we have continual exponential increase, and for exponent exactly at zero we have the ideal oscillator which only happens in theory or in passing.

 

[fouadalnoor]

Hmm, how do you get the (b-a) term for the impulse response?

I understand how to get the two exponential terms (you just need to do the inverse Laplace of the transfer function)

 

[MrAl]

Hi,

Do a partial fraction expansion of that first equation 1/((s+a)*(s+b)).

 

[fouadalnoor]

Hi,

Do a partial fraction expansion of that first equation 1/((s+a)*(s+b)).

Okay, so I have now worked through all the tasks and I got the following results:

Q1)
a) G(s) = 1/((s^2) + s(6+k) -4k + 25))
b) K=0, S = -3±4j

(I asked my lecturer and he said the frequency of oscillation in this question is the imaginary part of the poles, i.e. 4 as shown in the question)

c) When K = 4, S = -1, -9 and to get the impulse response we do Inverse laplace of the transfer function and get the response shown below:

y1(t) = Inv Laplace G1(t) = g1(t) = (e(^-t) - e^(-9t))/8

see here: https://www.wolframalpha.com/input/?i=y =(e^(-x) - e^(-9x))/8
When K = 6.25, S = 0, -12.25

y2(t) = Inv Laplace G2(t) = g2(t) = (1/(12.25))*(1-e^(-12.25t))

see here: https://www.wolframalpha.com/input/?i=y+=+(1/(12.25))*(1-e^(-12.25x))

Steady State error:

Unit step R(s) = 1/s

Since Y(s) = R(s)G(s)

When K = 4:

E(s) = R(s) - Y(s) = R(s)(1-G(s))

E(s) = (1/s)*(1-(1/s^(2)+10s+9))

**Using final value theorem**

e∞ = Lim s->0 sE(s) =(1-(1/9)) = 8/9

When K = 6.25

E(s) = (1/s)*(1-(1/s^(2)+12.25s))

e∞ = (1-(1/0)) = undefined

d)
system is critically damped when ζ = 1.

using the general characteristic equation:

s^(2)+ 2ζwns + wn^(2) = 0

It's critically damped when k = 2.12, -30.12

(if these values are correct then getting the poles are easy, since we only need to substiture them into the characteristic equation and solve for s)

e)
S(T,k) = k(s-4)/((s^2) + s(6+k) -4k + 25))

It's at this point I am unsure what to do... if I did the sensitivity correct, then what happens as K changes? do we look at S(T,k) as k->∞ and k->0?


Thanks for any comments.

Fouad.

 

[fouadalnoor]

Also I am now trying to answer Q2)

It says:

For each system, derive the simplest transfer function and determine the missing information in order to complete table ( see attached pic)



When attampting to fill in the values for the first system I get the pole
locations by using the general characteristic equation:

s^(2) + 2ζwn + wn^(2) = 0

ζ = 5/13 and wn = 13, we get s = -5±12j

In this case, how can the rank be 2?

Is'nt the rank just the number of poles - number of zeroes?

Shouldn't the rank be 0? if not, then wouldnt we need to get 4 poles in order to get a rank of 2 (since we already know that the number of zeroes is 2?)

Hope you can help!

Fouad.

 

[MrAl]

Hi,

Im not 100 percent sure why they are asking you to compute the sensitivity for the transfer function, unless maybe they want you to look at the sensitivity vs the pole positions to determine what the sensitivity function can tell you about the system. In that case you would look at the more important pole positions, or perhaps vary K over a range and see what happens to the roots, and compare that to the sensitivity function at those same values of K.
Normally a sensitivity function can tell you how much a certain quality about the main function changes when a given parameter (like K) is changed, which tells you how bad the system is going to turn out if that parameter varies over its known range. Thus you can figure out if a system is going to go unstable all of a sudden when the temperature goes up to 70 deg C for example, by estimating this at maybe 20 deg C for example. It tells you the answer to the question: "Is this system very sensitive to the parameter (probably very bad) or not very sensitive (probably very good)". Of course that's the usual application, but in a sensor examination it may be just the opposite, where you might hope for good (high) sensitivity instead of low (crappy) sensitivity.


The rank would be the rank of the matrix required if the characteristic equation was converted into a matrix. That would mean the rank would be 2 for a second order system.

 

[fouadalnoor]

Hi,

Im not 100 percent sure why they are asking you to compute the sensitivity for the transfer function, unless maybe they want you to look at the sensitivity vs the pole positions to determine what the sensitivity function can tell you about the system. In that case you would look at the more important pole positions, or perhaps vary K over a range and see what happens to the roots, and compare that to the sensitivity function at those same values of K.
Normally a sensitivity function can tell you how much a certain quality about the main function changes when a given parameter (like K) is changed, which tells you how bad the system is going to turn out if that parameter varies over its known range. Thus you can figure out if a system is going to go unstable all of a sudden when the temperature goes up to 70 deg C for example, by estimating this at maybe 20 deg C for example. It tells you the answer to the question: "Is this system very sensitive to the parameter (probably very bad) or not very sensitive (probably very good)". Of course that's the usual application, but in a sensor examination it may be just the opposite, where you might hope for good (high) sensitivity instead of low (crappy) sensitivity.


The rank would be the rank of the matrix required if the characteristic equation was converted into a matrix. That would mean the rank would be 2 for a second order system.

Okay, I will figure that out I guess...but does the rest look okay?

 

[MrAl]

Hi

In part 'c' why did you calculate the step response that way? If Ys is the transfer function, the output is the step Rs=1/s times Ys right?
That will give you a different result for the final value for K=4.