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Transfer Function from block notation?

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Hi Fouad,


You were right in that it should be a second order system. The transfer function is:
(12*p+360)/(p^2+12*p+180)

You should try this again really to see if you can get it.
This system has two feedback loops, an inner and an outer. Simplify the inner loop first then you are left with only one loop, then simplify that one. The rest is just multiplication. I can show a step by step procedure if you'd like to see it.
 
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Hi Fouad,

The solution is found by trying to find sections of the network diagram that can be reduced using the various rules for Block Diagrams. Sometimes it takes several steps, finding one section and reducing it and then replacing the reduced equation back into the network, then the next section, etc. This discussion will outline that procedure for one particular network.

The original network is shown on Page 1, Figure 1.
We first identify the inside feedback loop, shown in Figure 2.
We want to reduce this sub network. To do this, we use the general formula shown in Figure 3 as "Solution". G is the forward part, H is the feedback part. Note that there are two signs for the summing junction. One sign here is positive and one is negative. The positive sign is the feedforward sign, the negative sign is the feedback sign.
The feedfoward sign is usually positive, but if it is negative we just multiply everything by -1 later (ie a gain block of -1). The feedback sign however is reflected in the variable 'm'. m takes on the values as shown in figure 3.
In Figure 4, we use that feedback solution equation to form the equation for the new gain block which we will calculate to replace the feedback loop. By substituting the values of G and H and m into the solution formula we get the result shown in the upper part of Figure 4. This gain also reduces further as shown in the lower part of Figure 4.
Now that we have the replacement gain block calculated, we replace that feedback sub network with the new gain block, and this is shown in Figure 5.

Turning now to Page 2, Figure 6 shows the reduced network as we found so far.
We next note that the new gain block can easily combine with the gain block 1/p quite easily by simple multiplication, so we do that next as shown in Figure 7.
We reduce that slightly and we see that in Figure 8.
Now that we have reduced those two blocks, we replace them with the new block as shown in Figure 9.

Turning now to Page 3, we see the network as we have computed so far. We can draw this slightly differently to make the other feedback loop more apparent, and this is shown in Figure 11. Nothing here has changed, we just redrew it slightly.
Now that we can see another feedback loop, we want to reduce that one next. Figure 12 identifies this feedback loop and the values of G, H, and m for that loop. Note we use the same formula again the only difference is the different values of G, H, and m.
In the lower part of Figure 12 we replace the variables again with their respective values. This results in a more complex equation, but we can reduce that and this is shown in Figure 13. Of course we again replace the feedback loop with the new gain block we calculated, and that is shown in Figure 14. Note how much simpler the network appears now, and that there are no more feedback loops. Thus, we've reduced the network with two feedback loops to a network that is made up of only feedforward gain blocks. That is what we wanted to do in order to get this network reduced to only one gain block eventually.

Looking at the last page now, Page 4, in Figure 15 we have the network as we calculated so far. Since there are only forward gain blocks left all we have to do now is some multiplication and addition.
We can identify that more complicated network in the center and define that as the gain 'A', as shown in the upper part of Figure 16.
In the lower part of Figure 16, all we did was multiply all the gains in the top forward path (contains the gain block 'p' but not '30') and add that to all the gains multiplied in the lower forward path (contains the gain block '30' but not 'p'). The result is shown there in the lower part of Figure 16, keeping the math form simple by using the temporary variable 'A'.
Next we simply replace both occurrences of A with its true value, and we get two terms as shown in Figure 17. These two terms of course combine and we get the final equation shown in Figure 18.

The final solution is to replace the entire network with the equivalent gain block that we just calculated, and that is shown in Figure 19.
 

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Thank you!!! That explains it perfectly! I think I just struggled with the negative/positive feedback loops, the rest is pretty straight forward. Umm, now I'm trying to get the damped frequency of Oscillation, but get w= 13.45 rad/s instead of 12 rad/s as they said (though the poles are still 6+12j when solving the characteristic equation in the denominator?)

oh and I'm also stuck on part c) because I'M not sure how to expand the fraction into partial fractions and then use the unit step input response. (I get complex roots when trying to use partial fractions)

Thanks for any help!
 

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I think I figured out part 1 c) now.

I used partial fractions on the equation you guys gave and got:

(12*p+360)/(p^2+12*p+180) = ((6+12*j)/(p+6+12j)) + ((6-12*j)/(p+6-12j))



Is the step response going to be:

c(t) = (1-e^-(6+12j)t)r(t) ?

(of course we have another root at 6-12j so is it still necessary to put that in?)
 
Hello again,

you got the part correct:
(12*p+360)/(p^2+12*p+180)

If you factor the denominator (as you did) you can see you have complex roots. That indicates that there will be at least one sine (or cosine) term in your response. Also, im not sure what you did but when you do a partial fraction expansion you have to solve for A and B as in:
(12*p+360)/(p^2+12*p+180)=A/(p-r1)+B/(p-r2)
where r1 and r2 are the roots of the denominator on the left,
and then solve for the step function as per your Lecture notes, following those notes very carefully
Note you get two terms on the right, so you must later find two transforms that will involve both A and B.
Remember that the roots get subtracted from p in the denominators on the right, not added to p. You got that part right. To make sure you got the right solutions for A and B, simply recombine the right side of that above and see that you get the exact same equation on the left.
Alternately, we would do:
(1/p)*(12*p+360)/(p^2+12*p+180)=A/(p-r1)+B/(p-r2)+C/s
and solve for A, B, and C, then construct the time function by adding all the individual transforms and then simplifying.
 
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Yeah, I did get A and B from doing partial fractions. My A = 6+12*j and my B = 6-12*j (I checked that they are correct by substituting 3 into p and then seeing if both sides of the equation shown above is the same)

According to the notes (eqn 5.12) we have a unit step response for a "first order block" given by:

k(1 – e-t/T)U(t)

here k = numerator of the first order term and T = time constant.

As you know we get k and T by re-arranging our first order terms into k/(1+Tp), when I did that for the first term I got:

c(t) = (1-e^-(6+12j)t)r(t) (where r(t) is given in the question as the unit step response).

Also, I know that we get a sine/cosine term when we have a complex root, but I still dont quite know what the form will be.

Is the general response form for a complex root?

(Ae^−σt) sin (ωt + φ) as shown on page 4, part 4 of: https://www.electro-tech-online.com/custompdfs/2011/05/PoleZero.pdf
 
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Hi there Fouad,

You should be able to take the solutions you found for A and B and use them with the e^-at forms to form the (1-e^-at) forms as in 5.1 of your Lecture. That will give you the step response.

Alternately, try solving for A,B, and C in:
(1/p)*(12*p+360)/(p^2+12*p+180)=A/(p-r1)+B/(p-r2)+C/p

which you will note on the left is just the same equation we started with multiplied by 1/p, and the right is the partial fraction expansion before we solve for A, B, and C. Doing it this way you may find it a little easier to get the required solution.
You might note that the equation:
(12*p+360)/(p^2+12*p+180)
is the original response, but does not represent the step response until it is transformed via 5.1 of your Lecture, but once we multiply by 1/p as in:
(1/p)*(12*p+360)/(p^2+12*p+180)
the factor 1/p is called a 'step' input, so the response we get results in the step response for the original transfer function. The drawback is we have to solve for three variables instead of 2, but it's just a linear equation anyway so it's not too hard. The resulting partial fraction expansion is easier to transform that way, having two simpler terms and one very simple term which transforms to a constant. Try it that way too and you'll see how much simpler it turns out.

When we have a complex root we get the exponent from the denominator just as with a real root, and the gain we get from the numerator as well even if it is also complex. I think that's what you were asking about.
(a+b*j)/(s+A+B*j) => (a+b*j)*e^(-t*(A+B*j))
 
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Ah I didn't know you could just multiply by 1/p to get the unit step response!

oh and regarding

"When we have a complex root we get the exponent from the denominator just as with a real root, and the gain we get from the numerator as well even if it is also complex. I think that's what you were asking about.
(a+b*j)/(s+A+B*j) => (a+b*j)*e^(-t*(A+B*j))"


if you look at the notes on equation 4.13 and 4.14 you will see that he says when we have a complex root the response will be "x(t) = Acos(wot)"

I actually talked to him today and he said it will be x(t) = (Ae^pt)cos(wot)

I am now confused...

When it's x(t) = Acos(wot)

is that ONLY when its simple harmonic motion?

(p^2)+wo^2 = 0

p = +- jWo

I think the best way to learn is probably by doing questions and seeing a "Model solution". If I see how the exam questions are answered I can easily figure out how most things work.

If say we take Question 1 we have been doing on the exam paper (2009/2010) above and go through it fully then I should be able to understand it much more. I have attached my own solutions in this post so you can see what I'm doing and where I might be wrong (as you can see, I am doing okay until part b - i and then I mess up a bit on part c and d)

Once I understand that part fully I can easily do any past paper for practice because the questions are essentially the same..

We can then move on to part 2 (which I have also attempted, but not sure of my answers... )

I know it may be a lot to ask, but I feel like I'm close to understanding most of it, only a few bits and pieces left really...


Thanks for any help!

Fouad.
 

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Hi again,

They were talking about the simple harmonic motion there yes. The more general equation has both a sin and cos term, but sometimes the sin or cos can cancel out anyway leaving just the other.

Did you try following the procedure in that Lecture from 5.0 on? You should try to follow the example they give line by line first. Make sure you see what is going on with each new line in the Lecture. Notice how they use A and B once they are solved for, in particular equation 5.13, you need to get an equation like that and then reduce it.

Did you want to try the method that uses the step 1/p first though? Try that one and see how it works out for you.
 
Hello again,


I must have missed your solution pdf file, but i took a look at it now.
It's a little hard to read though, but it looked like maybe you did not consider the 'gain' part of the expressions, but then i looked again and it appears that maybe you did get the right result up to the exponential part, only having to use the Euler identities as the last step which i didnt see yet. We'll see i guess as im going to outline the whole procedure here and you can compare your results and let me know. I also added an extra transform checking step to make sure we got the right result and it checked out ok.

Notes to keep in mind:
A and B come from the numerators of the partial fraction expansion, while Td1 and Td2 come from the denominators.


We'll start with the transfer function:
(12*s+360)/(s^2+12*s+180)

and the denominator has two roots:
r1=-6+12*j
r2=-6-12*j

and this gives us the partial fraction expansion:
(12*s+360)/(s^2+12*s+180)=A/(s+6-12*j)+B/(s+6+12*j)

and after solving for A and B we get:
A/(s+6-12*j)+B/(s+6+12*j)=(6-12*j)/(s+6-12*j)+(6+12*j)/(s+6+12*j)

where we see:
A=6-12*j
B=6+12*j

and writing that equation a little differently:
A/(s+Td1)+B/(s+Td2)=(6-12*j)/(s+6-12*j)+(6+12*j)/(s+6+12*j)

and this gives us the two Td's:
Td1=6-12*j
Td2=6+12*j

and this gives us two partial responses:
res1=e^(-t*Td1)
res2=e^(-t*Td2)

and using 5.13 from the Lecture we get the total response:
f(t)=A/Td1*(1-res1)+B/Td2*(1-res2)

which after some substitutions comes out to:
f(t)=((1-e^(-(6+12*j)*t))*B)/(6+12*j)+((1-e^(-(6-12*j)*t))*A)/(6-12*j)

and after subt A and B and simplifying we get:

f(t)=2-e^(-(6+12*j)*t)-e^(-(6-12*j)*t)

and after using Euler's identities we get:
f(t)=2-2*e^(-6*t)*cos(12*t)

These results were checked carefully and appear to be accurate.
In fact, integrating:
f(t)*e^(-s*t)=(2-2*e^(-6*t)*cos(12*t))*e^(-s*t)
from t=0 to t=+inf gives us:

[LATEX]\int_0\ ^\infty f(t)\ e^{-s t}\,\mathit{dt} = \int_0\ ^\infty (\, 2\ -\,2 e^{-6 t}\, cos(12 t)\, )\, e^{-s t}\,\, \mathit{dt} = \frac{12 s+360}{s^3+12 s^2+180 s}[/LATEX]

which in text is:
(12*s+360)/(s^3+12*s^2+180*s)

which is the original transfer equation in s multiplied by the step (1/s)

[LATEX]\frac{12 s+360}{s^3+12 s^2+180 s} =\,\,\left(\frac {1}{s}\right)\,\, \frac{12 s+360}{s^2+12 s+180}[/LATEX]

so we know f(t) is correct.

Other notes:
This example was a little easier to do because A and B were the same as their denominators in the 5.13 form. If they were not the same it would be a little more difficult to simplify.
 
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Ah... I think I almost got it now.

I understand things up to:

"which after some substitutions comes out to:
f(t)=((1-e^(-(6+12*j)*t))*B)/(6+12*j)+((1-e^(-(6-12*j)*t))*A)/(6-12*j)"

When we look at the transfer function of the first order components:

we get:

1) k1/(1+T1p) (eqn 5.12) where T1 = 1/(6+12j)
2) k2/(1+T1p) (eqn 5.12) where T2 = 1/(6-12j)

*In our case k1 = 1 and k2 = 1

*because we can simplify our first order components from the transfer function to be:

12p+360/(p^2+12p+180) = 1/(1+(1/(6+12j))p) + 1/(1+(1/(6-12j))p)

If everything I have done above is correct, then the response to a step input would be as given by 5.13. But my case is much simpler as both my k-terms are 1 and I get:

(k1(1-e^-(t/T1)+k2(1-e^-t/T2))U(t) ---> Eqn 5.13

Substituting my values

((1-e^-(6+12j)t+(1-e^-(6-12j)t) U(t)

Now I only need to use Euler's identities to simplify that function right?

Thanks for the help!
 
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Hi again,

Yes, that looks right now :)
 
Cool!

Right, I have used Euler's identities to simplify the equation to the following form:

c(t) = (2-4e^(-6t)cos(12t))U(t)

And if the above is correct, then my steady state out-
put of the system for a unit step input U(t) is (given that U(t) = 0, t<0 and U(t) = 5, t>= 0 )

c(infinity) = 10.

Yes, yes? :)
 
Hello again,

Well, if you looked at my previous post you would have found this:
and after using Euler's identities we get:
f(t)=2-2*e^(-6*t)*cos(12*t)
When you compare the two, are they the same? I dont think so, so maybe you should try applying the Euler identities one more time :)
 
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Hello again,

Well, if you looked at my previous post you would have found this:
When you compare the two, are they the same? I dont think so, so maybe you should try applying the Euler identities one more time :)

Oh right! Sorry must have missed your answer! I shall try Euler's again but at least I know I'm on the right track.

Also, isn't there some sort of short cut we can take when we get the poles?
 
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Hello again!

I am have tried another past exam paper question (attached) but I can't seem to simplify the transfer function further than what I have shown on the solution (also attached) have I gone wrong somewhere?
also, am I missing some obvious simplification that may make the algebra easier?

Note: I am attempting the first part of the first question.

Also for your previous answer

"Hello again,

Well, if you looked at my previous post you would have found this:
and after using Euler's identities we get:
f(t)=2-2*e^(-6*t)*cos(12*t)"

don't you have to multiply with the step response (as r(t) = 5), t>=0?
 

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Hi again,


Are you saying that the input was not the unit step, but the unit step multiplied by 5? If so, then yes you have to multiply by 5 to get the final answer.

I'll take a look at your new circuit next.

The transfer function of the new circuit is:

Vout=(Vin*R4)/(s^2*L*C*R4+s*(C*R1*R4+L)+R1+R4)

where
R4=R2+R3

[LATEX]Vout=\frac{Vin R4}{s^2 L C R4+s (C R1 R4+L)+R1+R4}[/LATEX]

If that is what you got then it was correct :)
If not, try it again :)

You could of course divide top and bottom by the coefficient of s^2 to get s^2 alone on the bottom.
 
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