Simple Control Block diagram question

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fouadalnoor

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Hey guys,

I am confused about how my lecturer got the attached block diagram Q2, a)?

I understand that the open loop transfer function is: Ko/(1+ts), how does he get the 1/s part? isnt that just integral control?

Hope you can help.

Fouad.
 

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Hi,

The 1/s path factor is introduced in order to obtain Theta (the angular displacement) from the speed. In other words, the angle is the integral of the speed.
 
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MrAl said:
Hi,

The 1/s path factor is introduced in order to obtain Theta (the angular displacement) from the speed. In other words, the angle is the integral of the speed.
Click to expand...

Ah thanks! So I suppose I was supposed to already know that before I drew the block diagram? Because I didnt know that in order to get the angle you do the integral of the speed, i.e. 1/s) Is this just specific to this question or am I missing something more general?
 
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Hi,

It's a very general thing because the velocity is actually angular velocity dA/dt so to get the angle A you integrate. We're calling it speed here. It's that simple.
 
ah haha, okay that makes sense. The rest just follows from that so that should be fine. I will be working on Root Locus now I guess.
 
Hello,

Like MrAI already explained, it's a pure integrator: We integrate speed to get the angle.

It depends on what you actually want to control: If you wanted to control speed, you would simply have fed back speed (Omega), but since you want the angle (Theta), you integrated the speed.
 
Hi,
The 1/s path factor is introduced in order to obtain Theta (the angular displacement) from the speed. In other words, the angle is the integral of the speed.
Click to expand...

Mr Al is great man of this forum, helpful very much.
 
Okay, I am now doing a few other questions regarding drawing the root locus... I think I am getting the basics right. But can you check that I drew the one (attached) correctly?
 

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Hi again,

Im not sure i can read your drawing accurately. For example, what is the angle of departure for your plot at the pole drawn at the origin where you have the X right on (0,0) ?

Also, what is the approach point where the circular part of your drawing meets the x axis? It looks like -5/3 on your drawing, and if that's correct then did you go back and check that in your kGH equation to see if it actually has a root there or near there?

Also, did you calculate the max (or min) value of k for the gain where the plot crosses the j axis, and if so (you should have i think done this) did you check it in your kGH to see if it produced a root when k is equal to the found value (that would be on the j axis)?

I ask these questions for good reasons that i believe will help you, and i'll post my own root locus after you try some of the above.
 
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Hmm I think I messed that one up a bit. I did the second GH(S) below and tried to answer all the questions...I got stuck at a),b) and e).

I'm not sure exactly how to calculate the conditions for k for stability, but for now I just want to make sure I can draw the root locus properly. From that I can see how k's value changes the system etc.

I know the drawing is a little confusing, but can you please check of all calculations are correct? if their fine, then the only parts I did not know how to draw properly are the break in points..

Hope you can help!
 

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Hi,

So what did you finally get for the first problem which we were discussing first here?
 
Hi again,

Here are my root locus plots for the first problem. The first pic is the overall plot while the second pic is a close up of the plot near the imaginary axis to show how the roots cross over for small k. This tells us something very interesting about that first function. The red line is the imaginary axis.

BTW these plots were made with software that i wrote for drawing Root Locus plots.
 

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