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Control theory: block diagram, problem (detailed below)

  1. Sep 17, 2016 #1
    I´m taking a course on control engineering and I have a test next Tuesday so I need to study the basics which are: Laplace transform, simplification of block diagrams, and analysis of transient and steady state responses. Right now I am dealing with the second one.

    I know the basic rules of block diagram simplification and the specific case I want to deal in this thread is this one:

    NEGATIVE FEEDBACK LOOP
    Rule-6.gif
    Image taken from: http://www.msubbu.in/sp/ctrl/BD-Rules.htm

    In other words, the equivalent transfer function is the product of the feed-forward path divided by one plus the product of the transfer functions that make the loop.

    So I get stuck on this and the problem is that the author (the book is MODERN CONTROL ENGINEERING by KATSUHIKO OGATA 5th ed.) skipped steps explaining how to simplify this:

    This is the way the author simplified it:
    https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/14358649_1778386355779234_9131464410329635376_n.jpg?oh=e7cd4c25c4caafab8260b0e61c814aa2&oe=583D6FC2
    The author simplified from step c) to step d). The equivalent transfer function is not similar to what I get. I'll explain you what I did and I will compare the results:

    First of all let's start in step c). The blocks containing the transfer functions [itex]\frac{G_{1}G_{2}}{1-G_{1}G_{2}H_{1}}[/itex] and [itex]G_{3}[/itex] are in series, therefore the equivalent transfer function is [itex]\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}[/itex]. The reduced diagram is the following:
    https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/14355645_1778854725732397_2673141930938132618_n.jpg?oh=187ae0673bdb279bd12d64a28207f93c&oe=5874F734
    Here's now where my problem starts. As far as I see the loop formed by the second-to right summation point, the transfer function [itex]\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}[/itex], and the transfer function [itex]\frac{H_{2}}{G_{1}}[/itex] are forming the negative feedback loop configuration, when I did it the result was not the same as the author's. Trust me, I actually did it and I would put the proof below but my computer is failing so don't insist please, I'm trying to understand this topic but I keep stuck so I would want someone to tell me what the author exactly did step by step.
     
  2. jcsd
  3. Sep 17, 2016 #2
    I mean, how did he actually simplify to that?
     
  4. Sep 18, 2016 #3
    $$Y=\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}X -
    \frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{G_{1}}Y
    $$

    $$Y=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{G_{1}}}X
    $$

    $$Y=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{1G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{1}}\frac{1-G_{1}G_{2}H_{1}}{1-G_{1}G_{2}H_{1}}X
    $$

    $$Y=\frac{\frac{G_{1}G_{2}G_{3}}{1}}{1-G_{1}G_{2}H_{1}+\frac{1G_{2}G_{3}}{1}\frac{H_{2}}{1}}X
    $$

    You probably just mixed up a numerator and denominator somewhere

    edit: Earlier I had a '+' instead of '-' in that first equation. Fixed!
     
    Last edited: Sep 18, 2016
  5. Sep 18, 2016 #4
    • Poster has been reminded not to post in all capital letters (and post has been fixed)
    Please someone answer. I have an exam of this on Tuesday. Please, I'm desperate.

    I find this example everywhere and everyone skips the same damn steps. Please, I'm tired of getting stuck and confused. I did this following the definition of negative feedback, and I got the result I am showing in the photo which is completely different, and my teacher will just give us thirty minutes for the exam and I can't waste too much time with long steps.

    Please I need help. If I can't even do this simple problem, who thinks I'm going to do harder problems?

    https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/14368868_1779167525701117_2312698362951297674_n.jpg?oh=166ebb913302a2f873efc60a6c857bb5&oe=58811457
     
    Last edited by a moderator: Sep 21, 2016
  6. Sep 18, 2016 #5
    Take a deep breath. This part that you did was unnecessary.

    $$Y=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{G_{1}}}X=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{G_{1}G_{2}G_{3}H_{2}}{G_{1}^2-G_{1}G_{2}H_{1}}}X$$

    It's easier to simply eliminate the term ##\frac{1}{1-G_{1}G_{2}H_{1}}## from the numerator and denominator like so:

    $$Y=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{G_{1}}}X=\frac{\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}}{1+\frac{1G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{1}}\frac{1-G_{1}G_{2}H_{1}}{1-G_{1}G_{2}H_{1}}X=\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}+1G_{2}G_{3}H_{2}}X$$

    Does that make it more clear?
     
    Last edited: Sep 18, 2016
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