Control theory: block diagram, problem (detailed below)

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The discussion revolves around the simplification of a negative feedback loop in control engineering, specifically using Laplace transforms and block diagrams. The original poster is struggling with the steps taken in a textbook to derive the equivalent transfer function, which differs from their own calculations. They seek clarification on the simplification process, particularly how to properly apply the negative feedback definition. A response suggests simplifying the expression by eliminating common terms in the numerator and denominator for clarity. The conversation highlights the challenges of understanding control theory concepts and the need for clear step-by-step explanations.
thegreengineer
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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 \frac{G_{1}G_{2}}{1-G_{1}G_{2}H_{1}} and G_{3} are in series, therefore the equivalent transfer function is \frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}. 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 \frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}, and the transfer function \frac{H_{2}}{G_{1}} 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.
 
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I mean, how did he actually simplify to that?
 
$$Y=\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}X -
\frac{G_{1}G_{2}G_{3}}{1-G_{1}G_{2}H_{1}}\frac{H_{2}}{G_{1}}Y
$$

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

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

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

You probably just mixed up a numerator and denominator somewhere

edit: Earlier I had a '+' instead of '-' in that first equation. Fixed!
 
Last edited:
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:
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:
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