When I was taught in how to simplify block diagrams, the teacher used internal signals besides the Reference R(s) and the Output C(s). That seemed to be ok with my mind. But when he started to do the math, something bothered me.

This is the way he (and every book) calculated the expression C(s)/R(s):

Let’s define a new internal signal, the Error, the difference between the Reference and the measured Output; and the Output itself in terms of the Error:

1.- E(s) = R(s) - C(s)H(s)
2.- C(s) = E(s)G(s)

Substituting E(s) in the equation 2 gives us:

C(s) = ( R(s) - C(s)H(s) )G(s)

making the multiplications:

C(s) = R(s)G(s) - C(s)G(s)H(s)

grouping terms that have C(s) in common:

C(s) + C(s)G(s)H(s) = R(s)G(s)

taking C(s) as a common factor:

C(s) ( 1 +G(s)H(s) ) = R(s)G(s)

dividing both sides by ( 1 + G(s)H(s) ) yields:

C(s) = R(s)G(s) / ( 1 + G(s)H(s) )

and finally dividing both sides by R(s) finishes the process:

C(s)/R(s) = G(s) / ( 1 + G(s)H(s) )

The thing that I just couldn’t understand was the fact that E(s) and C(s) depended on each other. If you can write E(s) in terms of C(s) and C(s) in terms of E(s): How does the system know when to stop substituting one signal in the other? Or How does the system know which signal to substitute in which one?

I decided to make a mental simulation and write it with mathematical language. This is how I proceeded:

Imagine that the system has output zero and the reference is applied to the input. The signal E(s) is equal to R(s) because C(s) is zero. C(s) becomes R(s)G(s) by the time the signal reaches the output. Then C(s) is multiplied by H(s) and is substracted from the reference which is still there. Now E(s) is equal to R(s)-R(s)G(s)H(s). Let’s see how the signal C(s) evolves over time while the signal is looping:

As we can see, C(s) is taking a general form which can easily be deduced and is as follows:

C(s) = R(s)G(s)-R(s)G(s)2H(s)+R(s)G(s)3H(s)2-R(s)G(s)4H(s)3+R(s)G(s)5H(s)4-…+R(s)G(s)nH(s)n-1 as n goes to infinity.

Every loop the signal makes thru the system, produces a new term, and n can be seen as the number of these loops.

I feel more comfortable with that expression despite its complexity, but the question is: Is this expression the same as the one taught in every book? The answer is yes, I figured out by using a mathematical tool called “Taylor Series”.

Having C(s) in the latest form:

C(s) = R(s)G(s)-R(s)G(s)2H(s)+R(s)G(s)3H(s)2-R(s)G(s)4H(s)3+…+R(s)G(s)nH(s)n-1

we can take R(s)G(s) as a common factor:

C(s) = R(s)G(s) ( 1-G(s)H(s)+G(s)2H(s)2-G(s)3H(s)3+…+G(s)n-1H(s)n-1 )

dividing both sides by R(s) we obtain the desired function:

C(s)/R(s) = G(s) (1-G(s)H(s)+G(s)2H(s)2-G(s)3H(s)3+…+G(s)n-1H(s)n-1 )

Now, we make use of the Taylor series. In many books about Calculus you can find the Taylor series for many functions such as sin(x), e^x, ln(x), x/(1-x), etc…

If we take a look at the Taylor series for the function 1/(1+x), we will see that it corresponds to: 1 - x + x2 - x3 + x4 - x5 + x6 - x7 + x8 - x9 + … which works for -1 < x < 1
And if we substitute x by G(s)H(s) the expression turns to the one that is multiplying G(s) in the last equation for C(s)/R(s), being that expression equal to 1/( 1+G(s)H(s) ).

1/( 1+G(s)H(s) ) = (1-G(s)H(s)+G(s)2H(s)2-G(s)3H(s)3+G(s)4H(s)4-G(s)5H(s)5+…)

Finally, substituting properly, we get:

C(s)/R(s) = G(s) / ( 1 + G(s)H(s) )

This is the same equation given by our teachers and books. I said it was going to be a more complicated way to obtain the same result but now I feel satisfied and  I hope you feel the same way.