# total transfer function of a LTI System with feedback

I have calculated the transfer function of such systems many times, but now i feel like I`m going around in circles. I have drawn a few help-signals to help me solve it, but they are dependent of one another and no matter how many times I replace the values, I just finish where I started. I must calculate the total transfer function of the system from the attached image. The coloured lines are the help-signals that I added and their description is below. Any help is appreciated]1

• Please show us what you have come up with so far. Where exactly are you getting stuck? – Elliot Alderson Dec 14 '18 at 1:30
• I tried to replace the variables with what is after the equal sign. for example A with X+H2*C in the second and fourth relationship. The thing is, there is no end to this replacing. So let's say we replace A B would become X+H2*C-XH1-H1*H2*C. D would become H4*Y+XH1+H1*H2*C And C would become H4*Y+XH1+H1*H2*C+XH3+H2*H3*C-X*H1*H3-H1*H2*H3*C. Now i have a relationship between Y,X and C and i somehow need to find Y/X. – Newbie010 Dec 14 '18 at 1:46
• And if get Y/X out of that relationship with C, i would have to replace C once again, getting into another loop, because C is dependent on the other signals – Newbie010 Dec 14 '18 at 1:47
• C=Y/H5 since Y=C*H5 – Tony Stewart EE75 Dec 14 '18 at 3:05
• The best and classical way is: Block diagram reduction technique. Start to replace local feedback loops by one block only! – LvW Dec 14 '18 at 8:54

My answer comes a little late, but I think it should still be interesting.

I took the block diagram from you (@Newbie010) as a basis and added some labels which are helpful for the calculation. My method gets along here without any simplification.

At this point I ask you to check if what I have done is correct. I did not find any mistake in my calculation, but a critical consideration would be appropriate.

Block diagram with new labels:

Based on this picture, the following equations can now be determined:

$$X(s)+C(s)=B(s)=A(s) \tag{1}$$ $$D(s) = B(s) - H_1A(s) \tag{2}$$ $$E(s) = H_4Y(s)+H_1A(s) \tag{3}$$ $$F(s) = E(s) + H_3D(s) \tag{4}$$ $$Y(s) = H_5F(s) \tag{5}$$

Other important and useful equations:

$$F(s) = \frac{Y(s)}{H_5} \tag{6}$$ $$C(s) = H_2F(s) \tag{7}$$ $$C(s) = H_2\frac{Y(s)}{H_5} \tag{8}$$

The aim now is to have only inputs and outputs in its transfer function $H(s)$. Important here is this connection $H(s) = \frac{Y(s)}{X(s)}$

I now begin to generate $H(s)$ by starting with equation 5 and using the relations to other equations in turn. This is straight forward:

\begin{align} Y(s) &= H_5F(s) \\ &= H_5\left(E(s)+H_3D(s)\right) \\ &= H_5\left(H_4Y(s)+H_1A(s)+H_3D(s)\right) \\ &= H_5\left(H_4Y(s)+H_1A(s)+H_3B(s) - H_3H_1A(s)\right) \\ &= H_5\left(H_4Y(s)+H_1\left(X(s)+C(s)\right)+H_3\left(X(s)+C(s)\right) - H_1H_3\left(X(s)+C(s)\right)\right) \\ &= H_5\left(H_4Y(s)+H_1\left(X(s)+H_2\frac{Y(s)}{H_5}\right)+H_3\left(X(s)+H_2\frac{Y(s)}{H_5}\right) - H_1H_3\left(X(s)+H_2\frac{Y(s)}{H_5}\right)\right) \\ &= H_5H_4Y(s)+H_5H_1X(s)+H_5H_1H_2\frac{Y(s)}{H_5}+H_5H_3X(s)+H_5H_3H_2\frac{Y(s)}{H_5}-H_5H_1H_3X(s)-H_5H_1H_3H_2\frac{Y(s)}{H_5} \\ &=Y(s)\left(H_5H_4+H_1H_2+H_3H_2-H_1H_3H_2\right)+X(s)\left(H_5H_1+H_5H_3-H_5H_1H_3\right) \end{align}

Therefor:

$$Y(s)(1-H_5H_4-H_1H_2-H_3H_2+H_1H_3H_2) = X(s)(H_5H_1+H_5H_3-H_5H_1H_3)$$

It follows: $$H(s) = \frac{Y(s)}{X(s)} = \frac{H_5H_1+H_5H_3-H_5H_1H_3}{1-H_5H_4-H_1H_2-H_3H_2+H_1H_3H_2}$$

Now I wanted to test if what I have done can be confirmed by a simulation. Therefore I converted the block diagram in Matlab. Also I realized my generated transfer function as a block in Matlab, the results were the same!

For the simulation I used the following blocks: $$H_1 = 5, H_2 = \frac{1}{s}, H_3 = \frac{1}{s+2}$$ $$H_4 = 7, H_5 = \frac{1}{s}$$

If I now insert these values into my TF and let the overall expression be simplified once with the use of Wolframalpha I get the following for $H(s)$:

$$\frac{5s+6}{s^2-10s-20}, s\neq 0, s \neq -2$$

In Matlab I implemented your block diagram with the corresponding values and my simplified transfer function. Judging by the simulation, the results are the same.

I hope that my answer meets the requirements and hope that I could help a little bit with this. Thanks a lot!

Not a full answer; just some assistance....

Just one simple block diagram re-arrangement gets you an improvement (as per LvW's comment to your answer): -

Notice the added red block contans $$\1-H_1(s)\$$ and this removes the summing node just in front of $$\H_3(s)\$$. I'm assuming that was an intentional negative sign on the summing block I removed hence the negative sign in the new block I added.

Sometimes you have to add stuff to simplify but other folk may choose a different starting point.

The next thing to do (in my opinion) would be to re-arrange things like this: -

Hopefully I've not screwed up anywhere but I'm sure if i have there'll be someone eager to point it out.

Can you take it from here?