0
\$\begingroup\$

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![enter image description here]1

\$\endgroup\$
5
  • \$\begingroup\$ Please show us what you have come up with so far. Where exactly are you getting stuck? \$\endgroup\$ – Elliot Alderson Dec 14 '18 at 1:30
  • \$\begingroup\$ 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. \$\endgroup\$ – Newbie010 Dec 14 '18 at 1:46
  • \$\begingroup\$ 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 \$\endgroup\$ – Newbie010 Dec 14 '18 at 1:47
  • \$\begingroup\$ C=Y/H5 since Y=C*H5 \$\endgroup\$ – Tony Stewart EE75 Dec 14 '18 at 3:05
  • 1
    \$\begingroup\$ The best and classical way is: Block diagram reduction technique. Start to replace local feedback loops by one block only! \$\endgroup\$ – LvW Dec 14 '18 at 8:54
2
\$\begingroup\$

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:

enter image description here

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!

\$\endgroup\$
0
\$\begingroup\$

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): -

enter image description here

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: -

enter image description here

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?

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.