I am trying to find the transfer function for this basic block diagram

enter image description here

According to the book I am reading I should be able to derive the transfer function (given in the image above) from the block diagram, but I think I am doing it totally incorrectly... This is what I tried:

enter image description here

Could anybody correct me?

  • \$\begingroup\$ Either use a block reduction method; in which case, work from inside the system to the outside. Or start with the output signal and work backwards through the individual blocks until you reach the input signal. \$\endgroup\$
    – Chu
    Oct 29, 2017 at 10:57

5 Answers 5

  1. First simplify the innermost feedback loop.

    $$\frac{\frac{1}{s}}{1+\frac{1}{s}\ \frac{R}{L}}=\frac{L}{L s+R}$$

  2. Now simplify the blocks in series.

    $$\frac{1}{s}\frac{1}{L}\frac{L}{L s+R}=\frac{1}{s (L s+R)}$$

  3. Simplify the remaining feedback loop.

    $$ \frac{\frac{1}{s (L s+R)}}{1+\frac{1}{s (L s+R)}\frac{1}{C}}=\frac{C}{C s (L s+R)+1} $$

  4. Finally things are in series.

    $$\frac{1}{C}\frac{C}{C s (L s+R)+1} R=\frac{R}{C L s^2+C R s+1}$$

  • \$\begingroup\$ could you just explain a bit how you simplify the innermost feedback loop (step 1)? How did you come up with that fraction? My initial apporach isn't like that at all... I may not be familiar with that approach. Once that's clarified I'll accept your answer \$\endgroup\$ Oct 30, 2017 at 14:25
  • \$\begingroup\$ See the section 'Feedback Connection' at tutorialspoint.com/control_systems/…. I use that in steps 1 and 3. \$\endgroup\$ Oct 30, 2017 at 14:35
  • \$\begingroup\$ Look at step 1) in my post. Same result. \$\endgroup\$
    – LvW
    Oct 30, 2017 at 14:38

The procedure you are following is correct. But there is a mistake in the derivation. Your 3rd step is wrong. It has to be:


So if you proceed with that you will end up with something like $$V(s)=R\left(-x_2\frac{R}{Ls}+\left(\frac{U(s)}{Cs}-\frac{x_2}{Cs}\right)\frac{1}{Ls}\right)\tag1$$

What we want is \$\frac{V(s)}{U(s)}\$ but we have \$V(s)\$ in terms of \$U(s)\$ and \$x_2\$. So we have to represent \$x_2\$ either in terms of \$U(s)\$ or \$V(s)\$. Here, it is easy to represent \$x_2\$ in terms of \$V(s)\$ $$x_2=\frac{V(s)}{R}$$

I guess you can do the remaining. Substitute for \$x_2\$ in equation \$(1)\$, take all the \$V(s)\$ terms to one side and then find \$V(s)/U(s)\$. You will end up in the required expression.

  • \$\begingroup\$ And how do I continue after this? I feel like this is still very far away from the actual answer \$\endgroup\$ Oct 29, 2017 at 8:26
  • \$\begingroup\$ Is it possible that you made a mistake? I think you forgot the power of 2 in equation1, ie $\frac{R}{s^2L}$ I did something similar to what you suggested: imgur.com/MBQSX4n Yet, there is still something incorrect I think somewhere somehow... \$\endgroup\$ Oct 29, 2017 at 12:38
  • \$\begingroup\$ @LandonZeKepitelOfGreytBritn whatever you did is correct.. except that in last 2 step you missed negative sign for 1st term in right hand side of equation.. I don’t think i missed a power of 2. If you take out the \$s\$ that’s multiplied with C, it becomes \$s^2\$.. I think our expressions are same except for the negative sign \$\endgroup\$
    – nidhin
    Oct 29, 2017 at 12:51

1.) Start with the most right small loop and apply the classical feedback formula:

H1=Forward block/(1+loop gain)=(1/s)/[1+(1/s)(R/L)]=1/[s+R/L] .

2.) Replace the small loop with H1.

3.) Now you have three blocks in series forming the forward path for the large loop. All three block transfer functins are to be multiplied. Now - apply the same reduction principle for the large loop (feedback formula) as in step 1).

4.) The result of step 3) has to be multiplied with (1/C) and "R".

This is the final solution based on block diagram reduction techniques.


Here's another approach along the lines you started (where I replace \$X_2\$ with \$\frac{V_0}{R}\$ and get with the main equation).

$$ V_0=R \frac{1}{s}\left[\frac{1}{L}\frac{1}{s}\left(\frac{1}{C}U-\frac{1}{C}\frac{V_0}{R}\right)-\frac{R}{L}\frac{V_0}{R}\right]$$

After this, it's just algebraic manipulation.

Expand out the terms.

$$V_0=\frac{R U}{C L s^2}-\frac{V_0}{C L s^2}-\frac{R V_0}{L s}$$

Multiply throughout by \$C L s^2\$.

$$C L s^2 V_0=R U-V_0-R C s V_0$$

Collect the \$V_0\$ terms.

$$(C L s^2 +R C s +1)V_0=R U$$

And the answer follows.

$$\frac{V_0}{U}=\frac{R}{C L s^2 +R C s +1}$$


You can always try Block diagram reduction technique or the Signal flow graph methods. They make the derivation of the transfer function quite simple.

  • 3
    \$\begingroup\$ It might be helpful to provide an example of how to do this \$\endgroup\$
    – Voltage Spike
    Oct 29, 2017 at 5:36

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