I've been having trouble doing this school assignment. My professor has not taught the class how to read control block diagrams so I assumed that we were supposed to research on it. I've watched a few YouTube tutorials online and have been able to grasp the rough concept of it. But I'm stuck halfway through. Shown below is the question itself and my work.



I tried expanding everything out but I'm not able to get it into the form of the transfer function that the question is asking for.


2 Answers 2


You are almost there. Your last line in the second section, $D(s)=F(s)-X(s)$, is pointless; that information is already included in the above lines. Instead, you need to close the loop by making the replacement $E(s)\rightarrow \theta_r(s)-\theta(s)$ and then solve for the two transfer functions you are asked for. The final equation is $$ \theta(s)=B(s)\left\{D(s)+A(s)\left[\theta_r(s)+\theta(s)\right]\right\}, $$ which you can solve for both of the requested transfer functions.

  • $\begingroup$ I tried expanding it but I get Θ/Θr as a function of Θr. The final expression that I got was : Θ=(BD+BAΘr)/(1+BA) $\endgroup$
    – John
    Commented Nov 5, 2015 at 13:23
  • 1
    $\begingroup$ @John It's just a result of having multiple open inputs to the loop ($\theta_r$ and $D$); the resulting equation isn't going to solve cleanly. I get $G_r=\frac{BA}{1-BA}\left(1-\frac{D}{A\theta_r}\right)$. This shows that, if $D$ is small compared to $A\theta_r$, then the transfer function reduces to the standard closed loop transfer function. $\endgroup$ Commented Nov 5, 2015 at 13:37
  • $\begingroup$ But if the equation doesn't solve cleanly. How would I determine the order of the system? Since D(s) is not specified. $\endgroup$
    – John
    Commented Nov 5, 2015 at 14:02
  • $\begingroup$ @John You are probably expected to assume that the disturbance is small and ignore the second term in my expression above. $\endgroup$ Commented Nov 6, 2015 at 12:43

This is the equation you want to solve for $\theta$. $$\theta =\frac{\left(\theta _r-\theta \right) \left(K_d s+\frac{K_i}{s}+K_p\right)+D}{s^2-\alpha }$$

The solution is $$\theta =\frac{s^2 K_d \theta _r+D s+K_i \theta _r+s K_p \theta _r}{s^2 K_d+K_i+s K_p+s^3-\alpha s}$$.

To get $G_r$, assume $D=0$. You can do this because the system is linear and superposition holds.

$$G_r=\frac{\theta}{\theta_r}=\frac{s^2 K_d+K_i+s K_p}{s^2 K_d+K_i+s K_p+s^3-\alpha s}$$.

Again, to get $G_D$ assume $\theta_r=0$.

$$G_D=\frac{\theta}{D}=\frac{s}{s^2 K_d+K_i+s K_p+s^3-\alpha s}$$

The systems are third order.

  • $\begingroup$ I don't get the part about the system being linear and the superposition part. I understand how to apply superposition for different circuits, but not for control systems. Would it be possible for you to elaborate further on that part? $\endgroup$
    – John
    Commented Nov 7, 2015 at 2:59
  • $\begingroup$ The system consists of transfer functions, so it is linear. The consequence of the system being linear is $\theta = G_r \theta_r + G_D D$. Thus when $D =0$, $\frac{\theta}{\theta_r} = G_r$; and when $\theta_r =0$, $\frac{\theta}{D} = G_D$. $\endgroup$ Commented Nov 9, 2015 at 0:02

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