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.


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 Nov 5 '15 at 13:23
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    $\begingroup$ @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. $\endgroup$ – Chris Mueller Nov 5 '15 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 Nov 5 '15 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. $\endgroup$ – Chris Mueller Nov 6 '15 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 Nov 7 '15 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 $\theta_r =0$, $\frac{\theta}{D} = G_D$. $\endgroup$ – Suba Thomas Nov 9 '15 at 0:02

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