0
$\begingroup$

Hi Guys I am trying to evaluate the following transfer function below can anyone verify if this is correct

enter image description here

$$e_i = -d_{r,i} + d_i $$ $$d_i = q_{i-1}-q_i$$ $$d_{r,i} =H_iq_i$$ $$Therefore, e_i = -Hiq_i + q_{i-1} - q_i$$ $$q_i = e_iPDG_i$$ $$q_i = PDG_i(-H_iq_i+q_{i-1}-q_i)$$ $$q_i = -PDG_iH_iq_i+PDG_iq_{i-1}-PDG_iq_i$$ $$\frac{q_i}{q_{i-1}} = \frac{PDG}{1+HPDG+PDG}$$

Im thinking this is incorrect but I'm hoping someone can verify this for me?

$\endgroup$
1
$\begingroup$

Shouldn't the error equation have $-H_iq_i$ instead of $H_iq_i$? $$e_i=−d_{r,i}+d_i$$ $$d_i=q_{i−1}−q_i$$ $$−d_{r,i}=−H_iq_i$$ Therefore (modified equation): $$e_i=-H_iq_i+q_{i−1}−q_i$$ $$q_i=e_iPDG_i$$ $$q_i=PDG_i(-H_iq_i+q_{i−1}−q_i)$$ $$q_i=-PDG_iH_iq_i+PDG_iq_{i−1}−PDG_iq_i$$ $$\frac{q_i}{q_{i−1}}=\frac{PDG_i}{1+H_iPDG_i+PDG_i}$$

$\endgroup$
1
  • $\begingroup$ Yes was making edits and realized my mistake but thank you for the assistance @wicked stat $\endgroup$
    – KoolGuy
    Feb 13 at 4:54

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.