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Hi Guys wanted some help for breaking down a block diagram which can be seen below. I know that typically with feedback systems be it negative or positive you would employ using the following:-

Close Loop = $C.L.T = \frac{open loop} {closed loop}$

However the signal $q_{i-1}$ is confusing me typically without it, the closed loop would be

$\frac{q_i}{d_r,i} = \frac{PDG_i(s)}{1-PDG_{i}(s)}$ as a result of positive feedback I'm not sure how to treat that $q_{i-1}$ signal in the system can anyone help help derive the transfer function.

enter image description here

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Suppose we acknowledge $q_{i-1}$ as an independent signal, then using the block diagram, the following equations can be derived: $$e_i = d_i - d_{r,i} = - q_i+q_{i-1} - d_{r,i}$$ $$q_i = G_i(s)u_i = PDG_i(s)e_i$$ substitute the first in the latter: $$q_i = PDG_i(s)(- q_i+q_{i-1} - d_{r,i})$$ $$(1+PDG_i(s))q_i = PDG_i(s)(q_{i-1} - d_{r,i})$$ $$q_i = \frac{PDG_i(s)}{1+PDG_i(s)}(q_{i-1} - d_{r,i})$$ Which represents a transferfunction with $(q_{i-1} - d_{r,i})$ as input and $q_i$ as output. The troublesome part here is this $q_{i-1}$, of which I do not know its meaning (is $i$ a discrete time index, or is $q$ a set of systems where $q_i$ is an part of it). Even thought one could substitute $$q_{i-1} = \frac{PDG_{i-1}(s)}{1+PDG_{i-1}(s)}(q_{i-2} - d_{r,i-1})$$ into the equation, the problem is only shifted to $q_{i-2}$ and so on. As such, the complete transfer matrix (where the input is every instance of $d_{r,i-n}$ and the output is every instance of $q_{i-n}$) will be the full definition of your system. However, if $i$ is a discrete time index, you could try to convert the model and controller to the Z domain and describe $q_{i-1} = z^{-1}q_i $

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  • $\begingroup$ Hi @Petrus sorry for the late response $q_{i-1}$ represents the position of a vehicle in front while $q_i$ represents the position of a vehicle in the back so therefore q is a set of systems in which $q_i$ is part of it. $\endgroup$ – KoolGuy Feb 13 at 2:22
  • $\begingroup$ can you verify if i am mistaking a mistake engineering.stackexchange.com/questions/40341/transfer-function $\endgroup$ – KoolGuy Feb 13 at 4:14

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