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Control Systems Engineering - Signal Flow Graphs

  1. Feb 12, 2014 #1

    I seem to be having some issues going from state space representation of a system to signal flow graph representation. My troubles seem to be, if I have something like this

    [itex]\frac{d}{dt}x_{1}(t) = v_{1}(t)[/itex]

    In state space representation I list functions to which the derivatives are as a function of the state space variables.

    In signal flow graph model of a system am I allowed to draw transistors from one node to another node by multiplication of one in either direction?

    I don't if I'm explaining this well, but if I have a node for [itex]sx_{1}(t)[/itex] and a node for [itex]v_{1}(t)[/itex] am I able to to draw the transistor arrow from the nodes in either direction since they are equal to each other? That is a arrow coming out of the node that represents [itex]sx_{1}(t)[/itex] and going into [itex]v_{1}(t)[/itex] or the arrow coming out of the node representing [itex]v_{1}(t)[/itex] and into the node representing [itex]sx_{1}(t)[/itex]. Mathematically I don't see why the direction of the arrows from the nodes can be reversed as needed since they have a one-to-one equivalency. I just want to make sure. If I'm not able to do this and I'm not sure I can solve this problem I'm trying to solve.

    Thanks for any help.
  2. jcsd
  3. Feb 12, 2014 #2


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    Gold Member

    your output doesn't always have to be equal to 1 state variable


    A=[0 4
    7 -2]


    if you want y to equal [itex]\dot{X1}[/itex], you can say
    C=[0 4]

    Does that make sense?
  4. Feb 12, 2014 #3


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    Gold Member

    So that means for the signal flow graph that your output can directly be that sx1 node. You don't need to draw an arrow, that node can simply be sx1 and v1
    Last edited: Feb 12, 2014
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