Control - Block Diagram Question

Discussion in 'Homework Help' started by Fraser_Integration, Apr 4, 2011.

  1. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
    5
    Hello. Got this block diagram question the other day. Having a bit of trouble with the last part, and not sure of my answer to part b) either. I attach the question.

    a) E = R - Y
    but Y = GKR / (1 + GKH)

    then E = R - GKR/(1+GKH)

    For E = 0,

    R = GKR/1+GKH

    Rearrange for H = 1 - 1/KG

    b) I just worked backwards to achieve:

    Y = [ (0 - (1-1/KG)) + Td*P ] *G
    = 1 - KG + Td*P*G

    c) Not really sure about this one. Not come across Y(s) = 0 transfer functions before. Any general information would be appreciated.
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    I think you have done a mistake on b. You can disregard R and think of Td as the input of your feedback system.

    I 'll rewrite it for you:
    Y=\frac G {HK} \cdot P \cdot T_d\\<br />
Y=\frac G {\frac {GK-1}{GK} K} \cdot PT_d\\<br />
Y=\frac {G^2} {GK-1} PT_d

    Now the question is if it is possible for this to happen:
    \frac {G^2} {GK-1}=0
    I think you can answer that.
     
    Last edited: Apr 4, 2011
  3. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
    5
    Hi.

    I don't really understand where your answer comes from initially.

    If R = 0, then immediately after the first summing junction aren't we going to have 0 - Return Loop Gain?
     
    Last edited: Apr 4, 2011
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    If R=0, then it will only nullify its factor of the addition node. The feedback from the output is still there.

    The new system has P*Td as input, G in the forward branch and H*K in the feedback branch.
     
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