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Control Systems Engineering - Circuits

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the transfer function, [itex]G(s) = \frac{V_{0}(s)}{V_{i}(s)}[/itex] for each network shown in figure P2.3. [Section 2.4]

    http://imagizer.imageshack.us/v2/800x600q90/20/1ocq.png [Broken]

    2. Relevant equations



    3. The attempt at a solution

    When I try and solve this problem I Kirchhoff's Current Law and get

    [itex]\frac{V_{O}(t) - V_{i}(t)}{R} + \frac{1}{L}∫_{0}^{t}V_{0}(\tau)d\tau + \frac{V_{0}(t)}{R} = 0[/itex]

    This doesn't really seem to help.

    The solutions manual writes the node equation as

    [itex]\frac{V_{0} - V_{i}}{s} + \frac{V_{0}}{s} + V_{0} = 0[/itex]

    and then solves this for

    [itex]\frac{V_{0}}{V_{i}} = \frac{1}{s + 2}[/itex]

    I don't exactly see how they get the node equation they do. I know that the impedance of a resistor is just the resistance and that the impedance of the inductor is Ls. So to me it looks like it should be

    [itex]\frac{V_{0} - V_{i}}{R} + \frac{V_{0}}{Ls} + \frac{V_{0}}{R} = 0[/itex]
    [itex]\frac{V_{0} - V_{i}}{1} + \frac{V_{0}}{1s} + \frac{V_{0}}{1} = 0[/itex]
    [itex]V_{0} - V_{i} + \frac{V_{0}}{s} + V_{0} = 0[/itex]

    I don't see were the [itex]\frac{V_{0} - V_{i}}{s}[/itex] term comes from.

    Thanks for any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 20, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    It looks like a mistake in their solution; They have taken the first 1 Ω resistor (connected to Vs) and treated it as a 1H inductor. So either the circuit diagram is incorrect or their equation is.
     
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