1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Signals and Systems

  1. Sep 3, 2011 #1
    Hey guys, I was wondering if I could receive any help on a homework problem I have. I need to find the differential equation relating the input to the output. I've began working on it but feel like I've hit a brick wall in my work, any input?

    1. The problem statement, all variables and given/known data

    Attachment - Problem
    Working on part c

    2. Relevant equations

    Attachment - Work

    3. The attempt at a solution

    Attachment - Work
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2011 #2
    Take a look at my solution:
    http://www.mypicx.com/09032011/Solution/

    I leave the rearranging terms to you.. By the way how do you upload the picture like what is shown in this page. I don't know how to do it..
     
  4. Sep 3, 2011 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You solved for i1 in terms of i2 and its derivatives. Now plug that into the other equation to eliminate i1 completely and leave you with x(t) in terms of i2 and its derivatives. Then since y(t) = L di2/dt, you can write i2 and its derivatives in terms of y(t).
     
  5. Sep 3, 2011 #4
    The thing is I need to do it using differential equations and not in the s domain. All I do to upload pictures is scan a document to paint then save it as a jpeg file and upload it using the manage attachments tool here.

    Ok, I understand what you mean until you say to plug i1 into the other equation.. do you mean plug it into my x(dot) equation?? Where it is (i1-i2)?
     
  6. Sep 3, 2011 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, plug it into the x-dot equation anywhere i1 appears.
     
  7. Sep 4, 2011 #6
    I've done so and then solved for [itex]\frac{di1}{dt}[/itex] in equation 3 and also plugged that into the x(dot) equation. Do I now take the derivative of this equation to leave it in terms of x?
     

    Attached Files:

  8. Sep 4, 2011 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Equation 3 should be y(t)=L di2/dt.
     
  9. Sep 5, 2011 #8
    If equation 3 is y(t)=L di2/dt, then I'm not quite sure where it is plugged into in the other formulas
     
  10. Sep 5, 2011 #9
    I've been trying to restart the problem and look at it from a different approach but still no luck, any idea if I'm doing something wrong here? I've followed the same steps as taken in my notes from class and I can't seem to grasp this problem.

    Edit: Any thoughts as to if I should be using KCL rather than KVL since I can't seem to figure this problem out?
     
    Last edited: Sep 5, 2011
  11. Sep 5, 2011 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your original loop equations were
    \begin{align*}
    x(t) - R_1 i_1 - \frac{1}{C}\int (i_1-i_2)\,dt &= 0 \\
    \frac{1}{C}\int(i_2-i_1)\,dt - R_2 i_2 - L_1 \frac{di_2}{dt} &= 0
    \end{align*}
    (I think you had a sign error in the second equation.) The last term in the second equation is the voltage across the inductor, right? In other words, it's y(t). That's what your equation (3) should have been. I'm not sure why you used i1 there instead. In fact, when you differentiated equation (2), you correctly expressed the derivatives of i2 in terms of y(t).

    You then differentiated each equation and got
    \begin{align*}
    \dot{x}(t) - R_1 \frac{di_1}{dt} - \frac{1}{C}(i_1-i_2) &= 0 \\
    \frac{1}{C}(i_2-i_1) - R_2 \frac{di_2}{dt} - L_1 \frac{d^2i_2}{dt^2} &= 0
    \end{align*}
    You then solved the second equation for i1 and obtained[tex]i_1 = i_2 - R_2 C \frac{di_2}{dt} - L_1C \frac{d^2i_2}{dt^2}[/tex]
    Now just plug it into the first equation to get[tex]
    \dot{x}(t) -
    R_1 \frac{d}{dt} \left( i_2 - R_2 C \frac{di_2}{dt} - L_1C \frac{d^2i_2}{dt^2} \right) -
    \frac{1}{C} \left[ \left( i_2 - R_2 C \frac{di_2}{dt} - L_1C \frac{d^2i_2}{dt^2}\right) -i_2 \right] = 0[/tex]
    Note i1 is gone. Now simplify it and then use the fact that y(t) = L di2/dt to get rid of i2.
     
  12. Sep 5, 2011 #11
    Gotcha, my question is when you want to use y(t) = L di2/dt how in the world can you manipulate this to get rid of i2 throughout the xdot equation??
     
  13. Sep 5, 2011 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The same way you did it before after you solved for i1 after differentiating equation (2). What did you do back then?
     
  14. Sep 6, 2011 #13
    I ended up solving for a final answer of xdot = R1([itex]\frac{y}{L1}[/itex]+R2C[itex]\frac{ydot}{L1}[/itex]+y(2dot)C)+(R2[itex]\frac{y}{L1}[/itex]+ydot). Thank you guys for all of your help and input on this problem!
     
  15. Sep 6, 2011 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Sorry, as you apparently realized, your original equations were right, and it was I who made the sign error in equation (2). So the loop equations are
    \begin{align*}
    \dot{x}(t) - R_1 \frac{di_1}{dt} - \frac{1}{C}(i_1-i_2) &= 0 \\
    \frac{1}{C}(i_2-i_1) + R_2 \frac{di_2}{dt} + L_1 \frac{d^2i_2}{dt^2} &= 0
    \end{align*}
    which ultimately lead to the answer you found.
     
  16. Sep 6, 2011 #15
    No problem at all! I figured it out last night with one of my friends in my course and figured I would post my final answer. Thank you very much for your help and input!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Signals and Systems
  1. Signal and Systems (Replies: 1)

Loading...