1. Not finding help here? Sign up for a free 30min 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!

Interpreting a signal block diagram to form transfer funciton

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Given the following block diagram. Calculate the transfer function H(z) and find [itex]\theta (f=\frac{f_{s}}{6})[/itex]
    2.jpg

    2. Relevant equations

    [itex] H(z)=\frac{Y(z)}{X(z)}[/itex]

    3. The attempt at a solution
    Ok So I know i need to find my y(k) output equation from this. I can then rearange the equation so I have x terms and y terms on opposite sides and then take the Z transform of each side. Then rearange so that I end up with

    [itex]\frac{Y(z)}{X(z)}[/itex] Which should be my H(z). Then I can find [itex]\theta(f)[/itex] and evaluate at f= fs/6.

    I think thats the correct process anyway.


    However im not 100% on how to interpret the block diagram in this case. I have what looks like a feed back from x(k) i think i can handle that but not sure about the y(k) terms if any.

    This is what I have come up with so far.

    [itex]3x(k-1)-\frac{1}{3}x(k-1)+y(k)[/itex]

    However im not sure what I have here is 100% correct. Any hints would be much appreciated.

    Thanks,
     
  2. jcsd
  3. Dec 16, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi Evo8! :smile:

    I prefer to work in the z-domain for things like this.
    I do not know how you are taught these things, so perhaps the way I do it may be different from the way you are taught.
    The result however will be the same.


    So if you'll bear with me, then to the left, you have X(z), and to the right you have Y(z).

    Your problem is that you have a feedback loop that you need to solve.
    Let's give it a name.
    Let's say that in the middle you have V(z).

    This means that the left summing junction does:
    [tex]X(z) + 3z^{-1}V(z)=V(z)[/tex]

    Can you solve V(z) from this?
     
  4. Dec 16, 2011 #3
    If Ive done my algebra correctly I get

    V(z)=3zX(z)

    So would [itex]Y(z)=3zX(z)-\frac{1}{3}z^{-1}[/itex] maybe Im forgetting an X(z) in there?
     
  5. Dec 16, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    Your algebra appears to be a little off.
    How did you arrive at V(z)=3zX(z)?

    Could you say what the right summing junction does in terms of V(z) and Y(z)?
     
    Last edited: Dec 17, 2011
  6. Dec 17, 2011 #5
    I subtracted the x(z) term from either side moving it to the right. then divided both sides by v(z). so I would have

    [itex] 3z^{-1} V(z)=\frac{V(z)}{V(z)}-\frac{X(z)}{V(z)}[/itex]

    then errr wait. As Im typing this I realize my mistake. When I canceled out the v(z)/v(z) i didnt leave the 1.....Man my algebra skills are killing me here.

    In this case i guess im not really sure how to go about it. If I used the solver on my graphing calculator i get

    [itex]V(z)=\frac{X(z)z}{z-3}[/itex] Im not really sure how to come to this conclusion though. Or if it is correct.

    Ill have to play around with the algebra again and see if i can get it.
     
  7. Dec 17, 2011 #6
    I think the right summing junction would be V(z)-(1/3)z^-1?
     
  8. Dec 17, 2011 #7

    I like Serena

    User Avatar
    Homework Helper

    Suppose you had the equation 2+3v=v.
    How would you solve it?
     
  9. Dec 17, 2011 #8

    I like Serena

    User Avatar
    Homework Helper

    You appear to have dropped a V(z).
    When going down the first step is that V(z) gets multiplied by z-1, resulting in z-1V(z)...
     
  10. Dec 17, 2011 #9
    v=2/(1-3)

    So...

    [itex]V(z)=\frac{X(z)}{1-3z^{-1}}[/itex]

    Man its so easy too. I played with it for a little while and wasnt sure. Once i saw you question above I was like sure i guess I could factor the v out. This is assuming that what I have written now is correct.....*fingers crossed* I really should know my algebra better then this. I certainly used to anyway.
     
  11. Dec 17, 2011 #10

    I like Serena

    User Avatar
    Homework Helper

    Well you can uncross your fingers again, or push Ronald McDonald down the stairs, or both.
    You've got it right now! :)
     
  12. Dec 17, 2011 #11
    So would it be

    [itex] V(z)- \frac{1}{3}z^{-1}V(z)[/itex]?
     
  13. Dec 17, 2011 #12

    I like Serena

    User Avatar
    Homework Helper

    Yep. That is Y(z).

    Can you find H(z)=Y(z)/X(z) now?
     
  14. Dec 17, 2011 #13
    Ok not so bad. I think

    This is what I get for

    [itex] \frac{Y(z)}{X(z)}=H(z)=1-3z-\frac{1}{3}z^{-1}(1-3z)[/itex]



    Simplifies to [itex]H(z)=\frac{(-3z-1)^{2}}{3z}[/itex]

    I think anyway. The simplification is where I make the simple algebra mistakes. Which you already know. I found this last simplificaiton with my graphing calculator. I tried a few times to simplify, by distributing that -1/3z^-1 into the parenthesis and then clean things up a little. It never came out as clean as this though...:(
     
  15. Dec 17, 2011 #14

    I like Serena

    User Avatar
    Homework Helper

    Hmm, I get a different result... so I suspect your algebra needs some more polishing... (or mine does :shy:).

    You had [itex]Y(z)=V(z)- \frac{1}{3}z^{-1}V(z)[/itex].
    Can you substitute your result for V(z) into this?
     
  16. Dec 17, 2011 #15
    Looks like Ive jumped ahead too fast again.

    [itex]Y(z)=(\frac{X(z)}{1-3z^{-1}})-\frac{1}{3}z^{-1}(\frac{X(z)}{1-3z^{-1}})[/itex]

    Looking back agian I think i see where i went wrong.

    I wanted to get rid of the division in the V(z) term so i thought (I can see that it is wrong now) that I could do this.

    [itex] V(z)=\frac{X(z)}{1-3z^{-1}} = X(z)(1+3z).[/itex]

    Now im playing around with it to see if there is another way. Or if I can take the above Y(z) and just simplify it all together.
     
  17. Dec 17, 2011 #16

    I like Serena

    User Avatar
    Homework Helper

    You have 2 fractions with the same denominator.
    Perhaps you could add them?
     
  18. Dec 18, 2011 #17
    Indeed I do.

    So adding the fractions I would get.

    [itex]Y(z)=\frac{X(z)-\frac{1}{3}z^{-1}X(z)}{1-3z^{-1}}[/itex]

    If I factor out the X(z) I get

    [itex]Y(z)=\frac{X(z)(1-\frac{1}{3}z^{-1})}{1-3z^{-1}}[/itex]

    This is where i get nervous because im unsure if im breaking any rules or not

    I need to break that fraction up so I can isolate the Y(z) and X(z) on one side. My first thought is to rearange like this.

    [itex]Y(z)=X(z)(1-\frac{1}{3}z^{-1})(1-3z)[/itex] Is this ok? If so I can obviously bring my X(z) over to the other side and get my H(z)
     
  19. Dec 18, 2011 #18

    I like Serena

    User Avatar
    Homework Helper

    You added the fractions correctly and you also factored out X(z) correctly.
    But rearranging like that is not ok.

    Suppose you had the equation:

    y=3x/4

    How would you find y/x?
     
  20. Dec 18, 2011 #19
    You know its weird. I ask the same questions but with simple integers and variables and I mess with the algebra again and I usually find a method pretty quickly. I thought I tried that this time. I knew there was a way I just wasnt sure how to get there i guess. I will need to remember in the future to actually write down and replace sections wtih variables and try to simplify or rearrange that way to find a method.

    Now i get something like this

    [itex] \frac{Y(z)}{X(z)}=\frac{1-\frac{1}{3}z^{-1}}{1-3z^{-1}}[/itex]

    or I guess I could write it like this.

    [itex] \frac{Y(z)}{X(z)}=\frac{1-\frac{z^{-1}}{3}}{1-3z^{-1}}[/itex]

    Right....?
     
  21. Dec 18, 2011 #20

    I like Serena

    User Avatar
    Homework Helper

    Right! :smile:


    Btw, you're not the only one making mistakes with complex expressions.
    I know it helps to simplify them, and then suddenly the trees resolve into a small forest.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Interpreting a signal block diagram to form transfer funciton
Loading...