Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About H(jw) question

  1. Apr 26, 2016 #1
    擷取.PNG
    Determine the transfer function of the system H(jw)
    why the answer is not 1/[1+(jw/612)] but equal to 0.318/[1+(jw/612)]?
     
  2. jcsd
  3. Apr 26, 2016 #2

    LvW

    User Avatar

    What happens if you insert w=0 into the equation? Can you determine the remaining value?
     
  4. Apr 26, 2016 #3
    to which equation?Vi=Va sin(wt+θ)?
    i find that 0.318 is 1/π where the π come from?
     
  5. Apr 26, 2016 #4

    LvW

    User Avatar

    You have asked for the transfer function - thus, I mean H(jw), of course.
     
  6. Apr 26, 2016 #5
    but the equation of H(jw)=1/[1+jwRC] (from my lecture notes) 1/Rc=w0 (for w<0.1w0), hence i did not where 0.318 come from
     
  7. Apr 26, 2016 #6

    LvW

    User Avatar

    The magnitude plot shows for small w values (down to w=0) a level of -10dB. Can you calculate the corresponding absolute value?
    Then, you must compare this result with both transfer function for the case w=0. Which one is valid?
     
  8. Apr 26, 2016 #7
    20log(Vout/Vin)=-10
    丨Vout/Vin丨=10^(-0.5)=0.316? then what is the meaning of this number
    for the equation of 1/[1+jw/612] 丨Vout/Vin丨=0.999998 they are not the same,then how can i solve this problem?
     
  9. Apr 26, 2016 #8

    LvW

    User Avatar

    Don`t you see it in one of your transfer functions?
     
  10. Apr 26, 2016 #9
    NO,they are not the same ,for the equation of 1/[1+jw/612] 丨Vout/Vin丨=0.999998, but how can i make it become 0.316???
     
  11. Apr 26, 2016 #10

    LvW

    User Avatar

    I repeat: Don`t you see it in one of your transfer functions?
     
  12. Apr 26, 2016 #11
    sorry,still dont understand:frown: can you show me some step ?
     
  13. Apr 26, 2016 #12

    LvW

    User Avatar

    At the bottom of your first post I see TWO functions.
    Are you able to insert w=0 into BOTH functions - and decide, which one results in 0.318?
     
  14. Apr 26, 2016 #13
    from the graph (a) 0 to w0 ≅-10dB
    from the graph (b) phase differnet ≅-40 degree so the equation
    so Vi(t)=0.316Vi (sinθ-40o)then turn it to the H(jw) form?
     
  15. Apr 26, 2016 #14

    LvW

    User Avatar

    Sorry - I resign. Are you kidding me?
     
  16. Apr 26, 2016 #15

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    Sometimes teachers forget how strange it all seems to beginners.
    The mechanics of reading frequency response charts ( we called them Bode plots in my day)
    becomes so natural
    that your lecturer may have completely forgot to mention it,
    or his mention may have been so brief you missed it....



    Some of the filter's attenuation is not frequency dependent. Its "Gain", "Attenuation", "Transfer Function" , call it what you like ,
    has two parts, a constant part and a frequency dependent part
    so
    it is of the form K X f(frequency)
    and your graph shows you at its left edge that the constant term K is -10db not 0db
    and 10-0.5 = .316, not 1
    (if it started at zero db then K would be 1)

    That's how transfer functions work.

    I don't know how he got .318 instead of .316. You should ask him.
     
  17. Apr 26, 2016 #16

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    as explained above, it's K X f(frequency) ,, 0.316 X 1/[1+(jw/612)]
     
  18. Apr 26, 2016 #17
    First find the "DC gain" (for w = 0) for this transfer function 1/[1+(jw/612)]
    Next do the same for this one 0.318/[1+(jw/612)], then you need to compare the results. And you will see that only one of the results will fits the "graphical representation".
     
  19. Apr 26, 2016 #18
    thank for your,since my lecture notes just have the formula 1/[1+(jw/RC)],so i do not know that it have Dc gain K, but after search at google and read your comment, i got the point now, thank so much.
     
  20. Apr 26, 2016 #19

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    I think if you do that division and expand the polynomial you'll find in it terms that are not multiplied by jw

    whenever you see that , there's DC gain

    as you become accustomed to these exercises you'll come to recognize the "shape" of commonly occuring polynomials and your mind will leap immediately to an equivalent circuit.

    That one's a "first order lag"

    upload_2016-4-26_10-4-8.png

    R2/R1 is the "K" above.
     
  21. Apr 26, 2016 #20

    LvW

    User Avatar

    Hi Jim - perhaps you are right in some cases, however, after 25 years experience in teaching electronics, I know about the problems some beginners might have.
    But considering that
    * both possible transfer functions were given at the bottom of the first post
    * and reading again my several answers (in particular my post#6) and the OP`s answer (mentioning the magic value of 0.318)
    you certainly will come to the conclusion that I am not "guilty".

    In post#12 I have asked him to compare both given functions - he simply has ignored this hint.

    I simply have assumed that sombody who presents a magnitude plot together with two possible transfer functions (same denominator, only the constant factor is different) should be able to decide which function belongs to the graph.
    Or - do you really think I have overstrained the questioner?
     
    Last edited: Apr 26, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted