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How to get valid frequency for a transfer function?

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data
    transfer function: H(s) = 1/(s^2+s+1).
    question: compute half-power frequency

    2. Relevant equations


    3. The attempt at a solution
    I got 4 answers: -0.605i, 0.065i, -1.169, 1.169
    I know that only two of these are correct, one is 1.169 instead of -1.169 because frequency cannot be negative. Then I wrote the two imaginary answers into phaser form. I got 0.065<-90degree and 0.065<90 degree. Could someone tell me which phase I should choose and why?
  2. jcsd
  3. Sep 22, 2014 #2
    Did you come up with those answers yourself, or were they given to you in the problem statement? If they're your work, then it would be much easier to help you if you showed how you came up with them.

    The value of 's' is, in some contexts, called 'complex frequency', but I doubt that's what you're after. I think you probably just have to find the -3 dB point(s) of ##H(j \omega)##.
  4. Sep 22, 2014 #3

    I wrote the transfer function H(s) into H(w)=1/(-w^2+iw+1)
    so the magnitude would be |H(w)|=|1/(-w^2+iw+1)|=1/sqrt(1-w^2+w^4)
    |H(w)|max would be sqrt(4/3) when 1-w^2+w^4 is at minimum when w=+ or -1/sqrt(2)
    set |H(w)|=1/sqrt(1-w^2+w^4)=1/sqrt(2)*|H(w)|max=sqrt(2/3) and solve for w
    since this is a 4th order equation, I got 4 answers. 1 imaginary positive, 1 imaginary negative, 1 real positive, 1 real negative.
    Then I'm confused.

    What do you mean by -3dB point? I don't know about that.
  5. Sep 22, 2014 #4
    @Effy , i think the answers you calculated are wrong. Please go through it once again.
  6. Sep 22, 2014 #5
    Btw, the original question is a damper-spring-mass system. I got the transfer function from that and I'm pretty sure it's correct.
  7. Sep 22, 2014 #6
    -3db point is called as the cutoff frequency or half power frequency. It is a frequency at which magnitude of gain decreases by 3db from maximum gain.
  8. Sep 22, 2014 #7
    I've done that several times and I'm still try
    Does it mean that I shouldn't multiply the maximum magnitude by 1/sqrt(2) but subtract 3dB?
  9. Sep 22, 2014 #8
    I meant about the frequency values.
  10. Sep 22, 2014 #9
    I did it roughly .

    Attached Files:

  11. Sep 22, 2014 #10
    You can subtract 3 db only when the magnitude is also given in decibel .

    Attached Files:

  12. Sep 22, 2014 #11
    I got the same answers and finally found out what I did wrong. Thank you so much!!
    But only the +1.27 is a valid frequency, right?
  13. Sep 22, 2014 #12
    That depends on how pedantic your instructor is. You can interpret the real-valued signal ##\cos(\omega t)## as a complex-valued signal:
    \cos(\omega t) = \frac{e^{j \omega t} + e^{-j \omega t}}{2}
    with frequency components at ##-\omega## and ##\omega##, but I doubt that makes any difference for your assignment. The negative frequency components are redundant if you're only working with real-valued signals, so I think you can easily argue that it's correct to only include the positive frequency component in your answer.
    Last edited: Sep 22, 2014
  14. Sep 22, 2014 #13
    Got it. Thanks!
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