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Linear Phase system?

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A filter with the following impulse response
    h(k)=(-.75 .5 0 -.5 .75)

    find H(z)
    Is its a linear phase system?
    Is it a high pass filter?

    2. Relevant equations

    h=[c0, c1, c2, c1, c0] <--- I think this is valid for this problem anyway

    3. The attempt at a solution

    So if i understand this properly for my H(z) I get.

    [itex] H(z)=-0.75+0.52z^{-1}-0.5z^{-3}+0.75z^{-4}[/itex]

    Then to find out weather or not this is a linear phase system I find my H(f) frequency response?

    I get

    [itex]H(f)=-0.75+0.5e^{-j\theta}-0.5e^{-3j\theta}+0.75e^{-4j\theta}[/itex]

    First off, Am I on the right track here?

    Thanks for the help,
     
  2. jcsd
  3. Jan 7, 2012 #2
    Ok so I did a little more digging and I think Im going about things a little wrong here?

    The given values for h(k) are as follows
    H(0)=-.75
    H(1)=.5
    H(2)=0
    H(3)=-0.5
    H(4)=0.75

    I need to find the z transform of this sequence. I always have a hard time with this form. "transpose"

    do I do this form inspection? So I would end up with the following?

    [itex]H(z)=-0.75+0.5z^{-1}-0.5z^{-3}+0.75z^{-4}[/itex]

    This is what I had before though. Is my Z transform of the sequence above along the right lines?

    Thanks in advance..
     
  4. Jan 7, 2012 #3

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    Your z-transform is correct.

    TBH, I do not know whether this is a linear phase system or a high pass filter.
    I was hoping someone else would respond before now, but obviously no one else did.
    So I guess for now you're stuck with me.
    The wiki pages are not clear enough as far as I'm concerned and I haven't tried to delve too deeply into other sources yet.

    Do you have more information what the conditions are for both types of filters?
     
  5. Jan 7, 2012 #4
    I can dig up some information on the types of filters and add it to the thread.

    Right now Im a little stuck with the simplification of my H(f) frequency response.

    This is what I have right now from evaluating my H(z) at z=j/theta

    [itex] H(f)=-0.75+0.5e^{-j\theta}-0.5e^{-3j\theta}+0.75e^{-4j\theta}[/itex]

    When I use eulers identity to convert to sin and cosine it becomes a mess that only makes things worse. Im sure theres something I can do to simplify and factor before i apply euler's identity. I could factor out a 0.5 but i dont see how that would help. Since the exponents in the exponentials are different how can I combine them? Im not seeing something here.

    Any hints would be appreciated? As always Im stuck on the algebra.
     
  6. Jan 7, 2012 #5

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    You can take out ##e^{-2j\theta}##.
    That should leave you with 2 sines using Euler's formula.
     
  7. Jan 7, 2012 #6
    Man im not good at this. I dont fully see it?

    Something like this sort of?

    [itex]H(f)=-0.75+e^{-2j\theta}(0.5-0.5e^{-j\theta}+0.75e^{-2j\theta})[/itex]

    I dont think this is correct at all though. I feel like I have broken every rule in algebra even writing this? Im looking up basic algebra rules now to see if i can get this to make sense.
     
  8. Jan 7, 2012 #7

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    Hmm, that is "half" right.
    I don't quite understand how you did half correct and dropped stuff off in the other half...

    Anyway, it should be:
    $$
    e^{-2j\theta}(-0.75 e^{2j\theta}+0.5 e^{j\theta}-0.5 e^{-j\theta}+0.75 e^{-2j\theta})
    $$

    Can you recognize the sines now?
     
  9. Jan 7, 2012 #8
    Ok after looking at what you have written and played with the numbers a bit I see where i got lost. Again basic algebra skills that are missing.

    Ive used Euler's formula and but i end up with 2 sines and 1 cosine.

    The exponentials in the parenthesis simplify down to [itex]-1.5jsin(2\theta)+jsin(\theta)[/itex]

    When I apply Euler to the factored out [itex]e^{-2j\theta}[/itex] and add it to the above i get...

    [itex]cos(2\theta)-2.5jsin(2\theta)+jsin(\theta)[/itex]

    How do I get left with just 2 sines?

    Thanks again for your help on this.
     
  10. Jan 7, 2012 #9

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    Not missing. I'd say not fully developed yet, but I guess that won't be long now. ;)

    The expression you got is not right though. Apparently you did not properly eliminate the parentheses.

    But anyway, I intended:
    $$
    e^{-2j\theta}(-1.5j\sin(2\theta)+j\sin(\theta))
    $$
    Which you can rewrite as:
    $$
    e^{-2j\theta}e^{j{\pi \over 2}}(-1.5 \sin(2\theta)+ \sin(\theta))
    = e^{j({\pi \over 2} - 2\theta)}(-1.5\sin(2\theta)+\sin(\theta))
    $$
    I don't think there's much use in simplifying more.

    The 2 sines form a real number now.
    If you're interested in the argument of the complex number, you can read it off now.
    It's ##({\pi \over 2} - 2\theta)## if the sine expression is positive.
    Otherwise you need to add another pi.
     
    Last edited: Jan 7, 2012
  11. Jan 8, 2012 #10
    Your right I didnt evaluate the parenthesis properly. Looking back im not sure why I thought it was ok to just remove those after there were individual terms added or subtracted. I see where I went wrong. I understand now how to get the equation shown below.
    $$
    e^{j(\frac{\pi}{2}−2θ)}(−1.5sin(2θ)+sin(θ))
    $$

    This is actually in the correct form for me to determine weather or not this is a linear phase filter.

    Looking in the section about this in my book and following an example it talks about an:

    It then shows an example with similar coefficients unlike the problem that I have been working on through this thread. So now im not 100% sure how to go about it.

    If what I do have is a linear phase filter the [itex]\alpha=\frac{\pi}{2}[/itex] but im not sure what the [itex]\tau[/itex] would be...maybe [itex]\tau=-1.5T[/itex] I dunno.

    I still dont see anything in this chapter about high pass filters so Im looking into that.
     
  12. Jan 8, 2012 #11

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    Where you have written ##\theta## I think we should read ##2\pi f## (or something like that).

    And I think your formula should include "j":
    $$H(f)=A_{r}(f)e^{j(\alpha-2\pi\tau f)}$$

    As such your ##\alpha## would be ##\pi \over 2## and ##\tau## would be ##2## (or something like that), making it a linear phase filter.

    More specifically the phase changes linearly with f, with a fixed offset and a jump of pi somewhere.
    I think this constitutes a linear phase filter.
     
    Last edited: Jan 8, 2012
  13. Jan 8, 2012 #12
    Your correct I must have missed that j in there. and yes the [itex]\theta is 2\pi f[/itex]

    And after reviewing the examples again I see now where that tau comes from. I guess i didnt see or recognize that it was in the exponential. So my problem here is very similar to an example in the book (the end anyway) so the [itex]\tau=2T[/itex] I think. Plus the 4 in my exponential was throwing me off I suppose.

    Im still searching for high pass filter information. But the rest of this problem is MUCH more clearer now. Thank you as always for the help. MUCH appreciated.
     
  14. Jan 8, 2012 #13

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    I've helped you a couple of times now and I'd like to know a bit about who it is that I'm helping...?
    Could you for instance fill in a couple of fields in your profile?
     
  15. Jan 8, 2012 #14
    No problem. I filled in the "about me" section.

    Btw thanks for the Happy New year wishes on my profile! I guess its been a while since ive visited it.
     
  16. Jan 8, 2012 #15

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  17. Jan 8, 2012 #16
    Ill see if I can dig up a picture and post.

    Btw in case you were curious about determining weather if this filter is a high pass or not this is what I found out.

    [itex]Z=e^{j2\pi fT}[/itex] when [itex]f=0, z=1[/itex] and when [itex]f=\frac{F_{s}}{2} , Z=-1[/itex] So if I evaluate my H(z) for this filter at z=1 and z=-1 I get 0 for a response on both ends. So it cant be a highpass or a low pass for that matter. Band pass or something like that i would assume.

    so @f=0 H(z)=0
    @f=fs/2 H(z)=0

    Thought you might be interested in this.

    Thanks again!
     
  18. Jan 8, 2012 #17

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    Thanks for the info.
    It'll come in handy when next someone posts a question like this.

    Cheers! :smile:
     
  19. Jan 21, 2012 #18

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    Turns out, the information you posted comes in handy sooner than I expected! :wink:
     
  20. Jan 21, 2012 #19
    Thanks for pointing this out! I forgot we spoke about the phase response in here.
     
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