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Simplifying cosine and sine expressions

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    So im a little rusty on my basic math and I am trying to simplify this expression. In case you were curious the expression came from a signal block diagram that I calculated the transfer function H(z) to then found the frequency response H(f) and I think i have done that correctly I just need to reduce this further. I have attached a picture of what I have left. The theta stands for exp(j2pifT) if that maters to you.


    2. Relevant equations
    eq.jpg

    3. The attempt at a solution

    So this is where I get confused and cant remember the proper rules. These are super basic I know. Unfortunately I never learned them that well to begin with which is why im having trouble now..

    the terms that are j*sin(-2*theta) can I bring the -2 out of the sin function? Same with the cosine? I want to say no but part of me feels like you can..... If so im thinking this might help me cross out a few terms?

    Or if i break it up into sections i get something like this for the first few terms. -2/4 * cos(-theta)/cos(-theta). I then feel like i can cross out the cosine terms and be left with -1/2? Same with the -2jsin(theta)/4jsin(theta) etc.

    Am I going about this correctly? Some of it just doesn't seem right to me.

    I really appreciate any help here guys/girls.

    Thanks

    p.s. Sorry if my handwriting is messy. I tried my best to write it down as neat as possible. I figured it would be easier to see if i took a picture rather then type it out here and have you try to understand what im looking at.
     
    Last edited: Dec 8, 2011
  2. jcsd
  3. Dec 8, 2011 #2

    Fredrik

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    I don't see an attachment. There are formulas that tell you how to evaluate sin(2x) and cos(2x). If you remember the formulas for sin(x+y) and cos(x+y), you can use them with y=x. Also, keep in mind that sin is an odd function, and cos is an even function.
     
  4. Dec 8, 2011 #3
    Sorry about that i had a link to the picture but it must be just me that can see it because it was local or something like that. Ive uploaded it to a flikr type account and changed the link. You should be able to see it now.

    I saw some of these forumulas but im not 100% Im still looking at them as we speak.
     
  5. Dec 8, 2011 #4

    Fredrik

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    I can see it now. You can certainly use the formulas I mentioned, but my first thought is that it doesn't look like the result will be any simpler.
     
  6. Dec 8, 2011 #5
    Hmm maybe that's it then. The reason I said I want 100% was because I did try those formulas and it did not make the eq more simple so I suspected that I was doing something wrong. If it looks in simplest form to you then maybe that's the case.

    Thanks for taking a look at this for me.
     
  7. Dec 8, 2011 #6

    Fredrik

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    You should at least use the fact that sin is odd and cos even to get rid of the minus signs in front of the θ and 2θ.

    I had another look at this, and I think the formulas I mentioned can simplify this a little. The expression cos θ-j sin θ shows up in a couple of places. Is j2=-1? In that case, cos θ-j sin θ=e-jθ.

    The final result I got has 4(cos θ+1) as the denominator (after cancelling a factor of) e-jθ), and the numerator is at least a little bit simpler than what you started with.
     
  8. Dec 8, 2011 #7

    I like Serena

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    Hi Evo8! :smile:

    It seems a bit weird to me that you have a cosine with exp(j2pifT) as argument.
    And that for a transfer function H(z)?

    If you expand part of what you have, you get:
    [tex]\cos(-\theta)+j\sin(-\theta)=e^{-j\theta}=e^{-j e^{j2\pi f T}}[/tex]

    I don't know... it just doesn't look like part of a transfer function...
     
  9. Dec 8, 2011 #8
    I pretty sure that Evo8 meant to say that [itex]θ=2πfT[/itex], given the pattern of cos+j.sin combinations.

    [itex]\frac{-2cos(-θ)-2j.sin(-θ)+cos(-2θ)+j.sin(-2θ)}{2+4cos(-θ)+4j.sin(-θ)+2cos(-2θ)+2j.sin(-2θ)}[/itex]

    [itex]\ldots = \frac{-2e^{-jθ}+e^{-j2θ}}{2+4e^{-jθ}+2e^{-j2θ}}[/itex]

    [itex]\ldots = \frac{e^{-jθ}(e^{-jθ}-2)}{2(e^{-jθ}+1)^{2}}[/itex]

    Hmm, things behave badly when [itex]θ=π[/itex], or using my assumed identity above when [itex]2fT=1[/itex]
     
  10. Dec 9, 2011 #9
    Well for starters my transfer function was derived wrong...:( I derived it from a block diagram and now realize the mistake. So this is irrelivent but at least i get some more experience with this.

    Yes in this case i had theta=(2pifT). Its a method my book used so i tried mimicking it.

    I appreciate all the help guys. You have been super helpful.
     
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