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Math Help - A Quick Trig Identity Problem That I Need To Solve

  1. #1
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    A Quick Trig Identity Problem That I Need To Solve

    Hi,

    I'm looking to set up an equation involving the use of trig, just to help you understand take this example:

    H(z) = 1/(1+0.5z^-1)

    Substituting z=e^{j\omega}

    H(z) = 1/(1+0.5e^{-j\omega})

    And solves for:

    H(z) = 1/(1+0.5\cos(\omega)-j0.5\sin(\omega))

    So for my equation and substituting z=e^{-j\omega}:

    H(z) = (0.5 -0.75e^{-2j\omega})/(1e^{j\omega} + 0.25e^{-2j\omega})

    I'm trying to obtain mine but I am confused about the e^{-2j\omega} and the positioning of the subsquent 2 this time. I just can't be 100% of getting it right. Does anyone know?
    Last edited by CaptainBlack; March 28th 2010 at 01:21 AM.
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  2. #2
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    Quote Originally Posted by TomMUFC View Post
    Hi,

    I'm looking to set up an equation involving the use of trig, just to help you understand take this example:

    H(z) = 1/(1+0.5z^-1)

    Substituting z=e^{j\omega}

    H(z) = 1/(1+0.5e^{-j\omega})

    And solves for:

    H(z) = 1/(1+0.5cos(\omega)-j0.5sin(\omega))

    So for my equation and substituting z=e^{-j\omega}:

    H(z) = (0.5 -0.75e^{-2j\omega})/(1e^{j\omega} + 0.25e^{-2j\omega})

    I'm trying to obtain mine but I am confused about the e^{-2j\omega} and the positioning of the subsquent 2 this time. I just can't be 100% of getting it right. Does anyone know?
    I have corrected you LaTeX to the best of my ability, but it may still not be right. Also (not directly related to the LaTeX) use brackets to make your meaning clear.

    This is still confused, why not post the actual question.

    CB
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  3. #3
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    That's fine CB.

    Ok this is the question, you may remember my question regarding this transfer function I did with you:


     <br />
H(z) = \frac{(1-r) + (r^2-1)z^{-2}}{1-2rcos(\omega_cT)z^-1 +r^2z^-2}<br />

    So the question was:

    "Setting  \omega_cT = pi/2 and r = 0.5, compute the magnitude frequency response for the series of frequency values satisfying \omega_T = npi/4 where n = 0,1,2,3,4. Repeat this for r = 0.7 and r= 0.9. Using these calculations, sketch (or plot) - in one figure - the magnitude frequency responses for all three values of r."

    I was following an example used in a book, which for their example (following on from above) then obtained their magnitude frequency response to be:

    H(z) = \frac{1}{\sqrt{[(1+0.5cos(\omega))^2+(j0.5sin(\omega))^2]}}

    *Should be a modulus of H(z)

    Then they would produce a table and substitute values of 0, 0.25pi, 0.5pi, 0.75pi and pi all in radians (which obtained subsequent values of 0.670, 0.715, 0.894, 1.357, 2.000). And then would convert into dB's from which a graph was plotted. I assume this is the correct way of doing the quesiton. I have to do this for the three values of r.

    I would like to do the same with my question. I have substituted r = 0.5 and \omega_cT=pi/2 to get my equation the only problem is that I'm not sure what the step is then? Could you help?
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  4. #4
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    So basically to sum up I have this equation:

    H(z) = \frac{0.5 -0.75z^-2}{1z-1 + 0.25z^-2}

    Substitute z=e^{j\omega}:

    H(z) = \frac{0.5 -0.75e^{-2j\omega}}{1e^{-j\omega} + 0.25e^{-2j\omega}}

    I have heard the use of Euler's formular being used somewhere as well but I'm not sure.
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  5. #5
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    Quote Originally Posted by TomMUFC View Post
    So basically to sum up I have this equation:

    H(z) = \frac{0.5 -0.75z^-2}{1z-1 + 0.25z^-2}

    Substitute z=e^{j\omega}:

    H(z) = \frac{0.5 -0.75e^{-2j\omega}}{1e^{-j\omega} + 0.25e^{-2j\omega}}

    I have heard the use of Euler's formular being used somewhere as well but I'm not sure.
    As I understand it you want the amplitude response, from the frequency response:

    H(\omega) = \frac{0.5 -0.75e^{-2j\omega}}{1e^{-j\omega} + 0.25e^{-2j\omega}}

    Multiply top and bottom by the conjugate of the denominator, to give you something with a real denominator.

    H(\omega) = \frac{(0.5 -0.75e^{-2j\omega})(1e^{+j\omega} + 0.25e^{+2j\omega})}{(1e^{-j\omega} + 0.25e^{-2j\omega})(1e^{+j\omega} + 0.25e^{+2j\omega})}

    simplify the denominator:

    H(\omega) = \frac{(0.5 -0.75e^{-2j\omega})(1e^{+j\omega} + 0.25e^{+2j\omega})}{1+0.5\cos(\omega)+0.25}

    Now simplify the numerator and take the absolute value.

    CB
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  6. #6
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    Hi CaptainBlack, appreciate your response.

    I'm having difficulty simplifying the numerator here (0.5 -0.75e^{-2j\omega})(1e^{+j\omega} + 0.25e^{+2j\omega})

    When I do so I obtain:

    0.5e^{j\omega} + 0.125e^{2j\omega} - 0.75e^{-j\omega} - 0.1875

    To be honest I don't know if that's right, and if it is, I still don't know what to do with it.

    This is my last question of my coursework and I have to hand it in today at 4-5pm-ish (yeah I apologise) so if could help me solve the numerator I would very much appreciate it so I can crack on with the graphs. This thread has revealed how desperately I lacking I am in certain areas of maths, and I'm a 3rd year level Electrical and Electronic Engineering student!
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  7. #7
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    Quote Originally Posted by TomMUFC View Post
    Hi CaptainBlack, appreciate your response.

    I'm having difficulty simplifying the numerator here (0.5 -0.75e^{-2j\omega})(1e^{+j\omega} + 0.25e^{+2j\omega})

    When I do so I obtain:

    0.5e^{j\omega} + 0.125e^{2j\omega} - 0.75e^{-j\omega} - 0.1875

    To be honest I don't know if that's right, and if it is, I still don't know what to do with it.

    This is my last question of my coursework and I have to hand it in today at 4-5pm-ish (yeah I apologise) so if could help me solve the numerator I would very much appreciate it so I can crack on with the graphs. This thread has revealed how desperately I lacking I am in certain areas of maths, and I'm a 3rd year level Electrical and Electronic Engineering student!
    That looks correct for the numerator.

    Now you use Euler's formula and calculate the absolute values.

    CB
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  8. #8
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    Well ok so I'm assuming:

     <br />
-0.1875 -0.25cos(\omega) + 0.125e^{2j\omega}<br />

    You see the problem I have now is what to do with this 0.125e^{2j\omega} it was this that I couldn't understand initially during the question. So what exactly do I do with this now?
    Last edited by TomMUFC; March 29th 2010 at 03:27 AM.
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  9. #9
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    You say I can take the final answer by obtaining the absolute value? Can you confirm that is the final answer? As I can only see one section being broken down using Eular's formula.

    Can I now substitute my values of \omega in?
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