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Math Help - Finding a phase characteristic from a transfer function?

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    Finding a phase characteristic from a transfer function?

    Given a transfer function:
    H(s)=\frac{-1}{s+1}

    Show that the phase characteristic is \varphi(\omega)=\pi-Arctan \omega , \omega>0

    My text book only briefly covers this subject so I really have no idea what to do.
    I know the formula is:
    \varphi(\omega)=Arg H(i\omega) , \omega>0

    So I have that:
    H(i\omega)=\frac{-1}{i\omega+1}

    So I assume I need to find the argument of this fraction?
    Last edited by MathIsOhSoHard; October 15th 2012 at 05:46 AM.
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    Re: Finding a phase characteristic from a transfer function?

    My attempt:
    Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)

    I looked at this Wikipedia article under the section "Computation" where it says that:
    Arg(x+iy)=\pi/2-arctan(x/y)
    for y>0

    I'd assume this is the definition I need to look at because in the formula, it says that \omega>0 and the omega is the imaginary part.

    So that leaves me with:
    \varphi(\omega)=\pi/2-arctan(1/\omega)

    So I've figured out this much. I'm a bit confused as what to do next. What about Arg(-1)?
    Last edited by MathIsOhSoHard; October 15th 2012 at 07:34 AM.
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    Re: Finding a phase characteristic from a transfer function?

    Nevermind, I think I solved it on my own.
    Arg(-1)=pi
    Arg(ib+1)=arctan(b)

    So subtracting them with each other gives the answer.
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    Re: Finding a phase characteristic from a transfer function?

    Quote Originally Posted by MathIsOhSoHard View Post
    My attempt:
    Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)

    I looked at this Wikipedia article under the section "Computation" where it says that:
    Arg(x+iy)=\pi/2-arctan(x/y)
    for y>0

    I'd assume this is the definition I need to look at because in the formula, it says that \omega>0 and the omega is the imaginary part.

    So that leaves me with:
    \varphi(\omega)=\pi/2-arctan(1/\omega)

    So I've figured out this much. I'm a bit confused as what to do next. What about Arg(-1)?
    You can take the i out of the denominator by multiplying the numerator and denominator of  \frac{-1}{i\omega+1} by 1-i\omega to get,

    Arg\left ( \frac{-1}{1-\omega^2} + i \frac{\omega}{1-\omega^2} \right )=\frac{\pi}{2}-tan^{-1}(\frac{-1}{\omega})

    Using the properties of arctan,

    tan^{-1}(\frac{1}{x})=-\frac{\pi}{2}-tan^{-1}(x)
    and
    tan^{-1}(-x)=-tan^{-1}(x)

    results in the answer.

    EDIT:You've already done it. Well done
    Last edited by Krahl; October 15th 2012 at 08:16 AM.
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