# Finding a phase characteristic from a transfer function?

• Oct 15th 2012, 05:11 AM
MathIsOhSoHard
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?
• Oct 15th 2012, 07:21 AM
MathIsOhSoHard
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 $Arg(-1)$?
• Oct 15th 2012, 07:41 AM
MathIsOhSoHard
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.
• Oct 15th 2012, 08:12 AM
Krahl
Re: Finding a phase characteristic from a transfer function?
Quote:

Originally Posted by MathIsOhSoHard
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
$tan^{-1}(-x)=-tan^{-1}(x)$