Finding a phase characteristic from a transfer function?

Given a transfer function:

$\displaystyle H(s)=\frac{-1}{s+1} $

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

My text book only briefly covers this subject so I really have no idea what to do.

I know the formula is:

$\displaystyle \varphi(\omega)=Arg H(i\omega) , \omega>0$

So I have that:

$\displaystyle H(i\omega)=\frac{-1}{i\omega+1} $

So I assume I need to find the argument of this fraction?

Re: Finding a phase characteristic from a transfer function?

My attempt:

$\displaystyle Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)$

I looked at this Wikipedia article under the section "Computation" where it says that:

$\displaystyle Arg(x+iy)=\pi/2-arctan(x/y)$

for $\displaystyle y>0$

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

So that leaves me with:

$\displaystyle \varphi(\omega)=\pi/2-arctan(1/\omega)$

So I've figured out this much. I'm a bit confused as what to do next. What about $\displaystyle Arg(-1)$?

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.

Re: Finding a phase characteristic from a transfer function?

Quote:

Originally Posted by

**MathIsOhSoHard** My attempt:

$\displaystyle Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)$

I looked at

this Wikipedia article under the section "Computation" where it says that:

$\displaystyle Arg(x+iy)=\pi/2-arctan(x/y)$

for $\displaystyle y>0$

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

So that leaves me with:

$\displaystyle \varphi(\omega)=\pi/2-arctan(1/\omega)$

So I've figured out this much. I'm a bit confused as what to do next. What about $\displaystyle Arg(-1)$?

You can take the $\displaystyle i$ out of the denominator by multiplying the numerator and denominator of $\displaystyle \frac{-1}{i\omega+1}$ by $\displaystyle 1-i\omega$ to get,

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

Using the properties of arctan,

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

and

$\displaystyle tan^{-1}(-x)=-tan^{-1}(x)$

results in the answer.

EDIT:You've already done it. Well done