How to find the phase angle of the given transfer function

[SanjKrish]

Hello all,
While seeing a lecture about designing a simple digital high pass filter, I came to know its frequency response will of the form

H(e^j\[\omega\]) = je^{-j\[\omega\]/2}sin(\[\omega\]/2)

The author plotted the phase response graph as below.. :?:
View attachment phase.bmp

He explained that the phase at 0 will be \[\pi\]/2 because of the 'j' term and
beyond \[\omega\]=0 it will be \[\pi\]/2 - \[\omega\]/2 so it will linearly come down to 0 at \[\omega\] = \[\pi\]/2..

Please explain me how did he find out the phase response to be \[\pi\]/2 - \[\omega\]/2
I know tat the phase is given by tan^-1(img part/real part) and for me, for the given frequency response, it came out to be \[\omega\]/2
and not \[\pi\]/2 - \[\omega\]/2..:-(

 

[albbg]

It should be (to simplify I used x=w/2):

H(jx)=j•\[{e}^{-jx}\]•sin(x)=j•[cos(x)-j•sin(x)]•sin(x) that is:

H(jx)=\[{sin}^{2}\](x)+j•sin(x)•cos(x)

as you said the phase is φ=atan[Imag(H)/Real(H)] that means:

φ=atan[sin(x)•cos(x)/\[{sin}^{2}\](x)]

simplifying:

φ=atan[cotg(x)] = atan[tan(Π/2-x)] = Π/2-x

since I defined x=w/2 thus:

φ=Π/2-w/2

 

[SanjKrish]

Wow..
How did i miss it..
Thanks for your detailed explanation albbg..

Could you also reply for my other thread at
https://www.edaboard.com/threads/281292/