how does this phase change and phase plot work..

[SanjKrish]

This is a little tricky.. and I tried my best to explain my question..
I was watching a video in DSP posted by NPTEL in IIT video lectures..

He was explaining that....

A simple digital band stop filter's transfer function is given by
X(Z) = 1/2 (1+Z^-2)

Replacing Z by e^j\[\omega\] and reducing the expression we finally get the frequency response as
H(e^j\[\omega\]) = (1/2)e^-j\[\omega\] * 2\[\cos\]\[\omega\]

Now the author graphed a plot of cos(\[\omega\]) Vs \[\omega\] first as shown below


He said that cos(\[\omega\]) has 0 phase till \[\pi\]/2 and after \[\pi\]/2 the phase abruptly changes by \[\pi\]

He then came up with an equation for the phase angle, for the above mentioned frequency response equation, as
phase angle = -\[\omega\] + "angle of cos(\[\omega\])"

Now how did he get the phase angle like that.. I really don't understand the "angle of cos(\[\omega\])" part
what does that mean


He then plotted the phase plot of the digital filter like this


Please explain me how did he arrive at such a phase plot..
why does the the phase plot suddenly jump from -\[\pi\]/2 to +\[\pi\]/2

 

[zorro]

Hi SanjKrish,

The attachments can not be viewed. Please try to upload them again.

I really don't understand the "angle of cos(\[\omega\])" part
what does that mean

For any \[\omega\], cos(\[\omega\]) is a complex number whose real part ranges between -1 and 1, and the imaginary part is 0. The term "angle of" is equivalent to "argument of", so in the case of the complex number x+j0, its "angle" is 0 for x>0 and \[\pi\] for x<0.

If you upload the figures we can go further.
Regards

Z

 

[SanjKrish]

Hi zorro,

Thanks for your explanation
I uploaded the images again.. I think I kinda got it .. but could you please further expand on this sentence of what you said The term "angle of" is equivalent to "argument of", so in the case of the complex number x+j0, its "angle" is 0 for x>0 and for x<0.

you said that the imaginary part is 0. So argument should be tan^-1(img part/real part) which should always yield 0..

Also how does this have affect the phase plot in turn to have such spike like graph...

I know what a phase is.. if my function is cos(\[\omega\]+\[\pi\]/6) then \[\pi\]/6 is called the phase..
What about the term phase change..
Does cos(\[\omega\]) exhibit phase change at \[\pi\]/2..
I could only see that the value of it changes to negative..
How does negative value correspond to phase change of \[\pi\]...

 

[albbg]

You can proceed exactly in the same way as I did to answer your post "How to find the phase angle of the given transfer function":

H(jw)=\[{e}^{-jw}\]•cos(w)=[cos(w)-j•sin(w)]•cos(w) that leads into:

H(jw)=\[{cos}^{2}\](w)-j•cos(w)•sin(w)

Now, the phase will be:

φ=atan(-sin(w)/cos(w))=atan(-tan(w))=-w

Since the tan function repeats every Π, we can wrap the phase between -Π/2 and Π/2 having the drawing you post.

 

[zorro]

you said that the imaginary part is 0. So argument should be tan^-1(img part/real part) which should always yield 0..

...or \[\pi\] . Think in a pictorial view of vectors. The argument or angle (with respect to the real axis) of the complex number 1+j0 is 0. Which is the argument or angle of -1+j0 ?

I know what a phase is.. if my function is cos(\[\omega\]+\[\pi\]/6) then \[\pi\]/6 is called the phase..

Attention!!!
If you have a function of time t x(t)=cos(ω₀t+\[\pi\]/6), then the phase is \[\pi\]/6
But if you have a transfer function (a function of frequency ω) H(ω)=cos(\[\omega\]+\[\pi\]/6), that is real for any \[\omega\], then its phase is 0 or \[\pi\] (*), as shown above.

(*) to be complete, I should say "plus an arbitrary multiple of 2\[\pi\]" . We should remark that this H(ω) would have a complex (not real) impulse response, but this is another stuff.

regards

Z

 

[albbg]

But if you have a transfer function (a function of frequency ω) H(ω)=cos(\[\omega\]+\[\pi\]/6), that is real for any \[\omega\], then its phase is 0 or \[\pi\] (*), as shown above.
Z

Sorry, just a small clarification:

in the time expression sin[w*t+φ] w is an angular frequency and w*t is a phase (w=dφ/dt)
in the transfer function sin[w+φ] w is a phase itself (you cannot add different quantities)

 

[SanjKrish]

albbg,
you said

Since the tan function repeats every ?, we can wrap the phase between \[\pi\]/2 and \[\pi\]/2 having the drawing you post.

Since the final result has only \[\omega\] n is devoid of tan how can we say it will repeat every \[\pi\]..


zorrow,
Its difficult for me relate what u said with my question.. I agree cos repeats every \[\pi\], but why does ever appear in my phase angle \[\varphi\] of my frequency response function..
Zorrow, just tell me this..,
Is the phase angle of the frequency response function H(e^j\[\omega\]) = (1/2)e^-j\[\omega\] * cos(\[\omega\])

-\[\omega\]
OR
-\[\omega\]+

If it is -\[\omega\]+ tell me why...

 

[zorro]

...
Zorrow, just tell me this..,
Is the phase angle of the frequency response function H(e^j\[\omega\]) = (1/2)e^-j\[\omega\] * cos(\[\omega\])

-\[\omega\]
OR
-\[\omega\]+/cos(ω)
...

It's a very basic property of complex numbers!

arg(a*b)=arg(a)+arg(b)

So, if

H(e^jω) = (1/2)e^-jω * cos(ω)

then

arg{H(e^jω)} = arg{(1/2)} + arg{e^-jω} + arg{cos(ω)} = 0 - ω + arg{cos(ω)}


Z

 

[SanjKrish]

Hurray!!!:-D Thanks a lot zorrow...

Ur reply made everything clear..