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Offlineskydog
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Here's one for you
    #7709941 - 12/03/07 08:40 AM (16 years, 1 month ago)

angle(H(e^jw)) = arctan[sinw/cosw] - arctan[sinw/(cos(w) - 0.5)]

Ok, this is the formula for the phase response of a particular Discrete Time filter [H(z) = z/(z-0.5)].  My Signals & Systems professor says that, at w = pi/2 rad/s, angle(H(e^jw)) = -0.46. This is strange because if you plug pi/2 into the formula, the result is  2.68.

So I'm wondering, what is the secret step he took to get -0.46?  I know that arctan(sin(pi/2)/(cos(pi/2)-0.5)) - pi/2 = -0.46, but how did he change the formula so that it becomes arctan(sinw/cos(w)-0.5) - pi/2 at this particular frequency?

I have been playing with identities involving arctan such as arctan(x) = -arctan(-x) and the like, but to no avail.

Can anyone solve this mystery? :strokebeard:


btw, :crankey:


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OfflineTheCow
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Re: Here's one for you [Re: skydog]
    #7709977 - 12/03/07 08:57 AM (16 years, 1 month ago)

Seems to work for me


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Offlineskydog
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Re: Here's one for you [Re: TheCow]
    #7709994 - 12/03/07 09:09 AM (16 years, 1 month ago)

You got

angle(H(e^jw)) = w - arctan[sinw/(cos(w) - 0.5)] = -0.46

just from plugging in w=pi/2?

I get 2.68. :crankey:


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OfflineTheCow
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Re: Here's one for you [Re: skydog]
    #7710045 - 12/03/07 09:37 AM (16 years, 1 month ago)

Eh actually I can get either. This is what my calculator spits out at me:

"sign(0)*pi/2 + 1.10715"

So if its -pi/2 then you get his answer, if its positive pi/2 you get yours. I dont know what the fuck the sign function is, but I imagine my calculator doesn't like the divide by 0 thats going on.
Basically its just a difference of pi


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Offlineskydog
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Re: Here's one for you [Re: TheCow]
    #7710115 - 12/03/07 10:00 AM (16 years, 1 month ago)

Hmm you can get rid of the divide by zero since arctan(sinw/cosw) = arctan(tanw) = w.  If that's what you're referring to :awesome:


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OfflineTheCow
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Re: Here's one for you [Re: skydog]
    #7710146 - 12/03/07 10:11 AM (16 years, 1 month ago)

Yes but tan(pi/2) is |infinity| so you are taking the inverse tangent of |infinity| which can either be pos/neg pi/2


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Offlineskydog
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Re: Here's one for you [Re: TheCow]
    #7710164 - 12/03/07 10:16 AM (16 years, 1 month ago)

You are awesome :awesome:


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Offlinenobhdy
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Re: Here's one for you [Re: skydog]
    #7711044 - 12/03/07 01:40 PM (16 years, 1 month ago)

yeah, i heart math :awesome:


i have no idea whats happening, but its going into my siggy :mushroom2:


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[quote]Gumby said:
And if you are going to waste peoples time with your stupid questions, at least try to have grammar skills higher then that of a 7th grader.

READ DAMNIT! [/quote]


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