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Argand Diagram help

  • Thread starter Firepanda
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  • #1
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9abk87.jpg


For this I started with the 1st interval [0,pi/2] and said

arctan [4/ ... ] = [0 , pi/2]

but studying the tan graph this is all positive values of the y-axis

so surely the range of values for this is (0 , inf)?

If I've done that right, how do I do the other interval [-pi/2 , pi]?

Because using that method gives the result (-inf , 0), but then that would also be true for the quadrant [0 , -pi/2] surely, which brings me back to my first answer also being true for [pi/2 , pi]..

How do I attempt this because this is obviously wrong..

This was under the chapter about Laplace / Fourier Transforms

And if you could shed any light on how to maximise too I'd be greatful, because my notes havn't touched on this..

The y(t) = ... was formed from solving a differential equation using laplace transforms, I didn't include that because I didn't think it was relevant. =)
 

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  • #2
tiny-tim
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Hi Firepanda! :smile:

(have an omega: ω and a phi: φ and a pi: π :wink:)
For this I started with the 1st interval [0,pi/2] and said

arctan [4/ ... ] = [0 , pi/2] …
I don't really follow what you're doing :confused:

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan-1 :wink:
 
  • #3
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Hi Firepanda! :smile:

(have an omega: ω and a phi: φ and a pi: π :wink:)


I don't really follow what you're doing :confused:

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan-1 :wink:
Oh shoot I read the question wrong =)

I mean for [-π/2 , 0]

So I guess that's the values for (-inf , 0)..

AH yeah I see now I was doing something wrong in my method too

So how I'm interpreting the question is this:

Where does the angle given by arg[4/-ω2+5ωi+6] lie in the bottom right quadrant [-π/2 , 0] of an argand diagram?

So do I need to change [4/-ω2+5ωi+6] into the form x + iy?
 
  • #4
tiny-tim
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So how I'm interpreting the question is this:

Where does the angle given by arg[4/-ω2+5ωi+6] lie in the bottom right quadrant [-π/2 , 0] of an argand diagram?
No, it's asking what are the values of ω for which 4/3 lies in the bottom right (4th) quadrant? (and then same question for the 3rd quadrant).
So do I need to change [4/-ω2+5ωi+6] into the form x + iy?
Yes!! :smile: (or you'll end up in the wrong quadrant :rolleyes:)
 
  • #5
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No, it's asking what are the values of ω for which 4/3 lies in the bottom right (4th) quadrant? (and then same question for the 3rd quadrant).


Yes!! :smile: (or you'll end up in the wrong quadrant :rolleyes:)
Great ok,

I canceled it all down and got my

arctan 5ω/(ω2-6) (lets assume this is right, forget my working too much to type out :P)

then 5ω/(ω2-6) has to lie in (-inf,0) for my arctan 5ω/(ω2-6) to equal the interval (-pi/2 , 0)

so how do I find ω from this?
 
  • #6
tiny-tim
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This is getting unnecessarily complicated …

you don't actually need to consider tan-1 since all you are interested in is the quadrant, which means all you need to consider is the signs of x and y (in x + iy) …

just draw the four quadrants, and mark them ++ +- -+ and -- :wink:
 
  • #7
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This is getting unnecessarily complicated …

you don't actually need to consider tan-1 since all you are interested in is the quadrant, which means all you need to consider is the signs of x and y (in x + iy) …

just draw the four quadrants, and mark them ++ +- -+ and -- :wink:
now I got it! ty. Much simpler, and I have myself an answer

Any idea of how to maximise? because that part has me stumped
 
  • #8
tiny-tim
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Any idea of how to maximise? because that part has me stumped
just maximise x2 + y2 :smile:
 
  • #9
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just maximise x2 + y2 :smile:
So I use my x and y, and add up their squares..

Could you give me the basic method on how to do this, just so it clicks with me and I'm able to =)
 
  • #10
tiny-tim
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So I use my x and y, and add up their squares..

Could you give me the basic method on how to do this, just so it clicks with me and I'm able to =)
erm … I don't see what you think the difficulty is :confused:

add up the squares … what do you get?
 
  • #11
tiny-tim
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So I just add them up, then apply a value of ω from my previous intervals of ω that makes this the largest?
yes :smile:

(except I think the question is asking for any ω)
 
  • #12
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yes :smile:

(except I think the question is asking for any ω)
wouldnt that just be infinity then?

or do you mean I have to find another range of values?
 
  • #13
tiny-tim
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wouldnt that just be infinity then?
Nooo … what formula are you using for Aω? :confused:
 
  • #14
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Nooo … what formula are you using for Aω? :confused:
Well I was using my x+iy values

I then found x^2 + y^2

and I got ω^4 -12ω^2 +61

So I'm trying to find the value of ω where this is largest, and by the graph it zooms off the screen so I was assuming inf and -inf for my ω
 
  • #15
tiny-tim
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and I got ω^4 -12ω^2 +61
No, it's 1/(that) …

(except you've used 25 instead of 25ω2)
 
  • #16
tiny-tim
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Firepanda said:
About this thread

https://www.physicsforums.com/showthread.php?t=314510

I keep getting a page not found message whenever I try and reply in it..

So I just wanted to ask one more question

My x2+y2

= 16(ω4-12ω2+61)/(ω4+13ω2+36)

Do I just draw this graph?

It shows a maximum at ω=0

But as ω tends to -inf and +inf then the graph tends to +inf

So is my answer ω = +- inf

Or ω = 0? :)
Hi Firepanda! :smile:

I'm getting confused :confused:

|1/A| = 1/|A|, so isn't it just 1/(ω4+13ω2+36)?

(and you can factor that :wink:)
 
  • #17
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Hi Firepanda! :smile:

I'm getting confused :confused:

|1/A| = 1/|A|, so isn't it just 1/(ω4+13ω2+36)?

(and you can factor that :wink:)
Well you said this earlier

just maximise x2 + y2 :smile:
So that's why I was doing that

So I thought I was to compute x2 + y2, then find the maximum on the graph
 
  • #18
tiny-tim
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Well you said this earlier
I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from :confused:
 
  • #19
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I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from :confused:
EDIT: my working out was riddled with mistakes so ill jsut put the answer!

I seeeeee!!!!!

Woohoo

16/(ω4+36+13ω2)

I have the roots of the denominator, what do i do with those?
 
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  • #20
tiny-tim
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= 4(-ω2+6-5ωi)/(-ω2+6-5ωi)(-ω2+5ωi+6)

Ah I see now I didn't square the denominator :P

Woohoo

16/(ω4+36+13ω2)
he he :biggrin:

now you see you were going round in circles … |1/A| = 1/|A| ? :smile:
I have the roots of the denominator, what do i do with those?
hmm :rolleyes: … on second thoughts, factoring doesn't help … the roots are all imaginary, which is a teeny bit irrelevant since ω is real :redface:

 
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  • #21
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he he :biggrin:

now you see you were going round in circles … |1/A| = 1/|A| ? :smile:
Ah *facepalms* lol

hmm :rolleyes: … on second thoughts, factoring doesn't help … the roots are all imaginary, which is a teeny bit irrelevant since ω is real :redface:

Aww took me so long to find those roots too!

How do I complete the square with this? 13 is odd so when I do

2+b)2 = b2 + c

What is b?
 
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  • #22
tiny-tim
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try (ω2+b)2 + c, with b = 6.5 :wink:

… and now i'm off to bed … :zzz:
 

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