Solving Argand Diagram for [0,pi/2] & [-pi/2,pi] Intervals

[Firepanda]

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. =)

 

[tiny-tim]

Hi Firepanda!

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

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 …

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan^-1

 

[Firepanda]

Hi Firepanda!

(have an omega: ω and a phi: φ and a pi: π )I don't really follow what you're doing …

to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan^-1

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?

 

[tiny-tim]

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! (or you'll end up in the wrong quadrant )

 

[Firepanda]

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! (or you'll end up in the wrong quadrant )

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?

 

[tiny-tim]

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 --

 

[Firepanda]

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 --

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

 

[tiny-tim]

Any idea of how to maximise? because that part has me stumped

just maximise x² + y²

 

[Firepanda]

just maximise x² + y²

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 =)

 

[tiny-tim]

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

add up the squares … what do you get?

 

[tiny-tim]

So I just add them up, then apply a value of ω from my previous intervals of ω that makes this the largest?

yes

(except I think the question is asking for any ω)

 

[Firepanda]

yes

(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?

 

[tiny-tim]

wouldnt that just be infinity then?

Nooo … what formula are you using for A_ω?

 

[Firepanda]

Nooo … what formula are you using for A_ω?

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

 

[tiny-tim]

and I got ω^4 -12ω^2 +61

No, it's 1/(that) …

(except you've used 25 instead of 25ω²)

 

[tiny-tim]

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!

I'm getting confused …

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

(and you can factor that )

 

[Firepanda]

Hi Firepanda!

I'm getting confused …

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

(and you can factor that )

Well you said this earlier

just maximise x² + y²

So that's why I was doing that

So I thought I was to compute x² + y², then find the maximum on the graph

 

[tiny-tim]

Well you said this earlier

I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from

 

[Firepanda]

I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from

EDIT: my working out was riddled with mistakes so ill just put the answer!

I seeeeee!

Woohoo

16/(ω⁴+36+13ω²)

I have the roots of the denominator, what do i do with those?

 

[tiny-tim]

= 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

now you see you were going round in circles … |1/A| = 1/|A| ?

I have the roots of the denominator, what do i do with those?

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

sooo … https://www.physicsforums.com/library.php?do=view_item&itemid=107" instead! ​

 

[Firepanda]

he he

now you see you were going round in circles … |1/A| = 1/|A| ?

Ah *facepalms* lol

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

sooo … https://www.physicsforums.com/library.php?do=view_item&itemid=107" instead! ​

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

(ω²+b)² = b² + c

What is b?

 

[tiny-tim]

try (ω²+b)² + c, with b = 6.5

… and now I'm off to bed … :zzz:​