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

• Firepanda

#### 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. =)

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 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?

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 )

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?

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 -- 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

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

just maximise x2 + y2 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 =)

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?

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)

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?

wouldnt that just be infinity then?

Nooo … what formula are you using for Aω? Nooo … what formula are you using for Aω? 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

and I got ω^4 -12ω^2 +61

No, it's 1/(that) …

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

Firepanda said:

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 )

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

Well you said this earlier

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

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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 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! Last edited by a moderator:
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

2+b)2 = b2 + c

What is b?

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try (ω2+b)2 + c, with b = 6.5 … and now I'm off to bed … :zzz: