Question

How to convert 1-1/(jw+1)^2 to a real value and an angle?

Answer 1

j being complex

Answer 2

w=omega

Answer 3

\[1-\frac{1}{(1+jw)^2}\]
Let's simplify this first
\[\frac{(1+jw)^2-1}{(1+jw)^2}\]
We will get
\[\frac{-w^2-2jw}{1-w^2+2jw}\]
DO you get this part ?

Answer 4

I see how you did it but do we need it in that form?

Answer 5

yes, it's easy that way

Answer 6

Let's find the real value first ( or absolute value)

Answer 7

Okay I'm just having trouble like figuring how you find it from the form j

Answer 8

\[\frac{|-w^2-2jw|}{|1-w^2+2jw|} \]

This is the real value\[\frac{\sqrt{(-w^2)^2+(-2w)^2}}{\sqrt{(1-w^2)^2+(2w)^2}}\]

Answer 9

a+jb
Real value
\[\sqrt{a^2+b^2}\]

Answer 10

oh you do it for both top and bottom

Answer 11

oh I see so you try to get it in the form a+jb

Answer 12

yes :)

Answer 13

Yes, we can do it both for numerator and denominator

Answer 14

oh sweet okay how do you get the angle? is there some diagram we can see

Answer 15

like which is x and which is y

Answer 16

Yes, we can find the angle separately

Answer 17

\[\frac{-w^2-2jw}{1-w^2+2jw}\]
Angle get subtracted from numerator and denominator
\[{\tan^{-1} (\frac{-2w}{-w^2})}-{\tan^{-1}(\frac{2w}{1-w^2})}\]
Do you get this part ?

Answer 18

are you just dividing b/a?

Answer 19

Yes, but one more
\[a+ib=> \tan^{-1} \frac b a\]
only if a>0 and b>0 and a<0, b<0
if a>0,b<0 or a<0 b>0 =>\(180-\tan^{-1}{\frac ba }\)

Answer 20

There is a catch here, if w^2 <1 then
it'll be
\[{\tan^{-1} (\frac{2w}{w^2})}-{\tan^{-1}(\frac{2w}{1-w^2})}\]

if w^2>1
then
\[{\tan^{-1} (\frac{2w}{w^2})}-(180-{\tan^{-1}(\frac{2w}{w^2-1}))}\]

Answer 21

Oh whoa serious.. I am not given any info on w

Answer 22

but it's alright I'm going to assume it's greater one question though

Answer 23

For the real part though I am looking at my text and when trying your method for the real part it doesn't seem to work? They got 4.47 but if you do it your way is it not 4?

Answer 24

cool :)

Answer 25

it's easy
\[\frac{10\times \sqrt{2^2+4^2}}{\sqrt{6^2+8^2}}=10\times \sqrt {20}/10=\sqrt {20}\]
Did you also do the same thing?

Answer 26

oh you can't multiply the 10 into the brackets? to get 20 + j40?

Answer 27

You can
But it'll come out eventually so I kept it outside

Answer 28

oh but if you do it with 20+j40 can't you just assume a=20 and b =40?

Answer 29

ah sorry for all the questions

Answer 30

you can, it's right
No worries. I'm here to help you
questions are good :)

Answer 31

lol okay but if you do then you get sqrt of 1600 which is 40 / 10?

Answer 32

which gives you 4

Answer 33

\[20^2+40^2=400+1600=2000\]

Answer 34

Lol omg I'm so dumb so sorry for the dumb questions

Answer 35

yeah it works out haha thanks man

Answer 36

No problem :)

Answer 37

yeah seriously thanks for your help really got it now!!

Answer 38

Glad to help you :)

Answer 39

have a good night!

Answer 40

Night