Question

Hwo do you convert the product to polar form :
P = (12 +j5)(1<30 degrees)

Answer 1

Note 1<30 degrees is a phasor.. AND you CANNOT use a calculator
also its easy if you have a calculator but when you convert the rectangular part to its phase angle inverse tan = (5/12) , I dont know how the heck am I suppose to know the value in degrees . conversely converting the phasor to rectangular is difficult with out a calculator

Answer 2

phaser?
You can give me example of what that means exactly?

Answer 3

phasor is equivalent to polar coordinates

Answer 4

And I'm assuming your j is my i

that is, j is the actual imaginary unit

Answer 5

the work is obvious but NOT clean . thats the problem becuase we cannot use calculators on the exam , just want to clarify that

Answer 6

yes

Answer 7

so is that second factor just:
\[\cos(30)+j \sin(30)\]

Answer 8

or in the other form:
\[e^{j \frac{\pi}{6}}\]

Answer 9

(12 +j5)*(sqrt(3)/2 +j(1/2)) .. applying multiplication makes it a hair ball ,

Answer 10

\[P=(12+j5)e^{\frac{\pi}{6}j}\]
maybe we can also convert the other one to polar form
then find the product

Answer 11

ok but , inversse tan = (12/5)

Answer 12

how do you implement that

Answer 13

you cannot just leave theta as arctan(5/12)
also it is y/x inside tan

Answer 14

?

Answer 15

oh yeah my bad

Answer 16

\[P=\sqrt{12^2+5^2}e^{\arctan(\frac{5}{12})j} \cdot e^{\frac{\pi}{6} j} \\ P=\sqrt{169} e^{\arctan(\frac{5}{12})j}e^{\frac{\pi}{6} j} \\ P=13 e^{(\arctan(\frac{5}{12})+\frac{\pi}{6})j}\]

Answer 17

would that not work?

Answer 18

and just to be clear that factor rewrite was correct right?
1<30 degrees means e^(pi/6 i) ?

Answer 19

I got another method
hold on let me write it . i thhink it workds

Answer 20

oops j not i

Answer 21

(12+ j5)(cos(30) + sin(30)) = 12cos(30) -sin()30 +j (5cos(30)+12sin(30)).....can you help me simplify that

Answer 22

maybe if you conver cos to sin ..... sin (theta) = cos(theta - 90 ) ??? is that valid

Answer 23

no no its not never mind

Answer 24

\[(12+5j)(\cos(\frac{\pi}{6})+j \sin(\frac{\pi}{6})) \\ =12(\cos(\frac{\pi}{6})+j \sin(\frac{\pi}{6}))+5j(\cos(\frac{\pi}{6})+j \sin(\frac{\pi}{6})) \\ =12 \cos(\frac{\pi}{6})+12 j \sin(\frac{\pi}{6})+5j \cos(\frac{\pi}{6})+5j^2 \sin(\frac{\pi}{6}) \\ \text{ and } j^2=-1 \\ \text{ so we have } \\ =12 \cos(\frac{\pi}{6})+12 j \sin(\frac{\pi}{6})+5j \cos(\frac{\pi}{6})-5 \sin(\frac{\pi}{6}) \\\]
hmmm cos(x)=sin(pi/2-x)
so cos(pi/6)=sin(pi/2-pi/6)=sin(2pi/6)=sin(pi/3)
and sin(pi/6)=cos(pi/2-pi/6)=cos(pi/3)
hmm... So we have:
\[12e^{ \frac{\pi}{6} j} +5j \sin(\frac{\pi}{3})-5 \cos(\frac{\pi}{3})\]
is that what you wanted to do...

Answer 25

yes its close enough, you still have it in rectangular form - it still needs to be converted to polar . from what you have shown me - I cant imagine any other way to do this problem .

Answer 26

\[12e^{ \frac{\pi}{6} j} -5 \cos(\frac{\pi}{3})+5 j \sin(\frac{\pi}{3}) \\ 12e^{\frac{\pi}{6} j}-5(\cos(\frac{\pi}{3})-j \sin(\frac{\pi}{3}) ) \\ 12e^{\frac{\pi}{6} j } -5 (\cos(\frac{\pi}{3})+j \sin(-\frac{\pi}{3})) \\ 12e^{\frac{\pi}{6}j} -5(\cos(\frac{-\pi}{3})+j \sin(-\frac{\pi}{3}))\]
since cos(-x)=cos(x) and sin(-x)=-sin(x)

Answer 27

now that one part should be ready to go

Answer 28

Thank you , I have to double check with my professor . I will get back to you eventually

Answer 29

but you do see that one thing following the difference sign can be written in polar form right?

Answer 30

\[12e^{\frac{\pi}{6} j}-5e^{\frac{-\pi}{3} j}\]

Answer 31

can be written as a difference in polar form

Answer 32

To be honest I have never seen E raised to x/ a .. That is completely new to me

Answer 33

honestly though I kinda would have stuck with the above answer \[P=\sqrt{12^2+5^2}e^{\arctan(\frac{5}{12})j} \cdot e^{\frac{\pi}{6} j} \\ P=\sqrt{169} e^{\arctan(\frac{5}{12})j}e^{\frac{\pi}{6} j} \\ P=13 e^{(\arctan(\frac{5}{12})+\frac{\pi}{6})j}\]
But you are totally right
I don't know what goes in your course

Answer 34

oh what do you call polar form
\[e^{ i \theta} \text{ is the same as } \cos(\theta)+i \sin(\theta) \]

Answer 35

I call exp(i theta) polar form

Answer 36

http://web.eecs.umich.edu/~aey/eecs206/lectures/phasor.pdf
these use that imaginary unit j
so I assume this is the engineering version of the trig class I took

Answer 37

They call polar form the same thing I do

Answer 38

Thank you for the resourceful document - I will study this !!