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posted by  Remo on 7/3/2009 6:47:49 PM  |  status: Closed  |  Earned Karma: 165

Polar Notation

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Question Details:
Hi,
The polar notation for the function 1 + ej4 is:
a. cos(2) + 1
b. ej2[2cos(2)]
c. e-j4sin(2)
d. ej4sin(4)
Please show all basic steps.
Thanks, Remo
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cRequests for Clarification:

Apprentice
Karma Points: 165
posted by Remo on 7/3/2009 11:55:24 PM  |  status: Live
Clarification Details:
Your response:
 
We have ejθ = cosθ + jsinθ
Given
1+ ej4 = 1+ cos4 + jsin4
          =  1+ 0.99 + j0.07
          =  1 + 1/_4o
          =  1 + cos(4)
or
1+ ej4 = 1+ cos4 + jsin4
          =  1+ 0.99 + j0.07
          = 1.99 + j0.07
         =  2 /_2o
         = 2 cos (2)
Can you please explain you went from = 1.99 + j0.07
to the next step in both equations?
How does 1 + 0.99 + j0.07 tranlate to 1 + 1/_4o
Can you please explain the "/_" notation that you use?
Thank You
Thanks, Remo

AAnswers:

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Oracle
Karma Points: 25,678
posted by Shree_vani(MNK) on 7/3/2009 9:23:02 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
We have ejθ = cosθ + jsinθ
Given
1+ ej4 = 1+ cos4 + jsin4
          =  1+ 0.99 + j0.07
          =  1 + 1/_4o
          =  1 + cos(4)
or
1+ ej4 = 1+ cos4 + jsin4
          =  1+ 0.99 + j0.07
          = 1.99 + j0.07
         =  2 /_2o
         = 2 cos (2)
Oracle
Karma Points: 25,678
posted by Shree_vani(MNK) on 7/4/2009 1:28:53 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
/_ is the notation for angle of
1 + ei4
We have ejθ = cosθ + j sinθ
1+ej4 =  1 + cos 4 + jsin4
Using claulator or using log book
The value 0f cos4 = 0.99
The value of sin4 = 0.07
1+ej4 =  1 + 0.99 + j 0.07
         = 2.99 + j 0.07
We have the polar representation of complex number as
A + j B = r/_θ                 That is r angle of θ
Wher r = √A2 + B2
and θ  =tan -1 (B/A)
Using this
1+ej4
          
Which can be written as
          
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