Polar Form Conversion for Complex Numbers

  • Thread starter Thread starter shaltera
  • Start date Start date
  • Tags Tags
    Forms Polar
Click For Summary

Discussion Overview

The discussion revolves around the conversion of complex numbers to polar form and the discrepancies encountered when comparing manual calculations with results from MATLAB. Participants explore different methods for conversion and address potential errors in their calculations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation involving the polar form of complex numbers and seeks validation of their solution.
  • Another participant suggests that the original solution appears correct, allowing for multiple approaches to the problem.
  • Concerns are raised about discrepancies between the participant's results and those obtained from MATLAB, prompting further investigation into the calculations.
  • Participants discuss the importance of retaining precision in intermediate calculations to avoid rounding errors, particularly in angle calculations.
  • One participant acknowledges a mistake in their angle calculation and corrects it after comparing with MATLAB results.
  • Another participant highlights the potential for truncation errors when converting angles from radians to degrees.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the initial calculations, as discrepancies remain between manual results and MATLAB outputs. Participants express differing views on the accuracy of their computations and the impact of rounding errors.

Contextual Notes

Participants note that rounding and truncation errors can significantly affect the results, especially in angle calculations, and emphasize the need for careful handling of intermediate values.

shaltera
Messages
91
Reaction score
0

Homework Statement



A=1.5495<21.0363°x(22.1009<30.3658°/69.9667<9.1884°)

Homework Equations


The Attempt at a Solution


A=1.5495<21.0363x(22.1009/69.9667(30.3658-9.1884)=1.5495<21.0363(0.3159<21.1774)=(1.5495x0.3159)(21.0363+21.1774)=0.4895<42.2137°

Solution above is it correct or I have to convert polar forms to complex number, and final result to convert to polar form?

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
You can do it either way. Your solution looks fine.
 
why result is completely different compared with MatLAB
 
shaltera said:
why result is completely different compared with MatLAB

I couldn't say; I haven't seen your Matlab work.
 
IL=IN(ZTH/ZTH+RL)
IN=VTH/ZTH

FOR:
VTH=33.79310-j5.51724
ZTH=19.06896+j11.172413
RL=50

Here comes the tricky part:
I convert VTH and ZTH in polar forms I got
VTH=34.2460<-9.32954
ZTH=22.1009<30.3658
and
ZTH+RL=19.06896+j11.172413+50=69.06896+j11.172413=69.9667<9.1884

Then

IN=34.2460<-9.32954/22.1009<30.3658=1.1278<-39.6953

IL=1.1278<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.1278<-39.6953(0.3159<21.1774)=0.3563<-18.5169
 
shaltera said:
IL=IN(ZTH/ZTH+RL)
IN=VTH/ZTH

FOR:
VTH=33.79310-j5.51724
ZTH=19.06896+j11.172413
RL=50

Here comes the tricky part:
I convert VTH and ZTH in polar forms I got
VTH=34.2460<-9.32954
ZTH=22.1009<30.3658
and
ZTH+RL=19.06896+j11.172413+50=69.06896+j11.172413=69.9667<9.1884

Then

IN=34.2460<-9.32954/22.1009<30.3658=1.1278<-39.6953
That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?
 
  • Like
Likes   Reactions: 1 person
gneill said:
That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?

sorry
IN=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953
 
shaltera said:
sorry
IN=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953

Okay, that looks better.

I note that in your various conversions you are specifying several decimal places but the values after the first decimal place or so aren't always accurate... it looks like some truncation or rounding was done on some previous intermediate values, so the extra digits aren't helping (or useful). Make sure to keep extra digits in all intermediate values.

For example, you converted Vth from rectangular to polar as

33.79310-j5.51724 → 34.2460 ∠-9.32954

If I do the same conversion, being sure to retain all digits for intermediate steps, I obtain:

33.79310-j5.51724 → 34.24052 ∠-9.2726°

Angles are notorious sources of rounding/truncation errors if you skimp on digits, because the functions involved are not linear; in some regions a tiny difference can blow up into a big difference.
 
therefor

IL=1.5495<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.5495<-39.6953(0.3159<21.1774)=0.4895<-18.5179
 
  • #10
shaltera said:
therefor

IL=1.5495<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.5495<-39.6953(0.3159<21.1774)=0.4895<-18.5179

Yep. Again, rounding/truncation errors are suspect: I get 0.489 ∠-18.461° .
 
  • #11
interesting I calculate using a MatLAB, and that's what MatLAB gives me?
 
  • #12
fx>> A=33.79310-5.51724i
fx>> magn=abs(A)
fx>> magn=34.2405
fx>>ANGLE=angle(A)*180/pi
fx>>ANGLE=-9.2726
 
  • #13
Ohh, for calculation of Vth I used calculator
sqr(33.793102+5.517242)=34.2460
then tan-1(5.51724/33.79310)=-9.2726°

I don't know what I have done to get -9.32954 but obviously is not right :)
 
  • #14
shaltera said:
fx>> A=33.79310-5.51724i
fx>> magn=abs(A)
fx>> magn=34.2405
fx>>ANGLE=angle(A)*180/pi
fx>>ANGLE=-9.2726

Right. Compare these values with the ones you provided in post #5.
 
  • Like
Likes   Reactions: 1 person
  • #15
yes you are right angle is a bit off.And it has been corrected
 
  • #16
shaltera said:
Ohh, for calculation of Vth I used calculator
sqr(33.793102+5.517242)=34.2460
then tan-1(5.51724/33.79310)=-9.32954°

The inverse tan value looks suspect. Did you truncate a radian value before converting to degrees? My "calculator" gives 0.161837 radians for the arctan. Converting that to degrees gives 9.2726°.
 
  • #17
you are correct, I made a mistake.I copied the wrong value on the forum.I have 9.2726, and MatLAB confirmed it.Thanks
 
  • #18
shaltera said:
you are correct, I made a mistake.I copied the wrong value on the forum.I have 9.2726, and MatLAB confirmed it.Thanks

Excellent :smile:
 

Similar threads

Replies
22
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
10K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 24 ·
Replies
24
Views
13K
Replies
4
Views
2K
  • · Replies 28 ·
Replies
28
Views
7K