Polar Form Conversion for Complex Numbers

[shaltera]

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

 

[gneill]

You can do it either way. Your solution looks fine.

 

[shaltera]

why result is completely different compared with MatLAB

 

[gneill]

why result is completely different compared with MatLAB

I couldn't say; I haven't seen your Matlab work.

 

[shaltera]

I_L=I_N(Z_TH/Z_TH+R_L)
I_N=V_TH/Z_TH

FOR:
V_TH=33.79310-j5.51724
Z_TH=19.06896+j11.172413
R_L=50

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

Then

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

I_L=1.1278<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.1278<-39.6953(0.3159<21.1774)=0.3563<-18.5169

 

[gneill]

I_L=I_N(Z_TH/Z_TH+R_L)
I_N=V_TH/Z_TH

FOR:
V_TH=33.79310-j5.51724
Z_TH=19.06896+j11.172413
R_L=50

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

Then

I_N=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?

 

[shaltera]

That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?

sorry
I_N=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953

 

[gneill]

sorry
I_N=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.

 

[shaltera]

therefor

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

 

[gneill]

therefor

I_L=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° .

 

[shaltera]

interesting I calculate using a MatLAB, and that's what MatLAB gives me?

 

[shaltera]

fx>> A=33.79310-5.51724i
fx>> magn=abs(A)
fx>> magn=34.2405
fx>>ANGLE=angle(A)*180/pi
fx>>ANGLE=-9.2726

 

[shaltera]

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 :)

 

[gneill]

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.

 

[shaltera]

yes you are right angle is a bit off.And it has been corrected

 

[gneill]

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°.

 

[shaltera]

you are correct, I made a mistake.I copied the wrong value on the forum.I have 9.2726, and MatLAB confirmed it.Thanks

 

[gneill]

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