# phasor to complex

Discussion in 'Electronic Design' started by Wayne, Oct 16, 2004.

1. ### WayneGuest

Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
¦ ¦
¦--/\/\/\/\-----¦ ¦----¦
¦ Vs Vc
¦
---
GND

Vs^2=Vr^2+Vc^2 - 2VrVc Cos(theta1)
theta1=80deg
Vs=1v
Vr=0.25v

I am being dumb... but how do I get this into complex format e.g. a+jb.
Please give an example using my data.

Thanks

Wayne

2. ### ChrisGibboGibsonGuest

You could have at least made an attempt to pretend it wasn't your homework.

Gibbo

3. ### Glenn ElmoreGuest

Looks to me like a pretty bad homework problem if that's what it is. It
seems to be overspecified.

A source of 1 volt which develops 0.25V across a resistor connected to
any ideal capacitor can not yield an 80 degree phase shift. The current
is common to the two and the angle between Vr and Vc has to be 90
degrees, by definition. Unless I scribbled wrong, that means that |Vc| =
..968V and the angle = arctan (.25/.968)=14.5 degrees. The remaining
angle is the complement of this, 75.5 degrees.

I don't know where theta1 came from, but no angle in this situation is
80 deg. If you force the phase shift across the capacitor to be 80
degrees, you don't get 0.25V across the resistor. It takes a resistance
significantly higher in value than the reactance of the capacitor to
achieve this much phase shift. Such a resistor will have the bulk of the
voltage drop and much more than .25V across it.

I'll leave the details of that for a homework excercise.

Glenn

Gibbo wrote:

Homework ?

5. ### Paul BurridgeGuest

Use a Smith Chart. Piece of cake unless your teacher wants you to show
an absolute result with working.

6. ### Fred BloggsGuest

Theta1 would be the angle between Vr and Vc and this would be 90o. Vr
and Vc form a right angle. So there is no cosine term. The voltage
divider rule still applies with z=r-jx so that Vr=Vs*r/(r-jx) and
Vc=Vs*(-jx)/(r-jx) then use 1/(r-jx)=(r+jx)/(r^2+x^2) and Vs=Vr+Vc to
get the "format" you want.

7. ### Nicholas O. LindanGuest

Homework is \$20 a question, \$125/hour. Paypal or bank check ...

8. ### Don KellyGuest

----------------
You don't. It's not worth the bother and teaches you nothing.

theta1 is the angle of which voltage? With respect to what? I am guessing
that it is the phase of Vc with respect to the voltage Vr (check sign of
angle )

Look at the circuit- apply KVL in phasor form. ---done.

9. ### Don KellyGuest

--
How do you know that the "capacitor" doesn't have some loss component? I
suspect that this is the case.

Don Kelly

remove the urine to answer

10. ### tim gormanGuest

Wayne,

I'm not sure what this is supposed to represent. I think you meant to use
Vr under the resistor, not Vs. So it looks like you have a signal generator,
Vs, feeding a series combination of a resistor and a capacitor.

For what you are doing I am assuming the real part is along the x-axis and
the imaginary part is along the y-axis. That is the usual orientation for
the complex plane in doing electronics where Z = R + jX (X is positive for
inductive impedance and negative for capacitive impedance)

This gives an impedance with R = |Z|cos(theta) and X = |Z|sin(theta)

So you can write |Z|cos(theta)+j|Z|sin(theta) as the complex impedance.

|Z| = sqrt(R^2 + X^2) it also follows that tan(theta) = X/R

Since Vr = RVs/(R^2 + X^2)^(1/2) and you know Vr/Vs = .25 you can figure out
what X is.

R/(R^2 + X^2)^(1/2) = .25, R = .25(R^2 + X^2)^(1/2),
Squaring both sides you get R^2 = (1/16) (R^2 + X^2)
Solving for X^2 gives X^2 = 15R^2. So X = (R)(sqrt 15)

This gives a theta of 75deg not 80deg.

On the other hand, if your angle is correct, it looks like you should have
measured about .17v not .25v.

tim

11. ### WayneGuest

This is a psudo capasitace. It is an electrochemical cell and I am
simulating it as a RC network..

Some heavy home work!!!!!!!!!!!

Cheers

Wayne

12. ### Robert BaerGuest

Prolly; and he was asleep in simple Geometry...

13. ### GlennGuest

In that case I stand by my original posting but suggest that
a) your measurements are not correct
b) your model is not sufficient

Glenn

is.It > seems to be overspecified.
to > any ideal capacitor can not yield an 80 degree phase shift. The

14. ### WayneGuest

Tim

Your write, sorry, mis-place an 's'. The circuit should read:

Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
¦ ¦
¦--/\/\/\/\-----¦ ¦----¦
¦ Vr Vc
¦
---
GND

Vs^2=Vr^2+Vc^2 - 2VrVc Cos(theta1)
theta1=80deg
Vs=1v
Vr=0.25v

This is a psudo capasitace. It is an electrochemical cell and I am
simulating it as a RC network..

Some heavy home work!!!!!!!!!!!

15. ### peterkenGuest

"Theta1 would be the angle between Vr and Vc and this would be 90o"

ONLY 90o at resonance frequency remember....

--
How do you know that the "capacitor" doesn't have some loss component? I
suspect that this is the case.

Don Kelly

remove the urine to answer

16. ### peterkenGuest

have you taken the capacitance / resistance of the measurment tool into
account ?
at low C or high R they are important too....

In that case I stand by my original posting but suggest that
a) your measurements are not correct
b) your model is not sufficient

Glenn

is.It > seems to be overspecified.
to > any ideal capacitor can not yield an 80 degree phase shift. The

17. ### ChrisGibboGibsonGuest

What ?

Could you clarify that statement. In the hope that in doing so you will realise
how stupid it is and save someone else the bother of telling you.

Gibbo

18. ### WayneGuest

Yes, by subtraction.

Wayne

19. ### Fred BloggsGuest

Don Kelly wrote:

How do you know that the "capacitor" doesn't have some loss component? I
suspect that this is the case.

Don Kelly

remove the urine to answer

If that is the case then he uses Z=Vs/I which is again a measured phasor
for I- he will not be able to get away with just magnitude measurement.
Then he uses Zunknown=Z-R and there you have it. Zunknown= Ru-jXu.

20. ### WayneGuest

Don
I don't. I am going to try and work out what my electrochemical
cell looks like in terms of a RC network.

Wayne
Wa