phasor to complex

W

Wayne

Guest
Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
 
Homework ?



Wayne wrote:
Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
 
On Sat, 16 Oct 2004 17:51:27 GMT, "Wayne" <nospam-mail@wlawson.com>
wrote:

Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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.
Click to expand...
Use a Smith Chart. Piece of cake unless your teacher wants you to show
an absolute result with working.
--

"What is now proved was once only imagin'd." - William Blake, 1793.
 
Wayne wrote:
Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
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.
 
"Jan-Erik Söderholm" <aaa@aaa.com>
Wayne wrote:

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.
Homework ?
Click to expand...
Homework is $20 a question, $125/hour. Paypal or bank check ...

--
Nicholas O. Lindan, Cleveland, Ohio
Consulting Engineer: Electronics; Informatics; Photonics.
Remove spaces etc. to reply: n o lindan at net com dot com
 
"Wayne" <nospam-mail@wlawson.com> wrote in message
news:zcdcd.863$RL3.256@newsfe2-gui.ntli.net...
Hi

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

----------------
Click to expand...
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.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
 
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
"Wayne" <nospam-mail@wlawson.com> wrote in message
news:zcdcd.863$RL3.256@newsfe2-gui.ntli.net...
Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
 
Jan-Erik Söderholm wrote:
Homework ?

Wayne wrote:

Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
Prolly; and he was asleep in simple Geometry...
 
Yes, by subtraction.

Wayne
"peterken" <peter273@hotmail.com> wrote in message
news:XQAcd.281594$7k1.14337067@phobos.telenet-ops.be...
have you taken the capacitance / resistance of the measurment tool into
account ?
at low C or high R they are important too....


"Glenn" <this@isnt.it> wrote in message
news:dEwcd.17533$54.300468@typhoon.sonic.net...

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



Wayne wrote:
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

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
Click to expand...
 
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
dhky@peeshaw.ca
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.
 
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41730892.9060704@nospam.com...
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
dhky@peeshaw.ca
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.
Click to expand...
He has a magnitude and a phase angle listed. I suspect that the phase angle
is that of the voltage across the "capacitor" with respect to that across
the resistor (i.e. with respect to the current.) I don't recall him any
impedance values- just the voltages.
theta1=80deg
Vs=1v
Vr=0.25v
I asked Wayne for clarification as to what is what he has adrawn two Vs
Click to expand...
values and doesn't specify which voltage is associated with theta1. so the
problem is:
a) 1 @ 80 =0.25+jVc assuming a pure capacitor - this doesn't work as has
been pointed out.
b) 1 @ ?? =0.25 +|Vc| @80 does have a valid solution for |Vc| and for ??

Take your choice.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
 
Don Kelly wrote:
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41730892.9060704@nospam.com...


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
dhky@peeshaw.ca
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.


He has a magnitude and a phase angle listed. I suspect that the phase angle
is that of the voltage across the "capacitor" with respect to that across
the resistor (i.e. with respect to the current.) I don't recall him any
impedance values- just the voltages.
theta1=80deg

Vs=1v
Vr=0.25v

I asked Wayne for clarification as to what is what he has adrawn two Vs
values and doesn't specify which voltage is associated with theta1. so the
problem is:
a) 1 @ 80 =0.25+jVc assuming a pure capacitor - this doesn't work as has
been pointed out.
b) 1 @ ?? =0.25 +|Vc| @80 does have a valid solution for |Vc| and for ??

Take your choice.
Click to expand...
Right- well he does know R- so he measures magnitude and phase of Vr wrt
Vs and then I=Vr/R.
 
Wayne wrote:

Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
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
 
have you taken the capacitance / resistance of the measurment tool into
account ?
at low C or high R they are important too....


"Glenn" <this@isnt.it> wrote in message
news:dEwcd.17533$54.300468@typhoon.sonic.net...

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



Wayne wrote:
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

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
Click to expand...
 
--
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41719BBF.3000001@nospam.com...
Wayne wrote:
Hi

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

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

Don Kelly
dhky@peeshaw.ca
remove the urine to answer
 
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



Wayne wrote:
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

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
Click to expand...
 
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)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vr Vc
Ś
---
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!!!!!!!!!!!


"Wayne" <nospam-mail@wlawson.com> wrote in message
news:zcdcd.863$RL3.256@newsfe2-gui.ntli.net...
Hi

I have calculated the following network using the cosine law.
___SIGGEN (Vs)_
Ś Ś
Ś--/\/\/\/\-----Ś Ś----Ś
Ś Vs Vc
Ś
---
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
Click to expand...
 
"Theta1 would be the angle between Vr and Vc and this would be 90o"

ONLY 90o at resonance frequency remember....


"Don Kelly" <dhky@peeshaw.ca> wrote in message
news:chmcd.759540$gE.629871@pd7tw3no...


--
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41719BBF.3000001@nospam.com...
Wayne wrote:
Hi

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

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

Don Kelly
dhky@peeshaw.ca
remove the urine to answer
 
Wayne wrote:
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
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41730892.9060704@nospam.com...


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
dhky@peeshaw.ca
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.

Wa
Click to expand...
If you are modeling an electrochemical cell, i can guarantee losses...
 
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