LRC omplex Numbers (Try #2)

J

jalbers@bsu.edu

Guest
When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over correctly
or was omitted. Below is what I intended for everyone to see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j – 159j
Z = 50 – 96.2j or 108.417@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

I = V/Z or
I = 1@45 deg / 108.417@-62.53 deg = .0092@107.53 deg or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.

The voltage across R:
(50@0 deg) * (.0092@107.53 deg) = .46@107.53 deg or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage
source is at .707 volts, the voltage across R is at -.138 volts.

The voltage across L:
(62.8@90 deg) * (.0092@107.53 deg) = .577@197.53 deg or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage
source is at .707 volts, the voltage across L is at -.550 volts.

The voltage across C:
(159@-90) * (.0092@107.53 deg) = 1.4628@17.53 deg or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage
source is at .707 volts, the voltage across L is at 1.394 volts.

-----

Any comments or corrections would be greatly appreciated. Thanks
 
<jalbers@bsu.edu> wrote in message
news:8341e1c0-fdde-48ec-b452-f0e0889302d5@t54g2000hsg.googlegroups.com
When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over correctly
or was omitted. Below is what I intended for everyone to see.
Click to expand...
Check your numbers for the inductor.
 
jalbers@bsu.edu wrote:
When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over correctly
or was omitted. Below is what I intended for everyone to see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j – 159j
Z = 50 – 96.2j or 108.417@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

I = V/Z or
I = 1@45 deg / 108.417@-62.53 deg = .0092@107.53 deg or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.

The voltage across R:
(50@0 deg) * (.0092@107.53 deg) = .46@107.53 deg or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage
source is at .707 volts, the voltage across R is at -.138 volts.

The voltage across L:
(62.8@90 deg) * (.0092@107.53 deg) = .577@197.53 deg or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage
source is at .707 volts, the voltage across L is at -.550 volts.

The voltage across C:
(159@-90) * (.0092@107.53 deg) = 1.4628@17.53 deg or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage
source is at .707 volts, the voltage across L is at 1.394 volts.

-----

Any comments or corrections would be greatly appreciated. Thanks
Click to expand...
You should not be looking at just the real part of the voltage across
each part. You're using _complex_ currents, and _complex_ impedances;
why should the voltages not be complex also?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
<jalbers@bsu.edu> wrote in message
news:22836028-c0b3-47fb-b044-b1f04ca93468@z66g2000hsc.googlegroups.com
On Jun 23, 9:02 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:8341e1c0-fdde-48ec-b452-f0e0889302d5@t54g2000hsg.googlegroups.com

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted. Below is what I intended for everyone to
see.

Check your numbers for the inductor.


For some reason all of my complex numbers are bing treated as email
addresses. The numbers informt of the @ are being turned into ... .
I have replaced the @ symbols with an & symbol so that this will not
happen again. Everwhere yo see an & it should be an @ symbol.

Can you give me more of a hint as to what is wrong with the inductor.
Is the error in the instantaneous voltage across the inductor, the
value of the inductor, or the XL calculation?
Click to expand...
The voltage angle looks wrong. I make it out to be about
-160 degreees (I'll let you calculate an accurate value).
 
On Mon, 23 Jun 2008 05:45:35 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over correctly
or was omitted. Below is what I intended for everyone to see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j – 159j
Z = 50 – 96.2j or 108.417@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .
Click to expand...
Sounds like the source is 1 volt peak, not 0.707.


I = V/Z or
I = 1@45 deg / 108.417@-62.53 deg = .0092@107.53 deg or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.
Click to expand...
The RLC doesn't see "the real part" of the voltage, they see the real
voltage, namely 1 volt peak. The load has no idea that you consider
the applied voltage to be at 45 degrees, since it has no way of
knowing your time reference.

The same AC voltage can be named 1 at 0 degrees, 0.707+j0.707, or
0+j1.00; it just depends on where you're standing. A voltmeter will
read exactly the same for all cases, so the RLC can't tell the
difference. If you define the voltage source as having a non-zero
angle, you can *interpret* the resulting complex load current as
different angles, but the final magnitudes must be the same.

John
 
"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message news:9kfv54ladshberls4jmfdq762nq3ctlece@4ax.com
On Mon, 23 Jun 2008 05:45:35 -0700 (PDT), "jalbers@bsu.edu"
jalbers@bsu.edu> wrote:

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted. Below is what I intended for everyone to
see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j  159j
Z = 50  96.2j or 108.417@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

Sounds like the source is 1 volt peak, not 0.707.
Click to expand...
1V*(cos(q) + j*sin(q)) = (0.707 + j*0.707)V for q = 45 deg

It seems that he's looking for the instantaneous voltage
drops across the passive components when the driving
voltage is 45 degrees into its cycle.
 
On Jun 23, 9:02 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:8341e1c0-fdde-48ec-b452-f0e0889302d5@t54g2000hsg.googlegroups.com

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over correctly
or was omitted.  Below is what I intended for everyone to see.

Check your numbers for the inductor.
Click to expand...

For some reason all of my complex numbers are bing treated as email
addresses. The numbers informt of the @ are being turned into ... .
I have replaced the @ symbols with an & symbol so that this will not
happen again. Everwhere yo see an & it should be an @ symbol.

Can you give me more of a hint as to what is wrong with the inductor.
Is the error in the instantaneous voltage across the inductor, the
value of the inductor, or the XL calculation?

-----

Z = 50 + 62.8j – 159j
Z = 50 – 96.2j or 108.417&-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1&45 deg = .707 + .707j .

I = V/Z or
I = 1&45 deg / 108.417&-62.53 deg = .0092&107.53 deg or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage
source is at .707 volts, the current through L, R, and C is at -.0027
amps.

The voltage across R:
(50&0 deg) * (.0092&107.53 deg) = .46&107.53 deg or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage
source is at .707 volts, the voltage across R is at -.138 volts.

The voltage across L:
(62.8&90 deg) * (.0092&107.53 deg) = .577&197.53 deg or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage
source is at .707 volts, the voltage across L is at -.550 volts.

The voltage across C:
(159&-90) * (.0092&107.53 deg) = 1.4628&17.53 deg or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage
source is at .707 volts, the voltage across L is at 1.394 volts.
 
<jalbers@bsu.edu> wrote in message
news:06f5cdcd-29f5-4cb4-a3d2-6c2c301dbc1d@34g2000hsh.googlegroups.com
On Jun 23, 11:10 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:22836028-c0b3-47fb-b044-b1f04ca93468@z66g2000hsc.googlegroups.com





On Jun 23, 9:02 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:8341e1c0-fdde-48ec-b452-f0e0889302d5@t54g2000hsg.googlegroups.com

When I used cut and paste from a word processor on the first
post, I ddin't realize that some of the information didn't move
over correctly or was omitted. Below is what I intended for
everyone to see.

Check your numbers for the inductor.

For some reason all of my complex numbers are bing treated as email
addresses. The numbers informt of the @ are being turned into ...
. I have replaced the @ symbols with an & symbol so that this will
not happen again. Everwhere yo see an & it should be an @ symbol.

Can you give me more of a hint as to what is wrong with the
inductor. Is the error in the instantaneous voltage across the
inductor, the value of the inductor, or the XL calculation?

The voltage angle looks wrong. I make it out to be about
-160 degreees (I'll let you calculate an accurate value).- Hide
quoted text -

- Show quoted text -

+197.53 deg is the same as 162.47 deg which is close to your -160 deg.
Click to expand...
Okay.

Am I correct in regards to just looking at the real part of the
voltages if I am interested in an instantaneous voltages across X, L,
and R, since my calculations are based on the instance when Vsource > . 707V = 1&45 deg = .707 + .707j . Not that doing so has any
practical value. I just can't seem to find a "real" scope or DVOM
that displays imaginary numbers :)
Click to expand...
Sure.
 
On Jun 23, 11:51 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"John Larkin" <jjlar...@highNOTlandTHIStechnologyPART.com> wrote in
messagenews:9kfv54ladshberls4jmfdq762nq3ctlece@4ax.com





On Mon, 23 Jun 2008 05:45:35 -0700 (PDT), "jalb...@bsu.edu"
jalb...@bsu.edu> wrote:

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted.  Below is what I intended for everyone to
see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions.  I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j   159j
Z = 50   96.2j or 108....@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

Sounds like the source is 1 volt peak, not 0.707.

1V*(cos(q) + j*sin(q)) = (0.707 + j*0.707)V  for q = 45 deg

It seems that he's looking for the instantaneous voltage
drops across the passive components when the driving
voltage is 45 degrees into its cycle.- Hide quoted text -

- Show quoted text -
Click to expand...
Yes, that is "exactly" what I am trying to do. Can this be done the
way I am attempting to go about it?
 
On Jun 23, 11:10 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:22836028-c0b3-47fb-b044-b1f04ca93468@z66g2000hsc.googlegroups.com





On Jun 23, 9:02 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
jalb...@bsu.edu> wrote in message

news:8341e1c0-fdde-48ec-b452-f0e0889302d5@t54g2000hsg.googlegroups.com

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted.  Below is what I intended for everyone to
see.

Check your numbers for the inductor.

For some reason all of my complex numbers are bing treated as email
addresses.   The numbers informt of the @ are being turned into ... .
I have replaced the @ symbols with an & symbol so that this will not
happen again.  Everwhere yo see an & it should be an @ symbol.

Can you give me more of a hint as to what is wrong with the inductor.
Is the error in the instantaneous voltage across the inductor, the
value of the inductor, or the XL calculation?

The voltage angle looks wrong.  I make it out to be about
-160 degreees (I'll let you calculate an accurate value).- Hide quoted text -

- Show quoted text -
Click to expand...
+197.53 deg is the same as 162.47 deg which is close to your -160 deg.

Am I correct in regards to just looking at the real part of the
voltages if I am interested in an instantaneous voltages across X, L,
and R, since my calculations are based on the instance when Vsource = .
707V = 1&45 deg = .707 + .707j . Not that doing so has any practical
value. I just can't seem to find a "real" scope or DVOM that displays
imaginary numbers :)
 
On Mon, 23 Jun 2008 10:24:05 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:

On Jun 23, 11:51 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"John Larkin" <jjlar...@highNOTlandTHIStechnologyPART.com> wrote in
messagenews:9kfv54ladshberls4jmfdq762nq3ctlece@4ax.com





On Mon, 23 Jun 2008 05:45:35 -0700 (PDT), "jalb...@bsu.edu"
jalb...@bsu.edu> wrote:

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted.  Below is what I intended for everyone to
see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions.  I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j   159j
Z = 50   96.2j or 108....@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

Sounds like the source is 1 volt peak, not 0.707.

1V*(cos(q) + j*sin(q)) = (0.707 + j*0.707)V  for q = 45 deg

It seems that he's looking for the instantaneous voltage
drops across the passive components when the driving
voltage is 45 degrees into its cycle.- Hide quoted text -

- Show quoted text -

Yes, that is "exactly" what I am trying to do. Can this be done the
way I am attempting to go about it?
Click to expand...
Yes, but you can't throw away the imaginary part of the voltage, just
because it's imaginary!

John
 
On Jun 23, 9:09 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 23 Jun 2008 10:24:05 -0700 (PDT), "jalb...@bsu.edu"





jalb...@bsu.edu> wrote:
On Jun 23, 11:51 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"John Larkin" <jjlar...@highNOTlandTHIStechnologyPART.com> wrote in
messagenews:9kfv54ladshberls4jmfdq762nq3ctlece@4ax.com

On Mon, 23 Jun 2008 05:45:35 -0700 (PDT), "jalb...@bsu.edu"
jalb...@bsu.edu> wrote:

When I used cut and paste from a word processor on the first post, I
ddin't realize that some of the information didn't move over
correctly or was omitted.  Below is what I intended for everyone to
see.

-----

I am trying to understand XLR circuits and the use of complex numbers
to find solutions.  I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more
experience took a look at what I am doing.

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j   159j
Z = 50   96.2j or 108....@-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when
the power source is at .707V = 1@45 deg = .707 + .707j .

Sounds like the source is 1 volt peak, not 0.707.

1V*(cos(q) + j*sin(q)) = (0.707 + j*0.707)V  for q = 45 deg

It seems that he's looking for the instantaneous voltage
drops across the passive components when the driving
voltage is 45 degrees into its cycle.- Hide quoted text -

- Show quoted text -

Yes, that is "exactly" what I am trying to do. Can this be done the
way I am attempting to go about it?

Yes, but you can't throw away the imaginary part of the voltage, just
because it's imaginary!

John- Hide quoted text -

- Show quoted text -
Click to expand...
It sure is funny how the sum of the instataneous voltage drops (just
the "real" parts) any place in the AC cycle in my calculations always
adds up to the source which is a good thing because Kirchoff's voltage
law holds up.

Let V the voltage source be 1&x deg where 0<x<360 .
Let VR, VL, VC be the "instantaneous" voltage drops across R, L, and
C.

I = 1&x deg / 108.417&-62.53 deg

VL = [62.8&90 deg]*[.0092&(x+62.53) deg] = .57776&(x+152.53) deg
VC = [159&-90 deg]*[.0092&(x+62.53) deg] = 1.4628&(x-27.47) deg
VR = [50&0 deg]*[.0092&(x+62.53) deg] = .46&(x+62.53) deg

Taking just the "real" parts of VL, VC, and VR gives

Re[VL] = .5776 * Cos(x+152.53)
Re[VC] = 1.4628 * Cos(x-27.47)
Re[VR] = .46 * Cos(x+62.53)

It sure is a big coincidence that the graph of f(x) = .5776 * Cos(x
+152.53) + 1.4628 * Cos(x-27.47) + .46 * Cos(x+62.53) compares exactly
to g(x) = Cos(x) which is the power source.
 
jalbers@bsu.edu wrote:

It sure is funny how the sum of the instataneous voltage drops (just
the "real" parts) any place in the AC cycle in my calculations always
adds up to the source which is a good thing because Kirchoff's voltage
law holds up.
Click to expand...
Not funny at all, but exactly as expected. The sum of all
the imaginary voltages in the RLC also add up to the
imaginary part of the source. That is because the sum of
all the voltage around any loop has to be zero, regardless
of whether that voltage is expressed as a sum of components
at right angles (real and imaginary) or by any other
description.

Let V the voltage source be 1&x deg where 0<x<360 .
Let VR, VL, VC be the "instantaneous" voltage drops across R, L, and
C.

I = 1&x deg / 108.417&-62.53 deg

VL = [62.8&90 deg]*[.0092&(x+62.53) deg] = .57776&(x+152.53) deg
VC = [159&-90 deg]*[.0092&(x+62.53) deg] = 1.4628&(x-27.47) deg
VR = [50&0 deg]*[.0092&(x+62.53) deg] = .46&(x+62.53) deg

Taking just the "real" parts of VL, VC, and VR gives

Re[VL] = .5776 * Cos(x+152.53)
Re[VC] = 1.4628 * Cos(x-27.47)
Re[VR] = .46 * Cos(x+62.53)

It sure is a big coincidence that the graph of f(x) = .5776 * Cos(x
+152.53) + 1.4628 * Cos(x-27.47) + .46 * Cos(x+62.53) compares exactly
to g(x) = Cos(x) which is the power source.
Click to expand...
Now sum the imaginary components, then do the complex math
to sum the whole voltages around the loop. All versions
have to add up to zero.



--
Regards,

John Popelish
 
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