On 7/23/2015 7:58 AM, John S wrote:
On 7/22/2015 9:06 PM, rickman wrote:
On 7/22/2015 9:37 PM, John S wrote:
On 7/22/2015 8:23 PM, rickman wrote:
On 7/22/2015 8:00 PM, John S wrote:
On 7/10/2015 11:53 AM, John S wrote:

Generator----.---Xl---(1/4 wave 50+j0 line)-----.
| |
Xc 75+j0 load
| |

Assume ground in the obvious(?) places. Please let me know if I
need to
clarify.

1. The Generator produces 1W.
2. Xc is 0-j71.9
3. Xl is 0+j23.7
4. The line is loss-less.

Questions:

1. What is the power in the load?
2. What is the SWR at the load?
3. What is the SWR at the Generator?
4. What is the SWR in the line?
5. What impedance does the Generator see?
6. What impedance does the load see?

Answers:

1. 1W
2. In a 50 ohm environment it is 1.5:1
3. In a 50 ohm environment it is 1:1
4. In a 50 ohm environment it is 1.5:1
5. 50 ohms
6. Looking back into the line toward the generator, it is 0-j105 ohms.

Did I make a mistake anywhere?

I don't know. You didn't define the generator. Is it a zero ohm
voltage source? Is the Xc supposed to be the impedance of the
generator?


My apologies. It was meant to be a 0+j0 source impedance.

A voltage source with zero impedance will make Xc invisible.

Yes, when looking back at the generator from the junction of the
capacitor/inductor. The generator, however, sees the capacitor because
the generator supplies current to it.

The power to the load will be 1 W since there is nothing else that
dissipates power.

I knew you knew that.

I don't know the meaning of the question about SWR in the line. How is
that different from question 2?

Actually, it isn't. I probably should not have included it.

I don't know what impedance the generator sees, but I don't think it is
50 ohms real. Likewise I don't think the load sees a driving impedance
you state, but I don't know just what it is. Remember that Xc is
invisible since it is on a node with 0 ohms impedance to ground.
Consider that point to be shorted to ground for the purpose of
calculating source impedance at the load.

Xc is not invisible to the generator. The generator causes a current to
flow in the capacitor in accordance with the value of Xc.

Yes, the generator does supply current into the capacitor.


The 75 ohm load and 1/4 wavelength gets transformed to 33.33 ohms at the
input to the line. So now the generator sees:

------.--------Xl-----.
| |
Xc 33.33
| |

Doing the math yields 50.17+j0.66 for the given values.

As I was told in college, please show your work.

Rick

--

Rick

On 7/23/2015 10:10 AM, rickman wrote:
On 7/23/2015 7:58 AM, John S wrote:
On 7/22/2015 9:06 PM, rickman wrote:
On 7/22/2015 9:37 PM, John S wrote:
On 7/22/2015 8:23 PM, rickman wrote:
On 7/22/2015 8:00 PM, John S wrote:
On 7/10/2015 11:53 AM, John S wrote:

Generator----.---Xl---(1/4 wave 50+j0 line)-----.
| |
Xc 75+j0 load
| |

Assume ground in the obvious(?) places. Please let me know if I
need to
clarify.

1. The Generator produces 1W.
2. Xc is 0-j71.9
3. Xl is 0+j23.7
4. The line is loss-less.

Questions:

1. What is the power in the load?
2. What is the SWR at the load?
3. What is the SWR at the Generator?
4. What is the SWR in the line?
5. What impedance does the Generator see?
6. What impedance does the load see?

Answers:

1. 1W
2. In a 50 ohm environment it is 1.5:1
3. In a 50 ohm environment it is 1:1
4. In a 50 ohm environment it is 1.5:1
5. 50 ohms
6. Looking back into the line toward the generator, it is 0-j105
ohms.

Did I make a mistake anywhere?

I don't know. You didn't define the generator. Is it a zero ohm
voltage source? Is the Xc supposed to be the impedance of the
generator?


My apologies. It was meant to be a 0+j0 source impedance.

A voltage source with zero impedance will make Xc invisible.

Yes, when looking back at the generator from the junction of the
capacitor/inductor. The generator, however, sees the capacitor because
the generator supplies current to it.

The power to the load will be 1 W since there is nothing else that
dissipates power.

I knew you knew that.

I don't know the meaning of the question about SWR in the line. How is
that different from question 2?

Actually, it isn't. I probably should not have included it.

I don't know what impedance the generator sees, but I don't think it is
50 ohms real. Likewise I don't think the load sees a driving impedance
you state, but I don't know just what it is. Remember that Xc is
invisible since it is on a node with 0 ohms impedance to ground.
Consider that point to be shorted to ground for the purpose of
calculating source impedance at the load.

Xc is not invisible to the generator. The generator causes a current to
flow in the capacitor in accordance with the value of Xc.

Yes, the generator does supply current into the capacitor.


The 75 ohm load and 1/4 wavelength gets transformed to 33.33 ohms at the
input to the line. So now the generator sees:

+23.7
------.--------Xl-----.
| |
Xc=-71.9 33.33
| |

Doing the math yields 50.17+j0.66 for the given values.

As I was told in college, please show your work.

Rick

Okay. I assume you want numbers and not just algebra because I assume
you know how to work with complex variables.

Well, then we have the parallel connection of Xc and the series
connection of 33+j23.7. Yes? We will get the reciprocal of 33.33+j23.7
(admittance) and it is 0.020-j0.014. We will now add this to the
reciprocal of Xc (0-j71.9) which is 0+0.014. The sum is
0.020-j2.615e-004 (admittance) and to get the impedance we take the
reciprocal which is 50.174+j0.658 for the values given.

Does this help?

On 7/23/2015 11:04 AM, John S wrote:
On 7/23/2015 10:10 AM, rickman wrote:
On 7/23/2015 7:58 AM, John S wrote:
On 7/22/2015 9:06 PM, rickman wrote:
On 7/22/2015 9:37 PM, John S wrote:
On 7/22/2015 8:23 PM, rickman wrote:
On 7/22/2015 8:00 PM, John S wrote:
On 7/10/2015 11:53 AM, John S wrote:

Generator----.---Xl---(1/4 wave 50+j0 line)-----.
| |
Xc 75+j0 load
| |

Assume ground in the obvious(?) places. Please let me know if I
need to
clarify.

1. The Generator produces 1W.
2. Xc is 0-j71.9
3. Xl is 0+j23.7
4. The line is loss-less.

Questions:

1. What is the power in the load?
2. What is the SWR at the load?
3. What is the SWR at the Generator?
4. What is the SWR in the line?
5. What impedance does the Generator see?
6. What impedance does the load see?

Answers:

1. 1W
2. In a 50 ohm environment it is 1.5:1
3. In a 50 ohm environment it is 1:1
4. In a 50 ohm environment it is 1.5:1
5. 50 ohms
6. Looking back into the line toward the generator, it is 0-j105
ohms.

Did I make a mistake anywhere?

I don't know. You didn't define the generator. Is it a zero ohm
voltage source? Is the Xc supposed to be the impedance of the
generator?


My apologies. It was meant to be a 0+j0 source impedance.

A voltage source with zero impedance will make Xc invisible.

Yes, when looking back at the generator from the junction of the
capacitor/inductor. The generator, however, sees the capacitor because
the generator supplies current to it.

The power to the load will be 1 W since there is nothing else that
dissipates power.

I knew you knew that.

I don't know the meaning of the question about SWR in the line. How is
that different from question 2?

Actually, it isn't. I probably should not have included it.

I don't know what impedance the generator sees, but I don't think it is
50 ohms real. Likewise I don't think the load sees a driving impedance
you state, but I don't know just what it is. Remember that Xc is
invisible since it is on a node with 0 ohms impedance to ground.
Consider that point to be shorted to ground for the purpose of
calculating source impedance at the load.

Xc is not invisible to the generator. The generator causes a current to
flow in the capacitor in accordance with the value of Xc.

Yes, the generator does supply current into the capacitor.


The 75 ohm load and 1/4 wavelength gets transformed to 33.33 ohms at the
input to the line. So now the generator sees:

+23.7
------.--------Xl-----.
| |
Xc=-71.9 33.33
| |

Doing the math yields 50.17+j0.66 for the given values.

As I was told in college, please show your work.

Rick

Okay. I assume you want numbers and not just algebra because I assume
you know how to work with complex variables.

Well, then we have the parallel connection of Xc and the series
connection of 33+j23.7.

I probably stated that incorrectly. I will try again...

we have the connection of Xc in parallel with the series
connection of 33+j23.7.

On 7/23/2015 12:04 PM, John S wrote:
On 7/23/2015 10:10 AM, rickman wrote:
On 7/23/2015 7:58 AM, John S wrote:
On 7/22/2015 9:06 PM, rickman wrote:
On 7/22/2015 9:37 PM, John S wrote:
On 7/22/2015 8:23 PM, rickman wrote:
On 7/22/2015 8:00 PM, John S wrote:
On 7/10/2015 11:53 AM, John S wrote:

Generator----.---Xl---(1/4 wave 50+j0 line)-----.
| |
Xc 75+j0 load
| |

Assume ground in the obvious(?) places. Please let me know if I
need to
clarify.

1. The Generator produces 1W.
2. Xc is 0-j71.9
3. Xl is 0+j23.7
4. The line is loss-less.

Questions:

1. What is the power in the load?
2. What is the SWR at the load?
3. What is the SWR at the Generator?
4. What is the SWR in the line?
5. What impedance does the Generator see?
6. What impedance does the load see?

Answers:

1. 1W
2. In a 50 ohm environment it is 1.5:1
3. In a 50 ohm environment it is 1:1
4. In a 50 ohm environment it is 1.5:1
5. 50 ohms
6. Looking back into the line toward the generator, it is 0-j105
ohms.

Did I make a mistake anywhere?

I don't know. You didn't define the generator. Is it a zero ohm
voltage source? Is the Xc supposed to be the impedance of the
generator?


My apologies. It was meant to be a 0+j0 source impedance.

A voltage source with zero impedance will make Xc invisible.

Yes, when looking back at the generator from the junction of the
capacitor/inductor. The generator, however, sees the capacitor because
the generator supplies current to it.

The power to the load will be 1 W since there is nothing else that
dissipates power.

I knew you knew that.

I don't know the meaning of the question about SWR in the line. How is
that different from question 2?

Actually, it isn't. I probably should not have included it.

I don't know what impedance the generator sees, but I don't think it is
50 ohms real. Likewise I don't think the load sees a driving impedance
you state, but I don't know just what it is. Remember that Xc is
invisible since it is on a node with 0 ohms impedance to ground.
Consider that point to be shorted to ground for the purpose of
calculating source impedance at the load.

Xc is not invisible to the generator. The generator causes a current to
flow in the capacitor in accordance with the value of Xc.

Yes, the generator does supply current into the capacitor.


The 75 ohm load and 1/4 wavelength gets transformed to 33.33 ohms at the
input to the line. So now the generator sees:

+23.7
------.--------Xl-----.
| |
Xc=-71.9 33.33
| |

Doing the math yields 50.17+j0.66 for the given values.

As I was told in college, please show your work.

Rick

Okay. I assume you want numbers and not just algebra because I assume
you know how to work with complex variables.

Well, then we have the parallel connection of Xc and the series
connection of 33+j23.7. Yes? We will get the reciprocal of 33.33+j23.7
(admittance) and it is 0.020-j0.014. We will now add this to the
reciprocal of Xc (0-j71.9) which is 0+0.014. The sum is
0.020-j2.615e-004 (admittance) and to get the impedance we take the
reciprocal which is 50.174+j0.658 for the values given.

Does this help?

Yes, I had to look up the multiplicative inverse of a complex number.
It's been a number of years since I've done much with complex math.
Your numbers all look good. For now I'll have to take your word for the
transformation of the 75 ohm load to 33 ohms. I'll check Jim's
reference later. Right now I need to deal with some real estate things
and a PO a customer is trying to give me.

--

Rick

On 7/23/2015 2:03 PM, rickman wrote:
On 7/23/2015 12:04 PM, John S wrote:
On 7/23/2015 10:10 AM, rickman wrote:
On 7/23/2015 7:58 AM, John S wrote:
On 7/22/2015 9:06 PM, rickman wrote:
On 7/22/2015 9:37 PM, John S wrote:
On 7/22/2015 8:23 PM, rickman wrote:
On 7/22/2015 8:00 PM, John S wrote:
On 7/10/2015 11:53 AM, John S wrote:

Generator----.---Xl---(1/4 wave 50+j0 line)-----.
| |
Xc 75+j0 load
| |

Assume ground in the obvious(?) places. Please let me know if I
need to
clarify.

1. The Generator produces 1W.
2. Xc is 0-j71.9
3. Xl is 0+j23.7
4. The line is loss-less.

Questions:

1. What is the power in the load?
2. What is the SWR at the load?
3. What is the SWR at the Generator?
4. What is the SWR in the line?
5. What impedance does the Generator see?
6. What impedance does the load see?

Answers:

1. 1W
2. In a 50 ohm environment it is 1.5:1
3. In a 50 ohm environment it is 1:1
4. In a 50 ohm environment it is 1.5:1
5. 50 ohms
6. Looking back into the line toward the generator, it is 0-j105
ohms.

Did I make a mistake anywhere?

I don't know. You didn't define the generator. Is it a zero ohm
voltage source? Is the Xc supposed to be the impedance of the
generator?


My apologies. It was meant to be a 0+j0 source impedance.

A voltage source with zero impedance will make Xc invisible.

Yes, when looking back at the generator from the junction of the
capacitor/inductor. The generator, however, sees the capacitor because
the generator supplies current to it.

The power to the load will be 1 W since there is nothing else that
dissipates power.

I knew you knew that.

I don't know the meaning of the question about SWR in the line.
How is
that different from question 2?

Actually, it isn't. I probably should not have included it.

I don't know what impedance the generator sees, but I don't think
it is
50 ohms real. Likewise I don't think the load sees a driving
impedance
you state, but I don't know just what it is. Remember that Xc is
invisible since it is on a node with 0 ohms impedance to ground.
Consider that point to be shorted to ground for the purpose of
calculating source impedance at the load.

Xc is not invisible to the generator. The generator causes a current to
flow in the capacitor in accordance with the value of Xc.

Yes, the generator does supply current into the capacitor.


The 75 ohm load and 1/4 wavelength gets transformed to 33.33 ohms at
the
input to the line. So now the generator sees:

+23.7
------.--------Xl-----.
| |
Xc=-71.9 33.33
| |

Doing the math yields 50.17+j0.66 for the given values.

As I was told in college, please show your work.

Rick

Okay. I assume you want numbers and not just algebra because I assume
you know how to work with complex variables.

Well, then we have the parallel connection of Xc and the series
connection of 33+j23.7. Yes? We will get the reciprocal of 33.33+j23.7
(admittance) and it is 0.020-j0.014. We will now add this to the
reciprocal of Xc (0-j71.9) which is 0+0.014. The sum is
0.020-j2.615e-004 (admittance) and to get the impedance we take the
reciprocal which is 50.174+j0.658 for the values given.

Does this help?

Yes, I had to look up the multiplicative inverse of a complex number.
It's been a number of years since I've done much with complex math. Your
numbers all look good. For now I'll have to take your word for the
transformation of the 75 ohm load to 33 ohms. I'll check Jim's
reference later.

Sorry, I should have told you about the 1/4 wave transformation Zi =
Zo^2/Zl which, in this case is 50^2/75 or 2500/75 making Zi = 33.33.

To be complete, I should include all the imaginary parts, but in this
case all are zero and their inclusion makes it harder to read.