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