# Help me with Smith chart question

Status
Not open for further replies.

#### elvis0206

##### Junior Member level 1
The question may be a bit long....hope you guys can help me to solve the problem
Thanks very much.

A 2-GHz generator with internal impedance Zg=50ohm is connected to a load with impedance 100ohm. The reflection between the load and the generator is eliminated when the internal impedance equals the load impedance. To achieve this condition without changing Zg, the effective load impendance can be modified by adding a section of transmission line with length l and then a series open circuited line with length d.

Use Smith Chart to find length l such that the input resistance across A-A' is 50ohm?

Last edited:

#### WimRFP

Re: Smith Chart question

Hello,

I assume you understand the concept of the smith chart with the constant Re(Z) circles and constant Im(Z) circles. Note that when you add a line length of 10 degrees, the angle of RC changes -20 degrees (as both forward and reflected waves change).

100 Ohms is on the right of the center of the smith chart. When you increase "l", the RC travels clockwise (so the impedance gets a capacitive component). make "l" such that the impedance is on the circle with Re(Z) = 50 Ohms (that is 1 Ohm normalized).

Now your impedance has the form Z = 50 + jx. Depending on the length, jx can be capacitive or inductive. If you go for the inductive option (so your impedance is in the upper half of the smith chart), you need a capacitive series impedance that cancels the inductive component.

Your drawing shows an "open" line, so take a new chart and start fully right on the smith chart (RC = +1). Now increase the series stub length. The RC of the series stub travels clockwise at the edge of the chart. Make the length such that the impedance of the stub is on the required constant Im(Z) curve.

This stub in series with Z and cancels the jx component so the input impedance of your circuit is now 50 + j0 Ohms.

I hope this helps you.

Wim

elvis0206

V
Points: 2

### elvis0206

Points: 2

#### elvis0206

##### Junior Member level 1
Re: Smith Chart question

Wim, thanks for your help. But I still have some questions want to ask.

Does RC mean the REFLECTION COEFFICIENT and I'm not understand why it travels clockwise? If there is traveling clockwise or counter-clockwise on the Smith Chart, it must be the load which is moved toward or backward to the generator.

In the third paragraph, do you mean the capacitive and inductive should be in series so that they can cancel each other?

For the capacitive and inductive, how we can know their value such that we can find the length "l" of the transmission line?

When I use the Smith Chart to find the length "l" of the transmission line, I located the load impedance(2 ohm normalized) and then located the input resistance(1 ohm normalized). However, I don't know the imaginary part of Z so that I fail to find the length "l" of the transmission line.

Last edited:

elvis0206

### elvis0206

Points: 2

#### elvis0206

##### Junior Member level 1
Re: Smith Chart question

Hi biff44,
this is not a homework but an extra question for us to discuss after the lesson. Since I'm not good at the topic, so I may ask in a detailed way...Haha
Thanks very much! =]

#### WimRFP

Re: Smith Chart question

Wim, thanks for your help. But I still have some questions want to ask.

Does RC mean the REFLECTION COEFFICIENT and I'm not understand why it travels clockwise?

Yes, this is correct

If there is traveling clockwise or counter-clockwise on the Smith Chart, it must be the load which is moved toward or backward to the generator.

If you start with zero length (so you look directly into the load), RC=+0.333 (for 100 Ohms). When you add an increasing length of transmission line and you look into the combination of load and line, |RC| will not change, but the phase does. The phase of RC ( arg(RC) ) rotates clockwise (so will go in the capacitive region of the chart). Further length increase will move RC to the inductive part of the chart.

In the third paragraph, do you mean the capacitive and inductive should be in series so that they can cancel each other?

Correct

For the capacitive and inductive, how we can know their value such that we can find the length "l" of the transmission line?

When I use the Smith Chart to find the length "l" of the transmission line, I located the load impedance(2 ohm normalized) and then located the input resistance(1 ohm normalized). However, I don't know the imaginary part of Z so that I fail to find the length "l" of the transmission line.

When you rotate RC (clockwise) to the point where Re(Z) = 50 Ohms (1 Ohm normalized), yo can read the angle of RC and Im(Z) from the chart (don't forget to multiply with 50). If you need to rotate -30 degrees, you need a transmission line with 15 degrees electrical length between the load and the measuring point.

To get the right match, you need the series reactance to cancel Im(Z). The procedure for creating the right reactance is almost the same, except that you start from RC=+1 as you have an open series stub.

If you would use a shorted series stub, you start from RC=-1.

elvis0206

### elvis0206

Points: 2

#### elvis0206

##### Junior Member level 1
Re: Smith Chart question

Got it!
You are very helpful and thank you once again.

Status
Not open for further replies.