Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The textbook explains a situation in which when you have 2 unmatched transmission lines (different characteristic impedance), you can connect a new line in between such that the input impedance would match.

Say I have a line #1 with characteristic impedance \$Z_1=100 \Omega\$.

Line #1 is connected to Line #3 with \$Z_3=20 \Omega\$. There is mismatch and therefore reflection.

If I connect another transmission in between and its length is the quarter wavelength, then this new transmission line (we call it Line #2) would match the characteristic impedance of Line #1.

\$ Z_2=\sqrt{Z_1\times Z_3}=\sqrt{100 \times 20}=44.7 \Omega \$.

\$Z_{in}=\frac{Z_2^2}{Z_3}=\frac{44.7^2}{20}=100 \Omega\$.

Thus, overall input impedance at the junction of Line #1 and Line #2 \$Z_{in}\$ would be equal to \$Z_1\$. Therefore, no reflection at this junction.

But my question is: what about at the junction of Line #2 and Line #3? Since \$Z_2=44.7 \Omega\$ is not equal to \$Z_3=20\Omega\$, then there will be reflection at their junction? If that is the case, what is the point of using matching?

share|improve this question

Impedance matching is tricky, but the role of a quarter wave transmission line is to map from one impedance to another. The actual impedance of the line will not match either the input or the output impedance - this is entirely expected.

However at a given frequency, when a correctly designed quarter wave line is inserted with the correct impedance, the output impedance will appear to the input as perfectly matched. In your case, the transformer will make the \$20\Omega\$ impedance appear as if it is a \$100\Omega\$ impedance meaning no mismatch. Essentially it guides the waves from one characteristic impedance to another.

The easiest way to visualise this is on a Smith chart, plot the two points 0.4 (\$20\Omega\$) and 2 (\$100\Omega\$). Then draw a circle centred on the resistive/real axis (line down the middle) which intersects both points. You will find that this point is located at 0.894 (\$44.7\Omega\$) if your calculations are correct. This is shown below at \$500\mathrm{MHz}\$, but the frequency is only important when converting the electrical length to a physical length.

Smith Chart of Quarter Wave

What a quarter wave transformer does is rotate a given point by \$180^\circ\$ around its characteristic impedance on the Smith chart (that's \$\lambda/4 = 90^\circ\$ forward plus \$90^\circ\$ reverse).

Exactly why it does this is complex. But the end result of a long derivation is that for a transmission line of impedance \$Z_0\$ connected to a load of impedance \$Z_L\$ and with a length \$l\$, then the impedance at the input is given by:

$$Z_{in}=Z_0\frac{Z_L+jZ_0\tan\left(\beta l\right)}{Z_0+jZ_L\tan\left(\beta l\right)}$$

That is an ugly equation, but it just so happens if the electrical length \$\beta l\$ is \$\lambda/4\$ (\$90^\circ\$), the \$\tan\$ part goes to infinity which allows the equation to be simplified to:

$$Z_{in}=Z_0\frac{Z_0}{Z_L}=\frac{(Z_0)^2}{Z_L}\rightarrow Z_0=\sqrt{\left(Z_{in}Z_L\right)}$$

Which is where your calculation comes from.

With the quarter wave transformer in place, the load appears as matched to the source. In other words, the transformer matches both of its interfaces, not just the input end.

You can also see from this equation why the transformer only works for a single frequency - because it relies on the physical length being \$\lambda/4\$. You can actually (generally using advanced design tools) achieve an approximate match over a range of frequencies - basically a close enough but not exact match.

share|improve this answer
    
What tool did you use to simulate this please ? – Board-Man Feb 13 at 9:01
    
@Board-Man It's the "Smith-Chart" software by "Fritz Dellsperger". You can find it here. – Tom Carpenter Feb 13 at 15:05

There will be a reflection at the junction between lines 2 and 3. In fact, there has to be for this to work.

What you're doing is you're choosing the \$Z_0\$ of line 2 so that the reflection at the 2/3 transition after propagating back to the 1/2 junction, exactly cancels the reflection at the 1/2 transition. You are setting up destructive interference for the reflected waves and constructive interference for the transmitted waves.

Note that because line 2 has to be one quarter wavelength in length, this technique can only work perfectly at one specific frequency.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.