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Using the Smith Chart and a Single Stub to perform matching.

Often, there are reasons why using a discrete inductor or capacitor for matching is not such a good idea. Atthe high frequencies where matching is important, losses in both L or C mean that you don't get a good match, and most of thetime (except for some air-dielectric adjustable capacitors) it is hard to get just the value you want.

There is another approach though. A shorted or open transmission line, when viewed at its input looks like a purereactance or pure susceptance. With a short as a load, the reflection coefficient has unity magnitude Γ 1.0 and so we move around the very outside of the Smith Chart as the length of the line increases or decreases, and Z in Z 0 is purely imaginary. When we did the bilinear transformation from the Z s Z 0 plane to the r s plane, the imaginary axis transformed into the circle of diameter 2, which ended up being the outside circle whichdefined the Smith Chart.

Input impedance of a shorted line

Another way to see this is to go back to this equation . There we found:
Z s Z 0 Z L Z 0 β s Z 0 Z L β s
With Z L 0 this reduces to
Z s Z 0 β s
Which, of course for various values of s , can take on any value from to . We don't have to go to Radio Shack©and buy a bunch of different inductor and capacitors. We can just getsome transmission line and short it at various places!

Thus, instead of a discrete component, we can use a section of shorted (or open) transmission line instead . These matching lines are called matching stubs . One of the major advantages here is that with a line which has an adjustable short on the end ofit, we can get any reactance we need, simply by adjusting the length of the stub. How this all works will become obviousafter we take a look at an example.

A shortened stub

Let's do one. In we can see that, Z L Z 0 0.2 0.5 , so we mark a point "A" on the Smith Chart. Since we will want to put the tuning or matching stub in shunt across theline, the first thing we will do is convert Z L Z 0 into a normalized admittance Y L Y 0 by going 180 ° around the Smith Chart to point "B", where Y L Y 0 0.7 -1.7 . Now we rotate around on the constant radius, r s circle until we hit the matching circle at point "C". This is shown in . At "C", Y S Y 0 1.0 2.0 . Using a "real" Smith Chart, I get that the distance of rotation is about 0.36 λ . Remember, all the way around is λ 2 , so you can very often "eyeball" about how far you have to go, and doing so is a good check on making a stupid matherror. If the distance doesn't look right on the Smith Chart, you probably made a mistake!

Another load

Converting to normalized admittance

Converting to Y L Y 0

Moving to the matching circle

OK, at this point, the real part of the admittance is unity, soall we have to do is add a stub to cancel out the imaginary part. As mentioned above, the stubs often come with adjustable,or "sliding short" so we can make them whatever length we want .

Matching with a shortened stub

Our task now, is to decide how much to push or pull on thesliding handle on the stub, to get the reactance we want. The hint on what we should do is in . The end of the stub is a short circuit. What is the admittance of ashort circuit? Answer: , ! Where is this on the Smith Chart? Answer: on the outside, on the right hand side on the real axis. Now, if we start at ashort, and start to make the line longer than s 0 , what happens to Y s Y 0 ? It moves around on the outside of the Smith Chart. What we need to do is move away from the short until we get Y s Y 0 2.0 and we will know how long the shorted tuning stub should be . In going from "A" to "B" we traverse a distance of about 0.07 λ and so that is where we should set the position of the sliding short on the stub .

Finding the stub length

The matched line

We sometimes think of the action of the tuning stub as allowingus to move in along the Y s Y 0 to get to the center of the Smith Chart, or to a match . We are not in this case, physically moving down the line. Rather we are moving along a contour of constant real part because all the stub can do is change the imaginary part of the admittance,it can do nothing to the real part!

Moving with a stub

Moving along the Y s Y 0 1 circle with a stub.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4
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