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Chapter 25: Impedance Matching

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Chapter 25: Impedance Matching Chapter 25: Impedance Matching Chapter Learning Objectives: After completing this chapter the student will be able to: Determine the input impedance of a transmission line given its length, characteristic impedance, and load impedance. Design a quarter-wave transformer. Use a parallel load to match a load to a line impedance. Design a single-stub tuner. You can watch the video associated with this chapter at the following link: Historical Perspective: Alexander Graham Bell (1847-1922) was a scientist, inventor, engineer, and entrepreneur who is credited with inventing the telephone. Although there is some controversy about who invented it first, Bell was granted the patent, and he founded the Bell Telephone Company, which later became AT&T. Photo credit: https://commons.wikimedia.org/wiki/File:Alexander_Graham_Bell.jpg [Public domain], via Wikimedia Commons. 1 25.1 Transmission Line Impedance In the previous chapter, we analyzed transmission lines terminated in a load impedance. We saw that if the load impedance does not match the characteristic impedance of the transmission line, then there will be reflections on the line. We also saw that the incident wave and the reflected wave combine together to create both a total voltage and total current, and that the ratio between those is the impedance at a particular point along the line. This can be summarized by the following equation: (Equation 25.1) Notice that ZC, the characteristic impedance of the line, provides the ratio between the voltage and current for the incident wave, but the total impedance at each point is the ratio of the total voltage divided by the total current. We also use , the reflection coefficient, to indicate the ratio of the incident wave amplitude to the reflected wave amplitude. We also showed that the reflected term in the current is negative according to the transmission line equations. This equation can be simplified as follows: (Equation 25.2) Recalling the expression for the reflection coefficient from the previous chapter: (Copy of Equation 24.25) We can substitute this expression into Equation 25.2: (Equation 25.3) Multiplying numerator and denominator by ZL+ZC, we find: (Equation 25.4) 2 Rearranging this equation gives: (Equation 25.5) Next, we will apply two identities to simplify this equation. The identities are: (Equation 25.6) Substituting these into Equation 25.5 yields: (Equation 25.7) Both of the sine terms are negative because the exponents were in the opposite order from Equation 25.6. We have also reversed the terms in the denominator for clarity. If we now divide both numerator and denominator by 2cos(kz), we find: (Equation 25.8) This is the most compact and useful form of the expression for Z(z), the impedance at every point along a transmission line with a general load impedance. Example 25.1: A coaxial cable with characteristic impedance 100 and length 75m is attached to a signal generator set to 25MHz. The velocity of propagation along the wire is 2x108 m/s. The transmission line is terminated in a load impedance of 50. Calculate the impedance of the line at its input. 3 Let’s consider a transmission line of length L with characteristic impedance ZC and load impedance ZL, as shown in Figure 25.1. It would be very helpful if we could determine the input impedance of this line, which is simply the impedance at the input (left) end. Zin ZC ZL z=-L z=0 Figure 25.1. Input Impedance of a Transmission Line with a Load Impedance Recall from the previous chapter that we chose z=0 to be at the load, and everything to the left of that point has a negative value of z. Thus, since the transmission line has a length of L, the input impedance is Z(-L). (Equation 25.9) Inserting z=- L into Equation 25.8 (and remembering that tan(-x)=-tan(x), we find: (Equation 25.10) This is the general form for the input impedance of a line with length L and characteristic impedance ZC terminated in load ZL transmitting a signal with wave number k. Example 25.2: Calculate the input impedance of the transmission line described in Example 25.1. We have established that the impedance of the line varies from point to point along the line, as proven by the z that appears in Equation 25.8. It turns out that the impedance repeats every half-wavelength, as we can prove below: (Equation 25.11) Since =2/k: (Equation 25.12) 4 Simplifying gives: (Equation 25.13) Since tan(+)=tan(), we find: (Equation 25.14) Combined with Equation 25.10, this allows us to prove that: (Equation 25.15) It also turns out that the input impedance of a line that is one-quarter wavelength (/4) long has a special value. Beginning with Equation 25.10, we find: (Equation 25.16) Substituting k=2/, we find: (Equation 25.17) Simplifying gives: (Equation 25.18) But we know that tan(/2)=∞, so we can write: (Equation 25.19) (Technically, we shouldn’t write ∞ in an equation like this. It’s actually a limit.) The infinite terms in the numerator and denominator will overwhelm the non-infinite terms, giving: (Equation 25.20) Since both infinities represent the same function, they will cancel, leaving: 5 (Equation 25.21) This is a very special result, which we will use in upcoming calculations several times. Let’s also consider the input impedance of the three special cases of a transmission line termination: impedance matched, short-circuited, and open-circuited. If the transmission line is impedance matched (ZL=ZC), then Equation 25.10 becomes: (Equation 25.22) Since the numerator and denominator are identical, they cancel out, leaving: (Equation 25.23) Thus, for an impedance-matched transmission line, the input impedance is equal to the characteristic impedance. If the transmission line is short-circuited (ZL=0), Equation 25.10 becomes: (Equation 25.24) This reduces to: (Equation 25.25) This is the input impedance of a short-circuited transmission line. Example 25.3: Calculate the input impedance of the transmission line described in Example 25.1 if it is short-circuited rather than terminated with an impedance of 50. 6 Figure 25.2 shows the impedance of a short-circuited transmission line for various lengths. Note that this impedance is imaginary because of the j in Equation 25.25. Figure 25.2. Input Impedance of a Short-Circuited Transmission Line Two things are noteworthy about this figure. First, notice that the impedance does, in fact, repeat every half wavelength, just as we proved it would. Also notice that if you were asked to create any imaginary impedance between -∞ and +∞, you could do so just by selecting the length of a short-circuited transmission line. This will turn out to be very important later. Finally, we will consider the input impedance of an open-circuited transmission line. Substituting ZL=∞ into Equation 25.10 gives: (Equation 25.26) The non-infinite terms in the numerator and denominator can be neglected, and the infinities (which come from the same function) will cancel each other out, leaving: (Equation 25.27) This can be simplified by converting the tangent to a cotangent and by multiplying numerator and denominator by j: (Equation 25.28) This is the input impedance of an open-circuited transmission line. 7 Example 25.4: Calculate the input impedance of the transmission line described in Example 25.1 if it is open-circuited. We can also plot the input impedance of an open-circuited transmission line, as shown in Figure 25.3. Figure 25.3. Input Impedance of an Open -Circuited Transmission Line Again, notice that the impedance repeats every half wavelength, and that any imaginary impedance can be obtained simply by selecting the length of the open-circuited transmission line. 25.2 Introduction to Impedance Matching Now we have laid all the groundwork necessary to discover ways to correct an impedance mismatch between the line and the load. Doing so will eliminate reflections, which is very desirable for several reasons: 1. Typically, your goal is for the incoming signal to be fully absorbed by the load. If part of it is being reflected back along the line, then you are probably wasting energy and money. 2. When the reflected wave interacts with the incident wave, the maximum voltage can be higher than the amplitude of the incoming wave. Higher voltages require better cables, and better cables cost more money. Preventing reflections will allow you to use less expensive cables since the voltages will be lower. 3. The reflected signals will eventually return to the transmitter, which may be damaged by the reflections if it is not designed to handle them. 8 4. It is very likely that the reflected wave will also reflect off the left end of the wire, which will create an echo of the original signal some time later. This can create interference with new signals that are incoming at the same time. For these three reasons, we really want to reduce or eliminate reflections, which means the load impedance must match the characteristic impedance. But we don’t always control either of those quantities. Maybe the transmission line is very hard to replace, or it is prohibitively expensive to do so. Most of the time, the load impedance is set, and you can’t do anything to change it. So what can we do? There are three strategies we can apply. Each of them involves making a small change at or near the load end of the wire to make it look like the load impedance matches the line impedance, even when it doesn’t actually match.
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