# 2.11 Impedance Matching

A transmission line usually connects a generator circuit at one end to a load at the other. The load may be an antenna, a computer terminal, or any circuit with an equivalent input impedance Z L.

Since the primary use of a transmission line is to transfer power or transmit coded signals (such as digital data), a matched load ensures that all of the power delivered to the transmission line by the source is transferred to the load (and no echoes are relayed back to the source).

The simplest solution to matching a load to a transmission line is to design the load circuit such that its impedance Z L=Z 0. Unfortunately, this may not be possible in practice because the load circuit may have to satisfy other requirements. An alternative solution is to place an impedance-matching network between the load and the transmission line as illustrated in Fig. 2-32.

# Figure 2-32 The function of a matching network is to transform the load impedance Z L such that the input impedance Z in looking into the network is equal to Z 0 of the feedline.

Figure 2 32 Full Alternative Text

Within the matching network, reflections can occur at both terminals (A A′ and M M′), creating a standing-wave pattern, but the net result (of all of the multiple reflections within the matching network) is that the wave incident from the source experiences no reflection when it reaches terminals M M′. This is achieved by designing the matching network to exhibit an impedance equal to Z 0 at M M′ when looking into the network from the transmission line side. If the network is lossless, then all of the power going into it will end up in the load.

The matching network, which is intended to match a load impedance Z L=R L+j X L to a lossless transmission line with characteristic impedance Z 0, may be inserted either in series (between the load and the feedline) as in Fig. 2-33(a) and (b) or in parallel (Fig. 2-33(c) to (e)). In either case, the network has to transform the real part of the load impedance from R L (at the load) to Z 0 at M M′ in Fig. 2-32 and transform the reactive part from X L (at the load) to zero at M M′. To achieve these two transformations, the matching network must have at least two degrees of freedom (that is, two adjustable parameters).

# Figure 2-33 Five examples of in-series and in-parallel matching networks.

(a) If ZL is real: in-series λ/4 transformer inserted at AA′

Figure 2 33-a Full Alternative Text

(b) If ZL = complex: in-series λ/4 transformer inserted at d = dmax or d = dmin

Figure 2 33-b Full Alternative Text

(c) In-parallel insertion of capacitor at distance d1

Figure 2 33-c Full Alternative Text

(d) In-parallel insertion of inductor at distance d2

Figure 2 33-d Full Alternative Text

(e) In-parallel insertion of a short-circuited stub

Figure 2 33-e Full Alternative Text

If X L=0, the problem reduces to a single transformation, in which case matching can be realized by inserting a quarter-wavelength transformer (Section 2-8.5) next to the load (Fig. 2-33(a)).

The design procedure is outlined in Module 2.7. The in-parallel insertion networks shown in Fig. 2-33(c)–(e) are the subject of Examples 2-13 and 2-14.

# 2.11.1 Lumped-Element Matching

In the arrangement shown in Fig. 2-34, the matching network consists of a single lumped element, either a capacitor or an inductor, connected in parallel with the line at a distance d from the load. Parallel connections call for working in the admittance domain. Hence, the load is denoted by an admittance Y L, and the line has a characteristic admittance Y 0. The shunt element has an admittance Y s. At M M′, Y d is the admittance due to the transmission-line segment to the right of M M′. The input admittance Y in (referenced at a point just to the left of M M′) is equal to the sum of Y d and Y s:

Y in=Y d+Y s.(2.138)

# Figure 2-34 Inserting a reactive element with admittance Y s at M M′ modifies Y d to Y in.

Figure 2 34 Full Alternative Text

In general, Y d is complex, and Y s is purely imaginary because it represents a reactive element (capacitor or inductor). Hence, Eq. (2.138) can be written as

Y in=(G d+j B d)+j B s=G d+j(B d+B s).(2.139)

When all quantities are normalized to Y 0, Eq. (2.139) becomes

y in=g d+j(b d+b s).(2.140)

To achieve a matched condition at M M′, it is necessary that y in=1+j 0, which translates into two specific conditions, namely

The real-part condition is realized through the choice of d, which is the distance from the load to the shunt element, and the imaginary-part condition is realized through the choice of lumped element (capacitor or inductor) and its value. These two choices are the two degrees of freedom needed in order to match the load to the feedline.

# Example 2-13: Lumped Element

A load impedance Z L=25−j 50 Ω is connected to a 50 Ω transmission line. Insert a shunt element to eliminate reflections towards the sending end of the line. Specify the insert location d (in wavelengths), the type of element, and its value given that f=100 MHz.

Solution: The normalized load impedance is

z L=Z L Z 0=25−j 50 50=0.5−j 1,

which is represented by point A on the Smith chart of Fig. 2-35. Next, we draw the constant S circle through point A. As alluded to earlier, to perform the matching task, it is easier to work with admittances than with impedances. The normalized load admittance y L is represented by point B, which is obtained by rotating point A over 0.25 λ or equivalently by drawing a line from point A through the chart center to the image of point A on the S circle. The value of y L at B is

y L=0.4+j 0.8,

and it is located at position 0.115 λ on the WTG scale. In the admittance domain, the r L circles become g L circles, and the x L circles become b L circles. To achieve matching, we need to move from the load toward the generator a distance d such that the normalized input admittance y d of the line terminated in the load (Fig. 2-34) has a real part of 1. This condition is satisfied by either of the two matching points C and D on the Smith charts of Figs. 2-35 and 2-36, respectively, corresponding to intersections of the S circle with the g L=1 circle. Points C and D represent two possible solutions for the distance d in Fig. 2-34(a).

# Figure 2-35 Solution for point C of Example 2-13. Point A is the normalized load with z L=0.5−j 1; point B is y L=0.4+j 0.8. Point C is the intersection of the SWR circle with the g L=1 circle. The distance from B to C is d 1=0.063 λ.

Figure 2 35 Full Alternative Text

# Figure 2-36 Solution for point D of Example 2-13. Point D is the second point of intersection of the SWR circle and the g L=1 circle. The distance B to D gives d 2=0.207 λ.

Figure 2 36 Full Alternative Text

Solution for Point C (Fig. 2-35): At C,

y d 1=1+j 1.58,

which is located at 0.178 λ on the WTG scale. The distance between points B and C is

d 1=(0.178−0.115)λ=0.063 λ.

Looking from the generator toward the parallel combination of the line connected to the load and the shunt element, the normalized input admittance at terminals M M′ is

y in 1=y s 1+y d 1,

where y s 1 is the normalized input admittance of the shunt element. To match the feed line to the parallel combination, we need y in 1=1+j 0. Hence, we need y s 1 to cancel the imaginary part of y d 1; that is,

y s 1=−j 1.58.

The corresponding impedance of the lumped element is

Z s 1=1 Y s 1=1 y s 1 Y 0=Z 0 j b s 1=Z 0−j 1.58=j Z 0 1.58=j 31.62 Ω.

Since the value of Z s 1 is positive, the element to be inserted should be an inductor and its value should be

L=31.62 ω=31.62 2 π×10 8=50 nH.

The results of this solution have been incorporated into the circuit of Fig. 2-35.

Solution for Point D (Fig. 2-36): At point D,

y d 2=1−j 1.58,

and the distance between points B and D is

d 2=(0.322−0.115)λ=0.207 λ.

The needed normalized admittance of the reactive element is

y s 2=+j 1.58.

Hence,

Z s 2=−j 31.62 Ω,

which is the impedance of a capacitor with

C=1 31.62 ω=50 pF.

Figure 2-36 displays the circuit solution for d 2 and C.

# 2.11.2 Single-Stub Matching

The single-stub matching network shown in Fig. 2-37(a) consists of two transmission line sections: one of length d connecting the load to the feedline at M M′ and another of length l connected in parallel with the other two lines at M M′. This second line is called a stub, and it is usually terminated in either a short or open circuit; hence, its input impedance and admittance are purely reactive. The stub shown in Fig. 2-37(a) has a short-circuit termination.

# Figure 2-37 Shorted-stub matching network.

Figure 2 37 Full Alternative Text

Because at M M′ the stub is added in parallel to the line (which is why it is called a shunt stub), it is easier to work with admittances than with impedances. The matching procedure consists of two steps. In the first step, the distance d is selected to transform the load admittance, Y L=1/Z L, into an admittance of the form Y d=Y 0+j B when looking toward the load at M M′. Then in the second step, the length l of the stub line is selected so that its input admittance Y s at M M′ is equal to −j B. The parallel sum of the two admittances at M M′ yields Y 0, which is the characteristic admittance of the line. The procedure is illustrated by Example 2-14.

# Example 2-14: Single-Stub Matching

Repeat Example 2-13, but use a shorted stub (instead of a lumped element) to match the load impedance Z L=(25−j 50) Ω to the 50 Ω transmission line.

Solution: In Example 2-13, we demonstrated that the load can be matched to the line via either of two solutions:

1. d 1=0.063 λ, and y s 1=j b s 1=−j 1.58,

2. d 2=0.207 λ, and y s 2=j b s 2=j 1.58.

The locations of the insertion points at distances d 1 and d 2 remain the same, but now our task is to select corresponding lengths l 1 and l 2 of shorted stubs that present the required admittances at their inputs.

To determine l 1, we use the Smith chart in Fig. 2-38. The normalized admittance of a short circuit is −j∞, which is represented by point E on the Smith chart with position 0.25 λ on the WTG scale. A normalized input admittance of −j 1.58 is located at point F with position 0.34 λ on the WTG scale. Hence,

l 1=(0.34−0.25)λ=0.09 λ.

# Figure 2-38 Solution for point C of Example 2-14. Point A is the normalized load with z L=0.5−j 1; point B is y L=0.4+j 0.8. Point C is the intersection of the SWR circle with the g L=1 circle. The distance from B to C is d 1=0.063 λ. The length of the shorted stub (E to F) is l 1=0.09 λ.

Figure 2 38 Full Alternative Text

Similarly, y s 2=j 1.58 is represented by point G with position 0.16 λ on the WTG scale of the Smith chart in Fig. 2-39. Rotating from point E to point G involves a rotation of 0.25 λ plus an additional rotation of 0.16 λ or

l 2=(0.25+0.16)λ=0.41 λ.

# Figure 2-39 Solution for point D of Example 2-14. Point D is the second point of intersection of the SWR circle and the g L=1 circle. The distance B to D gives d 2=0.207 λ, and the distance E to G gives l 2=0.410 λ.

Figure 2 39 Full Alternative Text

# Example 2-15: λ/4 Transformer for Complex Load

Design a quarter-wavelength transformer to match a load with Z L=(100+j 100) Ω to a 50 Ω line.

Solution: The required transmission-line circuit is shown in Fig. 2-40(a). We need to insert a λ/4 transformer to eliminate reflections at M M′. To that end, we need to specify the insertion distance d and the characteristic impedance Z 0 2 of the λ/4 section so that Z in=Z 0 1.

# Figure 2-40 Solution for Example 2-15: (a) quarter-wave transformer and (b) Smith chart solution.

(a) Quarter-wave transformer

Figure 2 40-a Full Alternative Text

(b) Smith chart solution

Figure 2 40-b Full Alternative Text

A λ/4 section with characteristic impedance Z 0 2 transfers an impedance Z a at one end of the section to impedance Z b at the other end such that

Z a Z b=Z 0 2 2.(2.142)

In the present case (Fig. 2-40(a)), Z a=Z(d), the input impedance at B B′ looking towards the load, and Z b=Z in=Z 0 1=50 Ω. Since the λ/4 transformer is a lossless line, its impedance Z 0 2 is purely real. Consequently, to satisfy Eq. (2.142), it is necessary that Z(d) be purely real as well, which is the key to how we select the distance d. Given this rationale, we proceed as follows.

1. The normalized load impedance is

z L=Z L Z 0 1=100+j 100 50=2+j 2,

which is shown as point A in the Smith chart of Fig. 2-40(b).

2. Move on the SWR circle until z(d) becomes purely real, which occurs at points B and C.

(a) Point B:

d 1=(0.25−0.209)λ=0.041 λ,z(d 1)=4.27,Z(d 1)=4.27×50=213.5 Ω,Z 0 2=Z(d 1)Z 0 1=213.5×50=103.3 Ω.

(b) Point C:

d 2=d 1+λ 4=(0.041+0.25)λ=0.291 λ,z(d 2)=0.23,Z(d 2)=0.23×50=11.5 Ω,Z 0 2=Z(d 2)Z 0 1=11.5×50=24.2 Ω.

Concept Question 2-24: To match an arbitrary load impedance to a lossless transmission line through a matching network, what is the required minimum number of degrees of freedom that the network should provide?

Concept Question 2-25: In the case of the single-stub matching network, what are the two degrees of freedom?

Concept Question 2-26: When a transmission line is matched to a load through a single-stub matching network, no waves are reflected toward the generator. What happens to the waves reflected by the load and by the shorted stub when they arrive at terminals M M′ in Fig. 2-37?