# Introduction to Impedance Matching Utilizing Transmission Line Parts In a earlier article on this collection, we mentioned how lumped elements can be utilized to rework one impedance to a different desired worth. The frequency of utility of lumped matching networks is often restricted to a number of gigahertz as a result of, at excessive frequencies, the parasitic results of lumped elements turn into way more pronounced. There may be another choice, particularly transmission line-based matching networks, that higher fits high-frequency functions. On this article, we’ll discover this impedance-matching approach.

### Transmission Line Enter Reflection Coefficient

To raised perceive transmission line-based impedance matching, it’s vital to know how the enter impedance and reflection coefficient change as we transfer alongside a transmission line terminated in a given load impedance ZL. Take into account the next diagram in Determine 1. ##### Determine 1. A diagram displaying a transmission line of a load impedance and the reflection coefficient.

It may be proven that the reflection coefficient Γin(d) at a distance d from the load is given by:

[Gamma_in(d) = Gamma_0 e^-j2 beta d]

##### Equation 1.

The place:

[Gamma_0 = fracZ_L – Z_0Z_L + Z_0]

##### Equation 2.

Equation 1 exhibits how the reflection coefficient adjustments alongside the road—the magnitude of Γin(d) is fixed and equal to the magnitude of Γ0—nonetheless, its part angle varies linearly with distance (d) from the load.

Subsequent, let’s check out an instance.

### Instance 1: Discover the Enter Impedance and Reflection Coefficient

Discover the enter impedance and reflection coefficient of a 50 Ω line with βd = 71.585° terminated in a load impedance of Z= 100 + j50 Ω.

By making use of Equation 2, we first discover the reflection coefficient on the load finish:

[Gamma_0 = 0.4+j0.2 = 0.447 measuredangle 26.57^circ]

Determine 2 locates Γ0 on the Smith chart (the normalizing impedance is Z= 50 Ω). ##### Determine 2. A Smith chart displaying Γ0 and the normalizing impedance of Z0 = 50 Ω [click image to enlarge].

The βd parameter—referred to as the road’s electrical size—is 71.585° on this instance. From Equation 1, the part angle of Γin(d) is the part of Γ0 minus 2 × 71.585 = 143.17°. Due to this fact, the part angle of Γin(d) works out to 26.57° – 143.17° = -116.6°.

Determine 2 exhibits how a Smith chart can be utilized to graphically get hold of Γin(d) from Γ0; for a given Γ0, the reflection coefficient alongside the transmission line Γin(d) lies on the circle having a radius of |Γ0|. Usually, transferring away from a termination on a transmission line produces a clockwise rotation alongside the fixed |Γ| circle. A line with {an electrical} size of θ results in a rotation of 2θ on the Smith chart.

#### Two Necessary Observations About Impedance and Transmission Strains

Take into account a line with {an electrical} size of 180° (that corresponds to d = λ / 2). Such a line produces a full 360° rotation and strikes us again to the unique load impedance ZL the place we began. Which means that the impedance we observe alongside a transmission line repeats itself each half-wavelength.

Determine 2 additionally hints at an vital property of transmission traces; a transmission line can transfer us from one constant-resistance circle to a different. Within the above instance, a 71.585° lengthy line strikes us from the constant-resistance circle of r = 2 to the r = 0.5 circle. Which means that a transmission line can act as an impedance-matching part. We’ll shortly return to this instance and talk about transmission line-based impedance matching methods in additional element, however earlier than diving in too far, let’s first study concerning the wavelength scale of the Smith chart.

### Smith Chart Wavelength Scale

As mentioned above, the enter impedance of a transmission line may be discovered by a easy round movement on the Smith chart. Slightly than utilizing {the electrical} size of the road, we are able to additionally characterize the road utilizing its bodily size, which is usually expressed as a fraction of the wavelength.

The governing equation for this equivalence is (βd = frac2πdλ). For instance, βd values of 45°, 90°, 135°, and 180° may be respectively produced by bodily lengths of:

• λ / 8
• 2λ / 8 = λ / 4
• 3λ / 8
• 4λ / 8 = λ / 2

That’s why Smith charts usually present a wavelength scale alongside the perimeter of the chart, as proven in Determine 3. ##### Determine 3. Instance Smith chart with a wavelength scale round its perimeter [click image to enlarge].

The wavelength scale assigns a distance of d to every level on the Smith chart. For instance, level A within the above determine corresponds to a distance of about 0.312λ. Be aware that absolutely the worth proven on the wavelength scale shouldn’t be vital; we simply use this scale primarily to seek out the gap between two factors on a transmission line by way of the wavelength.

Beneath, we’ll take a look at an instance to make clear how the wavelength scale is used to unravel transmission line issues. Earlier than we transfer on, notice that since {an electrical} size of βd = 180° (akin to d = λ / 2) produces a full 360° rotation on the Smith chart, the wavelength scale of the Smith chart ranges from 0 to 0.5λ.

### Instance 2: Fixing Transmission Line Points Utilizing the Wavelength Scale

Assume that at a distance of l= 0.051λ from a load impedance ZLoad, the enter impedance is Z= 50 – j50 Ω (Determine 4 under). ##### Determine 4. Diagram displaying the distances and cargo and enter impedances of an instance transmission line.

Discover ZLoad in addition to the enter impedance Z2 at a distance of l= 0.074λ from Z1. Assume that the attribute impedance of each traces is 50 Ω.

Utilizing a normalizing impedance of fifty Ω, the normalized worth of Z1 is discovered as z= 1 – j, which is situated within the following Smith chart (Determine 5). ##### Determine 5. Smith chart displaying the normalized impedance of 50 Ω [click image to enlarge].

We all know that transferring alongside a transmission line results in a movement alongside a continuing |Γ| circle on the Smith chart. We solely want to find out the proper course of movement. Remember that transferring away from a termination produces a clockwise rotation (with respect to the termination), and getting near the termination results in a counterclockwise rotation.

As proven within the above determine, z1 corresponds to about 0.338λ on the wavelength scale. For z2, we’re transferring away from zLoad by one other 0.074λ. Due to this fact, z2 corresponds to 0.338λ + 0.074λ = 0.412λ on the wavelength scale, as illustrated in Determine 5. Then again, to find zLoad, we should always go in a counterclockwise course, which leads us to the purpose akin to 0.338λ – 0.051λ = 0.287λ. From the Smith chart, now we have ZLoad = 50 × zLoad = 100 – j50 Ω and Z= 50 × z= 25 – j25 Ω.

### Instance 3: Discovering ZLoad and Z2

Within the following diagram, now we have Z= 50 – j50 Ω, l= 0.051λ, and l= 0.574λ. Discover ZLoad and Z2. ##### Determine 6. Diagram displaying the necessity to discover ZLoad and Z2.

For the reason that enter impedance wanting right into a line repeats each half-wavelength, we are able to deduce {that a} 0.574λ line is equal to a 0.074λ line. Taking this under consideration and utilizing the outcomes of the earlier instance, now we have ZLoad = 50 × zLoad = 100 – j50 Ω and Z= 50 × z= 25 – j25 Ω.

### Wavelengths Towards Generator and Load Scales

Many textbooks, in addition to industrial Smith charts, embody two wavelength scales, one labeled “Wavelengths Towards Generator” and the opposite one “Wavelengths Towards Load” (Determine 7 under). ##### Determine 7. Instance Smith chart displaying the “Wavelengths Towards Generator” and the opposite one “Wavelengths Towards Load” labels [click image to enlarge].

The “Wavelengths Towards Generator” scale will increase in a clockwise course, whereas the “Wavelengths Towards Load” scale will increase in a counterclockwise course. Using the “Generator” and “Load” phrases can typically trigger confusion. To keep away from this kind of confusion, remember the fact that the precise perform of the termination—i.e. whether or not it’s a supply impedance —shouldn’t be vital. After we add a collection transmission line to a termination, the enter impedance and reflection coefficient of the road is discovered by a clockwise rotation on the Smith chart. Then again, after we cut back the size of the road and get nearer to termination, the rotation is counterclockwise.

### Transmission Strains as Impedance Matching Parts

We’re now in place to introduce transmission line-based impedance matching that we alluded to within the earlier sections. For example, assume that we have to remodel Z= 100 + j50 Ω to 50 Ω. The load impedance ZL is definitely the identical as the worth we utilized in Instance 1 above. The Smith chart of Determine 2, with some further particulars, is reproduced in Determine 9. ##### Determine 9. Smith chart from Determine 2 with further particulars [click image to enlarge].

On this instance, the size of the road is intentionally chosen in a method that strikes us from the load impedance zLoad to level A, which is on the g = 1 constant-conductance circle. {The electrical} size of the road is 71.585° which corresponds to a size of 0.412λ – 0.213λ = 0.199λ (from the wavelength scale).

Since we’re on the g = 1 circle, we are able to add a parallel part with an acceptable susceptance to the road’s enter and transfer alongside the g = 1 circle to the middle of the Smith chart (Determine 10). ##### Determine 10. Diagram displaying the addition of a parallel part.

From Determine 9, the normalized admittance at level A is discovered to be y= 1 + j. We’d like a parallel component with a normalized susceptance of -j to reach on the middle of the Smith chart. One possibility is to make use of a parallel inductor. If the frequency of curiosity is 1 GHz, the inductor worth is, as calculated under:

[frac1L omega Y_0=1 Rightarrow L=frac12 pi times 1 times 10^9 times 0.02= 7.96 text nH]

Slightly than utilizing a lumped component, we are able to make use of the impedance transformation property of transmission traces another time. For instance, a short-circuited line with {an electrical} size of 45°, which corresponds to a size of λ / 8, can produce the required normalized susceptance of -j. That is illustrated in Determine 11. ##### Determine 11. Smith chart displaying the correspondence of 45° to supply the required normalized susceptance [click image to enlarge].

Within the above determine, level B corresponds to a short-circuited load. A λ / 8 line produces a 90° clockwise rotation, transferring us to level C that has a normalized admittance of -j, as required. The ultimate matching circuits are proven in Determine 12. ### Transmission Line-based Impedance Matching Abstract

In contrast to lumped matching networks, transmission line-based impedance matching is healthier fitted to high-frequency functions. A correctly chosen mixture of a collection line together with a parallel stub can remodel an arbitrary impedance to a different desired worth. On this article, we launched this method by an instance. Within the subsequent article, we’ll proceed this subject and take a look at completely different strategies of making use of this method.