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**Dr. Sandra Cruz-Pol ECE Dept. UPRM**

Transmission Lines Dr. Sandra Cruz-Pol ECE Dept. UPRM

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Exercise 11.3 A 40-m long TL has Vg=15 Vrms, Zo=30+j60 W, and VL=5e-j48o Vrms. If the line is matched to the load and the generator, find: the input impedance Zin, the sending-end current Iin and Voltage Vin, the propagation constant g. Answers: ZL Zg Vg + Vin - Iin VL Zo=30+j60 g=a +j b 40 m

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**Transmission Lines TL parameters TL Equations**

Input Impedance, SWR, power Smith Chart Applications Quarter-wave transformer Slotted line Single stub Microstrips

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**Transmission Lines (TL)**

TL have two conductors in parallel with a dielectric separating them They transmit TEM waves inside the lines

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**Common Transmission Lines**

Two-wire (ribbon) Microstrip Stripline (Triplate) Coaxial

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**Other TL (higher order) [Chapter 11]**

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Fields inside the TL V proportional to E, I proportional to H

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**Distributed parameters**

The parameters that characterize the TL are given in terms of per length. R = ohms/meter L = Henries/ m C = Farads/m G = mhos/m

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**Common Transmission Lines**

R, L, G, and C depend on the particular transmission line structure and the material properties. R, L, G, and C can be calculated using fundamental EMAG techniques. Parameter Two-Wire Line Coaxial Line Parallel-Plate Line Unit R L G C

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TL representation

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**Distributed line parameters**

Using KVL:

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**Distributed parameters**

Taking the limit as Dz tends to 0 leads to Similarly, applying KCL to the main node gives

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**Wave equation Using phasors The two expressions reduce to**

Wave Equation for voltage

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TL Equations Note that these are the wave eq. for voltage and current inside the lines. The propagation constant is g and the wavelength and velocity are

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**Waves moves through line**

The general solution is In time domain is Similarly for current, I z

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**Characteristic Impedance of a Line, Zo**

Is the ratio of positively traveling voltage wave to current wave at any point on the line z

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Example: An air filled planar line with w=30cm, d=1.2cm, t=3mm, sc=7x107 S/m. Find R, L, C, G for 500MHz Answer d w

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Exercise 11.1 A transmission line operating at 500MHz has Zo=80 W, a=0.04Np/m,b=1.5rad/m. Find the line parameters R,L,G, and C. Answer: 3.2 W/m, 38.2nH/m, S/m, 5.97 pF/m

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**Different cases of TL Lossless Distortionless Lossy Transmission line**

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Lossless Lines (R=0=G) Has perfect conductors and perfect dielectric medium between them. Propagation: Velocity: Impedance

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**Distortionless line (R/L = G/C)**

Is one in which the attenuation is independent on frequency. Propagation: Velocity: Impedance

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Summary g = a + jb Zo =General Lossless Distortionless RC = GL

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Excersice 11.2 A telephone line has R=30 W/km, L=100 mH/km, G=0, and C= 20mF/km. At 1kHz, obtain: the characteristic impedance of the line, the propagation constant, the phase velocity. Solution:

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**Define reflection coefficient at the load, GL**

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**Terminated, Lossless TL**

Then, Similarly, The impedance anywhere along the line is given by The impedance at the load end, ZL, is given by

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**Terminated, Lossless TL**

Then, Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance. Recall for the lossless case, Then

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Example A generator with 10Vrms and Rg=50, is connected to a 75W load thru a 0.8l, 50W-lossless line. Find VL

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**Terminated, Lossless TL**

It is customary to change to a new coordinate system, z = - l , at this point. -z z = - l Rewriting the expressions for voltage and current, we have Rearranging,

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**Impedance (Lossy line)**

The impedance anywhere along the line is given by The reflection coefficient can be modified as follows Then, the impedance can be written as After some algebra, an alternative expression for the impedance is given by Conclusion: The load impedance is “transformed” as we move away from the load.

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**Impedance (Lossless line)**

The impedance anywhere along the line is given by The reflection coefficient can be modified as follows Then, the impedance can be written as After some algebra, an alternative expression for the impedance is given by Conclusion: The load impedance is “transformed” as we move away from the load.

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**Exercise : using formulas**

A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, ZL Zg Vg + Vin - Iin VL Zo g=j b 2 cm Voltage Divider:

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SWR or VSWR or s Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves. The (Voltage) Standing Wave Ratio - SWR (or VSWR) is defined as

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Power The average input power at a distance l from the load is given by which can be reduced to The first term is the incident power and the second is the reflected power. Maximum power is delivered to load if G=0

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**Three common Cases of line-load combinations:**

Shorted Line (ZL=0) Open-circuited Line (ZL=∞) Matched Line (ZL = Zo)

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**Standing Waves -Short Shorted Line (ZL=0), we had**

So substituting in V(z) Voltage maxima -z -l/4 -l/2 -l |V(z)| *Voltage minima occurs at same place that impedance has a minimum on the line

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**Standing Waves -Open Open Line (ZL=∞) ,we had So substituting in V(z)**

Voltage minima |V(z)| -z -l/4 -l/2 -l

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**Standing Waves -Matched**

Matched Line (ZL = Zo), we had So substituting in V(z) |V(z)| -z -l/4 -l/2 -l

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**Java applets http://www.amanogawa.com/transmission.html**

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The Smith Chart

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**Smith Chart Commonly used as graphical representation of a TL.**

Used in hi-tech equipment for design and testing of microwave circuits One turn (360o) around the SC = to l/2

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**What can be seen on the screen?**

Network Analyzer What can be seen on the screen?

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Smith Chart Suppose you use as coordinates the reflection coefficient real and imaginary parts. and define the normalized ZL: Gi |G| Gr

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**Now relating to z=r+jx After some algebra, we obtain two eqs.**

Similar to general equation of a circle of radius a, center at (h,k) Circles of r Circles of x

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**Examples of circles of r and x**

Circles of x

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**Examples of circles of r and x**

Circles of x

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**The joy of the SC Numerically s = r on the +axis of Gr in the SC**

Proof:

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**Fun facts about the Smith Chart**

A lossless TL is represented as a circle of constant radius, |G|, or constant s Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely. To generator

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**Fun facts about the Smith Chart**

One turn (360o) around the SC = to l/2 because in the formula below, if you substitute length for half-wavelength, the phase changes by 2p, which is one turn. Find the point in the SC where G=+1,-1, j, -j, 0, 0.5 What is r and x for each case?

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**Fun facts : Admittance in the SC**

The admittance, y=YL/Yo where Yo=1/Zo, can be found by moving ½ turn (l/4) on the TL circle

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**Fun facts about the Smith Chart**

The Gr +axis, where r > 0 corresponds to Vmax The Gr -axis, where r < 0 corresponds to Vmin Vmin Vmax (Maximum impedance)

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Exercise: using S.C. A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, Load is at S.C. Move .2l and arrive to .4179l Read ZL Zg Vg + Vin - Iin VL Zo g=j b 2 cm Voltage Divider:

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zL zin 0.2l

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**Exercise: cont….using S.C.**

A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, Distance from the load (.2179l) to the nearest minimum & max Move to horizontal axis toward the generator and arrive to .5l (Vmax) and to .25l for the Vmin. Distance to min= =.282l Distance to 2st voltage maximum is .282l +.25l=.482 See drawing ZL Zg Vg + Vin - Iin VL Zo g=j b 2 cm

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**Exercise : using formulas**

A 2cm lossless TL has Vg=10 Vrms, Zg=60 W, ZL=100+j80 W and Zo=40W, l=10cm . find: the input impedance Zin, the sending-end Voltage Vin, ZL Zg Vg + Vin - Iin VL Zo g=j b 2 cm Voltage Divider:

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**Distance to first Vmax:**

Another example: A 26cm lossless TL is connected to load ZL=36-j44 W and Zo=100W, l=10cm . find: the input impedance Zin Load is at S.C. Move .1l and arrive to .527l (=.027l) Read ZL Zg Vg + Vin - Iin VL Zo g=j b 26cm Distance to first Vmax:

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Exercise 11.4 A 70 W lossless line has s =1.6 and qG =300o. If the line is 0.6l long, obtain G, ZL, Zin and the distance of the first minimum voltage from the load. Answer The load is located at: Move to l and draw line from center to this place, then read where it crosses you TL circle. Distance to Vmin in this case, lmin =.5l-.3338l=

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**Java Applet : Smith Chart**

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**Applications Slotted line as a frequency meter Impedance Matching**

If ZL is Real: Quarter-wave Transformer (l/4 Xmer) If ZL is complex: Single-stub tuning (use admittance Y) Microstrip lines

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**HP Network Analyzer in Standing Wave Display**

Slotted Line Used to measure frequency and load impedance HP Network Analyzer in Standing Wave Display

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Slotted line example Given s, the distance between adjacent minima, and lmin for an “air” 100W transmission line, Find f and ZL s=2.4, lmin=1.5 cm, lmin-min=1.75 cm Solution: =8.6GHz Draw a circle on r=2.4, that’s your T.L. move from Vmin to zL

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**Quarter-wave transformer …for impedance matching**

ZL Zin Zo , g l= l/4 Conclusion: **A piece of line of l/4 can be used to change the impedance to a desired value (e.g. for impedance matching)

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**Single Stub Tuning …for impedance matching**

A stub is connected in parallel to sum the admittances Use a reactance from a short-circuited stub or open-circuited stub to cancel reactive part Zin=Zo therefore z =1 or y=1 (this is our goal!)

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Single Stub Basics We work with Y, because in parallel connections they add. YL (=1/ZL) is to be matched to a TL having characteristic admittance Yo by means of a "stub" consisting of a shorted (or open) section of line having the same characteristic admittance Yo

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Single Stub Steps First, the length l is adjusted so that the real part of the admittance at the position where the stub is attached is equal to Yo or yline = 1+jb Then the length of the shorted stub is adjusted so that it's susceptance cancels that of the line, or ystub= -jb

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Example: Single Stub A 75W lossless line is to be matched to a 100-j80 W load with a shorted stub. Calculate the distance from the load, the stub length, and the necessary stub admittance. Answer: Change to: =0.094l (1+j.96) or next intersection:0.272l, Short stub: =0.126l With ystub= -j.96/75 =-j.0128 mhos

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Example: Single Stub A 50W lossless T.L. is 20 m long and terminated into a 120+j220W load. To perfectly match , what should be the length and location of a short-circuited stub line. Assume an operating frequency of 10MHz. Answer: l=30m, zL= 2.4+j4.4 (circulito rojo). Trazo raya por el centro y leo yL al otro lado (circulito azul). La yL está en la posición 0.472l. Trazo círculo amarillo, esa es mi T.L. donde está la carga. Busco donde interseca el circulo de r=1(circulo turquesa). Lo interseca en 1+j3.2 Ese es mi 1+jb. Y está en la posicion 0.214l. Por tanto la distancia desde la carga al segmento(stub) = (debido a cambio de escala) +.214= .242 l. Ahora miro el círculo del segmento= circulo grande en el SC (donde Gamma =1).y busco donde jb= -j3.2 (abajo ver flecha roja). Para buscar su posición, trazo línea desde el centro hasta afuera y leo posición está en .2958l. El segmento empieza en carga de corto circuito (ver palabra roja que dice short) está en .25l. Por lo tanto el largo del segmento es = .0485l

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Microstrips

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**Microstrips analysis equations & Pattern of EM fields**

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**Microstrip Design Equations**

Falta un radical en eeff

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**Microstrip Design Curves**

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Example A microstrip with fused quartz (er=3.8) as a substrate, and ratio of line width to substrate thickness is w/h=0.8, find: Effective relative permittivity of substrate Characteristic impedance of line Wavelength of the line at 10GHz Answer: eeff=2.75, Zo=86.03 W, l=18.09 mm

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Diseño de microcinta: Dado (er=4) para el substrato, y h=1mm halla w para Zo=50 W y cuánto es eeff? Solución: Suponga que como Z es pequena w/h>2

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