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Home , Transmission coefficient, Transmission line

Chapter 2

Transmission Lines

ECE 130a

ZLocosbl + j Z sin bl ZLo+ j Z tan bl ZlZinaf-= o = Zo ZljZloLcosbb+ sin ZjZloL+ tan b Examples of Transmission Lines: (Chapter 2) Parallel Strip Line: metal conductors

d w

Coaxial Line: metal radius b metal radius a

dielectric

Parallel Wire Line: Air Line Dielectric Line D

ε r conductor radius a

Microstrip Line: w Dielectric Metal d

There is a simple way to view the guided on a . z Iztaf,

Vztaf, Iztaf, The potential difference () between the metal conductors with equal and opposite current flowing in them are circuit concepts, except they depend not only on time, but also on the distance z. So we describe the wave as voltage and current . itza , f + generator vtza , f load - Length L >>2πβ/ = λ Other guiding structures: 1. -- consist of a single hollow metal tube of various cross- sectional geometry. An EM wave propagates longitudinally inside the hollow structure.

The wave propagation in waveguides is not transverse (not TEM). That is, it has longitudinal field component(s). Transverse spatial depen- dence is fairly complicated. The β of a wave is not equal to that of plane waves, and velocity of propagation thus is not the same as . 2. Optical Fibers -- are used at optical frequencies, at , and at visi- ble . An consists of a very thin (50-300 mm) dielectric circular cross section cylinder. The material is usually glass or plastic. Inner and outer portions have different dielectric constants (index of refraction), as shown below.

core radius a, index n1

cladding, index n2

Optical fibers do not support TEM waves, like the hollow metallic guides. Their propagation constant and mode structures are even more compli- cated than for hollow metallic guides. Derivation of Transmission Line Equations (1-3) Let us consider a length of a transmission line at location z . The circuit model is clearly a series and resistance (since the whole line with a load at the end forms a loop) and a capacitance and shunt leak- age conductance (between the good conductors, across the dielectric which also has a small conductance). With the above circuit parameters being LRC,,, and G , the model is iztaf, L R iza +D zt, f + + vzta , f C G vza +D zt, f − −

A straight forward application of Kirchoff’s Loop Law gives ∂iztaf, vza +-DDD zt,,f vzta f =-LR z -zia zt, f , ∂t and Kirchoff’s Current Law at the upper gives ∂ vza +D zt, f izafafaf+-DDDD zt,, izt =-+-GC zvz zt , z . ∂ t Dividing through by ∆z and taking the limit ∆z→0 , we get: ∂v ∂i =-LR - i (1-5) ∂z ∂t ∂i ∂v =-CG - v (1-6) ∂z ∂t Sinusoidal Analysis of Transmission Lines (using phasors) ∂v =-aRL +jw fi (1) ∂z ∂ i =-afGC +jw v (2) ∂ z Differentiate (1) with respect to z : ∂ 2 vid =-afRL +jw ∂ z 2 dz Substitute from (2) ∂∂i/ z : ∂ 2v =+afafGCRLjjww + v ∂ z 2 Similarly, we get: ∂ 2i =+afafGCRLjjww + i ∂ z 2 The Lossless Transmission Line R = 0 (perfect conductor) G = 0 (lossless dielectric) This is not only instructive and simple, but also a good approximation of many real lines which are made of very good conductors (typically, copper) and nearly lossless dielectrics (e.g., teflon). For example, a commercial coaxial line is a nearly lossless line. ∂ 2 v = jjwwCLv ∂z 2 ◊ 2 ∂ 2 i ∂ v 2 2LC =-chw LC v Also, 2 =-chw i ∂ z 2 ∂ z ∂ 2 v ∂ 2 i =-b 2 v =-b 2 i ∂z 2 ∂z 2 ∂V ∂ I ∂ I ∂V =-L =-C ∂z ∂t ∂z ∂t The solutions are travelling waves: e ± jzβ (withe jtω already assumed)

jtafwb- z jtafwb+ z e ore or sum of both.

βω==LC ωµε

It will be shown later, when we consider transmission lines from the Electro- ’s point of view, that these voltage and current waves corre- spond to EM waves, also. It will be shown that: 1 LC ==µε , v where v is velocity of light in the medium between the conductors. We can get the relationship of V and I by substituting back into the original transmission line equations. jtafwb- z jtafwb- z v = Ve i = Ie (i.e., only +z going wave) dv di =−jωLi =−jω Cv dz dz −=−jjβωviL −=−jjβωivC

v ωL β == βω==LC ωµε, i β ω C v L + + ==Zo This is for z going wave only. i + C

Zo is called the of the line. For this lossless line,

Zo is real. It means that V and I are in phase. For example, power is prop- agated. None is absorbed. (Contrast this with ordinary circuits, where real impedance means power is absorbed and dissipated!) Power Propagated on a Lossless Line The power propagated can be calculated either from the electromagnetic wave or from the voltage-current wave. It is usual to use voltage and cur- rent for a transmission line. The power, of course, is an oscillating quantity. We are mostly interested, however, in the average power flow. If (real), then from the circuit picture: 2 ==1 1 Vo Pav Re vi* for lossless line 2 2 Zo If we have only a − z going wave, the same kind of derivation we did for the + z going wave gives:

v− =− − Zo (z going wave) i−

Circuits vs. Transmission Lines 1) Q: When does the “regular” circuit (E17) method hold correct; that is, only e jtω dependence (no propagation) and when does the transmission line method need to be used? A: Simple. When the is large compared to the dimensions, we can neglect transmission line concepts: π β =→2 zzλ 0 That is, whenever the distance scale (z) is small and/or the wavelength is long; that is, z << λ . = 1 2 2) For circuits: If is real (that is, ZR= ), PIRav 2 . This is power absorbed in the real impedance (resistance).

For a lossless transmission line: Zo is real. If we have only + going wave: 2 ==1 v+ 1 2 Pav i + Zo 2 Zo 2 This is not absorbed power. It is power propagating in the +z direction down the lossless line. = RZo ⇒ Power is absorbed in the load to Load where it does “something.” If we have - going wave: 2 ==−=−1 * 1 v− 1 2 Pav Re vi−− i − Zo 2 2 Zo 2 The negative sign means that power is propagating in the negative direc- tion. Unless otherwise stated, we always will mean average power. Transmission Line Circuits Transmission lines are used to carry signals from one location to another. Thus, they usually have a signal generator at one end and a load at another end. Let’s now examine the behavior of a transmission line with load. We use phasors and will assume lossless transmission line. 1) Transmission line terminated in its characteristic impedance. This is called “matched.” = Zo ZZLo Let’s assume we launched a + z going wave on this transmission line of

characteristic impedance Zo . Then, at any point z, we know that = = vi++/ Zo . At the load, viLL/ Z L . (This is definition of ZL .) Thus, if = = = = = ZZLo, then vvL + (ZL ) and iiL + (ZL ). Thus, there is no reflected wave (− z going wave). Rather, the wave continues into the load, where it gets dissipated. (In practical situations, the load is a whole circuit, for example, a detector, a , an oscilloscope. As long

as its is Zo , the foregoing discussion is true.) 2) Transmission line terminated in other impedance (1-7, text) ≠ ⇒ ZZLo Zo

Assume again we launched a signal at some frequency. Then, = = ≠ vi++/ Zo . But at the load, viLL/ Z L . If ZZLo , v L cannot be v+ and iL cannot be i + . The boundary condition at the load can only be satisfied if there is also a reflected wave vi−−, . This wave propagates backward in the − z direction. Then, at any point on the line: jtafwb- z jt af wb+ z vvv=+=+Ve+ Ve- +-

V+ jtafwb- z V- jt af wb+ z v− iii=-=+e e +- i− =− Zo Zo Zo Note the − sign on the current of the reflected wave! Now we can satisfy = the condition at the load vi/ ZL . Only one value of V− will satisfy. Let us define:

v− =Γ, the v+

Then:

vvv+++Γ = (1)

vv++v − Γ = (2) ZZZooL We solve for Γ to get: (Problem: Solve to G ) − Γ= ZZLo + ZZLo Γ Note: is complex, in general, because ZL may be complex. -£11G £ Fraction of the incident power reflected back:

1 1 221 Incident Power PZVZ= ReL vi* O = Re L v //O = iooNM2 ++QP NM2 +QP 2 + 1 1 Reflected Power P = ReL vi***O = Re L GG v iOa-1f r NM2 --QP NM2 - +QP 221 =-Re L GGvi* O =- P NM 2 ++QP i L1 ***O L1 O Total Power PTTT= Revi=+- Re a11GGf v++ej i NM2 QP NM2 QP L1 ***O =+--Re ej1 GG GGvi++ NM2 QP =−Γ 2 ≡ 1 Pi Transmitted power (into Load)

1 * Note: ejGG-=Im Gis purely imaginary. 2

3) Special Cases of Termination = A special case of the general load is ZL 0 , that is, a short circuited trans- mission line. (A transmission line may be shorted deliberately in order to achieve a certain result, or it may be shorted accidentally, something that has to be repaired.) For this case:

Zo Z Γ=−Zo =− 1 ZL = 0 Zo Z = 0 ~ We can say here that ΓΓ==eejjϕπ1 + here. If jtafwb- z, v + = Ae vafzAeAe=-- jzbb jz ce jtw understoodf 1 i afz =+Ae- jzbb Ae jz Zo Note that this time, the z dependence has two different parts. Thus, the z dependence cannot be lumped together with e jtω dependence, as it can be for a wave traveling in one direction only. Now, the entire V or I, including z dependence, is the phasor: vafzjAz=-2sinb jtw 2 A ce understoodγ iafz = cosb z Zo The real voltage and current are ( for AAe= jθ ) vztaf,sin=-Re mr2 jAejjtqwb ze =+2 Azsinbwq sinaf t

R2 Aejq U iztaf,cos= Re S b zejtw V T Zo W V 2 A =+cosbwqzt cosaf Zo z Conductor I

Both voltage and current oscillate sinusoidally in time with different maxi- mum values (amplitudes) at different locations. This is known as a --“standing,” because the amplitude remains the same at each location and the oscillating pattern is standing. Contrast this with “travelling” wave, in which a given point on the wave form progresses in distance with time. The amplitude of a travelling wave, however, is a constant value regardless of location. The instantaneous power flow is (must be calculated from real V, I): Pztaf,,,= vi afaf zt zt 4 A 2 =++sinbzz cos b sinafaf wq t cos wq t Zo A 2 =+sin22bwqzt sin af Zo The average power flow (we calculate it here direction from phasors) is: * L 1 * O L 1 2 A O ReM VIP =- Re M 2 jAsinbb z◊ cos zP N 2 Q NM 2 Zo QP ==Re 20jA2 sinββ z cos z

(purely imaginary) This makes physical sense, since no power flows into the short circuit. Thus, all the power must be reflected back, giving net power flow equal to zero (average). The same result, of course, could be derived from the field point of view. The short circuit is a perfectly conducting plate closing off the line. No fields penetrate into it. Thus, =−= PaveP z+− eP z 0

The Concept of Input Impedance ()Zin

Characteristic impedance Zo is the ratio of voltage to current of one wave direction.

==v+ v− Zo i + i − Since transmission lines are used in conjunction with lumped circuits, we must be able to treat them as regular circuits as well. The input impedance = of any circuit is ZVIin / at the terminals. Thus, for a transmission line, at any point on it:

= vT Zin , i T where vT and i T are the total phasor fields at that point. = Important: Since Zinvi T/ T , this is now the same as a regular circuit con- cept. So we can always replace a transmission line circuit with its input impedance! If there is only a single wave propagating in one direction (transmission line = = ≠ terminated in ZZLo ), then, of course, ZZin o . But, in general, ZZin o , and Zin is a function of both the load and of the position z on the line. Let’s measure as follows: Z = 0 at the load. Therefore, z is negative on the line zl=− . For the short circuited lossless transmission line: vaf-l --2 jAsin b af l Zl-= = z in a f 2 A iaf-l cosb af-l Zo zl=− z = 0 22p p f Za-= lf jZtanb l = jZ tan l = jZ tan l (for a shorted line) in o ol o v The input impedance of a short circuited lossless line is purely imaginary (as it should be, since it consists of distributed capacitance and inductance.) The values of the input impedance vary with the distance if the frequency is fixed or varies with frequency if the length is constant.

f

l

Reactance can be either capacitive or inductive, and its absolute value var- ies from zero to infinity. This property can be utilized in a number of ways. (a) Easily adjustable values of reactance. (b) Location of an accidental short on a line. (Problem)

(c) Matching of a load not equal to Zo (will study later)

Open Circuited Lossless Line In practice, we try to avoid open circuiting a transmission line. Reason: Theoretically, open circuit (i.e., open ends) means infinite load impedance =∞ (RL ). This would mean that: ∞− Γ= Zo =+ ∞+ 1 Zo That is, all of the wave is reflected, resulting in standing waves. However, in practice, this is not really true. The open end is not a truly infi- nite impedance. While no current flows beyond the open end, some of the electromagnetic wave goes out the open end and becomes a free space wave. Only a small fraction does, since the open circuited transmission line is “poorly matched” to space (i.e., it is a poor ). Yet, if not all the power is reflected, we obviously do not have infinite impedance.

Lossless transmission line terminated in a real load RLπ Z01 A = Z01 ZR02 L RL

Z01 RL A The above two cases are identical as far as T.L. 1 is concerned, since the impedance at point A is RL for both. We can always replace any transmission line by its equivalent input impedance. The fields up to that input point will be unchanged. ≠ For the above, since ZZ02 01 : − Γ=RZL 01 + RZL 01 Γ is real and its value varies from −1 to +1 . 1 A 2 = = ZR02 L ZRin A L

Line 1 has both + z and − z going wave. Line 2 has only +z going wave, since it is properly terminated. =+ Just to the left of junction A: vv11+− v = Just to the right of junction A: vv2+

These are obviously equal. So we replace at A by RL : 1

v + vv+−+ v − τ ==2 11=+111 =+Γ v1+ v1+ v1+ τ is called the . Thus:

= τ τ =+Γ =2 ZL vv21++ 1 + ZZLo τ can be greater than unity, but that does not mean more than 100% of power transmitted. += Power balance: PPPrti 02££t 2 2 ==vv1−+ΓΓ21 =2 Pr Pi 22ZZoo 2 τ 2 2 2 ==v2+ v1+ =τ 2 v1+ ⋅ Zo Pt 22ZZLL 2 Z oZL Substituting for Γ and τ and writing the sum of the reflected and transmit- ted power = incident: 2 2 LF I 2 F IO 2 vv1++ZZLo- 4 ZL Zo 1 MG J + 2 G JP = Z H ZZ+ K ZZ+ H ZZK o NM Lo b Log LoQP 22 ZL+- Z o24 ZZ Lo + ZZ Lo 2 = 1 bZZLo+ g 11= Q.E.D. Example: Z = 300Ω in z =0 Ω Ω = Ω 50 300 ZL 300 A Commercial coaxial line (Radio Shack) is 50Ω , flat antenna line is 300Ω . The above situation would happen if you connect them together without a matching transformer. (We shall learn later what a matching transformer is.) At the junction Azaf= 0 : − − Γ=ZZin o = 300 50 = + + 0714. ZZin o 300 50 τ =+11714Γ =. The fraction of power reflected is Γ 2 = 051. . The fraction of power transmitted to 300Ω line and on to the load is 1051049−=... General Transmission Line Now let’s return to a general case and find the total voltage and current on the line as a function of distance:

- jzbbbb+- jz jz2 jz v11=+Ve+ GG Ve 1+ = Veo ch1 + e (Let VV1+ = o )

V11+ - jzbbbbV + +-jz Vo jz2 jz i1 =-e GGe =eech1 - Z o Z o Z o jtω jtω e variation understood! We have dropped e .

The real fields, of course, vary sinusoidally in time. v1 and i1 are the com- plex phasor amplitudes. Since, for a complex voltage v = Ae jtω , the maxi- mum value (or sinusoidal amplitude) is A , and the sinusoidal amplitude of

V1 along the line is: - jzbb2 jz v1 =+=++Veo 1122GG e Vo afcosbb z j sin z (for G real)

2 22 2 =+VzzVzooa12GGcosbbf + sin 2 =++ 122 GG cos b Remember that Γ may be positive or negative. π β = 4 2 zzλ The minimum value of this is (if Γ is positive): v =+-=-VV12GG2 af 1 G 1 min oo For a general Γ : v =-V bg1 G 1 min o Similarly: v =++=+VV12GG2 b 1 Gg 1 max oo The location of consecutive minima and maxima are λ/2 apart. At the load, we have either a maximum or a minimum (max. for Γ positive). The shape of the function vs. z is not a pure sinusoid. The minima are not infinitely sharp (as they are for a short circuit).

Load

z An A.C. voltmeter, of course, does not measure the time variation, but only the amplitude (top solid line). (Actually the RMS value) This is called a partial standing wave. An important measurable quantity is the (SWR) = S. V 1+ Γ S ==max Vmin 1− Γ S can range from 1 to • . Example: For the previous example, 1714. S ==6 0286. Γ > We note also that for positive (ZZLo , ZL real), ZZ− 1 + Lo ZZ+ 2Z Z S = Lo==L L ZZ− 2Z Z 1 − Lo o o + ZZLo Γ < and for negative (ZZLo , ZL real), ZZ− 1 + oL ZZ+ 2Z Z S = oL==o o ZZ− 2Z Z 1 − oL L L + ZZoL This also agrees with our example: 630050= / If a transmission line is connected to an unknown load, the SWR can be measured and, from this, the load value found. But to fully understand how, we must consider a more general load.

Lossless Transmission Line Terminated in a General Complex Load ZL Most practical circuits/devices (whether the device be a transmitting antenna, a TV, a receiver, a , an oscilloscope, etc.) have complex input impedance. Thus, if such a device is fed by a transmission line, the transmission line sees a complex load. Without something special being done, the line is not matched to the load. There is a reflected wave. ZZ− Γ= Lo (complex, in general) + ZZLo =+− jzββΓ + jz vT Ve+ Ve+

=−V+ − jzββΓ V+ + jz iT e e Zo Zo =∠° = Assume for simplicity that v+ Vo 0 and that the load is at Z 0 . There- fore, on the transmission line, zl=− . Then, =+jlββΓ − jl vToVe Veo

=−Vo jlββΓ Vo − jl iT e e , Zo Zo where l is the distance measured backwards from the load. The input impedance at a point l distance back from the load is: eejlββ+ Γ − jl ZZ= in o eejlββ− Γ − jl 1+ Γ e − 2 jlβ = Z o 1− Γ e − 2 jlβ Substituting the value of Γ and rearranging, we get the alternate forms:

ZljZlLocosbb+ sin ZjZlLo+ tan b ZlZinaf-= o = Zo ZljZloLcosbb+ sin ZjZloL+ tan b = = When ZZLo , we get ZZin o . Now returning to the voltage and current:

− 2 jlβ ββ− 1 + Γ e v =+Vejl1 Γ e 2 jl ZZ= o in o 1 − Γ e − 2 jlβ V ββ− i =−o eejl1 Γ 2 jl another form Zo This is the same expression as before (zl=− ), except now Γ is complex Γ =∠Γ θ . It is easiest to see how this changes the standing wave by con- r sidering phasor diagrams. The amplitude of the quantity in square brackets determines the amplitude of the standing wave, since e jlβ = 1 . Definition of Generalized Reflection Coefficient - jlb Vl-() Ve- ()0 - 2 jlb GG()l = = + jlb = ()0 e zl=- Vl+() Ve+ ()0 In my notation GG= af0 , reflection coefficient at the load (complex quantity Γ~ in general) is equivalent to the book’s r .

Note: 1) Γ is real if load is real and line is lossless. Γ is complex if load is com- plex. 2) Gafl is always complex, except at certain points. These points have spe- cial significance! 3) GGalf ∫ , that is, moving about on the line only changes the phase angle of Gafl . Im At l = 0 (moving back, away from load) 1+Γ V min. Γ V max. Re −Γ Max. 1−Γ

It is obvious from this phasor diagram that the cases of real load and com- θ = plex load are not fundamentally different. r 0 or 180° for a real load, but can be anything for a complex load. The phasor rotates as l increases, making a complete cycle for 22βπl = ; that is, 42πλl/ = π , or l = λ /2 . +Γ =−Γ ΓΓ= Vmax is still 1 and Vmin 1 , where . Obviously, the shape of the standing wave is also the same as for a real load, except the maximum or minimum no longer occurs at the load.

Location of maxima and minima: Maximum occurs at: θβ−=±±ππ r 2024lmax ,,,etc. θβ−=±±ππ etc.Similarly, r 23lmin ,,etc.Similarly, Example: An RF signal is sent on a 300Ω parallel wire transmission line, ε = which has r 5 , to a receiver. A VSWR of S = 3 is measured on the line. The distance between two minima is 067..m and the nearest minimum to the load minimum is 045..m away. Find the frequency of the signal and the The’vnenin equivalent circuit of the receiver input. 067..m 045..m

Z = 300Ω o Z S = 3 L Solution: λ = λ = m /.2067 m 134. λλ==ε ⋅ = mr134.. 5 3m c 310× 8 f == ==108 100 MHz λ 3 π π β ==22 = π −1 λ 149. m m 134. −+=βθπ ∴+=ππθ 2 l rmin 298. lmr πθ==π = ° 2... 34r 0 34 61 2 1 + Γ S − 1 2 S = ∴=Γ ==05. 1 − Γ S + 1 4 Γ=0... 5 ∠ 61 2 °= 0 24 +j 0 . 44 − − Γ=ZZLo= Zl 300 + + ZZLoZL 300

ZL af1-=GG 300 af 1 +

ZjL a0.. 76-= 0 44f 300a 1 .. 24 + j 0 44f 124..+ j 044 132..∠° 195 Z = 300 = 300 =∠°450 49. 5 L 076..− j 044 088. ∠− 30 ° 342 − =+292j 342 Ω=RjL + ω Lh= =×05410. 6 210π × 8

Load is R = 292Ω Lh= 054. µ ANS. Looking at the definition of Zin and at the rotating phasor diagram of the pre- vious page, we also see that: At the location of a voltage maximum: 1 + Γ ZZ= = ZS in o1 − Γ o At the location of a voltage minimum: 1 - G Z ZZ= = o in o 1 + G S

= ZZSin o at voltage maximum Z Z = o at voltage minimum in S This is an important observation. It will be used soon for matching. Matching a transmission line to an unmatched load Reflections on a transmission line are undesirable for a number of reasons. For example, digital signals from PCM reflected pulses can bounce back again from the sending end if that end is also unmatched. Then false sig- nals can be detected. Q: How do we match an unmatched load? A: There are several methods. We now study the first one.

I. Quarter Wave Matching (λ /4 “Transformer”)

Suppose we have a transmission line terminated in a real load ZRLLbg∫

⇒ Z02 is intrinsic impedance. Zin Z02

ZL

2 = λ = Z02 If the length of this line is exactly l /4 , then Zin , since ZL 2π λπ sinβl =⋅== sin sin 1 λ 42 and cosβl = 0 This then offers a method to match this load to any lossless transmission line , intrinsic impendance to be matched to . real Z01 Z0 ZL ZL bgRL A 2 Z 01 1 Z02 ZL λ 4 λ = ⋅ We insert a transmission line of /4 , length (2 ), and having ZZZ02 01 L . Then, at (A ): Z 2 ZZ ==02 01 L = Zin Z 01 ! ZL ZL = Since ZZin A 01 , we are matched. No reflection, since at point A, we could replace the whole combination by Zin A . ZZ= Z 01 in 01

How is this possible physically? We won’t go into a detailed analysis of what really happens to the right of point A. But briefly, we do have (in actual physical fact) reflection of waves at point A and at ZL. The two reflections, however, cancel each other out on the line 1 .

Q: This obviously works if the load is real, since Z01 is also real, resulting in real Z02 . But if the load is complex, then what to do? A: Recall that the input impedance at a voltage minimum or maximum is real ZSo and ZSo / . Thus, these are the steps for matching a complex load

ZL to a line Z01 .

1) Add a section of Z01 line to the load. The length should be lmax or lmin of the standing wave that results. Say, we use lmax :

lmax

ZL = This is equivalent to a load of ZSZL 01 . 2) Now use λ /4 , matching on this equivalent load. For example, insert a line: == = λ ZZZSZZSZ02L 01 01 01 01, length /4

Z Z 01 B A 01 Zo λ /4

lmax The impedance at point B then is: SZ2 SZ2 ===01 01 Zin B 1 Z01 ZL SZ01 Thus we are matched. No reflection going back on line 1. Same com- ment applies as before. Actually, there are reflections at points A and B, but they are 180∞ out of phase and cancel out. The 50W - 300W “transformer” supplied for TV and video components is a lumped circuit of a λ /4 transformer! 2π λπ Important Note: l = λ /4 means βl ==⋅ λ 42 22π f π The matching is only valid at one frequency! Since β == , if f is ν λ changed, the wavelength is changed and βπl ≠ /2 for the same l. Example: (a) Match a 50 to a 300W real input impedance device at a frequency of 180 MHz. Assume all transmission lines used ε = have r 6 . (b) Find the reflection coefficient of the matched circuit at 90 MHz. (a) At 180 MHz: c 310× 8 λ == = 167..m o f 18. × 108 λ λ on the T.L., λ ==o 068..m m 6 ==⋅ Ω T.L. impedance needed Zans02 50 300 122.. 5

068. length needed ==017..mans . 4 50Ω 122. 5Ω = Ω ZL 300

λ = 017..m 4 We cannot readily buy a 122.5W transmission line. However, any transmis- sion line at these R.F. frequencies can be simulated by a proper lumped cir- cuit network. (In this class, we won’t study how.) (b) If the frequency is changed to 90 mHz: 2π π λ =×=068.. 2 136 βl =×=017. m 136 4. We have: π Ω βl = 50Ω 122. 5 300Ω 4 tanβl = 1 l = λ 8 300+×j 122. 5 1 324∠° 22. 2 Z = 122. 5 = 122. 5 inA 122. 5+×j 300 1 324∠− 90 22 =∠−°=−122.... 5 45 6 85 8j 87 5 Reflection coefficient at A is: −− − Γ=85.. 8j 87 5 50 = 33.. 5j 87 5 85.. 8−+j 87 5 50 135.. 8− j 87 5 937. ∠− 69 ° = =∠−°058... 362 ans 161.. 5∠− 32 8 ° Fraction of power reflected back at A is Γ 2 = 034. .