Experiment 3- Transmission Lines, Part 1

Transcription

1 Experiment 3- Transmission Lines, Part 1 Dr. Haim Matzner& Shimshon Levy. August, 2008.

2 Contents Introduction PrelabExercise BackgroundTheory CharacteristicImpedance TerminatedLosslessLine TheImpedanceTransformation ShortedTransmissionLine OpenCircuitLine LowLossLine CalculatingTransmissionLineparametersbyMeasuring,α,and β Calculating Coaxial Line Parameters by Measuring PhysicalDimensions Experiment Procedure RequiredEquipment Phase Difference of Coaxial Cables, Simulation and Measurement Wavelength of Electromagnetic Wave in Dielectric Material InputImpedanceofShort-CircuitTransmissionLine Impedance Along a Short - Circuit Microstrip Transmission Line FinalReport Appendix-1 - Engineering Information for RF Coaxial Cable RG v

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4 Introduction 0.1 Prelab Exercise 1. Define the terms: Relative permittivity, dielectric loss tangent (tan δ), skin depth and distributed elements. 2. RefertoFigure1,findthefrequencythatwillcausea2πradiansphase differencebetweenthetworg-58coaxialcables,assumeµ R =1,ɛ r =2.3. a. Find the relative amplitude of the two signals. b. Verify your answers in the laboratory. Signal generator ,00 MHz 0.7m RG-58 Oscilloscope 3.4m RG-58 Figure 1-Phase difference between two coaxial cables. 0.2 Background Theory Transmission lines provide one media of transmitting electrical energy between the power source to the load. Figure 2 shows three different geometry types of lines used at microwave frequencies. 1

5 2 CHAPTER 0 INTRODUCTION Coaxial cable Two paralel wire line Microstrip Line Figure 2- Popular transmission lines. Rectangular metal waveguide The open two-wire line is the most popular at lower frequencies, especially for TV application. Modern RF and microwave devices practice involves considerable usage of coaxial cables at frequencies from about 10 MHz up to 30GHzandhollowwaveguidesfrom1to300GHz. A uniform transmission line can be defined as a line with distributed elements, as shown in Figure 3. R =Seriesresistanceperunitlengthofline(Ω/m). Resistance is related to the dimensions and conductivity of the metallic conductors, resistance is depended on frequency due to skin effect. G =Shuntconductanceperunitlengthofline( /m). G isrelatedtothelosstangentofthedielectricmaterialbetweenthetwo conductors. ItisimportanttorememberthatG isnotareciprocal ofr. They are independent quantities, R being related to the various properties ofthetwoconductorswhileg isrelatedtothepropertiesoftheinsulating material between them.

6 0.2 BACKGROUND THEORY 3 First section Second section Third-section R/2 L/2 L/2 R/2 ZG G C G C G C Equivalent circuit of transmission line z=0 Figure 3-Transmission lines model L =Seriesinductanceperunitlengthofline(H/m). L - Is associated with the magnetic flux between the conductors. C =Shuntcapacitanceperunitlengthofline(F/m). C - Is associated with the charge on the conductors. Naturally, a relatively long piece of line would contain identical sections as shown. Sincethesesectionscanalwaysbechosentobesmallascomparedto the operating wavelength. Hence the idea is valid at all frequencies. The series impedance and the shunt admittance per unit length of the transmission linearegivenby: Z = R +JωL Y = G +JωC The expressions for voltage and current per unit length are given respectively by equations(1) and(2): dv(z) dz di(z) dz = I(z)(R +JωL ) (1) = V(z)(G +JωC ) (2) Wherethenegativesignindicatesonadecreaseinvoltageandcurrentasz increases. The current and voltage are measured from the receiving end; at z = 0 and line extends in negative z-direction. The differentiating equations, (3) and(4), associate the voltage and current:

7 4 CHAPTER 0 INTRODUCTION d 2 V(z) dz d 2 I(z) dz Where = (R +JωL ) di(z) dz = (G +JωC) dv(z) dz =(G +JωC )(R +JωL )V(z)=γ 2 V(z) (3) =(G +JωC )(R +JωL )I(z)=γ 2 I(z)(4) γ= ZY = (G +jωc )(R +jωl ) (5) The above equations are known as wave equations for voltage and current, respectively, propagating on a line. The solutions of voltage and current waves are: V(z) = V 1 e γz +V 2 e +γz (6) I(z) = I 1 e γz +I 2 e +γz These solutions are shown as the sumof two waves; the first term,v 1, indicates the wave traveling in positive z-direction, and is called the incident wave,whilethesecondterm,v 2,indicatesthewavetravelinginthenegative z-direction,andiscalledthereflectedwave. γ isacomplexnumberthatis called the propagation constant and can be defined as: γ=α+jβ (7) αiscalledtheattenuationconstantofthepropagatingwave,αistherealpart ofeq. (2.7)whileβ istheimaginarypartandiscalledthephaseconstant. Thus, propagation constant γ is the phase shift and attenuation per unit length along the line. Separating equation (2.5) into real and imaginary parts,wecanget: α= β= [ (G 2 +ω 2 C 2 )(R 2 +ω 2 L 2 )+(R G ω 2 L C ) [ (G 2 +ω 2 C 2 )(R 2 +ω 2 L 2 ) (R G ω 2 L C ) 2 2 ]1 2 ]1 2 (8) (9) αismeasuredinnepersperunitlengthofthetransmissionline(1neper = 8.686dB). β is measured in radians per unit length of tthe transmission

8 0.2 BACKGROUND THEORY 5 line. Thatmeansthatβ canbecalculatedas: β= 2π λ (10) Whereλisthewavelengthordistancealongthelinecorrespondingtoaphase changeof2πradians. Ifthewavelengthinfreespaceisdenotedbyλ 0,then: λ 0 = c = v p ɛr µ R =λ ɛ R µ f 0 f R (11) Wherecisthevelocityoflightinfreespaceandv p isthevelocityofelectric wave in a dielectric material Characteristic Impedance IfRFvoltageV isappliedacrosstheconductorsofaninfiniteline,itcausesa currentitoflow. Bythisobservation,thelineisequivalenttoanimpedance, whichisknownasthecharacteristicimpedance, : = V(z) I(z) TheexpressionforcurrentI,usingEqs. (2.1)and(2.6),isgivenby: Where 1 I= (R +JωL ) I 1 = V z = 1 (R +JωL ) ( γ)(v 1e γz +V 2 e +γz ) I(z)=I 1 e γz +I 2 e +γz V 1 (R +JωL ) (γ)and I 2= V 2 (R +JωL ) (γ) Infinitelinehasnoreflection,thereforeI 2 =V 2 =0and: V(z) I(z) = V 1 I 1 = (R +JωL ) (R +JωL )(G +JωC ) Thus the characteristic impedance of infinite line can be calculated by(12): = V(z) I(z) = (R +JωL ) (12) (G +JωC )

9 6 CHAPTER 0 INTRODUCTION Line Impedance Wecanrewritethecurrentandvoltangealongthelineas: V(z) = V 1 e γz +V 2 e +γz (13) I(z) = V 1 e γz + V 2 e +γz andthelineimpedance(theimpedanceatpointz)as: Z(z)= V(z) I(z) =(V 1 e γz +V 2 e +γz ) V 1 e γz V 2 e +γz (14) Terminated Lossless Line Consider a lossles line, length l, terminated with a load Z L, as shown in Figure 4. I(z) I L V(z) V L Z l z = l l z = 0 Figure 4- A transmission line, terminated with a load. Usingequation(2.13)andrecalling thatα=0, one candefine thecurrent andvoltageontheloadterminal(atz=0)as: V L I L =Z L (15) V(z = 0)=V 1 e jβ(0) +V 2 e jβ(0) =V 1 +V 2 (16) I(z = 0)= V 1 e jβ(0) V 2 e jβ(0) = V 1 + V 2

10 0.2 BACKGROUND THEORY 7 We can also add the boundary conditions: V(z = 0)=V L (17) I(z = 0)=I L Using equation(2.16) we can rewrite(2.15) as equation(2.18): Z L = V L = V(z=0) I L I(z=0) = V 1+V 2 V 1 (18) + V 2 Rearranging equation(2.18) one can rewrite: V 2 = Z L Γ V 1 Z L + Where Γ is the complex reflection coefficient. It relates to the magnitude and phaseofthereflectedwave(v 2 )emergingfromtheloadandtothemagtitude andphaseoftheincidentwave(v 1 ),or Equation(2.13) can be rewrite as: V 2 =ΓV 1 (19) V(z) = V 1 e jβz +ΓV 1 e jβz (20) I(z) = V 1 e jβz Γ V 1 e jβz The reflection coefficient at an arbitrary point along the transmission line is: Γ(z)= V 2e jβz =Γ(0)e 2jβz V 1 e jβz Whilethetimeaveragepowerflowalongthelineatpointzis: P av = 1 2 Re[V(z)I(z) ] (21) = 1 2 V 2 1 Re ( 1 Γ e j2βz +Γe j2βz Γ 2) WhenrecallingthatZ Z =2jImZ,equation(2.21)canbesimplifiedto: P av = 1 2 V 2 1 ( 1 Γ 2 ) (22) Which shows that the average power flow is constant at any point on the lossless transmission line. If Γ = 0(perfect match), the maximum power is delieveredtotheload,whileallthepowerisreflectedforγ=1.

11 8 CHAPTER 0 INTRODUCTION The Impedance Transformation According to the transmission line theory, in a short circuit line, the impedance become infinite at a distance of one-quarter wavelength from the short. The ability to change the impedance by adding a length of transmission line is a very important attribute to every RF or microwave designer. If we look at the transmission line(losseless line), as illustrated in Figure 5, anduseequation(2.20), theline impedance atz= l (inputimpedance) is: Z in = V(z= l) I(z= l) = ( ) e jβl +Γe jβl e jβl Γe jβl (23) l I L I (z) ZG Z in β V L ZL z = l z= 0 Figure5-TransmissionlineterminatedinarbitraryimpedanceZ L. IfweusetherelationshipΓ=(Z L )/(Z L + )withequation(2.23)we get: ( ( ZL e jβl +e jβl) ( + e jβl e jβl) ) Z in = (24) Z L (e jβl +Γe jβl ) (e jβl e jβl ) By recalling Euler s equations: e jβl = cosβl+jsinβl e jβl = cosβl jsinβl Wecanget: ( ) ZL cosβl+j sinβl Z in = cosβl+jz L sinβl Or: ( ) ZL +j tanβl Z in = +jz L tanβl Lets examine special cases: (25)

12 0.3 SHORTED TRANSMISSION LINE Shorted Transmission Line The voltage and current along a transmission line as a function of time is known as: [ ) ] [ ) ] V(z,t) = Acos ω (t zvp +θ +Bcos ω (t+ zvp +φ (26) I(z,t) = 1 [ ) ] [ ) ]} {Acos ω (t zvp +θ Bcos ω (t+ zvp +φ Wherez=0,atthegeneratorside,and d=0attheloadside,asshownin Figure 6. z=d z V I(z), β z z=0 Figure 6-Shorted transmission line. This expression can be rewrite in phasor form as: V(z) = V 1 e jβz +V 2 e jβz (27) I(z) = 1 ( V1 e jβz V 2 e jβz) WhereV 1 =Ae jθ istheforwardwave,v 2 =Be jϕ isthereflectedwaveand β =ω/v p.theboundaryconditionfortheshortcircuittransmissionlineat z=0isthatthevoltageacrosstheshortcircuitiszero: Using equations(27) and(28), we obtain: V(0)=0 (28)

13 10 CHAPTER 0 INTRODUCTION V(0) = V 1 e jβ0 +V 2 e jβ0 =0 (29) V 1 = V 2 InsertingEq(29)intoEq(28),weget: V(z) = V 1 e jβz V 1 e jβz =2jV 1 sinβz (30) I(z) = 1 ( V1 e jβz +V 1 e jβz) =2 V 1 cosβz WhichshowsthattheV =0 attheloadend,whilethecurrentisamaximum there. TheratiobetweenV(z)andI(z)istheinputimpedanceandisequal to: Z in =j tanβz (31) Which is purely imaginary for any length of z. The value of Z in(sc) vary between +j to j, every π/2 (λ/4),by changing z, the length of the line,orbychangingthefrequency. Thevoltageandcurrentasafunctionof time and distance are: V(z,t) = Re [ V(z)e jωt] (32) = Re ( 2e jπ/2 V 1 e jθ e jωt sinβz ) = 2 V 1 sinβzsin(ωt+θ) WhereV 1 = V 1 e jθ andj=e jπ/2. I(z,t) = Re [ I(z)e jωt] (33) ( = Re 2 V ) 1 cosβze jωt 2 = V 1 cosβzcos(θ+ωt)

14 0.3 SHORTED TRANSMISSION LINE 11 Figure 7-Voltage along a shorted transmission line, as a function of time and distance from the load. Frequency 1GHz. The result of the voltage on the shorted transmission line is shown in figure7,assumethatv 1 =sin2π10 9 t(λ=30cm),whichshowsthat: The line voltage is zero for βz = 0+nπ n = 0,1,2...for all value of time. Thevoltageateverypointofzissinusoidalasafunctionoftime. The maximum absolute value ofthe voltage is known as a standing wave pattern. Impedance magnitude Frequency difference 0 Wavelengths 1 (λ) 2 Figure 8- Input impedance of shorted low loss coaxial cable.

15 12 CHAPTER 0 INTRODUCTION Open Circuit Line- Forlosslesslineα 0and Y L =0,equation(2.25)isreduceto: or in impedance form: Y in =Y 0 tanβz Z in(os) = tanβz (34) UsingEqs. (2.24),(2.34)and(2.35), canbecomputedfromtheshortand open circuit, by the relation: = Z in(sc) Z in(os) Low Loss Line In many cases, the loss of microwave transmission lines is small. In these cases, some approximation can be made that simplify the expression of propagationconstantγ andcharacteristicimpedance.equation(5)canberearranged as: ( ) γ = (jωc jωl )(1+ R jωl ) 1+ G (35) jωc = jω ( ) R L C 1 j G ωl + R G ωc ω 2 L C IfweassumethatR <<ωl andg <<ωc forlowlossline,thanr G << ω 2 L C,thereforeEq. (36)isreducedto: γ=jω ( ) R L C 1 j G ωl + ωc (36) Byusing Taylorapproximation 1+x 1+ x, the propagation constant 2 can be approximated by: γ jω [ L C 1 j ( )] R G 2 ωl + ωc

16 0.4CALCULATINGTRANSMISSIONLINEPARAMETERSBYMEASURING,α, so Reγ = α is the attenuation constant and Imγ = β phase constant, therefore: ( α 1 ) C R L +G (37) 2 L C ( ) R = 1 2 +G =α c +α d Where α c is the conductor loss and α d is the dielectric loss. The phase constant is equal to: β ω L C (38) L Where =. C Note that the characteristic impedance of a low loss transmission line can be approximated by: = (R +JωL ) L (G +JωC ) C 0.4 Calculating Transmission Line parameters bymeasuring,α,and β Byknowingtheprimaryparameters,α,andβ,theequivalentparameters R,L,CandGcanbeextractedbymultiplyingthegeneralexpressionsof andγ: R +jωl = γ = (R 0 +jx 0 )(α+jβ) = (R 0 α x 0 β)+j(r 0 β+x 0 α) (39) WhereR 0 istherealpartof andx 0 isitsimaginarypart. ByEquating therealandimaginarypartweget: R =R 0 α x 0 β Ω m (40) and: L = R 0β+x 0 α ω H m (41)

17 14 CHAPTER 0 INTRODUCTION Bydividingthegeneralexpressionsof andγ,weget G +JωC = γ (42) = α+jβ R 0 jx 0 R 0 +jx 0 R 0 jx 0 = R 0α+x 0 β +j R 0β x 0 α R0 2+x2 0 R 2 0 +x2 0 By equating real and imaginary parts, we get: (43) and: G = R 0α+x 0 β R 2 0+x 2 0 C = R 0β x 0 α (R 2 0+x 2 0)ω (44) (45) Calculating Coaxial Line Parameters by Measuring Physical Dimensions ReferringtoFigure9,ifthecenterconductorischargedto+qandtheouter conductor(shielding) is charged to q, than the electric field lines will directe radially outward, while the magnetic lines will surround the inner conductor. Dielectric material z=0 l z=l magnetic field lines b E field lines a Figure 9- Geometry and elemagnetic fields lines of coaxial cable. By applying Gauss law to a cylindrical structure like a coaxial cable of length landradiusr,wherea< r>b: l 2π 0 0 Drdφdz=q

18 0.4CALCULATINGTRANSMISSIONLINEPARAMETERSBYMEASURING,α, Disindependentofzandφtherefore: D= q 2πrl ande= q 2πɛ 0 ɛ r rl The voltage difference between the conductors: V = b a E.dr= q 2πɛ 0 ɛ r l lnb a Hence: C= q V = 2πɛ 0ɛ r l (46) ln b a All dielectric materials have lossy at microwave frequencies. Hence one can look at a coaxial cable as a capacitor with parallel conductance or Y =G+jωC=jωC(1 j G ωc ) The quantity j G called material loss tangent and assign as tan δ. This ωc quantity indicates the relative magnitude of the loss component. It is often used to specify the loss properties of dielectrics: G=jωCtanδ= 2πɛ 0ɛ r l ωtanδ (47) ln b a Inductance of Transmission line The inductance of solenoid known as L= NΦ I (48) Flux φ Area- A l N turns r Current I Figure 10-Inductance of solenoid.

19 16 CHAPTER 0 INTRODUCTION wheren isthenumberofturns,listhelengthandi isthecurrentinthe solenoid. Applyingamperelawtoacircleradiusr(b<r>a)aroundany turnofthesolenoidwillresultin: H= I 2πr I r H= Magnetic field arround current carrying conductor I 2πr H Magnetic flux between the conductor Figure 11-Magnetic field around conductor and solenoid. Ifonewantstomeasurethetotalfluxbetweentheconductors,it stheflux enter the turn shown in Figure10 and Figure11. Therefore Φ= l b 0 a Bdrdz=µ 0 µ r l b a I 2πr dr= µ 0µ R li ln b 2π a Coaxial cable could be considered as a solenoid with one turn. By using equation 1.50, the expression for Φ and N=1 becomes: L= µ 0µ R l 2π lnb a (49) Resistance of Transmission Line The high frequency resistance of a coaxial cable is equal to the D.C. resistance of a equivalent coaxial cable composed of two hollow conductors with the radii, a and b, respectively, and with thickness equal to the skin depth penetration δ. Skin depth penetration is defined as δ skin = 1 πfµ0 µ R σ meters Where σ is the conductivity of the conductor. Therefore the resistance per unit length is: R = 1 2πaδ skin σ + 1 2πbδ skin σ = b+a 2πaδ skin σb (50)

20 Experiment Procedure 0.5 Required Equipment 1. Network Analyzer HP 8714B. 2. Arbitrary Waveform Generators(AW G)HP 33120A. 3. Agilent ADS software. 4. Termination-50Ω. 5. Standard 50Ω coaxial cable. 6. Open Short termination. 7. SMA short termination cm 50Ω FR4 microstrip line 0.6 Phase Difference of Coaxial Cables, Simulationand Measurement In this part of the experiment you will verify your answers to the prelab exercise. Signal generator ,00 MHz 0.7m RG-58 Oscilloscope 3.4m RG-58 Figure 1- Phase difference between two coaxial cables 17

21 18 CHAPTER 0 EXPERIMENT PROCEDURE 1. SimulatethesystemasindicatedinFigure2(seealsoFigure1). S-PARAMETERS S_Param SP1 Start=1.0 MHz Stop=100 MHz Step=1.0 MHz PwrSplit2 PWR1 S21=0.707 P_AC S31=0.707 PORT1 Num=1 Z=50 Ohm Pac=polar(dbmtow(0),0) Freq=freq COAX TL1 Di=0.9 mm Do=2.95 mm L=3400 mm Er=2.29 TanD= Rho=1 COAX TL2 Di=0.9 mm Do=2.95 mm L=700 mm Er=2.29 TanD= Rho=1 Term Term2 Num=2 Z=50 Ohm Term Term3 Num=3 Z=50 Ohm Figure 2- Phase Difference Simulation. 2. PlotS 21 (phase)ands 31 (phase)onthesamegraph,findthefrequency wherethephasesofthetwocablesareequal. Savethedata. Compare this frequency to the frequency you calculated in the prelab exercise. 3. Connect coaxial cables to a oscilloscope using power splitter, as indicatedinfigure1. 4. Set the signal generator to the frequency according to your simulation, and verify that the phase difference is near 0 degree. Save the data on magnetic media. 0.7 Wavelength of Electromagnetic Wave in Dielectric Material 1. Select a 50 cm length FR4 microstrip line, Width = 3mm, Height = 1.6mm,ɛ r =4.6, =50Ω,,v p =0.539c,ɛ eff =3.446.

22 0.7 WAVELENGTH OF ELECTROMAGNETIC WAVE IN DIELECTRIC MATERIAL microstrip length 50cm Figure3-50cmshortcircuitmicrostripline. 2. Calculate the frequency that half the wavelength in the dielectric materialisequalto50cm(λ/2=50cm). Recordthisfrequency. Verify your answer using ADS: Menu-> Tools-> LineCalc. MSub TRANSIENT PARAMETER SWEEP MSUB MSub1 H=1.6 mm Er=4.7 Mur=1 Cond=1.0E+50 Hu=1.0e+033 mm T=17 um TanD=0.01 Rough=0 mm Tran Tran1 StopTime=26 nsec MaxTimeStep=0.1 nsec Var Eqn VAR VAR1 x=26 ParamSweep Sweep1 SweepVar="x" SimInstanceName[1]="Tran1" SimInstanceName[2]= SimInstanceName[3]= SimInstanceName[4]= SimInstanceName[5]= SimInstanceName[6]= Start=0 Stop=50 Step=1 P_1Tone PORT1 MLIN Num=1 TL1 Z=50 Ohm Subst="MSub1" P=polar(dbmtow(0),0) W=3 mm Freq=161 MHz L=x cm I_Probe I_Probe1 MLIN TL2 Subst="MSub1" W=3 mm L=(50-x) cm Term Term2 Num=2 Z= mohm Figure 4- Simulation of short sircuit microstrip transmission line.

23 20 CHAPTER 0 EXPERIMENT PROCEDURE 3. Simulate a short circuit λ/2 microstrip line according to Figure 4. Drawagraphofthecurrentasafunctionoftime. Explain the graph. Find the distance where the current amplitude is minimum all the time. Findthetimewherethecurrentamplitudeisminimumateverypointon the microstrip line. Save the data. 4. Find the position where the current amplitude is maximum and change the simulation accordingly. Drawagraphofthecurrentasafunctionoftimetoproveyouranswer. Save the data. 5. Find the position where the current amplitude is minimum. Draw a graphof thecurrentasafunctionoftimetoproveyouranswer. Savethe data. 6. Connect the short circuit microstrip line to a signal generator, adjust the frequency to the calculated frequency and the amplitude to 0 dbm. 7. Measure the amplitude of the signal, using wideband oscilloscope at loadendandsignalgeneratorend,whytheamplitudeofthesignalatboth endsiscloseto0volt. 8. Measure the voltage along the line using the probe of the oscilloscope, verifythatthevoltageismaximumatthemiddleoftheline. 9. Calculate the frequency for λ = 50 cm in the dielectric material. Measure the voltage along the line, how many points of maximum voltage exist along the line? 0.8 Input Impedance of Short- Circuit Transmission Line 1. Simulate a transmission line terminated by a short circuit, as shown in Figure 5.

24 0.8 INPUT IMPEDANCE OF SHORT- CIRCUIT TRANSMISSION LINE21 S-PARAMETERS Zin N Zin Zin1 Zin1=zin(S11,PortZ1) Term Term1 Num=1 Z=50 Ohm Var Eqn COAX TL1 Di=2 mm Do=4.6 mm L=70 mm Er=1 TanD=0.002 Rho=1 VAR VAR1 X=1.0 S_Param SP1 Freq=642 MHz COAX TL2 Di=2 mm Do=4.6 mm L=X Er=1 TanD=0.002 Rho=1 PARAMETER SWEEP ParamSweep Sweep1 SweepVar="X" SimInstanceName[1]="SP1" SimInstanceName[2]= SimInstanceName[3]= SimInstanceName[4]= SimInstanceName[5]= SimInstanceName[6]= Term Start=0 Term2 Stop=30 Num=2 Step=0.25 Z= mohm Figure 5- Input impedance of a short circuit coaxial transmission line. 2. DrawagraphofZ in asafunctionoflengthx,savethedata. 3. If you use fix length of transmission line, which parameter of the simulation you have to change in order to get the same variation of Z in, prove your answer by simulation. 4. Connect line stretcher terminated in a short circuit to a network analyzer(see Figure 6).

25 22 CHAPTER 0 EXPERIMENT PROCEDURE R F O UT R F IN Coaxial line strecher Short Figure 6- Input impedance measurment of short circuit coaxial transmission line. 5. Calculate the frequency for the wavelength 1.5λ = 70cm(line stretcher at minimum overall length). 6. Set the network analyzer to smith chart format and find the exact frequency(aroundthecalculatedfrequency)fortheinputimpedancez in = 0.(line stretcher at minimum overall length), set the network analyzer to CW frequency equal to this measured frequency. 7. Stretchthelineandanalyzetheimpedanceinasmithchart. 0.9 Impedance Along a Short - Circuit Microstrip Transmission Line 1. Simulate a short circuit transmission line(see Figure 7) with length 50cm, width3mm,height1.6mmanddielectricmaterialfr4ɛ eff =3.446.

26 0.9 IMPEDANCE ALONG A SHORT- CIRCUIT MICROSTRIP TRANSMISSION LIN PARAMET ER SWEEP ParamSweep Sweep1 SweepVar=" x" SimInstanceName[1]=" SP1" SimInstanceName[2]= SimInstanceName[3]= SimInstanceName[4]= SimInstanceName[5]= SimInstanceName[6]= Start=0 Stop=50 Step=0.5 Var Eqn VAR VAR1 x=0 Term Term1 Num=1 Z=50 Ohm S-PARAMET ERS S_Param SP1 Freq=161 MHz MLIN TL2 Subst=" MSub1" W =3 mm L=x cm MLIN TL1 Subst=" MSub1" W =3 mm L=(50-x) cm Zin N Zin Zin1 Zin1=zin(S11,PortZ1) Term Term2 Num=2 Z=0.001 mohm Term Term3 Num=3 Z=100 MOhm MSub MSUB MSub1 H=1.6 mm Er=4.7 Mur=1 Cond=1.0E+50 Hu=1.0e+033 mm T=0 mm TanD=0.002 Rough=0 mm Figure 7- Simulation of a short and open circuit microstrip transmission lines. Pay attention that the simulation is parametric(x is the distance parameter). Verifythetforx λ/2theinputimpedanceisasuperpositionofashort and an open circuit transmission line connected in parallel and explain why isthisso. 2. Drawagraphoftheinputimpedanceasafunctionofthedistancex. Save the data. 3. Connect the microstrip to the network analyzer(as shown in Figure 8), set the network analyzer to impedance magnitude measurement.

27 24 CHAPTER 0 EXPERIMENT PROCEDURE 0 λ / 8 Open Short and Load Calibration RF OUT RF IN λ / 2 microstrip length 50cm 3 λ 8 λ / 4 Figure 8- Measurement of input impedance of short circuit microstrip transmission line 4. Set the network analyzer to Continuos Wave(CW), at the calculated frequency for λ/2 = 50cm.Calibrate the network analyzer to 1 port(reflection)attheendofthecable. 4. Whatistheexpectedvalueoftheimpedanceateachpoint? Measure the impedance magnitude at λ/2, 3λ/8, λ/4, λ/8, 0. Save the data on magnetic media Final Report 1. Attach and explain all the simulation graphs. 2. Using MATLAB, draw a 3D graph of the current as a function of timeanddistanceofashortcircuitmicrostriplineforafrequencyof1ghz (similartofigure7fromchapter1). Choosetherangesoftimeanddistance asyouwishandattachthematlabcode. 3. Using MATLAB, draw a graph of the input impedance,, as a functionofthelength,x,foralosslessshortcircuitcoaxiallineforafrequency of 161 MHz. Compare this graph to your measurement and ADS simulation. Attach the matlab code. 4. UsingthephysicaldimensionofacoaxialcableRG-58thatyoumeasured, find the following parameters versus frequency(if apply) and draw a graph of: a. Shunt capacitance C per meter. b. Series inductance L per meter. c. Series resistance R per meter. d. Shunt conductance G per meter.

28 0.11 APPENDIX-1- ENGINEERING INFORMATION FOR RF COAXIAL CABLE RG 0.11 Appendix-1 - Engineering Information for RF Coaxial Cable RG-58 Component Construction details Nineteen strands of tinned copper wire. Inner conductor each strand 0.18 mm diameter Overall diameter: 0.9 mm. Dielectric core Type A-1: Solid polyethylene. Diameter: 2.95 mm Outer conductor Single screened of tinned cooper wire. Diameter: 3.6mm. Jacket Typella: PVC. Diameter: 4.95mm. Engineering Information: Impedance: 50Ω Continuous working voltage; 1,400 V RMS, maximum. Operating frequency: 1 GHz, maximum. Velocity of propagation: 66% of speed light. Dielectric constant of polyethylene: 2.29 Dielectric loss tangents of polyethylene: tan δ = Operating temperature range: -40 C to +85 C. Capacitance: 101 pf. m

29 26 CHAPTER 0 EXPERIMENT PROCEDURE Attenuation of 100m Rg.-58 coaxial cable Attenuation (db) y = f f Frequency(MHz) Figure 9- Attenuation versus frequency of a coaxial cable RG-58.