Solution Manual for Microwave Circuit Design A Practical Approach Using ADS, 1st Edition

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Note:When Figures in the text are cited, it is denoted byFigureand figures in this End-ofchapter solutions are denoted by Figure.CHAPTER 1 PROBLEMS1.1Waveguide generally has lower line loss than microstrip. SIW(Substrate Integrated Waveguide)can be regarded as the planar version of waveguide. How SIW is configured using substrate?SolutionA post-wall waveguide (also known as substrate integrated waveguide (SIW) or a laminatedwaveguide) is a synthetic rectangular electromagnetic waveguide formed in a dielectric substrate bydensely arraying metallized posts or via-holes which connect the upper and lower metal plates ofthe substrate. The waveguide can be easily fabricated with low-cost and mass-production usingthrough-hole techniques where the post walls consists of via fences. The post-wall waveguide isknown to have similar guided wave and mode characteristics to the conventional rectangularwaveguide with equivalent guided wavelength. The structure is shown below. Since SIW provideslow loss, filters using SIW have studied by many researchers.Figure 1.1The structure of SIW1.2Find the microwave integrated circuit example built using LTCC in the web-site.SolutionIn the web site http://www. barryltcc.com/, there is a TR (Transmit and Receive) module for radar.1.3How the ALC inFigure1.7 is constructed?SolutionFigure 1.2 shows a block diagram for ALC. Usually PA module has the terminal for output powercontrol. If not, a driver amplifier should be designed to control the output power level. A smallportion of output power is sampled through a directional coupler as shown in Fig. 1.2 and the RFpower level is detected through a diode detector. The reference signal is a variable DC voltage

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source. Sometimes when the variable voltage source is difficult, it is generated using a lowpassfilter PWM (Pulse width modulation) signal generated from the controller. The average DC voltagecorresponds to the desired power level. Since it is a negative feedback loop, the detector voltage isequal to the reference voltage at the steady state. Thus the output power level can be controlled bychanging the reference voltage.Figure 1.2ALC block diagram1.4Refer to SA605 which is used to demodulate the FM signal. Explain its block diagram.SolutionFM demodulation in SA605 is based on quadrature FM demodulator shown in Fig. 1.3. In comingFM signal is split in two paths; one is directly applied to the mixer and the other is coupled to theresonant circuit centered at the FM carrier frequency by a high reactance capacitor. As a result, thesignal coupled to the high reactance capacitor suffers a 90 phase shift. The FM modulated signalsuffers additional phase shift proportional to the resonator slope and AM modulated waveformproportional to FM modulation appears at V2. As a result of mixing demodulated FM waveformappears.Figure 1.3Quadrature FM demodulator1.5Refer the web-site of PLL IC. Explain the synthesizer data bus shown in Fig. 1.7.SolutionRefer to Analog Devices, RF PLL Frequency Synthesizers ADF4110/4111/4112/4113, August,2012.

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1.6How the PLL can be modulated? Explain how to set the PLL loop bandwidth.SolutionFor FM signal generation at the VCO output, the modulation signal may be directly injected toVCOtuninginput. The fastvaryingsignalthanPLLloopbandwidthcanmodulateVCOsuccessfully, however the slow varying one will be compensated by the PLL and does not presentthe modulated output.Figure 1.4FM modulation using PLLCHAPTER 2 PROBLEMS2.1A company A fabricates a thin film resistor whose sheet resistivity is 50 ohm/square and in thecase of a company B, an identical thin film resistor has a sheet resistivity of 100 ohm/square. Whatis the difference between the processes of the two companies? Furthermore, given that thematerial’s volume resistivity isρand its thickness ist, find its sheet resistivity.SolutionThe thickness of the resistor material of the company A is two times thicker than that of thecompany B.The resistanceRwith a volume resistivity ofρand a thickness oftisSLLRRtWWρ==ThusSRtρ=

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2.2Given that the sheet resistivity is 50 ohm/(square) and we want to design a 100 ohm resistor. Ifthe current flowing in this resistor is 2 mA, find its minimum width. Its rated current per width is0.5 mA/μm.SolutionThe minimum widthWmintheta can sustain 2mA ismin2(mA)4(μm)0.5(mA/μm)W==TakingWmininto consideration, the width is determined as 10mm. Thus the length should be 20mmto yield 100 ohm resistor.2.3The dielectric material of a MIM capacitor is Silicon Nitride and its dielectric constant is 7.2. Ifthe area is 50μm2find the thickness of a 0.53 pF capacitor. In addition, find the capacitance perunit area in (F/μm2).Solution0.53( pFC=8.854( pF)orAtε ε==2/m)7.250(μm××)(μmt63)8.854107.2506.010(μm)6.0(nm)0.53t−−×××==×=The sheet capacitanceCSis220.53(pF)10.6 (fF /μm )50(μm )SC==2.4In the equivalent circuit of an inductor such as that inFig. 2.10, show that the approximatemagnitude of the impedance at resonance is()2maxoLZRω=1oLCω=SolutionThe impedance ofZ(ω) of the inductor equivalent circuit()()11RjLjCZRjLCωωωωω+=+−Near the resonanceZ(ω) can be approximated as()()()()()21111oooooRjLRjLjLLjCjCjCZRRRRjLCωωωωωωωωωω++=≅≅=+−

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2.5In this text, we have covered the method of extracting the equivalent circuit of a capacitor from|Z(ω)| characteristics obtained from impedance analyzer such as that shown inFig. 2.8. Theimpedance of the equivalent circuit of a capacitor is()1ZRjXRjLCωωω=+=+−Using this, the real and imaginary parts for frequency can be obtained by graphical representation.Then show that the approximate impedance near the resonant frequencyωois()()2oZRjLωωω≅+−Solution()()()()()()22222211112oooooooZRjXRjLRjLRjLCLCRjLRjLRjLRjLωωωωωωωωωωωωωωωωωωωωωωωω=+=+−=+−=+−−+−=+=+−+=+≅+−2.6Similar to Problem 2.5 above, for a parallel resonant circuit, show that the admittance near theresonant frequency,Y(ω) is given by()()2oYGj Cωωω≅+−SolutionThe admittance of a parallel resonant circuit can be expressed as()1YGjBRjCLωωω=+=+−This is similar toZ(ω) in problem 2.5. Thus,()()2oYGj Cωωω≅+−2.7(ADS Problem)In Example 2.2, the equivalent circuit of the 10 pF capacitor from Murata areobtained using the slope of the reactance versus frequency and resonance frequency such as4ofXLfπ∂=∂and1oLCω=.In another way, the equivalent circuit can be obtained using the impedance plot in log-log scale.Compute the equivalent circuit values using ADS.SolutionFig. 2.1 shows the simulation schematic for impedance plot.

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Figure 2.1Simulation schematic setup for impedance plot. Here log sweep is set to 30 points/decadeThe simulated impedance is shown in Fig. 2.2. Using the impedance values at markers, the seriesequivalent circuit values can be obtained. From the minimum value, the value ofRcan be obtainedandR=0.363Ω. The following equations forCandLare written in the Display window.EqnC=1/(2*pi*m2*indep(m2))EqnL=m3/(2*pi*indep(m3))The computedC= 9.9 pF andL=0.96 nH. The values ofR,C, andLare close to those computed inExample 2.2 but shows little differences. The error ofRis mainly due to the sparse sweep points.The differences of the other values are believed to come from different approximations in thecomputation of element values.Figure 2.2Simulated impedance for the Murata 10 pF capacitor.

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CHAPTER 3 PROBLEMS3.1In Example 3.1, a traveling sine wave propagating alongzdirection is given by cos(ωt–βz). Fora constantC, ifωt–βz=Cis chosen, what is the meaning? Also, when this constant is taken as thepeak of the sine wave, prove that its velocity isvp.SolutionWhenωt–βz=Cis chosen, the wave propagating alongzdirection becomes cos(C). This means thattime and space which yields constant phase is chosen. For example, letC=0. Then it represents awave at its peak value. This will propagate alongzdirection with time, which one usually observesthe wave velocity. Thus the phase velocity is{}0pddztzvdtdtωβωββ−=→==pvωβ=3.2In the special case of the transmission line whenθis small, find the values ofLandCusing theresults inFig. 3.25. In addition, show that the following equivalent circuits yield the same waveequation asθ→0.(a)(b)Figure 3.1Equivalent circuit of a small length transmission line: (a) Tee and (b) Pi shapeSolutionFor the circuit in Fig. 3.1(a), from Tee-equivalent circuit in section 3.6.3.2,1tan222acooopzjLZZjZjZjZvθθωω∆===≅=ReplacingZoandvpin the above equation byLandCgiven by (3.6) and (3.14),12LzL∆=This is phaysically obvious. Similar procedure leads to1CCz=∆For the circuit in Fig. 3.1(b), by inspection, one can obtain the following results as

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2LLz=∆212CCz=∆For the circuit in Fig. 3.1(a), definev(z,t) andi(z,t) as shown in Fig. 3.2 then()(){}()(),,11,,22i z ti zz tv zz tv z tLzLztt∂∂+ ∆−+ ∆−=∆+∆∂∂Figure 3.2Definition of voltages and currentsDividing both sides by∆z,()()()(),,,,1122v zz tv z ti z ti zz tLLztt+ ∆−∂∂+ ∆−=+∆∂∂In the limit∆z→0, the above equation becomes()(),,v z ti z tLzt∂∂= −∂∂This is equal to the result in (3.2). The other equation can be similarly derived. Thus, the waveequation in (3.4) is obtained irrespective of the equivalent circuit forms.3.3Considering the 2-port circuit in Fig. 3.3 as a transmission line, determine its equivalentcharacteristics impedance using open and short-circuit method presented in Section 3.2.4.Figure 3.3LC circuitSolutionThe open circuit impedance is

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1andopenshortZjLZjLCωωω=−=The equivalent characteristic impedance is221oopenshortoLZZZCωω=⋅=−Where21oLCω=3.4(1) If the electrical length at a frequency of 1 GHz is 90°, what is the electrical length at a frequencyof 3 GHz?SolutionThe electrical length is1GHz1GHz90fpflvωθ====°.Since the electrical length is proportional to frequency, the electrical length at 3 GHz is 270°.(2)Also, if the phase velocity is equal to the speed of light, what is its length?SolutionSincevp=c,1GHz1GHz1GHz902ffpfllvcωωπθ=======°Hence, the length is2clππω==83102 2π×930.1(m)7.5(cm)410=×=×(3)Finally, what is its propagation constant?SolutionSince the propagation constant is defined as the phase delay per length (rad/m), it can be computedas220.9(rad / m)7.5(cm)20.075clπωθπβ =====×3.5In the definition of the reflection coefficient shown below, when the real part ofZLis positive,prove that |ΓL|≤1.LoLLoVZZVZZ−+−Γ==+

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SolutionDefine the normalized impedancezLbyZoasfor>0Lzrjxr=+∞≥Then proving the following relation is the problem.11111LLLLLzzzz−Γ=≤→−≤++Squaring both sides and subtracting, one can find that{}2222(1)(1)40rxrxr++−−+=≥Thus |ΓL|≤1 and the equality is satisfied forr=0.3.6Figure 3.4 shows a slot line measurement. Voltage standing wave minimum points are shown bythe numbers for two measurements. When the load is replaced by a short, the first minimum occursat 90cm. The distancedminfrom the first minimum to the short is 15 cm (=90–75cm in Fig. 3.4) at afrequency of 375 MHz. Find the load impedance.Figure 3.4Slot line measurementsSolutionThis can be solved using Smith chart. Since VSWR=5, one can draw a circle in the Smith chartcorresponding to VSWR=5. Furthermore the half wavelength is found to be 40 cm (=75–35) fromFig. P3.3. Thus, the wavelength is 80cm. Thedminnormalized by wavelength is 0.1875 (=15/80).Thus the load impedanceZLis found by reading the impedance at point B shown in Fig. 3.5.ZLreads as56.594.5LZj=−

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Figure 3.5Slot line measurements3.7In Fig. 3.6,V+andV–represent incident and reflected phasors at load plane. Write voltageV1usingV+andV–. When the VSWR at the input is 2, find the magnitude ofΓinand the correspondingreturn loss. In above condition, settingZo=50 ohms, find the value of resistorZL.Figure 3.6Transmission line terminated by load.SolutionV1is composed of two voltage wavesVfandVb.Vfis propagating to the right and resulting inV+atthe load plane. The phase of the voltageVfadvances that ofV+byθ. This can be expressed asjfVV eθ+=The other propagating to the left from the load whose value isV–at the load plane. The phase ofVbis delayed byθcompared withV–.ThusjbVV eθ−−=

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As a result,1jjfbVVVV eV eθθ+−−=+=+Denote |Γin|=ρ,then111220 log9.54(dB)133VSWRRLρρρ+==→=→= −=−Sinceρ=1/3,1112 or132LoLLLLoLZZzzzZZzρ−−===→==++3.8Using the scale at the bottom of Smith chart, find VSWR and return loss corresponding toΓ=0.6using compass.SolutionThe scale in the bottom of the Smith chart is shown in Fig. 3.7. From Fig. 3.7, the return loss andVSWR corresponding toΓ=0.6 are about 4.4 dB and 4, respectively.Figure 3.7The scale in the bottom of the Smith chart.3.9For a transmission line withZoandθ, derive ABCD parameters defined as,1212VVABIICD=SolutionThe voltage and current of a transmission line atθcan be expressed as2jjVV eV eθθ+−−+=+()21jjoIV eV eZθθ+−−+=−.(1)From1VVV+−=+

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()11oIVVZ+−=−,V+andV–can be obtained as1111and22ooVZ IVZ IVV+−+−==.Substituting the aboveV+andV–into (1),1111211cossin22jjoooVZ IVZ IVeeVjZ Iθθθθ−++−=+=+11111211sincos22jjooooVZ IVZ IjVIeeIZZθθθθ−++−=+=+Thuscos,sin,andsinooADBjZCjYθθθ====.3.10Figure 3.8 shows an equivalent circuit of a distribution amplifier similar to that inFig. 3.1 inthe text. FindV1andV2.Figure 3.8Equivalent circuit of the output of a distribution amplifier.SolutionThe voltageV1andV2can be computed using superposition principle. The voltageVc1across thecurrent sourceIis computed to be12ocZ IV=This will propagates to both ends of the transmission lines and the resulting voltagesVo1andVo2are21, 12, 1and22jjooccZ IZ IVeVeθθ−−==Similarly for the current sourceIe-jθ231, 22, 2=and222jjjjoooccZ IeZ IZ IVeeVeθθθθ−−−−==Thus2211, 11, 22=2jjoccoZ IVVVeZ Ieθθ−−=+=×

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()322, 12, 22jjoccZ IVVVeeθθ−−=+=+3.11If a 50 ohm transmission line has a 1/4 wavelength length at 1 GHz and its end is short-circuited, can this be considered a parallel resonant circuit near the 1 GHz frequency? What then isthe capacitance of the parallel resonant circuit?SolutionThe transmission line of a quarter wavelength whose end is short-circuited can be approximated byparallel resonant circuit according to section 3.5.2 and the capacitance of the parallel resonantcircuit is9111==(nF)881050400ooCf Z=××3.12In Fig. 3.9, answer the following. Find the characteristic impedanceZxfor maximum powertransfer to load 2Zo. For this, find the time domain voltage waveformv1(t). Find also the timedomain voltage waveformv2(t).Figure 3.9Problem 3.12 circuit.SolutionFor the maximum power transfer, the impedance Zinshould be2222xxinoxoLoZZZZZZZZ===→=Since Zin=Zo,11( )cos2oovtEtω=This voltage propagates the transmission line and is delivered to the load 2Zo.whichis equal to thepower delivered to Zo. Thus221242122ooooVEVEZZ=→=The phase of V2should be–90° because the voltage propagates the quarter wavelength transmissionline. Thus

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221122jooVE ejEπ−== −And211( )Resin22ojtooovtjEeEtωω=−=3,13 (ADS Problem)The characteristic impedance of a transmission line can be found applying apulse signal to the transmission line. Fig. 3.10 shows a simulation set up to compute thecharacteristic impedance of the transmission line. By computing the current and voltage pulses andthe ratio of the pulse, show that the ratio is the characteristic impedance given by 50 ohms.Figure 3.10Simulation setup to compute the characteristic impedance using a pulseSolutionIn Fig. 3.10, the time of arrivaltdis81500.5sec310 [/ sec]dlmtcmm===×Fig. 3.11 shows the simulated pulse forVinandIin.i. From Fig. 3.11, it can be found that the pulseappears after 0.5msec. Also from the values of current and voltage, the characteristic impedance is50Ω.

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Figure 3.11The simulated voltage and current pulses.3,14 (ADS Problem)The standing wave can be simulated by sweeping the length of thetransmission line as shown in Fig. 3.12. Plot of mag(v1) vsl1 results in the standing wave pattern.Also plot the standing waveform with time as a parameter. It can be found that the minimum pointdoes not move despite of time change, which is why it is called “standing wave.”Figure 3.12Standing waveform simulation.SolutionAfter the simulation given in Fig. 3.12, v1 is plot for the length variable l1. Fig. 3.13 shows thesimulated standing waveform. Fig. 3.14 shows the time variation of the standing waveform. Thiscan be obtained using ts() function in ADS. As can be seen from Fig. 3.14, the minimum pointcalled “node” does not change with respect to time. This is the reason why we naming “standingwave.”

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Figure 3.13Standing waveformFigure 3.14Standing waveform change for time3.15 (ADS Problem)Using ADS, extract the equivalent circuit of corner-discontinuityMCORNwhich has a width of 50 ohm line microstrip width. Note that it can be represented by Teeequivalent circuit because the corner is a passive circuit. Here the substrate is a 10 mil thick aluminawithεr=9.6.SolutionFig. 3.15 shows the schematic setup for the simulation of the corner discontinuity. Also themicrostrip line length which has the corner discontinuity along the center line is also included forthe comparison. The variablew50 represents the 50 ohm microstrip line width which is computedusingLineCalcin ADS.

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Figure 3.15Schematic setup for the corner discontinuity simulationAfter the simulation, the transmission characteristics are compared in Fig. 3.16. From Fig. 3.16, onecan find that the length of the corner discontinuity has a shorter than that of the microstrip line.Figure 3.16TheS21of the corner discontinuity and microstrip lineGenerally, the corner discontinuity shows a shorter length than the microstrip line with the length ofthe corner discontinuity along the center line. Since the corner discontinuity can be represented by aTee-equivalent circuit shown in Fig. 3.12.

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Figure 3.17T-shape equivalent circuitThe values ofZAandZBcan be obtained from Z-parameters asEqnZA=Z11-Z12EqnZB=Z12EqnYB=1/ZBWriting the above equations in the Display window, their imaginary parts can be plotted as shownin Fig. 3.18. HereYB=1/ZBis plotted.(a)(b)Figure 3.18Imaginary parts ofZAandZBvs frequency.(a)YB(=1/ZB) and (b)ZAFrom Fig. 3.18(a),ZBis close to a capacitor. In Fig. 3.18(b)ZAis plotted with the microstrip lineZA. Since the microstrip lineZAistan2AoZjZθ=,it will increase linearly with frequency. On the contrary,ZAof the corner discontinuity decreasewith frequency and its value is negative. This phenomenon occurs because the length ofZAisnegative. Using the marker values, the inductance and capacitance corresponding toZAandZBcanbe computed inserting equations as

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Eqn LA=(m4-m3)/(indep(m4)-indep(m3))/(2*pi)Eqn CB=(m2-m1)/(indep(m2)-indep(m1))/(2*pi)The value ofCBandLAare33 fF and7.3 pHBACL== −.CHAPTER 4PROBLEMS4.1For a circuit in Fig. 4.1, calculate 2-port S-parameters normalized by the reference impedanceZo.Figure 4.1A two-port circuitUsing the previous results, calculate two-port S-parameters for the circuit where two transmissionlines of electrical lengthsθ1andθ2are added to the circuit in Fig. 4.1 as shown in Fig. 4.2.Figure 4.2A two-port circuit with transmission linesSolutionDue to symmetry,S12=S21andS11=S22.11221121112212, and33oooooooZEZZZZSSZZE−+== −==+Therefore12332133−= −SFor the circuit in Fig. 4.2,Scan be computed as

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1121222()1233()22133jjjjeeeeθθθθθθ−−+−+−−= −S,using the shift of reference plane explained in Section 4.1.5.4.2Calculate the S-parameters of the following parallel and serial resonance circuits.(a)(b)Figure 4.3(a) Parallel resonator and (b) the serial resonatorSolutionIn the case of Fig. 4.3(b), the S-parameters are presented in Example 4.5 in Section 4.1.5 and are1122112aSSjQδ−==+and2112212aajQSSjQδδ==+.HereQaandδare defined as2oaoLQZω=and=ooωωδω−In the case of Fig. 4.3(a), the S-parameters are()21122||||pooopZZSSZZZ==+and1122||||poooopZZZSSZZZ−==+.Zo||Zpis computed as1||and,=142ooopbooboZZZQZ CjQωωωδδω−≅=+SubstitutingZpintoS21()()21112||||12, and||12||12popooboopbopobZZZZZjQSSZZZjQZZZjQδδδ−−====++++4.3Explain how to obtain the circuit values of the above parallel resonant or series resonant circuitsusing |S21| and |S11| (Refer Example 4.5).Solution

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In the case of Fig. 4.3(b),112100 and1oSSωωωω→→ ∞== =Thus both |S21| and |S11| have the bell-shaped frequency response. From the peak value of |S21| and|S11|, one can find the resonance frequencies, which are equal toωo=2πfoas21oLCω=In addition,QaandQbof the circuits in Fig. 4.3(a) and (b) can be found from the 3-dB bandwidthBWas2oooafCZQBWω==and2oobofLQBWZω==UsingQs andωo, the values ofLandCin Fig 4.3 can be determined.4.4Find the maximum value of |S21| for the circuit shown in Fig. 4.4, and sketch the frequencyresponse (ωo=2πfo=2π×10 GHz).Figure 4.4A series resonant circuitSolutionThe transmission lines at the input and output have no effects on |S21|. Excluding the transmissionlines, one can form a new two port which yield the same |S21|. SinceS21can be expressed as2212VSE=|S21| approaches to 0 as frequency away from the resonance frequency 10 GHz. Thus, |S21| has abell-shaped frequency response centered at 10 GHz. The maximum value of |S21| appears at theresonance and the value is221210020.9532105ooZVSEZr====+4.5Find voltageVoutfor the circuit in Fig. 4.5.

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Figure 4.5Circuit for problem 4.5SolutionP=polar(dbmtow(0),0)inP_1Toneport means that, when a conjugate matched load is connectedto this port, the power of 0 dBm is delivered the conjugate matched load. Thus the power 0 dBm isdelivered to 50 ohm load. The voltageVoutis23211mW10W100.1(V)2outoutoutoVVVZ−−==→=→=Note that the valueVoutrepresents the peak value.4.6Given that the insertion loss in a passive lossless 2-port network is 3 dB, calculate the returnloss.SolutionSince lossless,221121111122SS=−=−=.Thus the return loss= 3dB.4.7Calculate the S-parameters of the attenuator circuit shown below. Determine whether |S21|2+|S11|2=1 is satisfied.Figure 4.6T-type 3 dB attenuatorSolutionS11andS22can be manipulated as1122{(508.56) ||141.88.56}500{(508.56) ||141.88.56}50SS++−===+++

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2112141.8 || (508.56)502 508.56141.8 || (508.56)508.562SS+==×++++=41.442×10050×10.70758.562==221121112SS+=<Since the circuit is lossy, |S11|2+|S21|2<1.4.8Prove equation (4.83)SolutionThe noise figure in (4.82) is()(){}22211ununsssssssGRGRFYYGGBBGGGGγγγ=+++=+++++ForBschange, minimum occurs atoptBBγ=The resulting noise figure is()()()222112122unsssunnnsnunnsGRFGGGGGRGR GR GR GG RR GGγγγγγ=++++=+++≥+++The minimum occurs at()()22ununsoptsnGRGGR GGGGRγγ+=→=+ThenFminis()()2min12212nunnnoptFR GG RR GRGGγγγ=+++=++UsingFmin,Gopt, andBopt, the noise figure relation can be re-arranged as()(){}22minnsoptsoptsRFFGGBBG=+−+−4.9Prove equation (4.84)SolutionFrom problem 4.8, the relation can be transformed into the equation using normalized admittancesas()(){}222minmin21nonsoptsoptsoptsosoR YrFFGGBBFyyGYgY=+−+−=+−Substituting the following

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11,and11optssoptsoptyy− Γ− Γ==+ Γ+ Γ,IntoF, the following equation can be obtained after manipulation.()2min22411nsoptsoptrFFΓ − Γ=+− Γ+ Γ4.10For the circuit in Fig. 4.7, findFminandΓopt. Use the definition in equation (4.68) aftertransforming the circuit into Norton equivalent circuit at port 2Figure 4.7A two-port resistor circuit.SolutionFrom Fig. 4.8, the Norton currentiTat the port221osTssosoYGiiiiGYGY=+−++Figure 4.8Equivalent Norton current computation.The contribution of the source noise to Norton current is,oT sssoYii GY=+Here

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()()()222124, and4sosooE ikT GE iE ikT Y===Thus()()222222211122TosososssoososooosssoosE iYGYGYGGG YYFGYGYYY GE iGYYG+++==++=++=+++Hencemin322F=+and215020.17221optoptRs−=→==+4.11The noise temperatureTeis defined in terms of the noise factor asTe=(F–1)To. Using thisrelationship, with the noise temperature of each stage known, derive the expression below byexpressing the noise temperature in terms of Frii’s formula.231112eeeeTTTTGG G=+++ ⋅⋅⋅SolutionThe noise temperature of a linear two-port network is often used in place of noise figure. Given thatthe linear two-port network has thetransducerpower gainG(which is defined as the ratio of theavailable power from the source to delivered power to a load and will be explained in detail inChapter 8). When the input termination with the temperatureTs. is connected to the linear two portnetwork, the delivered output powerPLcan be expressed as()LsePkG TTf=+∆.In the above equation,kTe∆frepresents the added equivalent thermal noise power density at theinput due to the noises of the two-port network. Thus the noisy linear two-port network can bemodeled as shown in Fig. 4.9.Figure 4.9Two port’s noise temperature conceptOne can find that from section 4.2.3,(1)(1)eoeokTfk FTfTFT∆=−∆→=−Now consider the following cascaded chains shown in Fig. 4.10.

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.(a)(b)Figure 4.10Cascaded block diagram (a) cascaded block and (b) its equivalent representationExcluding thermal noise source contribution toPL, the total output noise power excluding thecontribution of the source noise power can be related as{}1212122LTenenenenenPG TG GG TG GG TGG TG Tkf===+++∆Dividing both sides with total gainGT231112eeeeTTTTGG G=+++ ⋅⋅⋅4.12The attenuation with an attenuator isL(Lis not in decibel scale), given that its temperature isTo, determine the noise factor of the attenuator.SolutionWhen one end (source port) of the attenuator is terminated by 50 ohm termination, the impedanceseen from the other end (load port) is 50 ohm. Since the available power of a 50 ohm resistor iskTo,the delivered power to the 50 ohm load connected to the load port iskTo.On the other hand, the delivered power from 50 ohm termination connected to the source port of theattenuator iskTo/L. Thus, from the definition of the noise figure1ookTFLkTL==4.13When the attenuator in Problem 4.12 is connected next to a noise source having ENR1, what isthe resulting ENR2? Assume the cold temperature of the noise source isTo.

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SolutionDefining the hot and cold state noise temperaturesThandTc, the ENR of the noise source is1hccTTENRT−=Normally,Tc=To. After the attenuator, the corresponding output noise powers are1coNkTkT==2111hhoNkTkTkTLL==+−Thus212111111hoohoooTTTTTkTkTLLENRENRkTTLTL+−−−−====4.14 (ADS problem)It is explained thaticor(1,1) andicor(2,2) appearing after S-parameter noiseanalysis represent the short circuit noise current squares at ports 1 and 2 respectively. The shortcircuit noise currents can be obtained through AC simulation in ADS. The following figure showsAC noise simulation to obtain the short circuit noise currents. Since the noise currents cannot beobtained directly, the short circuit noise currents are computed using the node voltages across smallvalued resistors. Compute the short circuit noise currents through the AC noise simulation andcompare the results with theicor(1,1) andicor(2,2) obtained through the S-parameter noisesimulation.Figure 4.11AC noise simulation to compute the short circuit noise currentsSolutionAfter the AC noise simulation, the short circuit noise current squares are computed in the displaywindow using the equations asEqn Ii_sq=(Prob_15b..vi.noise/1e-3)**2

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Eqn Io_sq=(Prob_15b..vo.noise/1e-3)**2Figure 4.12AC noise simulation to compute the short circuit noise currentsIn the above equations, Prob_15b..vi.noise and Prob_15b..vo.noise represent the simulatednoise voltages in the dataset name Prob_15b. To represent the noise voltage, Ads uses theextension .noise after the node voltage name. The division factor 1e-3 appears tocompute thecurrent across 1mΩresistors. In addition, the S-parameter noise simulation for the same device isperformed. The computed two values are compared in Fig. 4.12. As can be seen in Fig. 4.12, theexact match can be found.CHAPTER 5 PROBLEMS5.1Figure 5.1 shows a simplified equivalent circuit of GaAs MESFET at low frequency.Figure 5.1A simplified FET equivalent circuit

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Figure 5.2A simplified FET equivalent circuit with a parallel feedback resistor(1) Calculate S-parameters setting the reference impedanceZo.SolutionSince input is open,S11=1. AlsoS12=0 by inspection. In the case ofS21,2211122momog Z EVSg ZEE−=== −When the gate is termed byZo, and source is applied to the drain,gmVgs=0. Thus the impedance seenfrom the drain is open, which means thatS22=1. Therefore1021mog Z= −S(2) SettingZo=50 ohms, computegmwhenS21=5∠180°.Solution212550 (mA/V)momSg Zg= −= −→=(3) When resistorfRis connected in parallel as shown in Fig. 5P.2, find the value ofRfthat makesS11=0.SolutionForS11=0,Zin=50. Applying KVL at the input1111()fmoVR IIg VZ=+−Thus the following condition can be obtained.2111foinofmomoRZVZZRg ZIg Z+===→=+5.2Figure 5.3 shows a simplified equivalent circuit of FET with source resistanceRsat lowfrequency. Due toRs, the equivalent trans-conductancegmeis lowered fromgmas explained inExample 5.3. Defining the equivalent trans-conductancegmeasDrain shorteddmegsigv=

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