射频工程师面试宝典

1、RFQuiz1.Whatistheimpedanceoffreespace? 377Ohm2.Whathappenstothenoisefigureofareceiverwhena10dBattenuatorisaddedattheinput?a)Noisefigureincreasesby10dBb)Noisefiguredecreasesby10dBc)NoisefiguredoesntchangeThe formula for cascaded noise figure is:NF(M stages)= 10*log nf1+ (nf2-1)/(gain1) + (nf3-1)/(gai

2、n1*gain2) + . + (nfM-1)/(gain1*gain2*.*gainM-1);where each nf and gain value is expressed as a ratio rather than in dB, and M is the total number of stages.The noise figure of an attenuator is equal to its insertion loss (10 dB in this case). Note that per the equation that the noise figure of the f

3、irst element in the chain is not modified by the gain of preceding stages - as are the subsequent stages noise figures. Therefore, any noise figure added to the front end adds directly to the overall system noise figure - in this case an increase of 10 dB.3.AnRFsystemhasalinearthroughputgainof+10dBa

4、ndanoutput3rd-orderinterceptpoint(OIP3)of+30dBm.Whatistheinput3rd-orderinterceptpoint(IIP3)?a)+20dBmb)+40dBmc)+30dBm4.Whichfiltertypehasthegreatestselectivityforagivenorder(i.e.,N=5)?a)Besselb)Chebychev(ripple=0.1dB)c)Butterworth5.Whichmixerspuriousproductisa5th-orderproduct?a)1*LO+5*RFb)6*LO-1*RFc)

5、3*LO-2*IF6.A2.8GHzoscillatorisphase-lockedtoa10MHzreferenceoscillatorthathasasingle-sidedphasenoiseof-100dBcat1kHzoffset.Whatisthesingle sidedphasenoiseofthe2.8GHzoscillatorat1kHzoffset?a)-48.6dBcb)-100dBcc)-51.1dBcWhen an oscillator (2.8 GHz in this case) is phase-locked (PLO) to a reference source

6、 (10 MHz in this case), the phase noise is increased in amplitude by an amount equal to 20*log (fPLO/fRef) + 2.5 dB, where the additional 2.5 dB (rule of thumb) is due to phase noise added by the phase locking circuitry. This explains why an extremely low phase noise reference oscillator is required

7、 when being used with a microwave frequency PLO.-100 dBc + 20*log (2800/10) + 2.5 dB = -48.6 dBc7.Whatisthepowerofa2Vpk-pksinewaveacrossa50ohmload?a)-20.0dBmb)+10.0dBmc)+19.0dBm8.Which2-portS-parameteriscommonlyreferredtoasreverseisolationinanamplifier?a)S21b)S22c)S12Common names for each of the fou

8、r 2-portS-parameters are:S11 : input return lossS21 : forward gainS12 : reverse isolation (or reverse gain)S22 : output return loss9.WhataretheminimumandmaximumcombinedVSWRlimitsataninterfacecharacterizedbya1.25:1VSWRanda2.00:1VSWR?a)1.75:1(min),2.25:1(max)b)1.60:1(min),2.50:1(max)c)0.75:1(min),3.25

9、:1(max)VSWR (max) = VSWR1 * VSWR2 : 1VSWR (min) = VSWR1 / VSWR2 : 1, where VSWR1 VSWR2.10.Anidealdirectionalcouplerhasadirectivityof25dBandanisolationof40dB.Whatisitscouplingvalue?a)65dBb)40dBc)15dBA directional coupler is characterized by five main parameters as follows:1. Frequency band of operati

10、on.2. Power coupling expressed as dB down from the input power level.3. Isolation of the coupled port from the output port (essentially coupling factor from the output port to the coupled port)4. Directivity, which is mathematically the difference between the magnitudes of the isolation and the coup

11、ling. If the coupler in this case had 0 dBm signals applied to both the input and output ports, the coupled port would see -15 dBm from the input port and -40 dB from the output port, hence, an isolation of 25 dB.Coupling = 40 dB - 25 dB = 15 dBRFQuiz#21.OnaSmithchart,whatdoesapointinthebottomhalfof

12、thechartrepresent?a)Aninductiveimpedanceb)Acapacitiveimpedancec)Powersaturation2.WhilewereonthesubjectofSmithcharts,whatistheimpedanceofthepointatthefarleftedgeofthecenterhorizontalline?a)Infiniteohms(opencircuit)b)Zeroohms(shortcircuit)c)50ohmmatch3.Asingle-conversiondownconverterusesahigh-sideloca

13、loscillator(LO)totranslatetheinputradiofrequency(RF)toanintermediatefrequency(IF).WillspectralinversionoccuratIF?a)Yes,alwaysb)No,neverc)SometimesSpectral inversion occurs when high frequencies within the input signal bandwidth are translated to low frequencies in the output bandwidth, and vice vers

14、a. Since a downconversion is being performed, the lower sideband of the mixing process is extracted, hence the difference between the LO frequency and the RF frequency is desired. Consider the following parameters and how spectral inversion occurs.RF input frequency band : fc = 1250 MHz, BW = 100 MH

15、z (1200 - 1300 MHz).LO frequency : 1600 MHz.IF output frequency band : fc = 350 MHz, BW = 100 MHz (300 - 400 MHz).When the lower frequency of the input band is subtracted from the LO frequency (1600 MHz - 1200 MHz = 400 MHz) a larger frequency is obtained than when the higher frequency of the input

16、band is subtracted from the LO frequency (1600 MHz - 1300 MHz = 300 MHz). This means that the output spectrum is the mirror image of the input spectrum.How to avoid spectral inversion? Always use a low-side LO (LO frequency below RF input frequency band) for mixing, or ensure that an even number of

17、spectral inversions are performed in the converter (i.e., two stages of conversion with high-side LOs).4.Whathappenstothenoisefloorofaspectrumanalyzerwhentheinputfilterresolutionbandwidthisdecreasedbytwodecades?a)20dBincreaseb)20dBdecreasec)40dBdecreaseThe input filter bandwidth determines the amoun

18、t of power that will be present at the detector circuitry. Since the detector performs a power integration function, it sums all of the incident power across the band. Decreasing the bandwidth by a factor of 100 (two decades) allows one one-hundredth of the amount of power to reach the detector, whi

19、ch in term of decibels is:10*log( 1/100) = -20 dB.5.Whatisaprimaryadvantageofaquadraturemodulator?a)LowLOpowerrequiredb)Fourseparateoutputsc)Single-sidebandoutput6.WhatismeantbydBiasappliedtoantennas?a)Isolationindecibelsb)Physicalsizerelativetointrinsicantennasc)Gainrelativetoanisotropicradiator7.W

20、hatisthepowerdynamicrangeofanideal12-bitanalog-to-digitalconverter(ADC)?a)36.12dBb)120dBc)72.25dBAn ideal 12-bit ADC can assume 212(4,096) unique voltage levels. Since power is proportional to the square of the voltage, the maximum power sample value is 40962(16,777,216) times the minimum power samp

21、le value. Therefore the dynamic range is 10*log (16,777,216) = 72.25 dB.A rule of thumb is 6 dB per bit.8.Anideal10dBattenuatorisaddedinfrontofaloadthathasa2.00:1VSWR.WhatistheresultingVSWRoftheload+attenuator?a)1.07:1b)2.10:1c)12.0:1VSWR is related to return loss (RL) according to VSWR = 10(RL/20)

22、+ 1 / 10(RL/20) - 1. It follows that increasing the return loss will result in a lower VSWR. The RL of a 2.00:1 VSWR is 9.542 dB. Add the 10 dB attenuator for a total RL of 2*10 dB + 9.542 dB = 29.542 dB. Convert back to VSWR using the given formula for a value of 1.07:1.Why add twice the attenuator

23、 value to the return loss? Return loss is the total decrease in signal strength in passing through the attenuator and being reflected back through the attenuator. Hence, the signal is decreased by twice the attenuator value.9.Whatisthethermalnoisepowerina1MHzbandwidthwhenthesystemtemperatureis15degr

24、eesCelsius(assumegainandnoisefigureare0dB)?a)-114.0dBm(ina1MHzbandwidth)b)-114.0dBmc)-114.0dBm/HzThermal noise power density is governed by the equation 10*log (k*T*B*1000) dBm, where k is the Boltzmann constant. T is the temperature in degrees Kelvin, and B is the bandwidth in Hertz. Multiplication

25、 by 1000 is to convert watts to milliwatts. A rule of thumb for temperatures near 15 C is to begin with a thermal noise density of -174 dBm/Hz, and scale accordingly (add 10 dB per decade of increased bandwidth).10.Twoequalamplitudetoneshaveapowerof+10dBm,andgenerateapairofequalamplitude3rd-orderint

26、ermodulationproductsat-20dBm.Whatisthe2-tone,3rd orderinterceptpointofthesystem?a)+40dBmb)+25dBmc)+20dBm2-tone, 3rd-order intermod products increase 3 dB in power for every 1 dB increase in tones that produce them. That means the intermods increase in power at a rate of 2 dB per 1 dB relative to the

27、 tone power. The 2-tone, 3rd-order intercept point is defined as the theoretical point where the two original tones and the two 3-rd-order products would have equal power (not possible in real systems due to saturation limits).If the two original tones have a power of +10 dBm and the 3rd-order produ

28、cts have a power of -20 dBm, then the intercept point will be at +10 dBm + (+10) - (-20)/2 dB = +10 dBm + 15 dB = +25 dBm.RFQuiz#31.Whatisaprimaryadvantagetousing90degree(quadrature)hybridcouplersinamplifierdesigns?a)Widerbandwidthpossibleb)Lownoisefigurec)Input/outputimpedancenotdependentondevicesa

29、slongasdeviceimpedancesareequalDue to the physical construction of the quadrature coupler, as long as the two devices between the couplers exhibit identical impedances the input and output impedances will exhibit the intrinsic coupler impedance. For example, if matched transistors with input impedan

30、ces of 12 - j5 are connected between to quadrature couplers that have an intrinsic impedance of 50 + j0 , then a 50 + j0 impedance would be exhibited at the circuit input (similar for the output).2.Whyisthereafrequencytermintheequationforfree-spacepathloss?a)Thereisnofrequencytermb)Atmosphericabsorp

31、tionc)AntennageometryrequiresitAntennas are an indispensable part of all wireless systems. There is no frequency dependency in the free-space power density equation as emitted from an isotropic radiator. Free-space power flux density decreases with distance due to energy being spread over the surfac

32、e of a sphere, hence:Pdensity = Ptransmitter / (4* d2) W / m2, where d is the distance in meters from the origin.However, the gain of the receiving antenna, including its effective area (Ae) is:G = Greceiver* l2/ (4)Total path loss = 20 * log (4* d / l) dB.3.Ifanamplifierhasanoisetemperatureof60K,wh

33、atisitsnoisefigureforanambienttemperatureof290K?a)8.0dBb)80dBc)0.82dBConversion from noise temperature to noise figure is a straightforward process.NF = 10 * log (NT / Ta) + 1 dB, where Tais the ambient temperature.4.Whatisaprimaryadvantageofoffset-quadrature-phase-shift-keying(OQPSK)overstandardQPS

34、K?a)Greaterdataratespossibleb)Greaterspectralefficiencyc)MoreconstantenvelopepowerOffset quadrature phase-shift keying (OQPSK) is a constant-envelope modulation that has no 180-deg phase shifts and, therefore, has a much higher spectral containment than non-offset quadrature phase shift keying (QPSK

35、) when transmitted over band-limited nonlinear channels.5.Amixerhasthefollowinginputfrequencies:RF=800MHz,LO=870MHz.Thedesiredoutputfrequencyis70MHz.Whatistheimagefrequency?a)940MHzb)1670MHzc)140MHzBy definition, the image frequency for any combination of input and LO frequencies is:fimage= 2 * fLO-

36、 finput.For any mixer, there are two input frequencies that, when mixed with the LO frequency, will generate the desired output frequency. In this example, the 70 MHz output can be generated either by taking 870 MHz - 800 MHz (desired), or by taking 940 MHz - 870 MHz (undesired).6.Whatisthespurious-

37、freedynamicrangeofasystemwithIP3=+30dBmandaminimumdiscerniblesignal(MDS)levelof-90dBm?a)80dBb)120dBc)60dBa) 80 dB.Spurious-free dynamic range (SFDR) is the maximum signal power above the minimum discernible signal (MDS) power level where two tones generate 2nd-order intermodulation products equal in

38、 power to the MDS. Input signals above that level will generate 2nd-order products that are greater in power than the MDS power level. MDS is generally defined as the noise power plus the minimum signal-to-noise ratio (SNR)One form of the equation is: SFDR = 2 / 3 * (IP3 - MDS) dB.7.Aspectrumanalyze

39、rdisplaysacomponentat10MHz0dBm,30MHz-10dBm,50MHz-14dBm,70MHz-17dBm,andalloftheotheroddharmonicsuntiltheydisappearintothenoise.Whatwasthemostlikelyinputsignalthatcausedthespectrum?a)A10MHzsquarewave(0Vdc)b)A10MHztrianglewave(0Vdc)c)A10MHzcosinewave(0Vdc)The Fourier series for a square wave with a 0 V

40、dc bias is the fundamental frequency and all of its odd harmonics. Amplitudes are scaled as the reciprocal of the harmonic number; in terms of power, the amplitudes are scaled according to 20 * log (1 / N) dB. A 10 MHz triangle wave also contains the odd harmonics, but amplitudes fall off according

41、to the reciprocal of the square of the harmonic number, 40 *log (1 / N) dB.8.OnwhichsideofarectangularwaveguideisanE-bendmade?a)Thelongdimensionb)Theshortdimensionc)TheinsideIn a rectangular waveguide, the E-plane is in the direction of the short dimension while the H-plane is in the direction of th

42、e long dimension. The type of bend is determined by which side is curved for the bend. A useful mnemonic is the short dimension is the Easy side to bend, while the long dimension is the Hard side to bend.9.Duringanetworkanalyzercalibration,whyarebothashortcircuitandanopencircuitused?a)Theyaverageto5

43、0ohmsinanRFsystemb)Todeterminethecharacteristicimpedanceofthemeasurementsystemc)Bothareeasytoproducetoathighaccuracy10.Whatisthefirstharmonicof1GHz?a)1GHzb)2GHzc)10GHzRFQuiz#41.Whichofthefollowingcancausefrequencyintermodulationproductsinasystem?a)Onlysemiconductorjunctionslikediodesandtransistors(a

44、mplifiers)b)Cableconnectors,boltedorrivetedantennapanels,isolatorsandcirculatorsc)Botha)andb)Intermodulation products are generated whenever currents of different frequencies flow in a nonlinear junction.2.Whatisthemeltingtemperatureofstandard60/40,tin/leadsolder?a)100C(212F)b)250C(482F)c)186C(386F)

45、3.Whatisthefrequencybandforthe900MHzGSMcellularband?a)Tx:880-915MHz/Rx:925-960MHzb)Tx:824-849MHz/Rx:869-894MHzc)Thereisno900MHzGSMband4.WhatdoesGSMstandfor?a)GeneralSystemforMobilephonesb)GreaterSpectrumforMobilephonesc)GlobalSystemforMobileCommunication5.WhatdoesPOTSstandfor(incommunications)?a)Pla

46、inOldTelephoneSystemb)PersOnalTelephonySystemc)PersonalOrbital&TerrestrialSatellite6.Whichofthesepairsofmaterialsinthetriboelectricserieshavethegreatestchargetransferpotential?a)Silk&Woolb)RubberBalloon&Celluloidc)Glass&HardRubber7.AlongwhichsideofrectangularwaveguideisanEbendmade?a)Longersideb)Shor

47、tersidec)Eitherside8.WhatisthelowestmodulationindexatwhichanFMcarrierissuppressed?a)2.40b)(3.1416)c)OnlyAMcarrierscanbesuppressedDepending on the modulation index (m) chosen, the carrier and certain sideband frequencies may actually be suppressed. Zero crossings of the Bessel functions,Jn(b), occur

48、where the corresponding sideband, n, disappears for a given modulation index, b. The carrier is the 0th sideband, so n=0. The next time the carrier disappears is for m=5.49.9.Howmuchcurrentisrequiredthroughthehumanbodytocauseanonsettomuscularparalysisduringelectrocution?a)Voltageistheculpritduringel

49、ectrocution,notcurrentb)21mAc)amp10.Atwhatfrequencyiselectromagneticenergymaximallyabsorbedduetooxygenintheatmosphere?a)22GHzb)Onlywaterintheatmosphereabsorbselectromagneticenergyc)63GHzRFQUIZ#5Thisquiztestsyourrecognitionoflogosfromcompanies*thatareleadersintheRFindustry.Insomecases,partofthestanda

50、rdlogowascroppedbecauseithadthecompanynameincluded.Printingoutthispagetofillintheblanksisprobablythebestwaytogoaboutthispuzzle.Hoveryourcursoroveralogotoseeahintastowhattypeofproductitmanufactures.AmericanTechnicalCeramics(ATC)PiconicsChannelMicrowaveMarkiMicrowaveScientificAtlantaAnadigicsNationalI

51、nstrumentsRemecAnalogDevicesDatelAndrewCorporationHittiteMicrowaveDeltaPowerCubeAgilentTechnologiesSynergyMicrowavePowerOneCypressSemiconductorConnecticutMicrowaveMiteqVicorNationalSemiconductorMauryMicrowaveCinchStanfordMicrodevicesHuber+SuhnerJohansonDielectricsSageLabsAnsoftAtlanticMicrowaveIntus

52、oftDallasSemiconductorMerrimacIndustriesAppliedRadioLabsTexasInstrumentsTrilithicRohde&SchwarzVoltronicsGeneralMicrowaveDielectricLaboratoriesRFMicroDevicesGHzTechnologyRFQuiz#6:WirelessCommunicationsFundamentals1.WhichofthefollowingWLANstandardsisonadifferentfrequencybandthantheothers?a)802.11ab)80

53、2.11bc)802.11gd)802.11n802.11b/g/n are all in the 2.4 GHz ISM band, 802.11a is in the 5 GHz ISM band.2.Whatdoesthetermruggednessrefertoinwirelesspoweramplifiers?a)Abilitytowithstandthermalstressb)Abilitytowithstandmechanicalstressc)Abilitytowithstandloadmismatchd)AlltheaboveA typical ruggedness spec

54、 is no damage into a 10:1 VSWR load3.WhichFCCregulationgovernstheunlicensedISMband?a)Part15b)Chapter11c)815.00d)Subsection114.InwhichsemiconductortechnologyarethemajorityofcellphonePAsmanufactured?a)Si/SiCMOSb)GaNc)GaAs/InGaPd)SiGe5.WhatisamajoradvantageofLowTemperatureCo-firedCeramic(LTCC)substrate

55、s?a)Denselyintegratedpassivecomponentsb)Betterthermaldissipationc)Lowerbillofmaterialsd)Alltheabove6.Whichphonestandardsupportsthehighestdatarate?a)iDENb)GPRSc)EDGEd)GSMiDEN=64kbps, GPRS=21.4kbps/time slot, EDGE=384kbps, GSM=14.4kbps7.Whichcomponentistypicallynotpartofafront-endmodule(FEM)?a)Poweram

56、plifierb)Filterc)Switchd)ControllerTransmit modules (TxM) incorporates a FEM + PA8.Whichtwosystemsaremostlikelytoexperienceconcurrentoperationproblems?a)WLAN+GSMb)Bluetooth+WLANc)GSM+W-CDMAd)Bluetooth+AMPSBluetooth and WLAN (802.11b/g/n) both operate on the 2.5 GHz ISM band.9.Anisolatoristypicallyre

57、quiredattheoutputofthePAforwhichtransmittersystema)EDGEb)Bluetoothc)GSMd)CDMA/W-CDMACDMA/W-CDMA power amplifiers, due to their highly linear region of operation, cannot tolerate a very high mismatch (VSWR) and must be protected with an isolator.10.WhatisthecommonlyclaimednominaloperationalrangeforBl

58、uetooth?a)10mb)32.8ftc)1.057x10-15lt-yrd)AlltheaboveRFQuiz#7:RadarFundamentals1.Whatisa“radarmile?”a)1nauticalmileb)1statutemilec)12.36sA radar mile is the time required for a signal leaving the antenna to go out and back one nautical mile (6076 feet). 2 * 6076 ft / 9.8356e8 ft/s = 12.36s2.Whichbest

59、describesabi-staticradar?a)Fixedtransmitterandfixedreceiveratdifferentlocationsb)Usesstaticelectricityforbothtransmittedandreceivedsignalsc)Onlydetectsfixed(non-moving)targetsMany bi-static systems use multiple receiving sites to be able to pick up weaker reflected signals and to correlate position

60、information with more certainty3.Whatisaradarcross-section(RCS)?a)Anengineeringisometricdrawingshowingaslicethroughtheequipmentrack(s)?b)Atargetsreflectioncoefficientrelativetoaperfectlyreflectingsphericalsurfaceof1m2cross-sectionc)Theplanethroughwhicharadarsignalpassesthroughatarget4.Whatarecommonu