1•Transformers and Impedance Matching•Transmission Lines•Fourier Waveforms•Transformers- VACPrimaryV2/2 center tapV1N1SecondaryV2N2V1/V2 = N1/N2 voltage ratio = winding ratioI1 V1 = I2 V2 conservation of energyI1/I2 = V2/V1 = N2/N1V=I Z Ohms AC LawZ1/Z2 = (V1/I1) / (V2/I2) =(I2/I1)(V1/V2)Z1/Z2 =(N1/N2)2 Impedance MatchingTransformer core enhances performance L~µFe(ω) µo x LairChapter 4 - AC Circuits IIFe core2Impedance matching - theoryP = I V =I2 R = [V/(r+R)]2 RMaximum power is delivered atwhat value of R?dP/dR= -2V2R(r+R)-3+V2/(r+R)-2 = 0-2V2R(r+R)-3+V2/(r+R)-2 =0R/(r+R)=1/2R=1/2r+1/2R --> R=r !Maximum power is transferredwhen Output impedance matchesInput impedance!riRVOutput impedanceInputimpedance3Impedance matching - w transformerZs=4900ΩZL=100Ω28VAC28VACWithout Transformeri = 28/(4900+100) = 5.6mAVL = (100/5000)28V =0.56VPL = i V2 = 3.1 mWZSource=4900ΩZLoad=100Ωii1i2With TransformerZ1/Z2 =(N1/N2)2N1/N2 = sqrt(4900/100)=7V2= (1/7)V1= (1/7) 28 = 4VI2 = 4V/100Ω = .040AP2 = (.040A)4V= 160mW !I1 = i2(N2/N1)=.0057AP1= (.0057A)28V = 160mW4Twisted pair (Z~100Ω)Parallel conductor (Z~90Ω)CoAxial (Z~50-100Ω) Zparallel=µµo!2""o# $ % & ' ( 1/ 2ln2da# $ % & ' ( Zcoax=µµo!2""o# $ % & ' ( 1/ 2lnba# $ % & ' ( 2d wire spacing2a wire diameter2d shield diameterhttp://www.eeweb.com/toolbox/wire-over-plane-inductance/Transmission Lines5Transmission Lines•We often use a cable or transmission line to transfer AC signals.•The cable can be modeled as a series L=L/l and parallelC=C/l to ground. Sometimes a series R=R/l is also included.•Since Vab ~ 1 for low frequencies (ω -> 0) generally signal propagation is only aconcern at higher frequency.•Resistance of the line will attenuate the signal at all frequencies!L/lC/lVoutVinabvVab=1 /!C(1 /!C)2+ (!L)2Vin!=!11 + (!2LC)2Vin!!!!!!!!!!!!Vout=11 + (!2LC)2"#$%&'NVin6Transmission Line Theory(1) Z1= R1/ ! + i!L / !!Z2= R2/ ! + i!C / !( ) 1)!!dV / dx!!= !!I ! Z1!!!!!!"!!!!!d2Vdx2= !! Z1!dIdx!=Z1Z2V !!!!!"!!!! !V (x) =! A!e!Z1Z2x!!+ !! B!e+Z1Z2x!!!!!!!!!!!!!!!!!2)!!!dI / dx = !VZ2!!!!!!"!!!!!!!!d2Idx2= !! Z1!dVdx!=Z1Z2I !!!!!!!!!"!!!!! I(x) =!C !e!Z1Z2x!!!!! D!e+Z1Z2x!!!! By!u sin g!1)!! I =!!1Z1dVdx!!!!"!!!!!! I(x) =!1Z1Z2(A!e!Z1Z2x!!!!! B!e+Z1Z2x)!!!where!!ZC= Z1Z2!!characteristic!impedance3)! Assume!we !ter min ate!the !transmission!at !! x = L!!line!with! ZC.!!!!!V(L) = ! A!e!Z1Z2L!!+ !! B!e+Z1Z2L!!and !!!!! I(L) =1ZC(!A!e!Z1Z2L!!!!! B!e+Z1Z2L)!!!!!!!!!Z(L) =V (L)I(L)!!= ZC!!!"!! B = 0!!!4)!!V(0) = VIN!!!!!!!!!!!!!!!"!!!!!! !V (x) =!VIN!e!Z1Z2x!and !!! I(x) =VINZC!e!Z1Z2x!!!!!!!!"!!!! ! ZIN=!V (0)I(0)=!ZCInput !Im pedance! = ! ZC!independent !of ! L!!!! "### $###I !!Z1!!!!!!!!!!dx!!!!!!!!!"VINI !!dIV ! dVdIZ2…………ZCVOUTL7•The characteristic impedance of a transmission line is•The input impedance of a transmission line is independent of L terminated if terminated in characteristic impedance is ZC. •The signal speed of propagation Transmission Line Theory(2)ZCZCharicteristic= ZC= Z1Z2 v = 1 / L'C' = c /µ!= c / n8Speed of Signal Propagation Vab=1 /!C(1 /!C)2+ (!L)2Vin!=!11 + (!2LC)2Vin!!!!!!!!!!!!Vout=11 + (!2LC)2"#$%&'NVinThe LC!circuit rings when !2LC = 1 !or != 1 / LC !!This is the optimum condition for signal transport.U sin g!the!ringing! period!as!!T = LC != 1 / LC !!!!!(!!!! !1 / v = LCT!/ l = (L / l)(C / l) v = 1 / L'C' = c /µ)!=!c / nL/lC/l VoutVinv9Reflection CoefficientIf a transmission line of characteristic impedance Zc is terminated with impedanceZ Zc a reflected signal will develop!K = A/Ao = [(Z-Zc)/(Z+Zc)] reflection coef (If Z=Zo then no reflection!)If T ~ 2(L/v) = transit time for 2L reflection then signal cancellation likely.If T<<To then no termination necessary.If T ~ To then cancellation or distortion and termination necessary.f =1/To =signal frequencyVoutLVinToWill a 100 MHz signal traveling down a 2 m RG58/U 50Ω cable need termination?To = 10 ns T= [2(2m)/2x108 m/s] = 20 ns : Yes cable should be terminated in50Ω!T!10Complex Signal and Fourier Decomposition V (t) =Ao2DC!+ An!sin(2n!!t / T )"!+ Bn!cos(2n!!t / T )"AC" #$$$$$$$$ %$$$$$$$$!!!!!!!!!#n= 2n!fFourier – Any!waveform!V(t)!can!be!decomposed!in!to!sin e!and!cosine!waves.Sawtooth!WaveAn=1n!!!n = 1, 3, 5,....!odd !!Bn= 0Square !WaveAn=4n!!!n = 1, 3, 5,....!oddBn= 011Fourier Decomposition- FFT• Fast Fourier Transform (FFT) is an algorithm to compute a discrete Fourier transform. This signal P(t) is decomposed in to a high frequency and low frequency P(ν). T-hiT-lo• Noise and signal can be separated in the frequency