ECE145a_218a_F11_HW4_Solns - ECE 145A/218A W09 HW3...

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ECE 145A/218A W09 HW3 SolutionsProblem #1Recall that a quarter wave transformer will transform any real impedance into another real impedance according to: ZIN(λ/4) = (Z0)^2 / ZL. Therefore when the frequency reaches a value such that the line length is an odd-multiple of a quarter-wavelength, the input impedance will cross the real axis with the impedance described in the preceding equation. In this case the load impedance is the 50Ω termination of port 2.We see above that for L=λ/4, ZIN= 3.4 + j*0. Therefore Z0= sqrt(ZIN* ZL) = 50 * sqrt(3.412 * 1) = 92.4ΩThis is confirmed in Line Calc:
To determine the propagation velocity vP, we use the fact that for a short-circuited transmission lineZSC= j*Z0*tan(βd), where β = 2π/λ = 2πf / vP. Since tangent goes to infinity for arguments equal to odd multiples of π/2, or where d is an odd multiple of λ/4, and we see below that there are peaks at odd multiples of roughly 105GHz, at that frequency 500μm must be λ/4. So, vP= f*λ = 105*109* (4*500μm) = 2.1*10^8 m/sCheck with εeff= 2.142 from Line Calc, vP= 3*10^8/sqrt(2.142) = 2.1*10^8 m/s.
Problem #2Part AWidth constraint: keep width below λ/2 at all design frequencies to prevent resonances. λ = c/(f*sqrt(εR)) = 609um. Wmax= 304.5um. Note: εeffectivecan be significantly less than εR, but because the line is going to be very wide the two will be very close. For narrow lines εEffectivewill approach 1.Length constraint: Keep impedance below infinity over design frequency range. Since impedance is infinite at DC (open circuit), quarter-wave will bring us to short circuit, and half-wavelength back to open circuit. So Lmax= λ/2 = 304.5um.Thus the low frequency capacitance C = εA/d = 371 fFPart CBecause the radial stub is narrow, then widens, we must use εEffective instead of just εR. But it changes over the entire length of the stub, since the relevance of the fringing effect depends on line width. As a quick, but reasonably accurate, approximation, take the average of the maximum: 2.7, and the minimum: 1, giving εEffective=1.85Lmax= λ/2 = c/(2*f*sqrt(εEffective)) = 368um. A = π*L^2/4. Thus, C = 423fFPart B & DPlots and schematic on next page. Z11 is rectangular line, Z22 is radial stub, Z33 is the capacitor from part A, Z44 is the capacitor from part C. We can see that with decent accuracy the impedance was kept finite up to 300GHz. The lower plots also show us that until the frequency increases such that the L = λ/2, for low frequencies there is close agreement between the imaginary components of impedance for the transmission line or stub, and the approximate capacitance calculated above. As frequency approaches the half-wavelength point, the discrepancy gets progressively larger.
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Top left: Rectangular Line Y and Z; Top right:Radial stub Y and Z; Bottom left:Rectangular line vs. capacitor; Bottom right:Radial stub versus capacitorMSUBMSub1T=0.25 umCond=1.0E+50Er=2.7H=6 umMSubS_ParamSP1Step=.1 GHzStop=320 GHzStart=1 MHzS-PARAMETERSCC2C=.423 pFRR3R=1 MOhmTermTerm4Z=50 OhmNum=4RR2R=1 MOhmCC1C=.371 pFTermTerm3Z=50 OhmNum=3RR1R=1 GOhmMRSTUBStub1Angle=90L=368 umWi=1 umSubst="MSub1"TermTerm2Z=50 OhmNum=2MLINTL1L=304.5 umW=304.5 umSubst="MSub1"TermTerm1Z=50 OhmNum=1

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