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.