Homework3_Solutions.pdf - Homework 3 Solutions ECE 333 Professor Amir Arbabi Problem 1 \u2202i \u2202v = R0(i L0 \u2202z \u2202t \u2202i \u2202v \u2212 = G0(i C 0 \u2202z \u2202t

Homework3_Solutions.pdf - Homework 3 Solutions ECE 333...

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SolutionsHomework 3ECE 333Professor Amir ArbabiProblem 1-∂v∂z=R0(i) +L0∂i∂t-∂i∂z=G0(i) +C0∂v∂tStarting with KVL, we havev(z, t)-R0Δ2i(z, t)-L0Δz2∂i(z, t)∂t-R0Δz2i(z+ Δz, t)-L0Δz2∂i(z+ Δz)∂t-v(z+ Δz, t) = 0Rearrange to getvand Δ terms on left side and take limit as Δz0:-[v(z, t)-v(z+ Δz, t)Δz] =R02[i(z, t) +i(z+ Δz, t)] +L02[∂i(z, t)∂t+∂i(z+ Δz)∂t]-∂v(z, t)∂z=R0i(z, t) +L0∂i(z, t)∂tFor KCL, at the central node (with 3 branches), we can insert temporary variablev(z+Δz2, t).Then we proceed as follows:i(z, t)-G0Δzv(z+Δz2, t)-C0Δz∂v(z+Δz2, t)∂t-i(z+ Δz, t) = 0Similar to approach with KVL, we bring current and Δzterms to one side and take the limitas Δz0-[i(z, t)-i(z+ Δz, t)Δz] =G0Δzv(z+Δz2, t) +C0Δz∂v(z+Δz2, t)∂t-∂i(z, t)∂z=G0v(z, t) +C0∂v(z, t)∂t1
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SolutionsHomework 3ECE 333Professor Amir ArbabiProblem 2f= 20GHza=0.912=.46mmb=2.972= 1.49mmσc= 5.8×107S/mμrc= 1r= 2.08μr= 1σ= 4×10-4S/mnote thatμc=μ0μrc= 4
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