ECE 391: Suggese Homework Problems Quarer an Half Wave T-line Behavior 1. receiver operaing a 451Mhz is being inerfere upon by a ransmier operaing a 461MHz. While no changes can be mae irecly o he receiver or anenna, he coaxial ransmission line beween anenna an receiver is accessible. In paricular, a coaxial ee fiing is accessible irecly where he coax is aache o he receiver. a) Design he a filer using coax cable, Ω, ɛ r = 2.3, ha will rasically reuce he 461.MHz signal wihou effecing 451Mhz signal. Show by rawing/schemaic iagram how your filer is connece o poin. Clearly sae require coax lengh an open or shor circuis. b) Design anoher filer using one 1pf capacior an an inucor ha will rasically reuce he 461MHz signal wihou effecing he 451Mhz signal. Show by rawing/schemaic iagram how your filer is connece o poin. Give he specific inucor value. nenna Zo= "ee" connecor receiver 2. Ω ransmission line which is λ 4 =75 Ohms long a MHz is riven by a 1MHz generaor. Wha m impeance oes he 1MHz generaor experience if he line is erminae C wih 1Ω? Zo=75 lpha=.1b/m 75 B G=6B R 2V 75 rms 1V @1Mhz Zo= E r = 1 Z =25 j25 z z=
3. The elecrical lengh of several ransmission Vin lines are given below. a frequency corresponing Vou o λ : V.5 ou, (a) raw he equivalen lumpe circui he line woul presen a is erminals (b) give he impeance a he of he line 1 1nH.5,75 Vin 1 1nH Vou C1 V ou,.,75 C1..125.125 R=Zo R=Zo P = 1mW in G=1B G=3B P Roa in= 1mW G=1B G=3B Roa.5 E r = Z o= 1.5 1 1 4sin(w) (vols) Zo= alpha=.5b/cm 12cm R Zo= alpha=.5b/cm 4sin(w) (vols) T6 4. For he ransmission line segmen below, raw Zin versus frequency12cm for frequencies from o 2 Ghz. Inicae he max value of Zin on he verical axis. R.25m. T6 E = r Z = 1 o DRWN BY.25m. 2
5. For he circui below, raw V ou an he curren for frequencies from o 2 Ghz. Use he same verical axis for volage an curren. The resonan frequency is foun a: f = 1 2π C Vin 1 1nH Vou V ou, C1. 6. For he circui below, (a) Skech he volage waveform a he o he ransmission line for f = 1Mhz, R=Zo f = 2Mhz, f = Mhz. Provie he unis an scale on he graph an raw a leas one full cycle. G=1B G=3B Roa (b) For eachp frequency = 1mWin (a), raw he value of V aa (). The line is sill.5λ long a 1Mhz. in 1 V () V()= sin(2pi * f) where f=1mhz 4sin(w) (vols).5 @ 1Mhz Z o= ohms Zo= alpha=.5b/cm 12cm V () R open circui. m. l_in.25 @ 5Mhz Z o= ohms 3 O
7. The elecrical lengh of several ransmission lines are given below. a frequency corresponing o λ : (a) Draw he equivalen lumpe circui he line woul presen a is erminals (b) Give he impeance a he of he line.5 R = Zo V()= sin(2pi * f) where f=1mhz.5 @ 1Mhz Z o= ohms open circui V () V ()?? R = Zo.25.5 Z = + j2 l_in.25 @ 5Mhz Z o= ohms c_ne 1.25 parallel resonaor pf 4
R = Z o.5 Z = + j2 25 Z = + j2 8. Consruc: (a) an ngspice simulaion o show ha he impeance of a parallel resonan circui is l_in high an he impeance of a resonan series circui is low. Use he parallel an series neworks below..5??.25 R = Zo R = Zo l_in.25 parallel @ 5Mhz resonaor Z o= ohms parallel resonaor pf.25 @ 5Mhz Z o= ohms pf (a) Parallel Resonan Circui V () V()= sin(2pi * f) where f=1mhz.5 @ 1Mhz open Z o= ohms circui (b) Series Resonan Circui (b) Consruc a ngspice simulaion showing he parallel resonan behavior a 5 Mhz for a shore,.25λ, Ω line as shown below. V ().5 Z = + j2 l_in.25 @ 5Mhz Z o= ohms 1.25 Figure 2: Transmission ine Resonaor parallel resonaor (c) Increase he resisor r in ha represens he series resisance of he inucor in he series an parallel circuis o 1 or 2 ohms. Commen on wha he effec is as seen in he simulaion. Do likewise for r cap excep ha you will lower pf is value o moel capacior leakage. () How oes he T-line resonaor compare wih he oher wo circuis? Which lumpe circui oes i behave mos like? 5