HW8-sol - Sahib“ ECE604 Homework #8 Out: Friday, March...

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Unformatted text preview: Sahib“ ECE604 Homework #8 Out: Friday, March 27, 2009 Due: Friday, April 3, 2009 These problems are from the text by Ramo, Whinnery, and Van Duzer (3rd edition, 1994) or are attached. 1) 2) 3) 4) 5) 6) 7) 8) attached attached attached 6.4b 6. 8c attached A planewave is incident at oblique angle from medium 1 onto medium 2, where 81 = 82 but m i uz. Show that there is an incident angle for which the reflection of the TE wave is zero. What is the angle? 6.14d 1) Consider a coaxial transmission line with 1 mm inner diameter and 10 mm outer diameter. The conductors are separated by an insulator with relative dielectric constant 8r =16. Assume u = no and that there is no loss. Assume the transmission line is driven at one end by a sinusoidal voltage source. At the other end a load resistor R = 5009 is connected. The transmission line is 1 meter long. For what range of frequencies (work out approximate numbers!) can the transmission-line and resistor combination be modeled as a (frequency—independent) lumped circuit equivalent? Draw the equivalent circuit for this frequency range, and explain your answer. 2) Consider a parallel plate transmission line made of perfectly conducting plates. The plates are 5mm wide and spaced by 250 um and the medium between the plates is air. The characteristic impedance of the transmission line is denoted by 20. At one end the transmission line is connected to a short circuit, while at the other end the transmission line is connected to a matched load (RL = 20). The transmission line is driven by a sinusoidal voltage source at a frequency of 10 GHz, separated a distance d from the short circuit. The voltage source has a generator impedance ZG = Z0 . The total length of the transmission line is 10 cm. perfectly conducting plates (extend 5‘ mm into paper) short circuit a) Where should the voltage source be placed to maximize the power delivered to the load resistor RL? (What is the value of d?) b) Now the short circuit at Z = O is replaced by an open circuit. Now where should the voltage source be placed to maximize the power delivered to the load resistor? 3) Consider an infinitely long coaxial transmission line made up of perfectly conducting electrodes with inner diameter a and outer diameter b. The spacer material between the plates is a poor conductor with permittivity 80, permeability no, and conductivity 6. The spacer conductivity, characterized by 0 << (1)8, is sufficiently poor that it does not significantly modify the electric and magnetic fields of the unperturbed transmission line (i.e., the case where the spacer material is a perfect insulator), except to introduce power loss. a) Assume a forward travelling wave on1y(+ z direction), with a sinusoidal temporal dependence at frequency (D, whose electric field amplitude is EO (voltage amplitude is V0) at some position zo. Give an expression for the time-average power (units Watts) carried by the wave at that position. b) Give an expression for the time—average power dissipated in propagating a distance Az (where AZ is taken to be very small). c) Show that the fractional time—average power loss per wavelength of propagation is small (much less than one). Also work out an expression for the exponential decay length of the voltage (length OVer which voltage drops to e'1 of initial value). 6) Write a computer program (e. g., in Matlab, or equivalent) to calculate the frequency- dependent reflectivity R in terms of power (R = loll) for a multi—layer mirror composed of the following layer sequence: (air)(HL HL HL) (substrate) \____——\,———————J I—_‘,—1 M pairs one extra "H" layer This means that the wave is incident from air, through a series of 2M + l dielectric layers, then onto the substrate. Take the substrate to be silica glass (index of refraction n z 1.45 ). The high- index (H) and low index (L) layers are made from ZnS (nH = 2.3) and Mng (nL =1.38) , respectively. The thicknesses of H and L layers are each chosen to be quarter—wave at a free— space wavelength of 1mm (frequency = 3x1014 Hz). That is, the thicknesses rig and dL are chosen such that anH =% and nLdL = 7V4, for 7t = lum. Using your program, plot the power reflectivity ‘pl2 as a function of frequency over the range 2-1014 Hz to 4.1014 Hz, for M = 7 and M = 9. Repeat assuming nH = 1.8 and HL = 1.38 and plot your results. What are the impacts of the number of layers and of the index contrast between the two materials? Problem 1 W W Lumped element model can be used forfl. >> I , likewise fll << 1. Note that fll is known as the electrical length. The wavelength and frequency are related as follows: C 1v f’ x/e—r Therefore, putting this all together we can get an expression for frequency that satisfies the short electrical length condition fll < 0.1. ,1: flL<0.1:>/1>10(2fl)L 6 _3-108 C <———#— =1.2MH2: xi 8" (207$)L‘l8r 807: f: The approximate range of frequency for which the circuit can be represented as a frequency independent lumped circuit equivalent is: 0 Hz< f< 1.2MHz. The equivalent circuit is as follows: a At these frequencies the transmission line is just a pair of wires, i.e. no transmission line effects. We can also look at the problem from the impedance transform: [2L cos(fll)—jZOsin(fll)] ZHFZ Zocos(,Bl)—jZLsin(fll) :Z L fll=0 We can expand this problem for small electrical lengths, using the small angle approximations we get: _ ~ ZL_jZUfll __ _' ( z< z>~ZU[———ZO_jZLfll]—ZO<ZL Jami—— Since fll << 1 this reduces further to: Zoz +(Zfl)2 o ‘ZL2 2 ZL2 =ZL+] fll— 1Z0fl1+JZL zZL+Jfll —Z0 20 Z0 Z 2 —Z(2 Z Z ZL >> I °r 'fil << lzgiéjl For our problem we have: Z0 = Jiln[2] 2—3—711n = 2179 8 a 4 1 17 <<___Z_LZ.(L_——(i)9)_(2__). : |ZL2 —z(fl _ 5002 —2172 Since fll << 1 anyway, we have that Z (—l) = ZL. Thus for fll < 0.1gives us the same range from before: 0 Hz< f< 1.2MHz. Problem 2 a) We can write the voltage and current on the left side as: K :‘ll+e—jkz +‘/l~ejkz Ii = W 6"“ +16“ Z Z 0 0 We can write the voltage and current on the right side as: Kl: : V2+e—jk(z—d) + I =V2 e~jk(z—d) 2 z 0 Note: For the voltage and current on the right side i used the fact that there is no reflection for a matched load, and it is also written as e‘Mz'd) to simplify the math. First we match the boundary conditions at the short circuit, i.e. V=O at z=0 hence: W“ = —V[ . This gives: V1 =—2ij sin(kz) = 2V: ‘ z 0 I cos(kz) At z=d, we require our voltage to be continuous giving: V2 (d) = Vl (d). V; = —2ij sin(kd) We can have maximum power when IV2+ is a maximum. This occurs when sin(kd) = 1-1. This occurs 7r 271' /l 1 for kd : —+m7r , where m is an integer. Since k = ——,d =——+m—. For a frequency of loGHz we 2 xi 4 2 get: 8 = 3 10 =3cm:>d =(0.75+1.5m)cm i f 1-101“ Another way to look at the circuit is having the excitation driving two parallel transmission lines: Length = 10 cm - d Using the wave impedance concept we can represent the short and load as impedance blocks Zeql and 2qu respectively. We can get the maximum current to Zeqz when the minimum current goes to Zeql. Zero current goes to Zeql when it looks like an open. From the wave impedance formula we get: Z . . . . d 221220 :20 ML) zjzotanw)” q Z(,cos(kd)+]ZLs1n(kd) ZOcos(kd) 7: This occurs when kd : 3+ m7: which is the same solution as before. b) Now we have an open instead of a short at 2:0 and following the discussion as before we just need to maintain that the left half of the circuit looks like an open. Since Zeql is periodic with M2 and it is terminated with an open we get: . . Z :20 =20 zjzocot(kd)=m Zocos(kd)+]ZL sm(kd) JZL $1n(kd) eql This occurs when: kd=m7r k=~21 /l d =m§=m(1.5c‘m) Where m is again an integer. > 04] 4mg 5 <Pdlrss> T 8% m (’3 3*) M m r: Ammv: °‘\Eo‘L abba~ ,L L pow”: CMVM‘H \vgfI'Q/xj (:IM‘E- W145 ‘5; mil ‘ (fiipm) v3 3““ "1 {JUCJ‘MM' F A % I ““ wwww vM-mxv-w-wM-«M WWMMM‘W‘MWMM W M “ MWM‘MWMMWW “MW—"MIWMWWMJ‘M..MMM‘ mmwvwmwvwwww Mom/M m 9- " . \ A o- ; Holga 32¢ <90M$$> \ 2 bio/60 ” w «660 I A P 7 D g, :- A O“ : 121’?" L32: 3 art > 4?; m ¢>~ = —. 49d?” > ___> - % Ai- .. ZP>= A\°(2¢°)>£&PC~0~@%), FMLM LVN/5+ flmw, J‘ fliflfigfowm '57 {6“ = \L:(; >03) 2. (0»80 Veq. and e' = 8\e°. these to be lndependent 0Q Freq’uencg- For -€ 2 good wnductor so that d a W. [5%. 6-4- (12)} At higher ¥requencjes , use .6-4—(5) Wm =£2; oz wfiflm-gr ~11 ’ Note 43mm um: plot “Uth extranelg high attenuation occurs 03: m‘torowowe Muencies (-9 _>_ \0“Hz> Plot 0% Mm 410!“ Seawater ‘0i Ht, aan oonSIder as (o4 105 IO“ (0“ (06 go“ J; (He) Treat 3: Rom Tame 94b , e.— = r2. For '33: coat‘mg, ‘p'rum b-BCb) , n}. = $1.13 or (u: mm, , "1‘ 47"" A. _ §x19____.5 = as CL dkdeo’cric . :: \. . 0-4-0 m For l0 GHE ‘. 2.1 = 4n; _ lo‘°x4.x l-eb X\o““’ __ For N, 2 (own 2 1L}, = 4xl-Bb ‘ "SAW" Reflectivity (power) L” i [4 c ‘ _( / w " l w 1%” gt: a . (/‘zwé’a )“PemiMW «’ MIC «7(5 eflogS“ (I ‘ n w 4;; + {M 4 (3445444 i rmxmwwa ’1‘37 (4 >‘\ /‘ (ficfivV/w’ U5 Mfwwfl/ U W 3 “£2 \ My: 1.: c; .0 oo 1 | .0 \l I I .0 0) .0 01 .O 4:. .0 o.) F3 N 0.1 ‘ __J | | g 2.6 2.8 3 3,2 3.4 Frequency (Hz) Reflectivity (power) if?!“ :1 i’ flh‘ M; (/F‘J‘Zfi/i’c’e A ) 1 | | | | l I I 0.999 — 0-998 - WW SEQ” <2 9 2 0 997 ~ g... WWW::: (N Z 0.996 — 0.995 — 0.994 — 0.993 — 0.992 — 0.991 e 0.99 ' I L. I I .1 I . 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 Frequency (Hz) X 10 F1?) C/LQ 6/5/‘w Csz f a , /2 w g ‘) a.» (M .9 M7 (“M/Q [M iii (.74? mg: I J V ’ i V I (3, +51 S'fi 52m. 4.}... ‘ ,-~ . t . :K r .5 {I V I . i. > (é. : ML“ r , If; 2’ ); By {R {3” iflbi gaff I’ J56 ‘ ALL/$6 by 513/) a) .0 (é) / (“new Z/j / .154 J” l [A]; /U [C mt“ (U g kg J24? IL“ I, ' ' W? L/é/g’x [/3 wa if: 621/ 5/17 ' 3! 4 gay/2,: [($33 NH 3’ A? 1 I ' ' /‘ (we: ("1": aéwf (Iain 75'1"”? ’V/ C I 1 | | | Reflectivity (power) 0 O O O C) Le. 01 '03 A: be I | l | | l .0 o.) .0 M 0.1 .1 | | 2.8 3 3.2 3.4 3.6 3.8 4 Frequency (Hz) X1014 Reflectivity (power) 0.99 0.98 .0 «a \l .0 no 0) 29 co 01 0.94 0.93 0.92 0.91 0.9 2.2 2.4 2.6 2.8 3 Frequency (Hz) 3.2 3.4 3.6 3.8 x10 \ “Z Lm ~ %Lt\ OE» :: “1%; “(f/Wu (MM-"a w / ? ’(fiw $1.“ "J/L ,\ 4,.“ (:LPVL V4 1/ u/‘Mgfi A 3 _ a A ‘ A KM" TIC?“ F9“ zwcfl Mcsf‘m L) W CM bum/k KC Ts wagde EauAdlo/L.‘ CM&¢~k—;LWJ) (JV-k M& «M m 'nguuusL k; 4-6 LN»; Saw l!“ K)“: / tLCt‘IC—Ys Cal-IQ 'UMS kx ,_ {‘A‘) kKC 2. hx.‘ TL’W‘ fn *Q°‘"Q" ) L :— (AZ/cue; "‘ka New +64% LACva GJ 6:“;0 [07“ -)-a koka 511M Hfié )lkuk €L=€Lf dflc‘lffi-f, ...
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This note was uploaded on 12/12/2010 for the course ECE 604 taught by Professor Staff during the Spring '08 term at Purdue.

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HW8-sol - Sahib“ ECE604 Homework #8 Out: Friday, March...

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