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ECE 350 Name:________________________ Section:_______________________ March 6, 2003, 7-8:30 pm FIRST EXAM NOTE:This is a closed book and closed notes exam. You are allowed one 8 ½” × 11” sheet with writing on both sides. Unless stated otherwise, do your work on the page of the problem and, if necessary, on the preceding blank page. Use only the Smith Charts provided. Additional copies of the Smith Chart, if needed, must be obtained by the proctor of the exam. All quantities are in the International System of Units, unless specified otherwise. For each problem you must show complete work and/or reasoning. No credit will be given if you do not show the complete work, even if the answer is correct. Also, numerical answers must be followed by units, whenever units exist. Write your name on all sheets, including the Smith Charts, and sign below on this page. Problem 1 Problem 2 Problem 3 Problem 4 TOTAL SIGNATURE________________________________ Signature__________________ PROBLEM 1. (50 points) a) A slotted coaxial transmission line with a characteristic impedance Z0 = 50 Ω is used to determine an unknown load impedance. We know that if the slotted line is terminated with a short-circuit, the voltage minima are found to be 20cm apart. One of these minima was used as a reference point. With the unknown impedance connected to the receiving end of the line, the VSWR was found to be 3 and the voltage minimum was found to be 5 cm from the reference point TOWARD THE LOAD. Find the value of the unknown load impedance. [20 pts] Solution: λ = 2 × 20 cm = 40 cm λ d min = − δ = 20 − 5 = 15cm 2 VSWR = 3 VSWR − 1 3 − 1 ΓR = = = 0 .5 VSWR + 1 3 + 1 4π π θ= d min − π = λ 2 ∴ ΓR = 0.5∠0.5π = 0.5∠90 = 0.5 j 1 + ΓR 1 + 0 .5 j 1.12 ∠ 26 .6 ZR = Z0 = 50 = 50 = 50 ∠53 .1 Ω 1 − ΓR 1 − 0 .5 j 1.12 ∠ − 26 .6 b) A loss-less transmission line is terminated in an unknown load, but measurements reveal that Zmax=75Ω and Zmin=25 Ω. What is the VSWR and characteristic impedance Z0? [10 pts] Solution: Z0 Z max = Z 0 × VSWR Z min = VSWR Z max Z 0 = Z max Z min = 43 .3Ω VSWR = = 1.73 Z min Signature___________________ c) The following problem is a classic example of a system with multiple reflections. Consider a source have ZERO internal resistance driving a λ/4 section of a lossless transmission line terminated by a SHORT CIRCUIT. If the source is Vs=50∠0o, find the current at the input to the transmission line I(d=λ/4).and the load current, I(d=0). [20 pts] Solution: There are a number of ways to answer this problem. For example, you can sum an infinite series of current reflections; or you can just apply the boundary conditions on the transmission line equations for the current and voltage. V ( d = λ / 4) = 50 = V + e jβd + V − e − jβd = jV + − jV − 0 = V + +V − ⇒ V + = − j 25V , V − = + j 25V , I ( d = λ / 4) = 1 Z0 ( ) ( V + e j β d − V − e − jβ d = 1 Z0 ) jV + + jV − = 1 50 ( 25V + 25V ) = 1 A I ( d = 0) = 1 Z0 ( V + −V − = )1 50 ( − j 25V − j 25V ) = − jA PROBLEM 2. (50 points) Signature___________________ The following is a problem on basic Smith chart properties. All answers must be obtained using the Smith Chart. You MUST show your work on the Chart for full credit. Assume that the characteristic impedance of a lossless transmission line is Z0=50Ω. The source voltage is 50∠0o and the source impedance is Zs=20Ω as illustrated by the figure below: (a) Assume that the length of the transmission line L=5λ/16 and that it is terminated in short circuit at d=0. Evaluate the input impedance of the transmission line at d=L. [20points] Solution: Z ( d = 5λ / 16 ) = − j 2 .4 ⇒ Z ( d = 5λ / 16 ) = − j120 Ω Z0 Signature___________________ (b) Indicate on the Smith chart the magnitude of the current ratio |I(d=0)|/(|V+|/Z0) and |I(d=L)|/(|V+|/Z0) and calculate the ratio K=|I(d=0)|/ |I(d=L)| [30points] Solution: I ( d = 5λ / 16 ) I ( d = 0) 2 + = 0.76 + =2 K= = 2.63 V / Z0 V / Z0 0.76 PROBLEM 3. [50 points] Signature___________________ Consider the transmission line circuit below. The tables under the circuit show a list of parameters measured on the line. Line 1 (Low-Loss) Line 2 (Loss-Less) Z01 VG Z1 Z02 Z2 α1 L1 L2 VG = 10.0∠0°V Z01 = 50 Ω Z02 = 300 Ω ZG = 50.0 Ω L1 = 5.0 λ L2 = 1.25 λ α 1 = 0.076 Ne/λ Z2 = 720 – j 360 Ω Z1 = j 250.0 Ω (a) Find the input impedance of Line 2. [10 points] L2 = 1.0 λ + 0.25 λ, therefore we can use Z2 2 Zin2 = 02 = 300 = 100 + j50 Ω Z R 720 − j360 (b) Find the total equivalent load of Line 1. [10 points] The equivalent load of line 1 is the parallel between Z1 and Zin2 −1 1 + 1 Z R1 = Z1 Zin2 = = 65 + j65 Ω j 250 100 + j50 Signature___________________ (c) Find the input impedance of Line 1. [15 points] Z − Z 65 + j65 − 50 Γ R1 = R1 0 = = 0.341+ j 0.3725 Z R1 + Z0 65 + j65 + 50 0.46766 1 −2α L1 −2 j β L1 1+ Γ R1e e 1+ (0.341+ j0.3725) e −2×0.076×5 e− j 2×2π ×5 Zin1 = Z0 = 50 1−Γ R1e−2α L1e−2 j β L1 1− (0.341+ j 0.3725)e−2×0.076×5 e− j 2×2π ×5 1 = 64.0717 + j 23.6415 Ω (d) Find the total time-average power delivered to the input of the network. [10 points] VG 10 Iin1 = = = 84.0 − j17.4 mA (ZG + Zin1) 50 + 64.07 + j 23.6415 Zin1 64.07 + j 23.6415 Vin1 = VG = 10 = 5.797 + j 0.871V (ZG + Zin1) 50 + 64.07 + j 23.6415 Pin1 = 1 Re Vin1Iin1 = 0.5*Re (5.797 + j0.871)(84.0 + j17.4) = 236.06 mW * 2 (e) Find the total time-average absorbed by Z1. [5 points] Z1 is purely imaginary, so it does not absorb time-average power. PROBLEM 4 [50 points] Signature___________________ This is a Smith Chart problem. All answers must be obtained using the Smith Chart. You must show your work on the Chart for full credit. Consider a transmission line with characteristic impedance Z0 = 50Ω, terminated by a load ZR=20.0+j10.0 Ω. Your task is to design a single stub matching network using a short circuited stub. (a) Determine the normalized load admittance on the chart. [5 points] Z 20.0 + j10.0 zR = R = = 0.4 + j0.2 Z0 50.0 yR = 2.0 − j1.0 (b) Find the location (in units of wavelength) closest to the load where a stub may be inserted to achieve impedance matching. [10 points] dstub1 = 0.0512λ (c) Determine normalized and actual input admittance of the stub. [10 points] y(dstub1) = 1.0 − j1.0 ⇒ ystub1 = + j1.0 y Ystub1 = stub1 = j 1.0 = 0.02[S ] Z0 5.0 (d) Determine the length of the stub in units of wavelength. [10 points] Lstub1 = 0.375λ ECE350 – Spring 2003 Exam 1 – Problem #4 ystub1 zR yR y(dstub1) Lstub1 dstub1