ECE Name ________________________ Section _______________________ March pm FIRST EXAM by benbrown

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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

								
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