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Test 11.a).Write Expression for E,E,H,H for a y-polarized UPW, motion in +x direction in anonmagnetic lossy dielectricb) The time instantaneous electric and magnetic fields of a z-polarized UPW waveat 12 Ghz traveling in the +y direction in a nonmagnetic, lossy dielectric at y=0 are given byEz(0,t)=200cos(wt+0.9pi)H(0,t)=-1.25sin(wt+0.34pi)

Write E(y,t), H(y,t),E(y),H(y)Find n,e"/e', alpha, and the percentage reduction in amp over distance of wavelength.E(x)=ayEo/e^(-gmax)H(x)=(Eo/n)e^(-gmax)

Ey(x,t)=Eoe^(-alphay)cos(wt-By+phi)H(y,t)=(Eo/n)e^(-alphay)cos(wt-Bx+phi-theta)

b)Ez(y,t)=200e^(-alphay)cos(wt-By+0.9pi)Ez(y)=200e^(j0.9pi)e^(-gmay)

Hx(y,t)=-1.25e^(-alphay)sin(wt-By+0.34pi)Hx(y)=-1.25(-j)e^(j0.34pi)e^(-gmay)=1.25e^(j0.84pi)e^(-gmay)

n=E/H=200e^(j0.9pi)e^(-gmay)/1.25e^(j0.84pi)e^(-gmay)=160e^(j0.06pi)=ne^(jtheta)

e"/e'=tan(2theta)=tan(2*.06)=0.396er={(120pi)/(160[1+(e"/e')^2]^(1/4)}^2=5.162alpha=110.99 (Np/m)beta=185.15pi (rad/m)calculated by f=12x10^9(2pif/c)*sqrt(er)/sqrt(2)

lamda=2pi/beta=0.0108(m)e^(-alpa(lamda))=0.3

thus 70% reduciton

Test 12.A dc source Vg=3(V) is turned on a tranmission line system at time t=0. The current at the two ends of the line (z=0 and z=l=150m) are shown. Find Zo, Rg,Rl,er,Iss,V(l,10us)

--Rg-----------

| Zo |Vg(t) er Rl| |

_______________

| I(0,t)|20mA|_____ 8mA| |____|-----| | |

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-----------------t(us)6 12

|I(l,t)| 10mA| _____| | |______| | |-------------------t(us)

3 6

I1+=0.02 (A)I1++I1-=0.01(A)I1++I1-+I2+=0.008(A)thus: I1-=-0.01(A)I2+=-0.002(A)F(0)=ref/inc=-I2+/I1-=-0.002/-0.01=-1/5F(l)=-I1-/I1+=-0.01/0.02=1/2

Rg=Zo[(1+F(0)/(1-F(0)]=2/3Zo

I1+=Vg/(Rg+Zo) plug in I1+, Vg, Rg, solve for ZoZo=90Rg=2/3Zo=60

Rl=Zo[(1+F(l)/(1-F(l)]=270 plug in Zo and F(l)

T=l/v=3usv=l/T=150/3us=0.5x10^8(m/s)er=(c/v)^2=36

Iss=Vg/(Rg+Rl)=9.1mAV(l,10us)=V1++V1-+V2++V2-=Zo(I1++I1-+I2++I2-)=2.43V

3. For the branched load arrangement shown,Vg=18V, Rl=150, Rg=100

a. Obtain reflection coefficient at AA,CC, BB, DD

b. Draw the bounce diagram for AA,BB, and DD

c. Use the bounce diagram to mark the graphs shown for Vaa dn Vdd. Mark the horizontal axes in terms of transit time and vertical axes for voltages (with numebrfor three levels shown)

Faa=0=Fcc=Fbb-Fdd=1Fbb(150line)=[100||300-150]/[100||300+150]=-1/3Fbb(300line)==[100||150-300]/[100||150+300]=-2/3

Vss=[Rl/(Rl+Rg)]Vg=10.8Vplug in Rl,Rg,Vg given values

Draw bounce diagram with three levels at F(0),Faa-Faa+, and Fdd

then label the V vs t graph accordingly. at nT is the sum of the voltage at thatpoint and previous voltage

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Test 2A load, Zl terminating a 100 ohm line produces the current SW pattern shown. Obtain the following...---------100ohm FS|

Zl---------|

| I(z)|nm---------------|

/ \ / \ / \|/ \ / \ / |

---------------|____|______|____|_z(mm)

-58.1 -20.6

3/4lamd0=58.1-20.6=37.5mm

lamda0=0.05m; f=c/lamdo=6Ghzdmin=20.6/50=0.412

Rotate 0.412 from Imin(Vmax) towards load and draw line. This point is zl, markIload=0.89

zl=1-1j; ZL=100-100jSWR=2.6

Read Fload=0.45e^(-j63) and |Vload|nrm=1.26|Vmax|norm=1.43|Imin|norm=0.55

2. For the arrangement shown, where Zl=15+j35d==0.25lmd givenZtaa=50-j50 and |Ibb|=1.12|Io|

Find the following

a. Find Ylaa ans SWR on the main line right of AAFind Ytaa and SWR on the Main line left of AAFind Ysaa and SWR on Stub

b. Find Zs, l/lmd(there are two soln, select Zs such that l is the longer length)

Find Ylaazl=Zl/Zo=.3+.7j=ylaa since d=.25 rotate yl to ylaaYlaa=ylaa/50

SWRzl=right=5.0Enter ztaa=1-1j

read ytaa=0.5+0.5j andYtaa=ytaa/50SWRaa=left=2.6Ysaa=Ytaa-Ylaaysaa=.4-.4j=Ysaa100SWR stub=3enter ysaa on SCmark |Ibb|=1.12 and readzs=(0.6-j.8 (for longer l)

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Zs=zs100=60-80j,l=0.3lmd

3. Fo rthe branched load system shown, calculate the input impedance, Zin usingthe analytical expression. Graphicl soln will not be accepted. Show all calculations by writing the expression for Zin for each transmission line wih the appropriate values given.

Zl=100+j100

y=0.5lmd A d=1.5ld____________|_________|Zin-> 100 | \ Zl

| \ || \ |

------------\--------- ...l=0.125lmd \ Zs=0 short\ |

Zin=Zo[(Zl+jZotan(Bl))/(Zo+jZltan(Bl))]

Zlaa=100[(Zl+j100(2pid))/(100+jZltan(2pid))]=100+100j

Zsaa=50[0+j50tan(2pi.125).....=50j

Ztaa=Zlaa||Zsaa=ZlaaZsaa/(Zlaa+Zsaa)=7.69+38.461j

Zin=100[Ztaa+j100tan(2piy)..........=7.69+38.4j