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### EE335 HW#4-Chp2_45-49

Course: EE 335, Fall 2008
School: Montana
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Word Count: 124

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45-49 2.45 EE335 HW#4 Chp2: Z0=50 l=0.6 ZL=50+j25 zL=(50+25j)/50=1+0.5j yL=0.8-0.4j translate 0.3 (0.395+0.3)=.695=.195 1.37+0.46j z=3.04+0.4j (0.492+0.3)=.792=.292 yr=50/30=1.67 zin=1.9-1.4j Zin=50*zin=95-70j z@0.3 z@0.6 2.46 Z0=50 ZL=75-20j zL=(75-20j)/50=1.5-0.4j yL=0.62+0.17j Solution#1 ys=-0.5j @ l=(0.426-0.25)=0.176 d=(.143-.042)=.1 Solution#2 ys=0.5j@(.25+.08)=0.33 d=(.355-.042)=.313 zload 2.47 = Z0...

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45-49 2.45 EE335 HW#4 Chp2: Z0=50 l=0.6 ZL=50+j25 zL=(50+25j)/50=1+0.5j yL=0.8-0.4j translate 0.3 (0.395+0.3)=.695=.195 1.37+0.46j z=3.04+0.4j (0.492+0.3)=.792=.292 yr=50/30=1.67 zin=1.9-1.4j Zin=50*zin=95-70j z@0.3 z@0.6 2.46 Z0=50 ZL=75-20j zL=(75-20j)/50=1.5-0.4j yL=0.62+0.17j Solution#1 ys=-0.5j @ l=(0.426-0.25)=0.176 d=(.143-.042)=.1 Solution#2 ys=0.5j@(.25+.08)=0.33 d=(.355-.042)=.313 zload 2.47 = Z0 50 ZL = 100+50j zL = 2+1j 0.375 D=0.2 yin Zload yload Yin2 0.125 0.375 d1=2, d2=0.375, l=0.125 l=0.375 2.48 Z1=50+50j z1=1+1j y1=.5-.5j z2=1-1j y2=.5+.5j Z2=50-50j Yt=y1+y2=3.9 yin=0.28+0.3j Zin=82.5-88j l=0....

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Montana - EE - 335
EE335 2.50HW#5Chp2: 50-56 oddRg = 100 Vg = 60 V Z0 = 50 ZL = 25 g = V1+ =Z g - Z0 Z g - Z0 Vg Z 0 Rg + Z 0= =100 - 50 50 1 = = 100 + 50 150 3 60(50 ) = 20 100 + 50L =25 - 50 - 25 1 = =- 25 + 50 75 3=1 L = = 5 10 -9 8 u 2 10
Montana - EE - 335
EE335 8.1 a) =HW#7Chp8: 1-25 odd 2 - 1 = 2 + 10 0r2 r1 - 0 r2 r1 r2 r1 + 0 r2 r1= r1 - r 2 r1 + r 2=1 - 25 1 + 25=-4 = -0.6667 6 = 1 + = 1 + -0.6667 = 0.3333 b)~ E1 1 + 1 + 0.6667 1.6667 = =5 = S = ~ max = 1 - 1
Montana - EE - 335
EE335 8.27HW#8Chp8: 27-43 odda) polarization: H is perpendicular to the plane of incidence incidence. b) angle of incidenceE is in the plane of- jk ( x sin i + z cos i ) = - j (8 x + 6 z )~ ^ H i = y 2 10 -2 e - j (8 x +6 z )z 0, r
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EE335 9.1HW#9 Hertzian DipoleL=Chp9: 1-13 odd, 14-17I 0 = 20( A) S max 15 I 02 = R2502R = 1000(m )215 I 02 L L -6 2 = = 7.54 10 W / m 1000 2 50 L ()9.3 120 degree cone centered on z-axis1 F ( , ) = 0 F D = max = 1
Montana - EE - 335
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;* ;* Program Name: Lab#1 - Example of Indexed Addressing ;* Author Name: Chris Davenport ;* Date: 02/01/2007 ;* Description: This example shows how indexed addressing can be used ;* to cycle through a table of constants. Specifically, ;* this exampl
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Montana - EE - 465
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54321VCC Note: Can be implemented using pushbuttons on SLK VCC R5 4700kDVCC Check the resistor values to the I2C specification R14 C1 30k R15 30k R18 10k U1 8 1 Check the resistor values to the I2C specification R9 1k VCC VCC 16 U6 1 2 3
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54321Note: Can be implemented using pushbuttons on SLK VCC R5 4700kD DC1 RESET 1 2 0.1u U1 C41 PIN 4 ON J6 RESET 2 PIN6 ON J6 BKGD 3 +3.3V C2 10u GND .1u C3 D3 D2 D1 D0CRESET BKGD VDD VSS PTB7/SCL PTB6/SDA PTB5 PTB4PTA0 PTA1 PTA2/
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Requirements Specification for EE465 Lab Project 6: Digital Temperature Sensor with I2C Serial Two-Wire InterfaceLab project goal: Read the temperature from an LM92 digital temperature sensor via I2C communication and display the latest value on the
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