Notes 19 Spring 2005

Notes 19 Spring 2005 - Notes 19 Spring 2005.doc Example: I....

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Notes 19 Spring 2005.doc Example : I. Let us assume that an antenna is feeding a television at a frequency of where we consider the antenna to be the source for the television, which we will consider the load. Connecting the two is a 12-meter length of lossless coaxial line. We use a Thevenin equivalent to represent our antenna, which has a Thevenin source voltage of MHz 6 75 ( ) V t 10 12 cos 10 6 6 4 Th × π × = V and a Thevenin output resistance of . Let us further assume that the impedance of the receiver is real and has a value of as shown in the figure below. = 75 R Th 75 = 75 R Th in i L i = 75 Z o + The phase velocity on the line is resulting in a wavelength of c 85 . 0 m 5 . 42 10 6 c 85 . 0 6 = × = = λ f v p on the transmission line. The line length of 12 meters then represents 0.282 wavelengths or we can say that since λ π = β 2 the electrical length of the line is ( ) ° = π = = β 6 . 101 radians 565 . 0 12 5 . 42 π 2 l . We note that the source is perfectly matched to the line and delivers maximum power to the line, and since the line is lossless, is also delivered to the load. The input voltage to the line can, through an application of voltage division, be seen to be half of the source voltage, : Th V ( ) ( ) V t 10 12 cos 10 3 75 75 75 t 10 12 cos 10 6 R Z Z 6 4 6 4 Th o o Th in × π × = + × π × = + = V V . Since there is no attenuation (a lossless line) and no reflections from the load (matched condition) we can expect that the magnitude of the voltage at the load is also but will be delayed by , i.e., V 10 3 4 × ° = β
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This note was uploaded on 01/20/2012 for the course EE 4460 taught by Professor Czarnecki during the Fall '10 term at LSU.

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Notes 19 Spring 2005 - Notes 19 Spring 2005.doc Example: I....

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