# test1 - ECE 303 Sum2014 1 Test 1 w/Answers 1 Let vp =...

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ECE 303 - Sum2014 Test 1 w/Answers 1 1. Let v p = 2(10) 8 m/s and T = 0 . 3 μ sec, and let a waveform be defined by the math form g ( u ) = braceleftBigg 2 - 10 7 u, 0 u T 0 , else Let time t 1 = 0 . 5 μ sec and define the x -domain function v ( x ) by v ( x ) = g parenleftBig t 1 - x v p parenrightBig . (a) Compute the simplest math form for v ( x ). (b) Plot v ( x ) over 0 x 300. DETAILED SOLUTION: (a) Compute the simplest math form for v ( x ) . Substitute u = t - x v p into the equation for g ( u ) in the problem statement: g parenleftBig t - x v p parenrightBig = 2 - 10 7 parenleftBig t - x v p parenrightBig , 0 t - x v p T 0 , else ( a ) Since the problem statement requests a plot on the x -axis, we need to rewrite the bounded inequality in ( a ) as an inequality on x . One way to do this is to separate the bounded inequality into the two separate inequalities shown below, and then work on each inequality to obtain inequalities for x : 0 t - x v p t - x v p T x v p t t - T x v p x v p t v p ( t - T ) x The two inequalitites in the last row above are equivalent to the single inequality shown below: v p ( t - T ) x v p t ( b ) Using ( b ), the relation in equation ( a ) can now be expressed as g parenleftBig t - x v p parenrightBig = 2 - 10 7 parenleftBig t - x v p parenrightBig , v p ( t - T ) x v p t 0 , else ( c ) We can now substitute the appropriate values of T , v p , and t into ( c ) to compute the math answer, after which we can plot it on the x -axis. Substitute v p = 2(10) 8 m/s and T = 0 . 3 μ sec from the problem statement and t = t 1 = 0.5 μ sec into the terms in the inequalities in equation ( c ): v p ( t 1 - T ) = 2(10) 8 parenleftBig 5(10) - 7 - 3(10) - 7 parenrightBig = 100 - 60 = 40 ( d ) v p t 1 = 2(10) 8 × 5(10) - 7 = 100 ( e )

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ECE 303 - Sum2014 Test 1 w/Answers 2 Problem 1. (cont.) The math expression on the left side of ( c ) can be computed as 2 - 10 7 parenleftBig t 1 - x v p parenrightBig = 2 - 10 7 parenleftBig 5(10) - 7 - x 2(10) 8 parenrightBig = - 3 + 0 . 05 x ( f ) Note that the substitution t = t 1 has removed the t dependences and we are left with the space- domain function v ( x ). Thus, substitute the results of ( d ), ( e ), and ( f ) into ( c ) to obtain the desired form for the simplest math answer: v ( x ) = braceleftBigg 3 - 0 . 005 x, 40 x 100 0 , else ( e ) (b) Plot v ( x ) over 0 x 300 . A plot of equation ( e ) over the region 0 x 3 is shown below: ( ) v x 0 40 100 2 x (m)
ECE 303 - Sum2014 Test 1 w/Answers 3 2. At t = 0, a source with R s = 200 Ω produces a rectangular pulse which begins to propagate on a lossless transmission line. The line is terminated with a 80 Ω load at x = 5 meters. The pulse v 0 ( t ) initially (at x = 0 + ) on the line is a 3 volt, 0.04 μ sec pulse. The line has characteristic resistance R 0 = 120 Ω and v p = 2(10) 8 m/sec. Plot the voltage on the line at t = 60 nanosec. DETAILED SOLUTION: We need to determine the Bounce Diagram, from which it is easy to draw the voltage on the line. From the problem statement, an equivalent math form for v 0 ( t ) is v 0 ( t ) = braceleftBigg 3 , 0 t 40 nsec 0 , else ( a ) Equation ( a ) means the pulse voltage on the first transit (toward the load) is A = 3 volts.

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• Spring '08
• ALEXANDER
• Problem Statement, Transmission line

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