311ps5sol

311ps5sol - ECE311 Autumn 2011 Problem Set 5 Solutions...

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Unformatted text preview: ECE311 Autumn 2011: Problem Set 5 Solutions Problem 1 : a) First, this is definitely a transient problem because there is a particular time ( t = 0) when things change, and because we’re asked about time domain voltages and currents. Note that the time required for one trip down the line is distance velocity = . 6 3 × 10 8 √ 2 . 3 = 3 . 0332 nsecs. Thus at t = 2 nsecs, the initial voltage wave has not reached the end of the line, but rather traveled only 3 × 10 8 √ 2 . 3 (2 × 10- 9 ) = 0 . 3956 m or 0.2044 m from the load end. The line initially looks like a resistor of Z ohms before any reflections. We simply have a V + wave of (2)(50 k 110) = 68 . 75 V propagating down the line. Sketch is V(z,t=2nsec) (v) z (m) 68.75-0.6-0.2044 z (m) V+ (v) z (m) 68.75-0.6-0.1913 V- (v) z (m) Z=-0.6-62.77 Total current (A) z (m) 0.63-0.6 1.20 b) At t = 4 nsecs, the pulse has propagated 2 × as far as in part (a) = 0.7913 m which is down the line and 0.1913 m way back up. At the load end of the line, a reflected V- wave of amplitude Γ V + = 5- 110 5+110 (68 . 75) =- 62 . 77 V is generated. The V + and V- waves and the current look like: V(z,t=2nsec) (v) z (m) 68.75-0.6-0.2044 z (m) V+ (v) z (m) 68.75-0.6-0.1913 V- (v) z (m) Z=-0.6-62.77 V(z,t=2nsec) (v) z (m) 68.75-0.6-0.2044 z (m) V+ (v) z (m) 68.75-0.6-0.1913 V- (v) z (m) Z=-0.6-62.77 Total current (A) z (m) 0.63-0.6 1.201....
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This note was uploaded on 11/11/2011 for the course ECE 311 taught by Professor Johnson,j during the Fall '08 term at Ohio State.

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311ps5sol - ECE311 Autumn 2011 Problem Set 5 Solutions...

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