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Wpower_solutions - Solutions Set 7 Power Flow on...

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ECE 303 - Sum2014 Solutions Set 7: Power Flow on Transmission Lines 1 7-1. Let the voltage across a 100 Ω resistor be v ( t ) = 5 cos(2 π 1000 t ) and let the current be i ( t ). (a) Compute and plot the current i ( t ) over 0 t 3 msec. (b) Compute and plot the instantaneous power p ( t ) = v ( t ) i ( t ) over 0 t 3 msec. (c) Compute P avg , the time-average power, using the time-domain integral given by eq. (3) in the notes. DETAILED SOLUTION: (a) Compute the current i ( t ) : The current is easily computed using Ohm’s Law: i ( t ) = v ( t ) R = 5 cos(2 π 1000 t ) 100 ( a ) Simplifying ( a ) then gives the desired result : i ( t ) = 0 . 05 cos(2 π 1000 t ) ( b ) (b) Compute the instantaneous power p ( t ) = v ( t ) i ( t ) : The instantaneous power is given by the product of current and voltage : p ( t ) = v ( t ) i ( t ) ( c ) Substitute into equation ( c ) the voltage given in the problem statement and the current from ( b ) : p ( t ) = 5 cos(2 π 1000 t ) × 0 . 05 cos(2 π 1000 t ) ( d ) Performing the multiplication in ( d ) then gives the desired result : p ( t ) = 0 . 25 cos 2 (2 π 1000 t ) ( e ) (c) Compute P avg , the time-average power, using the time-domain integral. From the notes, the time-domain integral definition of the time-average power is given by P avg = 1 T p integraldisplay T p t =0 p ( t ) dt ( f ) The period T p is given by the inverse of the frequency of the sinusoid. With f = 1000 Hz from the problem statement, we therefore have T p = 1 f = 1 1000 = 0 . 001 ( g ) Substituting ( e ) and ( g ) into ( f ) then gives the following integral : P avg = 1 0 . 001 integraldisplay . 001 t =0 0 . 25 cos 2 (2 π 1000 t ) dt = 250 integraldisplay . 001 t =0 cos 2 (2 π 1000 t ) dt ( h )
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ECE 303 - Sum2014 Solutions Set 7: Power Flow on Transmission Lines 2 Prob. 7-1 (cont.) To evaluate ( h ), use the following trigonometric identity : cos 2 α = 1 2 + 1 2 cos2 α ( i ) Use ( i ) in ( h ) with α = 2 π 1000 t . This gives P avg = 125 integraldisplay . 001 t =0 bracketleftBig 1 + cos(2 π 2000 t ) bracketrightBig dt ( j ) Expanding ( j ) into the two integrals then gives P avg = 125 integraldisplay . 001 t =0 dt + 125 integraldisplay . 001 t =0 cos(2 π 2000 t ) dt ( k ) The first integral in ( k ) evaluates to 0.001 and the second integral (containing the cosine) in ( k ) evaluates to zero. This provides the following answer for the time-average power : P avg = 0 . 125 ( l )
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ECE 303 - Sum2014 Solutions Set 7: Power Flow on Transmission Lines 3 7-7. A lossy transmission line operating at f 0 = 200 KHz has per kilometer propagation constant γ = 0 . 8 + j 5. The characteristic impedance Z 0 = 226 - j 35 Ω. A transmitter voltage waveform v 0 ( t ) = A cos(2 πf 0 t ) enters the line at y = 0 + . The time-average power received at a distance 700 meters from the transmitter must be at least 10 milliwatts. Compute the minimum value of A which will provide this power. DETAILED SOLUTION: Here is the constraint expressed in watts as a mathematical relation: P avg 0 . 01 ( a ) The expression for P avg has been found from the notes to be P avg = A 2 e - 2 αx 2 | Z 0 | cos φ ( b ) where | Z 0 | and φ are obtained from the polar form for Z 0 : Z 0 = | Z 0 | e j φ = 228 . 7 e - j 0 . 154 ( c ) It should also be recognized that since α is given in km - 1 units then the distance x must be in km units. Therefore, we need to use x = 0 . 7 km ( d ) in this problem. Now substitute the expression for P avg from ( b ) into the left side of ( a ), and
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