The procedure on page 7 of notes set 2 also

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The procedure on page 7 of Notes Set 2 also simplifies the math function in equation ( p 1) of the problem statement, giving v 1 ( t ) = braceleftBigg A sin parenleftBig 2 π f c t - Θ L parenrightBig , t L t t L + 2 T p 0 , else ( e ) where Θ L , the phase “length” of the line, is given by eq. (15) from Notes Set 2: Θ L = 2 π x L λ ( f ) In equation ( f ), λ = v p /f c is the wavelength of the sinusoid on the line.
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ECE 303 - Sum2018 Turn-In Assignment 1: 25 May 2018 5 Problem 2. (cont.) With f c = 250 MHz = 2 . 5(10) 8 Hz, the time period T p is given by T p = 1 f c = 1 2 . 5(10) 8 = 4(10) 9 ( g ) Using ( d ) and x L = 0 . 4 m from the problem statement, the time length of the line is given by t L = 0 . 4 2(10) 8 = 2(10) 9 = 2 ns ( h ) The wavelength of this sinusoid is λ = v p f c = 2(10) 8 2 . 5(10) 8 = 0 . 8 meters ( i ) The ratio of the length of the line to the wavelength for this case is therefore x L λ = 0 . 4 0 . 8 = 0 . 5 ( j ) Substituting ( j ) into ( f ) gives the phase length of the line as Θ L = 2 π (0 . 5) = π rad = 180 ( k ) With A = 5 and f c = 2 . 5(10) 8 from the problem statement, substituting ( g ), ( h ), and ( k ) into equation ( e ) then gives the following for the voltage at the end of the line: v 1 ( t ) = braceleftBigg 5 sin parenleftBig 2 π 2 . 5(10) 8 t - π parenrightBig , 2 ns t 10 ns 0 , else ( l ) Substituting the same parameters into equation ( p 1) for the source signal gives v 0 ( t ) = braceleftBigg 5 sin(2 π 2 . 5(10) 8 t ) , 0 t 8 ns 0 , else ( m ) A plot of v 1 ( t ) from equation ( l ) and v 0 ( t ) from equation ( m ) is shown in the figure below: x = L 0.4, 0 t = 2 t = t = 8 10 t = t, ns v t ( ) 0 v t ( ) 1 5 250 MHz f = The phase length Θ L = π radians in ( l ) is significant compared to one sinusoidal cycle, which is 2 π 6 . 283 radians. This means that the signal v 1 ( t ), at the load end of the line, is different from v 0 ( t ), the signal at the beginning of the line. This signifies that circuit theory will not provide a
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