s Line E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 1 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. If fact signals travel at arond half the speed of light (c = 30 cm/ns). E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. If fact signals travel at arond half the speed of light (c = 30 cm/ns). Reason: all wires have capacitance to grond and to neighboring condctors and also self-indctance. It takes time to change the crrent throgh an indctor or voltage across a capacitor. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. If fact signals travel at arond half the speed of light (c = 30 cm/ns). Reason: all wires have capacitance to grond and to neighboring condctors and also self-indctance. It takes time to change the crrent throgh an indctor or voltage across a capacitor. A transmission line is a wire with a niform goemetry along its length: the capacitance and indctance of any segment is proportional to its length. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. If fact signals travel at arond half the speed of light (c = 30 cm/ns). Reason: all wires have capacitance to grond and to neighboring condctors and also self-indctance. It takes time to change the crrent throgh an indctor or voltage across a capacitor. A transmission line is a wire with a niform goemetry along its length: the capacitance and indctance of any segment is proportional to its length. We represent as a large nmber of small indctors and capacitors spaced along the line. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Lines s Line Previosly assme that any change inv S (t) appears instantly atv R (t). This is not tre. If fact signals travel at arond half the speed of light (c = 30 cm/ns). Reason: all wires have capacitance to grond and to neighboring condctors and also self-indctance. It takes time to change the crrent throgh an indctor or voltage across a capacitor. A transmission line is a wire with a niform goemetry along its length: the capacitance and indctance of any segment is proportional to its length. We represent as a large nmber of small indctors and capacitors spaced along the line. The signal speed along a transmisison line is predictable. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 2 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t = v t. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t Capacitor eqation: C v t = v t. = i(x,t) i(x+δx,t) = i x δx E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t Capacitor eqation: C v t Indctor eqation: L i t = v t. i = i(x,t) i(x+δx,t) = = v(x,t) v(x+δx,t) = v x δx x δx E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t Capacitor eqation: C v t Indctor eqation: L i t Transmission Line = v t. i = i(x,t) i(x+δx,t) = = v(x,t) v(x+δx,t) = v x δx x δx SbstitteC 0 = C δx andl 0 = L δx wherec 0 andl 0 are capacitance and indctance per nit length. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t Capacitor eqation: C v t Indctor eqation: L i t Transmission Line SbstitteC 0 = C δx andl 0 = L δx wherec 0 andl 0 are capacitance and indctance per nit length. = v t. i = i(x,t) i(x+δx,t) = = v(x,t) v(x+δx,t) = v x δx x δx v C 0 t = i x i L 0 t = v x E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Transmission Line s Line Short section of lineδx long. v(x,t) andi(x,t) depend on both position and time. Small δx ignore 2nd order derivatives: v(x,t) t = v(x+δx,t) t Capacitor eqation: C v t Indctor eqation: L i t Transmission Line SbstitteC 0 = C δx andl 0 = L δx wherec 0 andl 0 are capacitance and indctance per nit length. = v t. i = i(x,t) i(x+δx,t) = = v(x,t) v(x+δx,t) = v x δx x δx v C 0 t = i x i L 0 t = v x General soltion to two simltaneos partial differential eqations will inclde two arbitrary fnctions (instead of the arbitrary constants that yo get with ordinary differential eqations). E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 3 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x L 0 i t = v x E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x General soltion: v(t,x) = f(t x )+g(t+ x ) i(t,x) = f(t x ) g(t+ x ) where = 1 L 0 C 0 and = L 0 i t = v x L 0 C 0. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x General soltion: v(t,x) = f(t x )+g(t+ x ) i(t,x) = f(t x ) g(t+ x ) where = 1 L 0 C 0 and = L 0 i t = v x L 0 C 0. is the propagation velocity and is the characteristic impedance. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x General soltion: v(t,x) = f(t x )+g(t+ x ) i(t,x) = f(t x ) g(t+ x ) where = 1 L 0 C 0 and = L 0 i t = v x L 0 C 0. is the propagation velocity and is the characteristic impedance. f() and g() can be any differentiable fnctions. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x General soltion: v(t,x) = f(t x )+g(t+ x ) i(t,x) = f(t x ) g(t+ x ) where = 1 L 0 C 0 and = L 0 i t = v x L 0 C 0. is the propagation velocity and is the characteristic impedance. f() and g() can be any differentiable fnctions. Verify by sbstittion: i x = ( f (t x ) g (t+ x ) 1 ) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Soltion to Transmission Line s Line Transmission Line : C 0 v t = i x General soltion: v(t,x) = f(t x )+g(t+ x ) i(t,x) = f(t x ) g(t+ x ) where = 1 L 0 C 0 and = L 0 i t = v x L 0 C 0. is the propagation velocity and is the characteristic impedance. f() and g() can be any differentiable fnctions. Verify by sbstittion: i x = ( f (t x ) g (t+ x ) 1 ) = C 0 ( f (t x )+g (t+ x )) = C 0 v t E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 4 / 13
Forward Wave s Line = 15 cm/ns E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) 0 2 4 6 8 10 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) f(t-90/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. Atx = 90 cm [ ],v R (t) = f(t 90 ); now delayed by6ns. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) f(t-90/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. Atx = 90 cm [ ],v R (t) = f(t 90 ); now delayed by6ns. Waveform at x = 0 completely determines the waveform everywhere else. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) f(t-90/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. Atx = 90 cm [ ],v R (t) = f(t 90 ); now delayed by6ns. Waveform at x = 0 completely determines the waveform everywhere else. Snapshot att 0 = 4ns: the waveform has jst arrived at the point t = 4 ns f(4-x/) 0 20 40 60 80 x = t 0 = 60 cm. Position (cm) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) f(t-90/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. Atx = 90 cm [ ],v R (t) = f(t 90 ); now delayed by6ns. Waveform at x = 0 completely determines the waveform everywhere else. Snapshot att 0 = 4ns: the waveform has jst arrived at the point t = 4 ns f(4-x/) 0 20 40 60 80 x = t 0 = 60 cm. Position (cm) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward Wave s Line = 15 cm/ns Atx = 0 cm [ ], v S (t) = f(t 0 ) f(t-0/) f(t-45/) f(t-90/) Atx = 45 cm [ ], v(45,t) = f(t 45 ) 0 2 4 6 8 10 f(t 45 ) is the same asf(t) bt delayed by 45 = 3 ns. Atx = 90 cm [ ],v R (t) = f(t 90 ); now delayed by6ns. Waveform at x = 0 completely determines the waveform everywhere else. Snapshot att 0 = 4ns: the waveform has jst arrived at the point t = 4 ns f(4-x/) 0 20 40 60 80 x = t 0 = 60 cm. Position (cm) f(t x ) is a wave travelling forward (i.e. towards+x) along the line. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 5 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 f(t-0/) g(t+0/) 0 2 4 6 8 10 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 f(t-0/) g(t+0/) Atx = 0 cm [ ], v S (t) = f(t)+g(t) 0 2 4 6 8 10 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 f(t-0/) g(t+90/) f(t-90/) g(t+0/) Atx = 0 cm [ ], x=90 v S (t) = f(t)+g(t) 0 2 4 6 8 10 Atx = 90 cm [ ],g starts att = 1 andf starts att = 6. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 f(t-0/) g(t+90/) f(t-45/)+g(t+45/) f(t-90/) g(t+0/) Atx = 0 cm [ ], x=90 v S (t) = f(t)+g(t) 0 2 4 6 8 10 Atx = 45 cm [ ],g is only1ns behindf and they add together. Atx = 90 cm [ ],g starts att = 1 andf starts att = 6. x=45 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 f(t-0/) g(t+90/) f(t-45/)+g(t+45/) f(t-90/) g(t+0/) Atx = 0 cm [ ], x=90 v S (t) = f(t)+g(t) 0 2 4 6 8 10 Atx = 45 cm [ ],g is only1ns behindf and they add together. Atx = 90 cm [ ],g starts att = 1 andf starts att = 6. x=45 Snapshots: Att = 2 nsf andg have not f(2-x/)+g(2-x/) yet met. 0 20 40 60 80 t = 2 ns Position (cm) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 x=45 f(t-0/) g(t+90/) f(t-45/)+g(t+45/) f(t-90/) g(t+0/) Atx = 0 cm [ ], x=90 v S (t) = f(t)+g(t) 0 2 4 6 8 10 Atx = 45 cm [ ],g is only1ns behindf and they add together. Atx = 90 cm [ ],g starts att = 1 andf starts att = 6. Snapshots: Att = 2 nsf andg have not yet met. f(2-x/)+g(2-x/) 0 20 40 60 80 t = 2 ns Position (cm) Byt = 5 ns,f andg have jst crossed. f(5-x/)+g(5-x/) 0 20 40 60 80 t = 5 ns Position (cm) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Forward + Backward s Line Similarlyg(t+ x ) is a wave travelling backwards, i.e. in the x direction. v(x,t) = f(t x )+g(t+ x ) x=0 x=45 f(t-0/) g(t+90/) f(t-45/)+g(t+45/) f(t-90/) g(t+0/) Atx = 0 cm [ ], x=90 v S (t) = f(t)+g(t) 0 2 4 6 8 10 Atx = 45 cm [ ],g is only1ns behindf and they add together. Atx = 90 cm [ ],g starts att = 1 andf starts att = 6. Snapshots: Att = 2 nsf andg have not yet met. f(2-x/)+g(2-x/) 0 20 40 60 80 t = 2 ns Position (cm) Byt = 5 ns,f andg have jst crossed. f(5-x/)+g(5-x/) 0 20 40 60 80 t = 5 ns Position (cm) Magically, f and g pass throgh each other entirely naltered E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 6 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z 1 0 (f x (t) g x (t)). E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x ) andgy (t) = g x ( t+ y x ). E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t)= Z 1 0 (f x (t)+g x (t))(f x (t) g x (t)) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t)= Z0 1 (f x (t)+g x (t))(f x (t) g x (t)) = f2 x (t) g2 x (t) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t)= Z0 1 (f x (t)+g x (t))(f x (t) g x (t)) = f2 x (t) g2 x (t) f x carries power into shaded area andg x carries power ot independently. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t)= Z0 1 (f x (t)+g x (t))(f x (t) g x (t)) = f2 x (t) g2 x (t) f x carries power into shaded area andg x carries power ot independently. Power travels in the same direction as the wave. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Power Flow s Line Definef x (t) = f ( t x ) andgx (t) = g ( t+ x ) to be the forward and backward waveforms at any point, x. i is always measred in the +ve x direction. Then v x (t) = f x (t)+g x (t) and i x (t) = Z0 1 (f x (t) g x (t)). Note: Knowing the waveformf x (t) org x (t) at any positionx, tells yo it at all other positions: f y (t) = f x ( t y x Power Flow ) andgy (t) = g x ( t+ y x ). The power transferred into the shaded region across the bondary atxis P x (t) = v x (t)i x (t)= Z0 1 (f x (t)+g x (t))(f x (t) g x (t)) = f2 x (t) g2 x (t) f x carries power into shaded area andg x carries power ot independently. Power travels in the same direction as the wave. The same power as wold be absorbed by a [ficticios] resistor of vale. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 7 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L v x = f x +g x i x = Z 1 0 (f x g x ) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z 1 0 (f L (t) g L (t))r L v x = f x +g x i x = Z 1 0 (f x g x ) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z0 1 (f L (t) g L (t))r L From this: g L (t) = R L R L + f L (t) v x = f x +g x i x = Z 1 0 (f x g x ) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z0 1 (f L (t) g L (t))r L From this: g L (t) = R L R L + f L (t) v x = f x +g x i x = Z 1 0 (f x g x ) We define the reflection coefficient: ρ L = g L(t) f L (t) = R L R L + = +0.5 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z0 1 (f L (t) g L (t))r L From this: g L (t) = R L R L + f L (t) v x = f x +g x i x = Z 1 0 (f x g x ) We define the reflection coefficient: ρ L = g L(t) f L (t) = R L R L + = +0.5 Sbstittingg L (t) = ρ L f L (t) gives v L (t) = (1+ρ L )f L (t) andi L (t) = (1 ρ L )Z 1 0 f L(t) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z0 1 (f L (t) g L (t))r L From this: g L (t) = R L R L + f L (t) v x = f x +g x i x = Z 1 0 (f x g x ) We define the reflection coefficient: ρ L = g L(t) f L (t) = R L R L + = +0.5 Sbstittingg L (t) = ρ L f L (t) gives v L (t) = (1+ρ L )f L (t) andi L (t) = (1 ρ L )Z 1 0 f L(t) v 0 (t) 0 2 4 6 8 10 12 14 16 18 At sorce end: g 0 (t) = ρ L f 0 ( t 2L ) i.e. delayed by 2L = 12 ns. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflections s Line From Ohm s law atx = L, we have v L (t) = i L (t)r L Hence(f L (t)+g L (t)) = Z0 1 (f L (t) g L (t))r L From this: g L (t) = R L R L + f L (t) v x = f x +g x i x = Z 1 0 (f x g x ) We define the reflection coefficient: ρ L = g L(t) f L (t) = R L R L + = +0.5 Sbstittingg L (t) = ρ L f L (t) gives v L (t) = (1+ρ L )f L (t) andi L (t) = (1 ρ L )Z 1 0 f L(t) v 0 (t) i 0 (t) 0 2 4 6 8 10 12 14 16 18 0 2 4 6 8 10 12 14 16 18 ( ) At sorce end: g 0 (t) = ρ L f 0 t 2L i.e. delayed by 2L Note that the reflected crrent has been mltiplied by ρ. = 12 ns. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 8 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R +1 ρ depends on the ratio R. R ρ 3 +0.5 v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 R -1 Comment E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ +1 ρ depends on the ratio R. R ρ v L (t) f(t) 3 +0.5 1.5 ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 R -1 Comment E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R ρ v L (t) f(t) ρ i L (t) f(t) 3 +0.5 1.5 0.5 1 0-1 0 1 2 3 4 5 R -1 Comment E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R ρ v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 R -1 Comment 3 +0.5 1.5 0.5 R > ρ > 0 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R ρ v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 R -1 Comment 3 +0.5 1.5 0.5 R > ρ > 0 1 3 0.5 0.5 1.5 R < ρ < 0 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R ρ v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 R -1 Comment 3 +0.5 1.5 0.5 R > ρ > 0 1 0 1 1 Matched: No reflection at all 1 3 0.5 0.5 1.5 R < ρ < 0 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 ρ Comment +1 2 0 Open circit: v L = 2f,i L 0 3 +0.5 1.5 0.5 R > ρ > 0 1 0 1 1 Matched: No reflection at all 1 3 0.5 0.5 1.5 R < ρ < 0 R -1 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 ρ Comment +1 2 0 Open circit: v L = 2f,i L 0 3 +0.5 1.5 0.5 R > ρ > 0 1 0 1 1 Matched: No reflection at all 1 3 0.5 0.5 1.5 R < ρ < 0 0 1 0 2 Short circit: v L 0,i L = 2f R -1 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 ρ Comment +1 2 0 Open circit: v L = 2f,i L 0 3 +0.5 1.5 0.5 R > ρ > 0 1 0 1 1 Matched: No reflection at all 1 3 0.5 0.5 1.5 R < ρ < 0 0 1 0 2 Short circit: v L 0,i L = 2f R -1 Note: Reverse mapping isr = v L il = 1+ρ 1 ρ E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Reflection Coefficients s Line ρ = R R+ = R Z 1 0 R v L (t) f(t) = 1+ρ i L (t) f(t) = 1 ρ +1 ρ depends on the ratio R. R v L (t) f(t) ρ i L (t) f(t) 1 0-1 0 1 2 3 4 5 ρ Comment +1 2 0 Open circit: v L = 2f,i L 0 3 +0.5 1.5 0.5 R > ρ > 0 1 0 1 1 Matched: No reflection at all 1 3 0.5 0.5 1.5 R < ρ < 0 0 1 0 2 Short circit: v L 0,i L = 2f R -1 Note: Reverse mapping isr = v L il = 1+ρ 1 ρ Remember: ρ { 1, +1} and increases with R. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 9 / 13
Driving a line s Line From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Driving a line s Line v x = f x +g x i x = f x g x From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. Sbstittingv 0 (t) = f 0 +g 0 andi 0 (t) = f 0 g 0 f 0 (t) = R S + v S (t)+ R S R S + g 0 (t) leads to: E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Driving a line s Line v x = f x +g x i x = f x g x From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. Sbstittingv 0 (t) = f 0 +g 0 andi 0 (t) = f 0 g 0 leads to: f 0 (t) = R S + v S (t)+ R S R S + g 0 (t) τ 0 v S (t)+ρ 0 g 0 (t) E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Driving a line s Line v x = f x +g x i x = f x g x From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. Sbstittingv 0 (t) = f 0 +g 0 andi 0 (t) = f 0 g 0 leads to: f 0 (t) = R S + v S (t)+ R S R S + g 0 (t) τ 0 v S (t)+ρ 0 g 0 (t) Sof 0 (t) is the sperposition of two terms: (1) Inptv S (t) mltiplied byτ 0 = R S + which is the same as a potential divider if yo replace the line with a [ficticios] resistor. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Driving a line s Line v x = f x +g x i x = f x g x From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. Sbstittingv 0 (t) = f 0 +g 0 andi 0 (t) = f 0 g 0 leads to: f 0 (t) = R S + v S (t)+ R S R S + g 0 (t) τ 0 v S (t)+ρ 0 g 0 (t) Sof 0 (t) is the sperposition of two terms: (1) Inptv S (t) mltiplied byτ 0 = R S + which is the same as a potential divider if yo replace the line with a [ficticios] resistor. (2) The incoming backward wave,g 0 (t), mltiplied by a reflection coefficient: ρ 0 = R S R S +. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Driving a line s Line v x = f x +g x i x = f x g x From Ohm s law atx = 0, we havev 0 (t) = v S (t) i 0 (t)r S wherer S is the Thévenin resistance of the voltage sorce. Sbstittingv 0 (t) = f 0 +g 0 andi 0 (t) = f 0 g 0 leads to: f 0 (t) = R S + v S (t)+ R S R S + g 0 (t) τ 0 v S (t)+ρ 0 g 0 (t) Sof 0 (t) is the sperposition of two terms: (1) Inptv S (t) mltiplied byτ 0 = R S + which is the same as a potential divider if yo replace the line with a [ficticios] resistor. (2) The incoming backward wave,g 0 (t), mltiplied by a reflection coefficient: ρ 0 = R S R S +. ForR S = 20: τ 0 = 100 20+100 = 0.83 and ρ 0 = 20 100 20+100 = 0.67. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 10 / 13
Mltiple Reflections s Line ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) = ( ) f 0 (t)+g L t L E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L v L (t) = f 0 ( t L ) +gl (t) f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L v L (t) = f 0 ( t L ) +gl (t) f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 v L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L v L (t) = f 0 ( t L ) +gl (t) f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 v L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Mltiple Reflections s Line Each extra bit off 0 is delayed by 2L (=12ns) and mltiplied byρ L ρ 0 : f 0 (t) = i=0 τ ( ) 0ρ i L ρi 0v S t 2Li ( ) g L (t) = ρ L f 0 t L v 0 (t) = ( ) f 0 (t)+g L t L v L (t) = f 0 ( t L ) +gl (t) f 0 (t) ρ 0 = 2 3 ρ L = 1 2 v x = f x +g x 0 5 10 15 20 25 30 g L (t) 0 5 10 15 20 25 30 v 0 (t) 0 5 10 15 20 25 30 v L (t) 0 5 10 15 20 25 30 E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 11 / 13
Transmission Line s Line Integrated circits & Printed circit boards High speed digital or high freqency analog interconnections 100Ω, 15 cm/ns. Long Cables Coaxial cable ( coax ): naffacted by external fields; se for antennae and instrmentation. = 50 or75ω, 25 cm/ns. Twisted Pairs: cheaper and thinner than coax and resistant to magnetic fields; se for compter network and telephone cabling. 100Ω, 19 cm/ns. When do yo have to bother? Answer: long cables or high freqencies. Yo can completely ignore transmission line effects iflength Adio (< 20 khz) never matters. Compters (1 GHz) sally matters. Radio/TV sally matters. freqency = wavelength. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 12 / 13
Smmary s Line Signals travel at arond 1 2 c = 15 cm/ns. Only matters for high freqencies or long cables. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: f x (t) = f 0 ( t x ) and ( ) g x (t) = g 0 t+ x E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x Terminating line withratx = L links the forward and backward waves: backward wave isg L = ρ L f L whereρ L = R R+ E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x Terminating line withratx = L links the forward and backward waves: backward wave isg L = ρ L f L whereρ L = R R+ the reflection coefficient,ρ L { 1,+1} and increases withr E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x Terminating line withratx = L links the forward and backward waves: backward wave isg L = ρ L f L whereρ L = R R+ the reflection coefficient,ρ L { 1,+1} and increases withr R = avoids reflections: matched termination. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x Terminating line withratx = L links the forward and backward waves: backward wave isg L = ρ L f L whereρ L = R R+ the reflection coefficient,ρ L { 1,+1} and increases withr R = avoids reflections: matched termination. Reflections go on for ever nless one or both ends are matched. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13
Smmary s Line Signals travel at arond 1 2c = 15 cm/ns. Only matters for high freqencies or long cables. Forward and backward waves travel along the line: ( ) ( ) f x (t) = f 0 t x and g x (t) = g 0 t+ x Knowing f x andg x at any singlexposition tells yo everything Voltage and crrent are: v x = f x +g x andi x = f x g x Terminating line withratx = L links the forward and backward waves: backward wave isg L = ρ L f L whereρ L = R R+ the reflection coefficient,ρ L { 1,+1} and increases withr R = avoids reflections: matched termination. Reflections go on for ever nless one or both ends are matched. f is infinite sm of copies of the inpt signal delayed sccessively by the rond-trip delay, 2L, and mltiplied byρ Lρ 0. E1.1 Analysis of Circits (2015-6973) Transmission Lines: 17 13 / 13