Issue link: https://resources.pcb.cadence.com/i/1480209
Rs Source L1 L2 C1 C2 Ln V2' Cn Transmission line consisting of "n" LC segments Step injected back into transmission line due to reflection at load Total energy injected into the transmission line during time 2Td from step reflected at load side = Total energy stored in the LC network during this time + Total energy dissipated in the source resistance Energy injected during 2T d =V*I*t=V 2 'I 2 '2n√(LC) Original voltage source shorted out to apply superposition principle Load end Source end Figure 9: Effective circuit Starting right from the reflected step input application instant from the load side, for a time duration = the round trip transmission line delay time = 2T d =2n√(LC), the energy injected by the reflected step into the transmission line is V*I*t=V2'I2'2n√(LC), where V2',I2' is V2-V1 and (V2-V1)/Z0, respectively. In other words, the amplitude of the voltage step injected back by reflection at the load end, and the effective current injected back due to this reflected step (this current is actually the delta between the current drawn by the output load and the current initially drawn by the transmission line from the main source = V/(Z0+Rs), due to the effect of load reflection), n is the number of LC segments, and L and C are the unit inductance and capacitance values per segment. By the principle of energy conservation, this is the total available energy to be stored in the inductors and capacitors within the transmission line and to be dissipated in the source resistance Rs with which the transmission line is terminated. In this consideration, we are neglecting the energy stored within the LC network due to the main step from the source side. We are only considering the energy freshly injected into the LC network by the reflected load step, for applying the principle of superposition. Let the new voltage at the source end of the transmission line (due to the reflected load step alone, ignoring the effect of the main source by shorting it out) be V3'. Following the same approach we used before: • 3' *n+ *n√(LC) 2 1 2 1 V 2 'I 2 '2n√(LC) = { L( ) 2 3' 2 + CV 3 ' 2 } • We know I 2' =V 2' /Z0. Thus, 3' 2n√(LC)={ *n√(LC) *n+ 2 1 2 1 L( ) 2 2 3' 2' 0 2 + CV 3 ' 2 } • Dividing both sides by n√(LC), 3' *2= *Z0 + + 2 1 2 1 ( ) 2 2 3' 2 3' 0 0 2' 2 • That is, *2 = V 3 ' 2 + 2 0 1 + 1 2 2 0 0 2' 2 * { 2 } • 2ܴܼܵ 0 2 *2 = V 3 ' 2 + + ܼ0 ܸ 2' 2 * { } 2ܴܵ 3 4ܴܵ 2 ܼ0 4ܴܵ 3 ܼ0 • ܼ 0 2 + + * { } ܴܵ 2 2ܴܼܵ0 4ܴܵ 2 V 2 ' 2 = V 3 ' 2 • (ܼ0 + * { } ܴܵ) 2 4ܴܵ 2 V 2 ' 2 = V 3 ' 2 • (ܼ0 + * { } ܴܵ) 2ܴܵ V 2 ' = V 3 ' -IV • Or, (ܼ0 + * { } ܴܵ) 2ܴܵ V 2 ' = V 3 ' , where V 2 '=V 2 -V 1 -V This is the expression for new voltage at the source end of the transmission line (due to the reflected load step alone, ignoring the effect of the main source by shorting it out) after the first delay Td, i.e. after the load reflected step first reaches the source end. Superimposing the effect of the actual source, too, the actual voltage at the source end after the first source reflection is given by V3 =V3'+V1, where V1 is VZ0/(Z0+Rs) (V being the amplitude of the original step from the main source). The reflection coefficient at the source side is given by the ratio of the amplitude of the reflected wave from the source end (V3-V2) to the incident wave (V2-V1), i.e. reflection coefficient at the source end ρ = (V3-V2) / (V2-V1). • Setting (ܼ0 + * ܴܵ) 2ܴܵ (ܼ0 + ܴܵ) 2ܴܵ = = = ( V 3 V 2 -V 1 )* + V 1 + V 1 + V 1 V 3 ' V 2 ' • Thus, (ܼ0 + * ܴܵ) 2ܴܵ (ܼ0 + ܴܵ) 2ܴܵ - - V 3- V 2 = -(V 2 -V 1 )+(V 2 -V 1 )* = (V 2 + V 1 V 2 V 1 ) • So, reflection coefficient at the source end (ܼ0 + ܴܵ) 2ܴܵ ρ = (V 3 -V 2 ) / (V 2 -V 1 ) = -1+ www.cadence.com 7 Accurately Modeling Transmission Line Behavior with an LC Network-based Approach