4.4. Time Domain Reflectometry

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Cuthbert Ward
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1 4.4. Time omain Reflectometry Task. lossless line 4 km long has characteristic impedance of 6 Ω and is terminated at the far end with 6 Ω. Exactly in the middle of the line there is an impedance of 3 Ω between the wires. For testing purposes a voltage square impulse of 5, µs long is sent through the line. raw lattice diagram of signal propagation through the line. ssume propagation speed of, km/s and output impedance of the transmitter of 3 Ω. raw lattice diagram of signal propagation in case the test signal is Heaviside step function having 5 amplitude. Solution The scheme of the line in question is shown in Fig.. source Figure. Schematics of the line oefficients of division and reflection at node are source ρ k source source source 6 Ω 6 Ω 3 Ω 3 3 Ω 6 Ω 3 Ω 6 Ω 3. oefficients of reflection at node are: ρ' ρ 4.4 -

2 4.4. Time omain Reflectometry In, the single reflection coefficient is: Ω 6 Ω '. Ω 6 Ω.3 The respective lattice diagram is shown in Fig. in case of a single square pulse. 4 km ρ = ρ' = -/ km ρ = -/ ρ = -/3 km µs µs 3 µs 4 µs 5 µs 6 µs Figure. Lattice diagram of the propagation the single square pulse through the line In Fig. 3 is shown the lattice diagram of this same line but in the case of the propagation of the Heaviside s step function signal. 4. -

3 4.4. Time omain Reflectometry 4 km ρ = ρ' = -/ km ρ = -/ ρ = -/3 km µs µs 3 µs 4 µs 5 µs 6 µs 5.98 Figure 3. Lattice diagram of the propagation the step pulse through the line Task. For a lossless line, shown in Fig. 4 determine reflections in all legs using lattice diagram and sketch voltage signals at the reception node. Noise level in the channel is ± m, propagation delay is 4 μs/km, width of the square test impulse is μs, and its amplitude is 5. Output impedance of the TR is 5 Ω and the output impedance is Ω. haracteristic impedance of all segments is the same: Ω, and the 5 m outlet is terminated with a = kω impedance. Repeat the procedure using Heaviside step function with an amplitude of 5. m 5 m out, in 5 m Figure 4. Lossless line

4 4.4. Time omain Reflectometry Solution oefficients of division and reflection at node are out,in ρ k in in out Ω Ω 5 Ω 3 Ω Ω Ω Ω. t node : ρ ρ' ρ'' 5 ' ''. 5 3 t : kω Ω 9.3 kω Ω nd at :.4 The respective lattice diagram in case of a single square pulse is shown in Fig. 5, while in Fig. 6 a voltage signal at is sketched

5 4.4. Time omain Reflectometry.5 km ρ = 9/ ρ' = -/3 km ρ = -/ below the noise level ρ = km 4 μs 8 μs μs 6 μs μs 4 μs.5 km ρ = ρ'' = -/3 km ρ = -/ below the noise level 4 μs 8 μs μs 6 μs μs 4 μs Figure 5. Lattice diagram of the propagation the single square pulse through the line μs 6 μs.7 μs 4 μs 8 μs -.99 μs 4 μs Figure 6. oltage signal at node Similarly, Fig. 7 and 8 represent propagation of the signal in case the line is driven by a Heaviside s step function and the respective shape of the signal at

6 4.4. Time omain Reflectometry.5 km ρ = 9/ ρ' = -/3 km ρ = -/ below the noise level ρ = km μs 8 μs μs 6 μs μs 4 μs.5 km ρ'' = -/3 km ρ = -/3 ρ = below the noise level 4 μs 8 μs μs 6 μs μs 4 μs Figure 7. Lattice diagram of the propagation the Heaviside impulse through the line μs 4 μs 8 μs μs 6 μs μs 4 μs Figure 8. oltage signal at node

7 4.4. Time omain Reflectometry Task 3. Figure 9 shows a line with longitudinal losses of α = 3 d/km. The line is divided in two sections having different characteristic impedances = Ω and = Ω. Propagation delay of the line is 5 μs/km, and the instrument emits a square impulse of -, μs wide. etermine reflections in all legs using lattice diagram and sketch voltage signals in all discontinuity nodes, and. Noise level in the channel is ± m. TR - 5 l m Figure 9. Line with longitudinal losses Solution oefficients of division and reflection at are ρ k Ω Ω Ω Ω Ω Ω Ω 3. t node they are: ρ ρ' ' 3. t node, the reflection coefficient is: SS Ω Ω 3.3 Ω Ω SS

8 4.4. Time omain Reflectometry Longitudinal attenuation is defined as P U P U log out log out log out 3.4 in in U U in So, for a line segment km long, between nodes and, the absolute attenuation of the voltage signal will be U out U in km.78u in 3.5 For the line segment laying between nodes and that is 5 m long, the signal will attenuate for the absolute amount U out U in.5 km.84u in 3.6 ased on all these coefficients the lattice diagram is obtained shown in Fig...5 km ρ = m - m m m km ρ' = / ρ = -/ m -7 m m m m m 9 m m m ispod nivoa šuma ρ' = km μs -,46 3,5-58 m 7 m 5 μs μs 5 μs μs 5 μs Figure. Lattice diagram 3 μs

9 4.4. Time omain Reflectometry The respective sketches of signals at nodes, and are shown in Fig.. μs 5 μs μs 5 μs μs 5 μs 3 μs μs 5 μs m μs -39 m μs 5 μs 3 μs m 7 m μs 5 μs μs 5 μs -58 m 5 μs 3 μs -. Figure. oltage signals at nodes, and Task 4. For a line configuration shown in Fig. sketch the curve that will appear on a reflectometer connected to the node if its impedance is adjusted to the characteristic impedance of the first line leg. Propagation velocity as well as characteristic impedances of each line leg are denoted as general values in Fig.. It is known that > and RT <. TR emits Heaviside step function impulse with amplitude. l l l, L v, v, v L RT Figure. Line under test Solution Taking into account that the TR s output impedance is matching the characteristic impedance of the first line segment the amplitude of the initial pulse at the start of the reflectogram will be divided by two

10 4.4. Time omain Reflectometry When the rising edge ω of this impulse arrives at node the inductivity L will appear as an infinite impedance resulting in an initial total reflection back towards the node. ut as this rapid change is short lived, the inductance will start filling with the magnetic energy allowing the passage of the current through it. t the flat level of the step function ω =, the inductivity L will zero impedance and will not affect the signal any further. fter long enough time the reflection will be determined only by the difference of two facing impedances and. It will be positive because >. 4. fter l/v time elapses this positive reflection will arrive at node and will be indicted as a positive increase in voltage. Hence, the voltage at will jump to l v 4. l/v What doesn t reflects back to will continue towards node. t it meets capacitance which briefly acts as a short circuit between the wires. This will initiate a negative reflection back towards, but it will be short lived again as the condenser immediately begins to charge. fter it is fully charged, the reflection coefficient in the steady regime will be 4.3 ack at the instrument at node this change will be noted after l/v + l/v. ut in order to arrive to, the voltage has to pass through node once more where it will be attenuated by a secondary reflection. This reflection has an opposite sign from the direct one -ρ and hence the transmission coefficient will be ρ. 4. -

11 4.4. Time omain Reflectometry l v l v 4.4 direct imuls arrivingat impact of the return reflection at l/v l/v Whatever is left of the original incident impulse will continue to travel towards where it will face a complex impedance first seen as an open circuit due to inductivity L resulting in an initial overshot of the voltage. IT will be immediately followed by an undershot caused by the capacitor. In a steady regime after both the inductor and the capacitor are charged they can be replaced with a short circuit and open circuit respectively and the coefficient of the reflection at will be determined only by the resistivity RT which is smaller than : RT 4.5 R T fter l/v + l/v + l/v this disturbance will be visible at node. ut not before it passes first through and then again where they will be attenuated by additional backscatter. The actual value of the signal at will be l v l v l v theincident signalarrivng at impact of the impact of the secondpass secondpass through through 4.6 l/v l/v l/v Fig. 3 illustrates the propagation of the signal from to and back again to where only asymptotic values of the voltage are taken into account. 4. -

12 4.4. Time omain Reflectometry 4. - Figure 3. Propagation of a ramp function through the line Remark: In this task all secondary and multiple reflections have been neglected because they were impossible to take into account without knowing exact delays of each line segment. In reality they would eventually reach at node and some of them may arrive before some of the direct reflections. Task 5. For the reflecting curve shown in Fig. 4 obtained using a TR with an output impedance of 5 Ω reverse engineer all possible line configurations. How many possible configurations are there? The initial impulse is Heaviside step function with amplitude. Figure 4. Reflecting curve obtained at the TR s screen Solution Each point in Fig. 4 indicating any change in voltage even a temporary one followed by a long straight line indicates a discontinuity point or a node. Let s denote these points with capital letters of the alphabet,, etc. as shown in Fig ' ' ' ' Neglected Neglected

13 4.4. Time omain Reflectometry E Figure 5. Identifying points of discontinuity t node there is a division of voltage by factor meaning that the output impedance of the TR matches the characteristic impedance of the first line segment. This also means that the reflection coefficient for all returning signals is zero. So there is only one possibility for node and it shown below. 5 Ω = 5 Ω t node, considering the rise of the voltage, there are at least two possibilities. First, the increase of the impedance could be a simple joint of two line segments where the continuing segment has larger characteristic impedance. > nother option is that there is a serial resistance added to the line, e.g. due to a poor or a decaying joint. The assumption is valid as long as the sum of that resistance and the characteristic impedance of following segment is greater than the characteristic impedance of the previous segment. R t node a short decline of the signal is visible indicating an increased capacity between the lines. fter that there are several possibilities. The first and the simplest is that the line is continued with the segment having identical characteristic impedance =. =

14 4.4. Time omain Reflectometry The second option is that the parallel capacitance is followed by a resistance in a series and a segment of the line with the characteristic impedance lower such that the condition R + = is satisfied. R < The third possibility assumes a combination of a parallel resistance and characteristic impedance such that the condition R = is satisfied. > 3 R The second and the third possibilities are highly unlikely considering a very strict equality that needs to be satisfied. ctually, there are many more options and possibilities but they are all too complicated to even be considered. t node, another short-lived fall of the voltage indicates a capacitive drain. It is then followed by a negative reflection, which is most likely the result of a continuing segment with characteristic impedance 3 being lower than. 3 < 3 Some further possibilities are shown below, including a branching line. > R R n

15 4.4. Time omain Reflectometry t the ending node E, a capacitive line termination is evident, since in a steady state is behaves like an open circuit. 3 E E 4 So there are at least 3 x 3 x = 8 different configurations with varying complexity. To determine coefficients of reflections and characteristic impedances of all segments one concrete configuration has to be adopted. Engineering empirical logic states that more complicated explanations are less likely to be correct. Therefore, we will always assume the simplest line configuration, which in this case will be ----E. It is shown in Fig Ω 3 E 3 4 Figure 6. The simplest straightforward line configuration is also the most probable For the configuration from Fig. 6 the following series of equations can be set representing voltage values at node along the lattice diagram shown in Fig end

16 4.4. Time omain Reflectometry E Neglected ' 3 Neglected Neglected 3 Neglected 4 9 Neglected ' Neglected end Figure 7. Lattice diagram of signal propagation From the above system of equations it is easy to obtain reflection coefficients and the required unknown characteristic impedances of all segments Ω Ω Ω 5. 3 The asymptotic value of the ending voltage that can never reach will actually be 7 end 5. 9 Remark: gain, all secondary reflections have been cancelled to simplify the solution of this task

17 4.4. Time omain Reflectometry Task 6. For a lossless branching line shown in Fig. 8 determine reflection coefficients at all nodes and sketch the lattice diagram of signal propagation through all legs. Using the lattice diagram sketch voltage signals in time across each of the nodes,,, and E. The initial impulse sent to the line from node is a Heaviside step function with amplitude 3. Lengths and characteristic impedances of all legs, as well as resistances of all drains and terminations are designated in Fig. 8. Propagation delay of all legs is the same and has a value of 4. μs/km. km km 5 Ω 5 Ω 3 5 Ω Ω Ω Ω E 5 m Ω Figure 8. ranching lossless line with a parallel drain and a series ohmic discontinuities Solution t the division and reflection coefficients are 5 Ω ρ 5 Ω 5 Ω k 5 Ω 5 Ω 5 Ω 5 Ω 5 Ω 5 Ω

18 4.4. Time omain Reflectometry t there are 3 directions from which an impulse can arrive. They all have to be considered and they may all result in different reflection and transmission coefficients. ρ ρ' 5 Ω 5 Ω Ω Ω Ω E ρ'' When the signal is arriving at from, we obtain : back to Ω Ω 5 Ω 6,9 Ω 5 Ω Ω Ω 5 Ω 6,9 Ω 5 Ω 35 Ω 5 Ω 6. Ω 5 Ω : forwarded to , E: forwarded to E across Ω 4 4 E 6.4 Ω Ω 35 7 When the signal is arriving at from, the following will be true : back to : forwarded to , E: forwarded to E across Ω 4 4 E 6.7 Ω Ω 35 7 Finally, when a signal arrives at travelling from E it has to pass across the node and we obtain

19 4.4. Time omain Reflectometry E E: back to E Ω Ω 5 Ω 5 Ω Ω 4 Ω Ω 3 Ω Ω 5 Ω 5 Ω Ω 4 Ω Ω E, : forwarded to across Ω 5 Ω 5 Ω Ω Ω Ω 5 Ω 5 Ω Ω Ω 7 7 E, : forwarded to across 6. 7 t the reflection coefficient is: 5 Ω ρ Ω 5 Ω 6. Ω 5 Ω While in E there will be no reflection since: Ω E Ω Ω Ω E 6. Ω Ω Having calculated all coefficients it is possible to draw the lattice diagram shown in Fig. 9. The signals at,, and E are shown in Fig

20 4.4. Time omain Reflectometry 3 km ρ = ρ' = -/35 km ρ = -/ ρ = km μs 6.8 μs 5. μs 33.6 μs 4. μs 5.4 μs ρe = E.5 km km ρ'' = -3/ μs 6.8 μs 5. μs 33.6 μs 4. μs 5.4 μs Figure. Lattice diagram of the signal propagation 4. -

21 4.4. Time omain Reflectometry E μs 8.4 μs 6.8 μs 5. μs 3.6 μs 4. μs 5.4 μs μs 6.8 μs 5. μs 3.6 μs 4. μs 5.4 μs μs 6.8 μs 5. μs 3.6 μs 4. μs 5.4 μs μs 6.8 μs 5. μs 3.6 μs 4. μs 5.4 μs Figure. Signals at nodes,, and E 4. -

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