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The energy injected into the node by the incident pulse and the total energy of the scattered pulses are correspondingly
: <math>E_I = vi\,\Delta t = 1
: <math>E_S = \left[0.5^2+0.5^2+0.5^2+(-0.5)^2\right](\Delta t/Z) = \Delta t/Z</math>
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The next scattering event excites the neighbouring nodes according to the principle described above. It can be seen that every node turns into a secondary source of spherical wave. These waves combine to form the overall waveform. This is in accordance with Huygens principle of light propagation.
In order to show the TLM schema we will use time and space discretisation. The time-step will be denoted with <math>\Delta t</math> and the space discretisation intervals with <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math>. The absolute time and space will therefore be <math>t = k\,\Delta t</math>, <math>x = l\,\Delta x</math>, <math>y = m\,\Delta y</math>, <math>z = n\,\Delta z</math>, where <math>k=0,1,2,
: <math>\Delta t=\frac{\Delta l}{c_0}.</math>
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The figure on the right presents a structure, referred to as a ''series node''. It describes a block of space dimensions <math>\Delta x</math>, <math>\Delta y</math> and <math>\Delta z</math> and consists of four ports. <math>L'</math> and <math>C'</math> are the distributed inductance and capacitance of the transmission lines. It is possible to show that a series node is equivalent to a TE-wave, more precisely the mesh current ''I'', the ''x''-direction voltages (ports 1 and 3) and the ''y''-direction voltages (ports 2 and 4) may be related to the field components <math>H_z</math>, <math>E_x</math> and <math>E_y</math>. If the voltages on the ports are considered, <math>L_x = L_y</math>, and the polarity from figure holds, than the following is valid
: <math>-V_1+V_2+V_3-V_4 = 2L'\,\Delta l\frac{\partial{I}}{\partial{t}}</math>
where <math>\Delta x = \Delta y = \Delta l</math>.
: <math>\left(V_3 - V_1\right)-\left(V_4-V_2\right) = 2L'\,\Delta l\frac{\partial I}{\partial t}</math>
: <math>\left[E_x(y+\Delta y)-E_x(y)\right]\,\Delta x-[E_y(x+\Delta x)-E_y(x)]\Delta y = 2L'\,\Delta l\frac{\partial{I}}{\partial{t}}</math>
and dividing both sides by <math>\Delta x \Delta y</math>
: <math>\frac{E_x(y+\Delta y)-E_x(y)}{\Delta y}-\frac{E_y(x+\Delta x)-E_y(x)}{\Delta x} = 2L'\,\Delta l\frac{\partial{I}}{\partial{t}}\frac{1}{\Delta x \, \Delta y}</math>
Since <math>\Delta x = \Delta y = \Delta z = \Delta l</math> and substituting <math>I = H_z \,\Delta z</math> gives
: <math>\frac{\Delta E_x}{\Delta y} - \frac{\Delta E_y}{\Delta x} = 2L'\frac{\partial H_z}{\partial t}</math>
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