Multiple reflection in a terminated line
A transmission line is terminated by R_{S }and R_{T} at the source and termination end respectively.
The voltage and current at the source end are:
The voltage and current at the termination end are:
where Ń is the propagation delay and Ą_{S},Ą_{T }are the reflection coefficients:
Assuming the input voltage v_{0}(t) is a rising step:
The voltage and current at the source end are:
The voltage and current at the termination end are:

We will derive a special solution for a terminated line.
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We assume the boundary conditions at the source and termination end as:
Laplace transformed boundary conditions are:
Inserting general solutions
into the boundary conditions, we obtain the equations in terms of V_{a} and V_{b}:
Combining like terms:
Using the propagation delay Ń and reflection coefficients Ą_{S},Ą_{T} instead of l and R_{S,}R_{T,}
then:
Eliminating V_{b}, we obtain V_{a}
Here we expand 1/(1x) to move the factor exp(2Ńs) from the devisor to dividend. This geometric series represents the multiple reflection.
Substituting the V_{a} and V_{b} into the general solutions, we obtain the special solutions in the 's' domain:
Noting that multiplying exp(Ńs) in the 's' domain is equivalent to shifting by Ń(>0) in the time domain:
,then we obtain the special solutions in the time domain:
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Noting l/u=Ń, the voltage and current at the source end (x=0) are:
These solutions satisfy the boundary condition:
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The voltage and current at the termination end (x=l) are:
These solutions satisfy the boundary condition:
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We assume that the input voltage v_{0}(t) is a rising step:
Using the identity:
, the voltages can be expressed in terms of the amplitude in nŃ< t <mŃ: v_{0}(tnŃ)v_{0}(tmŃ):
At the source end:
At the terminaton end:
or, explicitly
At the source end:
At the terminaton end:
Moreover expressing the amplitude in terms of R_{S},R_{T} and Z_{0},
At the source end:
At the terminaton end:
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So far we have assumed the initial condition as v(x,0)=0,i(x,0)=0. When the source voltage V_{SS} or termination voltage V_{TT} is not zero,
we may just superpose the DC circuit.
Then the solution is just shifted by constant:
For example, a transition from HIGH(V_{SS}) to LOW(0V) is equivalent to a transition from 0V to negative voltage.