Multiple reflection in a terminated line
A transmission line is terminated by RS and RT 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 ĄS,ĄT are the reflection coefficients:
Assuming the input voltage v0(t) is a rising step:
The voltage and current at the source end are:
The voltage and current at the termination end are:
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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 Va and Vb:
Combining like terms:
Using the propagation delay Ń and reflection coefficients ĄS,ĄT instead of l and RS,RT,
then:
Eliminating Vb, we obtain Va
Here we expand 1/(1-x) to move the factor exp(-2Ńs) from the devisor to dividend. This geometric series represents the multiple reflection.
Substituting the Va and Vb 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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The voltage and current at the source end (x=0) are:
These solutions satisfy the boundary condition:
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By noting l/u=Ń, 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 v0(t) is a rising step:
Using the identity:
, the voltages can be expressed in terms of the amplitude in nŃ< t <mŃ: v0(t-nŃ)-v0(t-mŃ)
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 RS,RT and Z0,
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 VSS or termination voltage VTT is not zero,
we may just superpose the DC circuit.
Then the solution is just shifted by constant:
For example, a transition from HIGH(VSS) to LOW(0V) is equivalent to a transition from 0V to negative voltage.