Multiple reflection in a terminated line

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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 ƒŃ is the propagation delay and ƒĄS,ƒĄT 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:

We will derive a special solution for a terminated line.

***

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 ƒŃ and reflection coefficients ƒĄS,ƒĄT instead of l and RS,RT,

then:

Eliminating Vb, we obtain Va

Here we expand 1/(1-x) to move the factor exp(-2ƒŃs) 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(-ƒŃs) in the 's' domain is equivalent to shifting by ƒŃ(>0) in the time domain:

,then we obtain the special solutions in the time domain:

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Noting l/u=ƒŃ, 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 v0(t) is a rising step:

Using the identity:

, the voltages can be expressed in terms of the amplitude in nƒŃ< t <mƒŃ: v0(t-nƒŃ)-v0(t-mƒŃ):

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.


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