High-Speed Digital System Design ~ A Handbook of Intercon..

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where t is time and V
input
is the input voltage. If V
input
is assumed to be 1 V, the time it takes
for the voltage V to rise to 0.9 V, or 90% of the signal swing, is calculated as
(C.3)
The signal swing to 0.1 V, or 10% of the maximum swing, is
(C.4)
Subsequently, the time it takes a step to transition from 10% to 90% of the signal swing is
(C.5)
The frequency response of a network is with a time constant of (i.e., RC)
(C.6)
If
(C.6)
is inserted into
(C.5)
, the result is
(C.7)
Equation (C.7)
is the resulting 10 to 90% rise or fall time of a perfect step function driven
through a network with a time constant of . Obviously, this does not apply to real systems
because realistic edge rates are finite. However, the F
3 dB
frequency is still a reasonable
approximation of the spectral envelope of a digital pulse because below this frequency is
where most of the spectral energy is contained (as depicted in
Figure 10.3
).
When a signal passes through a network with a time constant of , the edge rate will be
degraded because of the filtering effects. The degradation of a finite edge rate can be
approximated by the equation
(C.8)
where t
input
is the input rise or fall time and t
step
is the result from
equation (C.7)
.
The reader should note that the relationships above between the edge rate and the spectral
content are only approximations, however, they tend to be extremely useful for back-of-the-
envelope calculations and for determining the capability of laboratory equipment. For
example, an oscilloscope with an input bandwidth of 3 GHz will be able to measure edge
rates correctly only when they are greater than approximately 0.35/3 × 10
9
117 ps. If it
were attempted to measure a 50-ps rise time on such a scope, for example, the waveform
on the screen would have an edge rate of approximately
ps.

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Summary :

If V input is assumed to be 1 V, the time it takes for the voltage V to rise to 0.9 V, or 90% of the signal swing, is calculated as (C.3) The signal swing to 0.1 V, or 10% of the maximum swing, is (C.4) Subsequently, the time it takes a step to transition from 10% to 90% of the signal swing is (C.5) The frequency response of a network is with a time constant of (i.e., RC) (C.6) If (C.6) is inserted into (C.5) , the result is (C.7) Equation (C.7) is the resulting 10 to 90% rise or fall time of a perfect step function driven through a network with a time constant of .

Tags : time,input,edge,signal,rise,swing,rate,step,because,network,equation,frequency,spectral
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