bitseq()
Returns the bit sequence at specified time point as a real number
Syntax
bitseq(time, ClockFreq, Rise, Fall, Vlow, Vhigh, BitSeq)
Arguments
| Name | Description | Range | Type | Default | Required |
|---|---|---|---|---|---|
| time | program variable time | [0, ∞) | real | yes | |
| ClockFreq | Clock frequency of the signal | (0, ∞) | real | Fstop † | no |
| Rise | Rise time of pulse | [0, ∞) | real | Tstep † † | no |
| Fall | Fall time of pulse | [0, ∞) | real | Tstep † † | no |
| Vlow | Minimum voltage level | (-∞, ∞) | real | 0 V | no |
| Vhigh | Maximum voltage level | (-∞, ∞) | real | 1 V | no |
| BitSeq | Bit sequence | [0, 1]† † † | string | "1101010100101" | no |
| † Fstop is 1/Transient StopTime or 1/Envelope Stop. † † Tstep is Transient MaxTimeStep or Envelope Step. † † † sequence of 0s and 1s. | |||||
Examples
This example assumes that a transient simulation is performed using:
StartTime = 0, StopTime = 2*BitPeriod*NumBits,
MaxTimeStep = BitPeriod*NumBits/pow(2,NumBits)
where:
BitRate = 500MHz, BitPeriod = 1/BitRate,
NumBits = length(BitSeq)-1, BitSeq = "110101110011"
This expression creates a bit sequence which repeats every 24 nsec:
value = bitseq(time, BitRate, 0.1nsec, 0.1nsec, 0, 5, BitSeq)
Notes/Equations
The bitseq() function can be used to vary the waveform of a pulse, an arbitrary bit pattern such a 110101110011. When the end of the sequence is reached, the sequence is repeated.
The transient stop time (Tstop) should be exactly one bit cycle for good results. For the example given above, BitPeriod=1/BitRate = 1/500MHz = 2nsec . For BitSeq="110101110011", Tstop=NumBits * BitRate = 12bits * 2nsec = 24 nsec .
The bitseq() Function
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