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BPSK Waveform

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by Tapajit Mondal, Sep 3, 2017.

  1. Tapajit Mondal

    Tapajit Mondal New Member

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    Hello,

    The following is a Data waveform modulated as shown in the image. Can someone guide me how to generate the waveform using a MCU / FPGA. I am trying to figure out how to jump between the voltage rails quick enough to generate an acceptable waveform.
     

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  2. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    The only 'slight' difficulty is the three level waveform, centred about zero volts - you will need a little circuitry to create that, and probably two I/O pins from a processor to produce the three levels.

    The waveform itself is simple enough, and easily done in assembler with simple loop delays - with pulse widths of 8uS a 4MHz PIC could easily generate it.
     
  3. Tapajit Mondal

    Tapajit Mondal New Member

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    I am looking some help in the circuitry only. The digital part I should be able to figure out.
     
  4. dave

    Dave New Member

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  5. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    Tapajit, where the heck did you find that old waveform? It is not very BW efficient and not exactly called BPSK either. It is BPSK + RZ (return to zero) which doubles the bandwidth consumed.

    It can be easily generated from the master clock, and dividers with an XOR gate + a couple gates each with resistor divider network.

    What data rate? 125 kbps as shown?
     
  6. Dick Cappels

    Dick Cappels Member

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    He needs it for backwards compatibility.
     
  7. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Quick idea on an envelope. lol

    From computer or FPGA. (A, B) Digital data 0/5V could be 0/3.3.
    Transistor changes 0/5V to = -5V/0V. Level shifter.
    IC1 is simple buffer (could be inverter)
    IC2 is also simple buffer BUT the supply is connected to 0V and the "ground pin" is connected to -5V. (output is 0 and -5V)
    I chose resistors to get near 50 ohms. Could choose higher values. (if 3.3V supply you will need to change the 680 resistors.
    AB=00= -0.35V if I did the math right.
    AB=11= +0.35V
    AB=01 or 10 you will get 0V.

    upload_2017-9-4_21-43-6.png
    There are other ways. Maybe 30 different ways. You will need to update the outputs every 2uS.
    p.s. looks like I did not do the math right. (voltage divider, 5V to 0.35V, simple math for some one awake)(resistor to +5V and resistor to ground, the third resistor is in parallel with the 56 ohm resistor)
     
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  8. Tapajit Mondal

    Tapajit Mondal New Member

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    Thank you Mr.Simpson. Ideally I would like to have a ZERO output when the inputs are open or at logic low. Could you please advice how can I do it differently ?

    Moreover I would be putting Opto-isolators at the digital input, and unit gain buffer at the analog side. Any suggestion on the selection on the op-amp ??
     
  9. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Here is a part.
    Many isolators only pull down. This type pulls up and down! I think you can pull 1 mA with out a problem.
    The delay through the part is about 50nS. The output rise and fall time is 3nS.
    You should drive the LED with 2 to 6mA.
    Power supply pins should have a 0.1uf cap.
    When the LED is on the output is low. When there is no LED current the output is high.
    upload_2017-9-10_14-35-15.png
    upload_2017-9-10_14-51-37.png
    On the input side:
    (top) If input = 0 then the LED current is 3.5mA and the output is low.
    If the input = 1 then the LED current is 0 and the output is high.
    .................
    (bottom) If input = 0 the LED current is 0 and the output is high.
    If the input = 1 the LED current is 3mA and the output is low.

    On the output side: (about the same as last time)
    There is a +5V, 0V and -5V supply
    One isolator has an output of 0 or 5V.
    The other isolator has an output of -5V or 0V.

    Input 0,0 output = 0V
    Input 1,0 output = +0.35V
    Input 0,1 output = -0.35V
    Input 1,1 output = about 0V (not used)
    -----edited-----
    On the left side (logic) that +5V and ground is not the same as the +5V, gnd, -5V on the right side. It sounds like you want isolation.
     

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