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I am trying to understand following signal trace picture that is shown on the wiki page of 1553 Bus. The Bus has balanced differential pair wires.

1553 Bus differential signal trace

As the 1553 bus has differential signals on the twisted pair of wires it means that the physical arrangement of the oscilloscope probe was that the probe SIGNAL and GND pins would be connected on the differential wire-pair. As there is no 'ground' wire in the 1553 bus so the oscilloscope probe GND will not be connected to any 'ground' potential but one of the two differential wires will act as the reference potential point (ground) for the oscilloscope input signal.

In the picture I can see 3 voltage levels, Low, Mid and High. My understanding is that when the Bus is quiet then both lines are at 0 volts w.r.t their signal ground so their difference will also be 0 volts. It looks like this is the 'Mid' signal level case.

I could not understand how the Logic '0' and '1' is translated as 'Low and 'High' in the above signal trace on the oscilloscope? If I want to draw the individual signals on the differential lines w.r.t. to the signal ground then what would be those 2 signal traces that produces the trace as shown in the above picture?

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2 Answers 2

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This bus is Manchester encoded. This means that one symbol (bit) is encoded as a low transitioning to a high, or a high transitioning to a low. These correspond to 0 and 1 respectively. Therefore, if you want to send: 1 0 0 1 0 1 1 1 the sequence is: HLLHLHHLLHHLHLHL

This encoding moves the center frequency up to the bit rate, eliminating the low frequency content, and makes clock recovery easier.

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  • \$\begingroup\$ If it connect oscilloscope probe on the differential lines then if I see H->L transition on oscilloscope that will mean Logic '0'? \$\endgroup\$
    – alt-rose
    Commented Jan 8, 2020 at 14:23
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    \$\begingroup\$ The opposite...look at the display you posted. L->H is zero, H->L is one. Note that there is no transition between bits if they're different bits, but there needs to be a transition between them if they're the same...so you see above that 0x03 (0b00011) goes L H L H L H H L H L. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Jan 8, 2020 at 14:32
  • \$\begingroup\$ I understand a little bit now. Please tell more about "0x03" as shown in the above posted picture.. Will I get this trace when I connect my oscilloscope probe (SIGNAL, GND) pins to the differential lines pair of the 1553 bus? If this is the case then what traces will I get if I connect my two probes of oscilloscope to the individual lines of the 1553 bus w.r.t. the system ground? That is to say that one probe connect to Line 1 and second probe to Line 2. The oscilloscope Ground is connected to the system ground. Assume the data is "0x03" on differential lines. \$\endgroup\$
    – alt-rose
    Commented Jan 9, 2020 at 5:39
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    \$\begingroup\$ Typically you'd do that using the second connection method, and use the math function on your scope to do A-B; that accurately represents the differential content of the signal. What you will see is similar to the trace you posted above. Look at that trace, and you'll see a field which is annotated to be 0x03, and is 5 bits long. It decodes exactly as I described above. Also see en.wikipedia.org/wiki/Manchester_code for more detail. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Jan 9, 2020 at 13:14
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the 1553 bus is mainly used in the space industry, a low signal is equivalent to a negative voltage not zero, if you have a look onto your specification you will see what's is acceptable as the range could be quite wide. Same for a high signal it should be a positive voltage within a specific range. the peak to peak voltage must be >18V

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