PAMn Capabilities

IBIS BIRD (Buffer Issue Resolution Document) 213 adds supports to IBIS-AMI models for any level of signaling from PAM2 (NRZ) to upwards, collectively known as PAMn. You can simulate PAMn models for upto n = 32 with specific mapping using the Serial Link Designer or Parallel Link Designer apps.

There are some specific terms defined for PAMn signaling:

• Payload — Binary bits to be transmitted using PAMn voltage levels.

• Message — PAMn symbols.

• Mapping — A look-up table used to translate PAMn voltage levels to binary bits.

• Encoding — The entire process of converting binary payload to PAMn symbols sent.

• Decoding — The entire process of converting received PAMn symbols to binary bits.

The number of Modulation Levels is selected with the Modulation_Levels AMI Parameter in the solution space.

To define PAMn mappings, in the Stimulus editor, select the parameters Number Modulation Levels and Mapping. The mapping is dependent on the modulation scheme selected.

Mapping Binary Payload to PAMn Symbols

For a system that transmits and receives PAMn symbols, the ability to reliably map between binary and PAMn symbols is crucial.

For NRZ or PAM2 modulation, the mapping is trivial.

Binary Payload	Message Symbol	Symbol Voltage
0	0	-0.5
1	1	0.5

For PAM4, the mapping can be done in 24 ways. One of the possible ways can be:

Binary Payload	Message Symbol	Symbol Voltage
00	0	-0.5
01	1	-0.1666
10	2	0.1666
11	3	0.5

For PAM2, PAM4, PAM8, and other PAMn, where n is a power of 2, the mapping is straightforward in that the message symbol sequence is of length 1. For PAM3 the mapping is not as clean since the number of bits per symbol is not an integer. For example, the PAM3 symbol mapping for the IEEE 100Base-T1 standard is:

Binary Payload	Message Symbol	Symbol Voltage
000	0 1	-0.5 -0.5
001	0 1	-0.5 0
010	0 2	-0.5 0.5
011	1 0	0 -0.5
100	1 2	0 0.5
101	2 0	0.5 -0.5
110	2 1	0.5 0
111	2 2	0.5 0.5

Here, the eight possible binary payload sequences are assigned a unique message symbol sequence. But the 1 1 message sequence is not specified. As such, if a 1 1 is detected at the receiver, a simple decoding of the symbol message back to binary is not possible.

To analyze this mapping further, the set size of the payload and message is needed. The binary payload length is 3 and therefore the binary payload set size is 2³ = 8. The message length is 2 and with PAM3, the symbol message set size is 3² = 9. Therefore, the coverage of this PAM3 mapping is 8/9 = 88.88%. The IEEE 100Base-T1 standard could have chosen a different symbol sequence for the missing message. So there are many different possible mapping schemes. A PAM3 mapping with greater coverage is one with a binary payload length of 11 and a message length of 7. This has a coverage of 2¹¹/3⁷ = 93.64% and has 139 missing messages.

The question becomes, how to map the binary payload to the PAMn symbols and where to locate the missing messages. One possible approach can be:

• Represent the binary payload as a base-10 integer x.

• Scale x from the binary payload range of [0, 2^{PayLoadLength-1}] to the range of [0, n^{MessageLength-1}] to obtain y.

• Convert y to a base-n number to obtain the PAMn message sequence.

This mapping algorithm has the benefit of distributing the missing symbol sequences uniformly throughout the range of message sequences.

The complete list of mapping depending on the number of modulation levels is shown:

Number of Modulation Levels	Mapping
2	Default
3	UNIFORM_3_2
	UNIFORM_11_7
	UNIFORM_19_12
	USB4_V2
	ETH_100BASE_T1
4	PAM4_0132
	PAM4_0123
	PAM4_0213
	PAM4_0231
	PAM4_0312
	PAM4_0321
	PAM4_1023
	PAM4_1032
	PAM4_1203
	PAM4_1230
	PAM4_1302
	PAM4_1320
	PAM4_1302
	PAM4_2013
	PAM4_2031
	PAM4_2103
	PAM4_2130
	PAM4_2301
	PAM4_2310
	PAM4_3012
	PAM4_3021
	PAM4_3102
	PAM4_3120
	PAM4_3201
	PAM4_3210
5	UNIFORM_9_4
	UNIFORM_16_7
	UNIFORM_23_10
	UNIFORM_30_13
	UNIFORM_37_16
	UNIFORM_44_19
6	UNIFORM_5_2
	UNIFORM_18_7
	UNIFORM_31_12
7	UNIFORM_14_5
	UNIFORM_5_2
	UNIFORM_8_3
	UNIFORM_11_4
8	UNIFORM_3_1
9	UNIFORM_3_1
	UNIFORM_19_6
10	UNIFORM_3_1
	UNIFORM_13_4
	UNIFORM_23_7
	UNIFORM_33_10
	UNIFORM_43_13
11	UNIFORM_17_5
	UNIFORM_3_1
	UNIFORM_10_3
	UNIFORM_24_7
	UNIFORM_31_9
	UNIFORM_38_11
12	UNIFORM_7_2
	UNIFORM_3_1
	UNIFORM_25_7
	UNIFORM_43_12
13	UNIFORM_11_3
	UNIFORM_3_1
	UNIFORM_7_2
	UNIFORM_37_10
14	UNIFORM_15_4
	UNIFORM_7_2
	UNIFORM_11_3
	UNIFORM_19_5
15	UNIFORM_31_8
	UNIFORM_11_3
	UNIFORM_15_4
	UNIFORM_19_5
	UNIFORM_23_6
	UNIFORM_27_7
	UNIFORM_35_9
	UNIFORM_39_10
16	UNIFORM_4_1
17	UNIFORM_4_1
	UNIFORM_49_12
18	UNIFORM_25_6
	UNIFORM_4_1
19	UNIFORM_21_5
	UNIFORM_4_1
	UNIFORM_38_9
20	UNIFORM_17_4
	UNIFORM_4_1
	UNIFORM_30_7
	UNIFORM_43_10
21	UNIFORM_13_3
	UNIFORM_4_1
	UNIFORM_35_8
22	UNIFORM_22_5
	UNIFORM_4_1
	UNIFORM_13_3
	UNIFORM_31_7
	UNIFORM_40_9
	UNIFORM_49_11
23	UNIFORM_9_2
	UNIFORM_4_1
24	UNIFORM_9_2
	UNIFORM_4_1
	UNIFORM_32_7
25	UNIFORM_9_2
	UNIFORM_4_1
	UNIFORM_23_5
	UNIFORM_37_8
26	UNIFORM_14_3
	UNIFORM_4_1
	UNIFORM_9_2
	UNIFORM_47_10
27	UNIFORM_19_4
	UNIFORM_9_2
	UNIFORM_14_3
28	UNIFORM_24_5
	UNIFORM_9_2
	UNIFORM_14_3
	UNIFORM_19_4
29	UNIFORM_34_7
	UNIFORM_9_2
	UNIFORM_14_3
	UNIFORM_19_4
	UNIFORM_24_5
	UNIFORM_29_6
30	UNIFORM_49_10
	UNIFORM_9_2
	UNIFORM_14_3
	UNIFORM_19_4
	UNIFORM_24_5
	UNIFORM_29_6
	UNIFORM_34_7
	UNIFORM_39_8
	UNIFORM_44_9
31	UNIFORM_64_13
	UNIFORM_29_6
	UNIFORM_34_7
	UNIFORM_39_8
	UNIFORM_44_9
	UNIFORM_49_10
32	UNIFORM_5_1

PAM6 Mapping Using UNIFORM_5_2 Standard

For PAMn = 6, a payload length of 5 bits and a message length of 2 symbols gives 88.88% coverage with 4 missing symbol sequences. This example illustrates how each possible payload is mapped to a symbol sequence and the missing symbol sequences are distributed uniformly throughtout.

Define the PAMn level, payload length, and message length.

n = 6;                  %PAMn Levels
PayloadLength=5;        %Number of binary bits to encode
MessageLength=2;        %Number of PAMn symbols to encode each payload into

Calculate the number of gaps between the number of possible messages and the number of possible payload values.

missingSymbolSequenceCount = n^MessageLength - 2^PayloadLength;

To loop over every possible payload value, first represent the binary payload as a base-10 integer Payload. Scale this value from the binary payload range of [0, 2^PayloadLength-1] to the range of [0, N^MessageLength-1] to get the ScaledPayload parameter. Then convert the ScaledPayload parameter to a base-N number. This approach uniformly distributes any missing message sequences throughout the message space. You can visualize the integer and binary payload, scaled payload, and resulting PAMn symbol values.

for Payload = 0:2^PayloadLength-1

    ScaledPayload = round(Payload*(1 + missingSymbolSequenceCount/(2^PayloadLength)));
    ScaledPayload1 = ScaledPayload; %Store initial value for report

    %Convert the integer Payload value to a base N number
    SymbolsOut=zeros(1,MessageLength);
    for jj=1:MessageLength

        %Determine the remainder after division to determine the next
        %least-significant base N value and store output symbols big-endian
        SymbolsOut(MessageLength+1-jj) = (mod(round(ScaledPayload),n));

        %Remove contribution of least-significant base N value and shift by
        %division (analogous to how divide by 2 is a binary bitshift).
        ScaledPayload = (ScaledPayload - SymbolsOut(MessageLength+1-jj)) / n;
    end

    %Report out payload (integer and binary), scaled payload and resulting
    %PAMn symbol values

    if Payload==0
        %Print out header of report
        fprintf('%13s -->  %6s --> %s\n','Payload   ','Scaled ','Output Symbols');
    end

    fprintf('%4g: %7s --> %8.6g --> %s\n',...
        Payload,...                  %Input payload value in integer form
        dec2bin(Payload,PayloadLength), ...      %Input payload value in binary form
        ScaledPayload1,...                  %Scaled payload value in double form
        num2str(round(SymbolsOut)))     %Output PAMn message symbol values
end

   Payload    -->  Scaled  --> Output Symbols

   0:   00000 -->        0 --> 0  0
   1:   00001 -->        1 --> 0  1
   2:   00010 -->        2 --> 0  2
   3:   00011 -->        3 --> 0  3
   4:   00100 -->        5 --> 0  5
   5:   00101 -->        6 --> 1  0
   6:   00110 -->        7 --> 1  1
   7:   00111 -->        8 --> 1  2
   8:   01000 -->        9 --> 1  3
   9:   01001 -->       10 --> 1  4
  10:   01010 -->       11 --> 1  5
  11:   01011 -->       12 --> 2  0
  12:   01100 -->       14 --> 2  2
  13:   01101 -->       15 --> 2  3
  14:   01110 -->       16 --> 2  4
  15:   01111 -->       17 --> 2  5
  16:   10000 -->       18 --> 3  0
  17:   10001 -->       19 --> 3  1
  18:   10010 -->       20 --> 3  2
  19:   10011 -->       21 --> 3  3
  20:   10100 -->       23 --> 3  5
  21:   10101 -->       24 --> 4  0
  22:   10110 -->       25 --> 4  1
  23:   10111 -->       26 --> 4  2
  24:   11000 -->       27 --> 4  3
  25:   11001 -->       28 --> 4  4
  26:   11010 -->       29 --> 4  5
  27:   11011 -->       30 --> 5  0
  28:   11100 -->       32 --> 5  2
  29:   11101 -->       33 --> 5  3
  30:   11110 -->       34 --> 5  4
  31:   11111 -->       35 --> 5  5

Create and print out the summary string.

strout = sprintf(['PAMn=%g, Payload Length=%g, Message Length=%g\n',...
    'Missing %g symbol sequences for coverage of 2^{%g}/%g^{%g} = %g%%'],...
    n,PayloadLength,MessageLength,...
    missingSymbolSequenceCount,...
    PayloadLength,n,MessageLength,2^PayloadLength/n^MessageLength*100)

strout = 
    'PAMn=6, Payload Length=5, Message Length=2
     Missing 4 symbol sequences for coverage of 2^{5}/6^{2} = 88.8889%'

Identify and visualize the gaps in message symbol sequences.

Payload = 0:2^PayloadLength-1;
ScaledPayload = Payload + (n^MessageLength-2^PayloadLength)/(2^PayloadLength)*Payload;

isaGap = ones(1,n^MessageLength-1);
isaGap(round(ScaledPayload)+1) = 0;
 
stem(0:length(isaGap)-1,isaGap)
axis([0 n^MessageLength-1 0 1.1])
title({'Location of the missing message symbol sequences for ',strout})
xlabel('Message (Decimal)')