This question is derived from my previous SO question's commends.
I am confused with PLC's interpretation of BCD and decimal.
In a PLC documentation, it somehow implies BCD = decimal:
The instruction reads the content of D300, 0100, as BCD. Referring to Cyber Slueth Omega's answer and online BCD-Hex converter, 0100 (BCD) = 4 (Decimal) = 4 (Hex), but the documentation indicates 0100 (BCD) = 100 (Decimal).
Why?
BCD is HEX
BCD is not binary
HEX is not binary
BCD and HEX are representations of binary information.
The only difference is in how you decide to interpret the numbers. Some PLC instructions will take a piece of word memory and will tell you that "I, the TIM instruction, promise to treat the raw data in D300 as BCD data". It is still HEX data, but it interprets it differently.
If D300 = [x2486] --> the timer (as example) will wait 248.6 seconds. This even though HEX 2486 = 9350 decimal. You can treat hex data as anything. If you treat hex data as encoded BCD you get one answer. If you treat it as a plain unsigned binary number you get another, etc.
If D300 = [x1A3D] --> TIM will throw an error flag because D300 contains non-BCD hex digits
Further, the above example is showing HEX digits - not BINARY digits. It is confusing because they chose [x0100] as their example - only zeroes and ones. When you are plugging this into your online converter you are doing it wrong - you are converting binary 0100 to decimal 4. Hexadecimal is not binary - hex is a base16 representation of binary.
Anatomy of a D-memory location is this
16 Bits | xxxx | xxxx | xxxx | xxxx | /BINARY/
---> | | | |
4 bits/digit D4 D3 D2 D1 /HEX/
example
D300 = 1234 | 0001 | 0010 | 0011 | 0100 |
----> 1 2 3 4
example
D300 = 2F6B | 0010 | 1111 | 0110 | 1011 |
----> 2 F 6 B
example (OP!)
D300 = 0100 | 0000 | 0001 | 0000 | 0000 |
----> 0 1 0 0
A D-memory location can store values from x0000 -> xFFFF (decimal 0-65535). A D-memory location which is used to store BCD values, however, can only use decimal digits. A->F are not allowed. This reduces the range of a 16-bit memory location to 0000->9999.
Counting up you would go :
Decimal BCD HEX
1 0001 0001
2 0002 0002
3 0003 0003
4 0004 0004
5 0005 0005
6 0006 0006
7 0007 0007
8 0008 0008
9 0009 0009
10 0010 000A
11 0011 000B
12 0012 000C
13 0013 000D
14 0014 000E
15 0015 000F
16 0016 0010
17 0017 0011
18 0018 0012
19 0019 0013
20 0020 0014
...etc
Going the other way, if you wish to pass a decimal value to a memory location and have it stored as pure hex (not BCD hex!) you use the '&' symbol.
For example -> [MOV #123 D300]
This moves HEX value x0123 to memory location D300. If you use D300 in a future operation which interprets this as a hexadecimal number then it will have a decimal value of 291. If you use it in an instruction which interprets it as a BCD value then it will have a decimal value of 123.
If instead you do [MOV &123 D300]
This moves the decimal value 123 to D300 and stores it as a hexadecimal number -> [x007B]! If you use D300 now in a future operation which interprets this as a hexadecimal number it will have a decimal value of 123. If you try to use it in an instruction which interprets it as a BCD value you will get an ERROR because [x007B] contains the hex digit 'B' which is not a valid BCD digit.
Binary-coded decimal is encoded as hex digits that have a limited range of 0-9. This means that 0x0100 should be read as 100 when BCD is meant. Numbers with hexadecimal digits from A to F are not valid BCD numbers.
Related
Hi I have a question about the bit representation in python
when I use bit operation 1<<31, then we can see the bits are
1000 0000 0000 0000 0000 0000 0000 0000
python will print this value as 2147483648
but when I give a variable value like a = -2**31
the bits are also
1000 0000 0000 0000 0000 0000 0000 0000
but python will print -2147483648
so if the bits are the same , how python decide to use 2147483648 or -2147483648 ?
In python integers do not have a limited precision. Which means among other things, that the numbers are not stored in twos compliment binary. The sign is NOT stored in the bit representation of the number.
So all of -2**31, 2**31 and 1<<31 will have the same bit representation for the number. The sign part of the -2**31 is not part of the bitwise representation of the number. The sign is separate.
You can see this if you try this:
>>> bin(5)
'0b101'
>>> bin(-5)
'-0b101'
The representation isn't really the same. You can use int.to_bytes to check it:
(1 << 31).to_bytes(32, 'big', signed=True)
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x80\x00\x00\x00'
(-2 ** 31).to_bytes(32, 'big', signed=True)
b'\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\x80\x00\x00\x00'
Also, be careful about the - operator, which have the lower priority here:
-2 ** 31 == -(2 ** 31)
I was working on a problem for converting base64 to hex and the problem prompt said as an example:
3q2+7w== should produce deadbeef
But if I do that manually, using the base64 digit set ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/ I get:
3 110111
q 101010
2 110110
+ 111110
7 111011
w 110000
As a binary string:
110111 101010 110110 111110 111011 110000
grouped into fours:
1101 1110 1010 1101 1011 1110 1110 1111 0000
to hex
d e a d b e e f 0
So shouldn't it be deadbeef0 and not deadbeef? Or am I missing something here?
Base64 is meant to encode bytes (8 bit).
Your base64 string has 6 characters plus 2 padding chars (=), so you could theoretically encode 6*6bits = 36 bits, which would equal 9 4bit hex numbers. But in fact you must think in bytes and then you only have 4 bytes (32 bits) of significant information. The remaining 4 bits (the extra '0') must be ignored.
You can calculate the number of insignificant bits as:
y : insignificant bits
x : number of base64 characters (without padding)
y = (x*6) mod 8
So in your case:
y = (6*6) mod 8 = 4
So you have 4 insignificant bit on the end that you need to ignore.
I am a bit lost on understanding the implementation of a specific command.
In this example, there is a command passed 0x00c6ba23 which is 0000 0000 1100 0110 1011 1010 0010 0011 in binary
I am attempting to find the ALU control unit’s inputs for this instruction.
From this I can see
opcode = 0100011
imm[4:0] = 10100
funct3 = 011 (incorrect...)
rs1 = 01101
rs2 = 01100
imm[11:5] = 0000000
I am using this image to decode it
My question is how do I get the ALU control bits and ALUOp control bits for this function? And why is the function SD, even though the funct 3 is showing 011 instead of 111?
... why is the function SD, even though the funct 3 is showing 011 instead of 111?
011 is correct. The funct3 bits must be 011 in order for this to be an SD instruction. According to page 105 of https://content.riscv.org/wp-content/uploads/2017/05/riscv-spec-v2.2.pdf the SD instruction has the format:
| imm[11:5] | rs2 | rs1 | 011 | imm[4:0] | 0100011 |
If the funct3 bits were 111 then this instruction would not be SD.
... how do I get the ALU control bits and ALUOp control bits for this function?
Since this is an SD instruction, you can read those bits straight out of the SD line of the lower table in the diagram that you referenced in your question.
I'm having a topic here which is from "Number Systems" in the subject of "Introduction to Computer Organisation & Architecture"
Then i came across this topic,"Self complementing Codes"
There are 3 parts of it which are as follows:
i)Excess-3 (I understand this part as it requires us to add 3 to BCD)
ii)84-2-1 (I don't understand)
iii)2*421 (I don't understand)
I hope someone could explain how the part ii & iii works.
Thanks alot.
I think this part of "Digital Design" book from Morris Mano will answer your question:
BCD and the 2421 code are examples of weighted codes. In a weighted code, each bit
position is assigned a weighting factor in such a way that each digit can be evaluated by
adding the weights of all the 1’s in the coded combination.
Four Different Binary Codes for the Decimal Digits
_____________________________________________________________
Decimal BCD 2421 Excess‐3 8, 4, -2, -1
Digit 8421
_____________________________________________________________
0 0000 0000 0011 0000
1 0001 0001 0100 0111
2 0010 0010 0101 0110
3 0011 0011 0110 0101
4 0100 0100 0111 0100
5 0101 1011 1000 1011
6 0110 1100 1001 1010
7 0111 1101 1010 1001
8 1000 1110 1011 1000
9 1001 1111 1100 1111
_____________________________________________________________
1010 0101 0000 0001
Unused 1011 0110 0001 0010
bit 1100 0111 0010 0011
combi- 1101 1000 1101 1100
nations 1110 1001 1110 1101
1111 1010 1111 1110
The 2421, the excess‐3 and the 84-2-1 codes are examples of self‐complementing codes. Such
codes have the property that the 9’s complement of a decimal number is obtained
directly by changing 1’s to 0’s and 0’s to 1’s (i.e., by complementing each bit in the pattern). For example, decimal 395 is represented in the excess‐3 code as 0110 1100 1000.
The 9’s complement of 604 is represented as 1001 0011 0111, which is obtained simply
by complementing each bit of the code (as with the 1’s complement of binary numbers).
Digital Design-Fifth edition-By Morris Mano
First of all 84-2-1 & 2421 code are "weighted code" and as well as "self-complementing code" both (because the necessary condition for a code to be self-complementing is that the sum of all of its weight must be equal to 9) i.e. 84-2-1(8+4-2-1=9) and, 2421(2+4+2+1=9).
So, constructing binary equivalent of decimal of number you to need to make sure one thing:
The number's code and its 9's complement's code should have complementary relationship (i.e. the number & its 9's complement code are complements of each other)
For example let's take a 84-2-1 System.
Decimal 0=0000(in 84-2-1 system) & (9's complement of 0=9) than 9(in 84-2-1) should have 1111. Hence 0 & its 9's complement i.e. 9 maintain their self-complement relationship.
Let's take another example:
Decimal 1=0111(in 84-2-1 system) & (9's complement of 1=8 ) that implies (84-2-1) equivalent of 8 should be 1000,hence again the number & its complement preserve their self-complementary relationship.
Similarly, as in 2421 system, all you need to construct the code by making sure that the number and its 9's complement should have maintained their self-complementing relationship.
I'm decoding jpeg file. I have generated huffman tables, and quantization tables, and I have reach the point where I have to decode DC and AC elements. For example lets say I have next data
FFDA 00 0C 03 01 00 02 11 03 11 00 3F 00 F2 A6 2A FD 54 C5 5F FFD9
If we ignore few bytes from SOS marker, my real data is starting from F2 byte. So lets write it in binary (starting from F2 byte):
1111 0010 1010 0110 0010 1010 1111 1101 0101 0100 1100 0101 0101 1111
F 2 A 6 2 A F D 5 4 C 5 5 F
When decoding, first element is luminance DC element so let's decode it.
[1111 0]010 1010 0110 0010 1010 1111 1101 0101 0100 1100 0101 0101 1111
F 2 A 6 2 A F D 5 4 C 5 5 F
So 11110 is Huffman code (in my case) for element 08. This means that next 8 bits are my DC value. When I take next 8 bits the value is:
1111 0[010 1010 0]110 0010 1010 1111 1101 0101 0100 1100 0101 0101 1111
F 2 A 6 2 A F D 5 4 C 5 5 F
DC element value is -171.
Here is my problem: next is luminance AC value, but I don't really understand standard in a case when is AC non zero? Tnx!
The DC values, as you've seen, are defined as the number of "extra" bits which specify the positive or negative DC value. The AC coefficients are encoded differently because most of them are 0. The Huffman table defines each entry for AC coefficients with a "skip" value and an "extra bits" length. The skip value is how many AC coefficients to skip before storing the value, and the extra bits are treated the same way as DC values. When decoding AC coefficients, you decode values from 1 to 63, but the way the encoding of the MCU ends can vary. You can have an actual value stored at index 63 or at if you're at index > 48, you could get a ZRL (zero run length = 16 zeros), or any combination which takes you past the end. A simplified decode loop:
void DecodeMCU(signed short *MCU)
{
int index;
unsigned short code, skip, extra;
MCU[0] = decodeDC();
index = 1;
while (index < 64)
{
code = decodeAC();
skip = code >> 4; // skip value
extra = code & 0xf; // extra bits
index += skip;
MCU[index++] = calcACValue(extra);
}
}
The color components can be interleaved (typical) or stored in separate scans. The elements are encoded in zigzag order in each MCU (low frequency elements first). The number of 8x8 blocks of coefficients which define an MCU varies depending on the color subsampling. For 1:1, there will be 1 Y followed by 1 Cr and 1 Cb. For typical digital camera images, the horizontal axis is subsampled, so you will get 2 Y blocks followed by 1 Cr and 1 Cb. The quality setting of the compressed image determines the quantization table used and how many zero AC coefficients are encoded. The lower the quality, the more of each MCU will be zeros. When you do the inverse DCT on your MCU, the number of zeros will determine how much detail is preserved in your 8x8, 16x8, 8x16 or 16x16 block of pixels. Here are the basic steps:
1) Entropy decode the 8x8 coefficient blocks, each color component is stored separately
2) De-zigzag and de-quantize the coefficients
3) Perform inverse DCT on the coefficients (might be 6 8x8 blocks for 4:2:0 subsampling)
4) Convert the colorspace from YCrCb to RGB or whatever you need