{"id":40,"date":"2008-12-14T14:00:57","date_gmt":"2008-12-14T14:00:57","guid":{"rendered":"https:\/\/blogs.mathworks.com\/seth\/2008\/12\/13\/representing-numbers-integers-and-fixed-point\/"},"modified":"2017-04-27T08:47:40","modified_gmt":"2017-04-27T13:47:40","slug":"representing-numbers-integers-and-fixed-point","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/simulink\/2008\/12\/14\/representing-numbers-integers-and-fixed-point\/","title":{"rendered":"Representing Numbers: Integers and Fixed-Point"},"content":{"rendered":"\r\n\r\n

In embedded systems, computation time and memory are\r\ncritical resources.\u00a0 Floating-point calculations require special floating-point\r\nunits for efficient computation and this translates to processors that are more\r\nexpensive.\u00a0 For this reason, designers of embedded systems often use\r\nfixed-point numbers.\u00a0 In this post, I want to introduce the basic concepts of\r\nfixed-point number representation.\u00a0 I also want to share with you an article\r\nfrom the MATLAB Digest about Converting\r\nModels from Floating Point to Fixed Point for Production Code Generation.<\/p>\r\n\r\n

Integers<\/strong><\/p>\r\n\r\n

Integers are the most basic way to store numbers in binary.\u00a0\r\nFor integers, each bit represents a different number in powers of two.\u00a0 To\r\nvisualize this, I want to look at a toy three-bit number system.\u00a0 The values of\r\nthe bits in this toy system are 22<\/sup> (4), 21<\/sup> (2), and 20<\/sup>\r\n(1).\u00a0 This three-bit system can represent all integer values between 0 and 7\r\nusing the following bit patterns:<\/p>\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n
\r\n

22<\/sup><\/p>\r\n <\/td>\r\n

\r\n

21<\/sup><\/p>\r\n <\/td>\r\n

\r\n

20<\/sup><\/p>\r\n <\/td>\r\n

\r\n

Value<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 0<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 1<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 2<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 3<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 4<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 5<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 6<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 7<\/p>\r\n <\/td>\r\n <\/tr>\r\n<\/table>\r\n\r\n

 <\/p>\r\n\r\n

With the addition of a sign bit to keep track of positive or\r\nnegative numbers, our three-bit unsigned<\/em> integers become four-bit signed\r\nintegers.\u00a0 Just putting a sign bit at the beginning of the number creates an\r\nintriguing problem of +0 and -0 representations.\u00a0 There are different ways to\r\ndeal with signed\r\nnumber representations<\/a>. \u00a0Using the two's complement<\/a> method, each negative\r\nnumber is the bitwise NOT of the positive number.\u00a0 Therefore, if one is 0001, negative\r\none is 1110.\u00a0 With two's complement numbers, negative zero is the most negative\r\nnumber, 1000 = -8.\u00a0 The signed four-bit patterns look like this:<\/p>\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n
\r\n

Sign<\/p>\r\n <\/td>\r\n

\r\n

22<\/sup><\/p>\r\n <\/td>\r\n

\r\n

21<\/sup><\/p>\r\n <\/td>\r\n

\r\n

20<\/sup><\/p>\r\n <\/td>\r\n

\r\n

Value<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 7<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 6<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 5<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 4<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 3<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 2<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 1<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 0<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= -1<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= -2<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= -3<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= -4<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= -5<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= -6<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= -7<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= -8<\/p>\r\n <\/td>\r\n <\/tr>\r\n<\/table>\r\n\r\n

 <\/p>\r\n\r\n

This extends the range from -8 to +7, and the numbers are\r\nevenly spaced over that range.\u00a0 Integer representations are really just a\r\nspecial case of fixed-point numbers.<\/p>\r\n\r\n

Fixed-Point Numbers: Binary Point<\/strong><\/p>\r\n\r\n

Fixed-point numbers use the same integer representations,\r\nbut they assign a different meaning to the bits.\u00a0 We can introduce the idea of fractional\r\nbits and a binary point.\u00a0 With integers, your smallest spacing between numbers\r\nis 1.\u00a0 If we decide to let a bit represent \u00bd we now have closer spacing in our\r\nsystem.\u00a0 The number of bits to the right of the binary point is the fraction\r\nlength.\u00a0 Using our toy three-bit system, we can assign a fixed binary point\r\nafter the first bit.\u00a0 \u00a0\u00a0This is then a three-bit system with a fraction length\r\nof two.\u00a0 The values of the bits are now 1, \u00bd, and \u00bc.\u00a0 A number 0.10 equals \u00bd.\u00a0 This\r\nthree-bit fixed-point system can represent the following range of numbers:<\/p>\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n
\r\n

20<\/sup><\/p>\r\n <\/td>\r\n

\r\n

2-1<\/sup><\/p>\r\n <\/td>\r\n

\r\n

2-2<\/sup><\/p>\r\n <\/td>\r\n

\r\n

Value<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 0<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= \u00bc<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= \u00bd<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= \u00be<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 1<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 1 \u00bc<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 1 \u00bd<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 1 \u00be<\/p>\r\n <\/td>\r\n <\/tr>\r\n<\/table>\r\n\r\n

 <\/p>\r\n\r\n

This allows more precision over a smaller range of numbers.\u00a0\r\nThe location of the binary point determines the range of the numbers and the\r\nspacing between them.<\/p>\r\n\r\n

What if we wanted to move the binary point to be after the 5th<\/sup>\r\nbit in our three-bit system?\u00a0 In this case, the fraction length would be -2.\u00a0 Instead\r\nof fractional bits, the least significant bit equal to 4.\u00a0 In a system like this,\r\nthe bits would be 16, 8, and 4.<\/p>\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n
\r\n

24<\/sup><\/p>\r\n <\/td>\r\n

\r\n

23<\/sup><\/p>\r\n <\/td>\r\n

\r\n

22<\/sup><\/p>\r\n <\/td>\r\n

\r\n

Value<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 0<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 4<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 8<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 12<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 16<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 20<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 24<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 28<\/p>\r\n <\/td>\r\n <\/tr>\r\n<\/table>\r\n\r\n

 <\/p>\r\n\r\n

This representation gives us a larger range, but greater spacing\r\nbetween bits.\u00a0 Imagine an engine controller that has to represent values\r\nbetween 0 and 10,000 RPMs.\u00a0 Unsigned eight-bit integers have a maximum value of\r\n255.\u00a0 A sixteen-bit unsigned int can hold values over the range 0 to 65535.\u00a0 If\r\nwe choose an eight-bit number with fraction length of -6, we could store these\r\nnumbers using only 8 bits.\u00a0 The range of such a system is 0 to 16320 with a\r\nspacing of 64.<\/p>\r\n\r\n

Fixed-Point Number: Slope and Bias<\/strong><\/p>\r\n\r\n

A fixed-point number system can also encode the real-world\r\nvalue using an arbitrary slope and bias.\u00a0 Regular integers have a slope of 1\r\nand a bias of 0.\u00a0 When using slope and bias, the binary representation stores\r\nan integer that is used to calculate the real-world value.\u00a0 The calculation is\r\nthe ever familiar line equation, y = m ∙ x + b<\/em>.\u00a0 This representation\r\nshifts the range and scaling of the numbers represented.\u00a0 Using our toy\r\nthree-bit integers, we can define a slope of 1.25 and a bias of 20. The bits still\r\nrepresent 4, 2, and 1, but the value computed is 1.25 ∙ x + 20.<\/p>\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n
\r\n

22<\/sup><\/p>\r\n <\/td>\r\n

\r\n

21<\/sup><\/p>\r\n <\/td>\r\n

\r\n

20<\/sup><\/p>\r\n <\/td>\r\n

\r\n

Integer<\/p>\r\n <\/td>\r\n

\r\n

Real World Value<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 0<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 0 + 20 = 20<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 1<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 1 + 20 = 21.25<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 2<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 2 + 20 = 22.50<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 3<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 3 + 20 = 23.75<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 4<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 4 + 20 = 25.00<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 5<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 5 + 20 = 26.25<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

0<\/p>\r\n <\/td>\r\n

\r\n

= 6<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 6 + 20 = 27.50<\/p>\r\n <\/td>\r\n <\/tr>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

1<\/p>\r\n <\/td>\r\n

\r\n

= 7<\/p>\r\n <\/td>\r\n

\r\n

1.25 ∙ 7 + 20 = 27.75<\/p>\r\n <\/td>\r\n <\/tr>\r\n<\/table>\r\n\r\n

 <\/p>\r\n\r\n

There are more intricacies to fixed-point representations,\r\nand the Simulink\r\nFixed-Point documentation<\/a> gives a great overview.<\/p>\r\n\r\n

Challenges of Working in Fixed-Point<\/strong><\/p>\r\n\r\n

The challenges of using fixed-point add complexity to\r\nembedded system development.\u00a0 You have to make sure you have the correct\r\nscaling and bias for your application in order to avoid overflow or underflow.\u00a0\r\nIf you have Simulink\r\nFixed-Point<\/a> you can use fixed-point data types in your Simulink model and\r\nget bit true simulation of your design.\u00a0 The MATLAB Digest article Converting\r\nModels from Floating Point to Fixed Point for Production Code Generation\r\nexplains the workflows that help you get this right in your Simulink design.<\/p>\r\n\r\n

Now it\u2019s your turn<\/strong><\/p>\r\n\r\n

Do you develop fixed-point code?\u00a0\u00a0 What did you learn from\r\nthis overview of fixed-point number representation?\u00a0 Leave a comment here<\/a> and share\r\nyour fixed-point story.<\/p>","protected":false},"excerpt":{"rendered":"

\r\n\r\nIn embedded systems, computation time and memory are\r\ncritical resources.\u00a0 Floating-point calculations require special floating-point\r\nunits for efficient computation and this translates to... read more >><\/a><\/p>","protected":false},"author":40,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[67,76],"tags":[68,441],"_links":{"self":[{"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/posts\/40"}],"collection":[{"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/users\/40"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/comments?post=40"}],"version-history":[{"count":6,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/posts\/40\/revisions"}],"predecessor-version":[{"id":6509,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/posts\/40\/revisions\/6509"}],"wp:attachment":[{"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/media?parent=40"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/categories?post=40"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/simulink\/wp-json\/wp\/v2\/tags?post=40"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}