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introduction_to_digital_systems:number_systems [2023/03/27 10:47]
mexleadmin 		introduction_to_digital_systems:number_systems [2023/11/16 01:09] (aktuell)
mexleadmin
Zeile 1: Zeile 1:
-====== 2. Number Systems ======+====== 2 Number Systems ======
  
 ===== 2.1 Types of Number Systems ===== ===== 2.1 Types of Number Systems =====
Zeile 28: Zeile 28:
  
 Besides this representation of quantities also the position of the symbol in the **numeral** was important: Besides this representation of quantities also the position of the symbol in the **numeral** was important:
-  * In general: the letters have to be arranged decreasing from left to right. For example $MDCI = 1601$+  * In general: the letters have to be arranged decreasing from left to right. For example $\rm MDCI = 1601$
   * There are deviations of this rule: When up to three of the lower symbols are written to the left, these have to be subtracted. Sounds complicated? \\ A kind of.... for example $\rm \color{blue}{M}\color{green}{CCD}\color{red}{L}IV = \color{blue}{1}\color{green}{3}\color{red}{5}4$. \\ Luckily, we do not have to learn this, but by this trick, the length of the numeral could be shortened,    * There are deviations of this rule: When up to three of the lower symbols are written to the left, these have to be subtracted. Sounds complicated? \\ A kind of.... for example $\rm \color{blue}{M}\color{green}{CCD}\color{red}{L}IV = \color{blue}{1}\color{green}{3}\color{red}{5}4$. \\ Luckily, we do not have to learn this, but by this trick, the length of the numeral could be shortened, 
  
Zeile 151: Zeile 151:
 As an example the Byte4 shall be written in decimal: As an example the Byte4 shall be written in decimal:
  
-$ Z_{10}(Byte4)= Z_{10}(19_{16}) \\ =\sum_{i=0}^1 z_i \cdot 16^i = \\ 1 \cdot 16^1 + 9 \cdot 16^0 = 16 + 9 = 25_{10}$+$ Z_{10}({\rm Byte}4)= Z_{10}(19_{16}) \\ =\sum_{i=0}^1 z_i \cdot 16^i = \\ 1 \cdot 16^1 + 9 \cdot 16^0 = 16 + 9 = 25_{10}$
  
 === Octal === === Octal ===
Zeile 200: Zeile 200:
  
 ^each position in decimal     ^ $\color{red}{1}$ ^ $\color{green}{12}$ ^ $\color{blue}{4}$ ^ ^each position in decimal     ^ $\color{red}{1}$ ^ $\color{green}{12}$ ^ $\color{blue}{4}$ ^
-|each position in hexadecimal | $\color{red}{1}$ | $\color{green}{C}$  | $\color{blue}{4}$ |+|each position in hexadecimal | $\color{red}{1}$ | $\color{green}{\rm C}$  | $\color{blue}{4}$ |
  
-The result is $\color{red}{1}\color{green}{C}\color{blue}{4}_{16}$+The result is $\color{red}{1}\color{green}{\rm C}\color{blue}{4}_{16}$
 </callout> </callout>
  
Zeile 216: Zeile 216:
  
 ^each position in decimal     ^ $\color{blue }{1}$ ^ $\color{green}{14}$ ^ $\color{brown}{11}$ ^  $\color{red  }{8}$ ^  $\color{grey }{5}$ ^  $\color{blue }{1}$ ^ $\color{green}{14}$ ^ $...$ | ^each position in decimal     ^ $\color{blue }{1}$ ^ $\color{green}{14}$ ^ $\color{brown}{11}$ ^  $\color{red  }{8}$ ^  $\color{grey }{5}$ ^  $\color{blue }{1}$ ^ $\color{green}{14}$ ^ $...$ |
-|each position in hexadecimal | $\color{blue }{1}$ | $\color{green}{\rm E}$  | $\color{brown}{B}$  |  $\color{red  }{8}$ |  $\color{grey }{5}$ |  $\color{blue }{1}$ |  $\color{green}{\rm E}$  |     |+|each position in hexadecimal | $\color{blue }{1}$ | $\color{green}{\rm E}$  | $\color{brown}{\rm B}$  |  $\color{red  }{8}$ |  $\color{grey }{5}$ |  $\color{blue }{1}$ |  $\color{green}{\rm E}$  |     |
  
-The result is $0.\overline{ \color{blue }{1} \color{green}{E} \color{brown}{B} \color{red  }{8} \color{grey }{5} }$+The result is $\rm 0.\overline{ \color{blue }{1} \color{green}{E} \color{brown}{B} \color{red  }{8} \color{grey }{5} }$
 </callout> </callout>
  
Zeile 301: Zeile 301:
                +672 \\                +672 \\
 \hline  \hline 
-1\overset{\color{red}{2}}{1}\overset{\color{red}{1}}{4}\overset{}{2}+1\overset{\color{red}{2}}{2}\overset{\color{red}{1}}{4}\overset{}{2}
 \end{align*} \end{align*}
  
Zeile 387: Zeile 387:
 \begin{align*} \begin{align*}
 \begin{array}{lll} \begin{array}{lll}
-{ +{a) \\ 
-a) \\ +\color{white}                       {+}5_{16} \\ 
-\color{white}{+}5_{16} \\ +                                     + 3_{16}\\
-+3_{16}\\+
 \,\overline{\color{white}{+}\overset{}{8_{16}}} \,\overline{\color{white}{+}\overset{}{8_{16}}}
-}&{ +}&{ b) \\ 
-b) \\ +\color{white}                           {+}7_{16} \\ 
-\color{white}{+}7_{16} \\ +                                         + 3_{16}\\ 
-+3_{16}\\ +\,\overline{\color{white}{+}\overset{}{\rm A_{16}}} 
-\,\overline{\color{white}{+}\overset{}{A_{16}}} +}&{ c) \\ 
-}&{ +\color{white}                       {+}\,    1_{16} \\ 
-c) \\ +                                     + {\rm D}_{16}\\ 
-\color{white}{+}\,1_{16} \\ +\,\overline{\color{white}{+}\overset{}{{\rm E}_{16}}} 
-+D_{16}\\ +}&{ d) \\ 
-\,\overline{\color{white}{+}\overset{}{E_{16}}} +\color{white}                             {+} {\rm E}_{16} \\ 
-}&{ +                                           +  {\rm A}_{16}\\
-d) \\ +
-\color{white}{+}E_{16} \\ +
-+A_{16}\\+
 \;\;\overline{\overset{\color{red}{1}}{1}\overset{}{8_{16}}} \;\;\overline{\overset{\color{red}{1}}{1}\overset{}{8_{16}}}
 }  } 
Zeile 412: Zeile 408:
  
 a) This one is simple: looks like a decimal formula.. \\ a) This one is simple: looks like a decimal formula.. \\
-b) Here, the summands look like decimal numerals, but the result $7_{10} + 3_{10} = 10_{10}$ is still within the range of the base. The correct symbol would be $10_{10} = A_{16}$ \\ +b) Here, the summands look like decimal numerals, but the result $7_{10} + 3_{10} = 10_{10}$ is still within the range of the base. The correct symbol would be $10_{10} = \rm A_{16}$ \\ 
-c) Now, the summands are a "bit more hexadecimally". The easiest way is: "convert the single digit from hex to decimal, do the operation, and re-convert to hex". For the given example: $1_{10} + 13_{10} = 14_{10} = D_{16}$ \\ +c) Now, the summands are a "bit more hexadecimally". The easiest way is: "convert the single digit from hex to decimal, do the operation, and re-convert to hex". For the given example: $1_{10} + 13_{10} = 14_{10} = \rm D_{16}$ \\ 
-d) Also for this calculation the described way is beneficial: $E_{16} + A_{16} = 14_{10} + 10_{10} = 24_{10}$. The result is larger than the base, and therefore the value has to be separated in more digits: $24_{10} = 16_{10} + 8_{10} = 10_{16} + 8_{16} = 18_{10}$ \\+d) Also for this calculation the described way is beneficial: $\rm E_{16} + A_{16} = 14_{10} + 10_{10} = 24_{10}$. The result is larger than the base, and therefore the value has to be separated in more digits: $24_{10} = 16_{10} + 8_{10} = 10_{16} + 8_{16} = 18_{10}$ \\
  
 For a hexadecimal value with more digits, the carry of the calculation before has to be added - otherwise, every step remains the same.  For a hexadecimal value with more digits, the carry of the calculation before has to be added - otherwise, every step remains the same. 
Zeile 426: Zeile 422:
 \begin{align*} \begin{align*}
 \begin{array}{lll} \begin{array}{lll}
-{ +{ a) \\ 
-a) \\ +\color{white}                       {-}0_2 \\ 
-\color{white}{-}0_2 \\ +                                     - 0_2 \\
--0_2\\+
 \,\overline{\color{white}{-}\overset{}{0_2}} \,\overline{\color{white}{-}\overset{}{0_2}}
-}&{ +}&{ b) \\ 
-b) \\ +\color{white}                       {-}1_2 \\ 
-\color{white}{-}1_2 \\ +                                     - 1_2 \\
--1_2\\+
 \,\overline{\color{white}{-}\overset{}{0_2}} \,\overline{\color{white}{-}\overset{}{0_2}}
-}&{ +}&{ c) \\ 
-c) \\ +\color{white}                       {-}1_2 \\ 
-\color{white}{-}1_2 \\ +                                     - 0_2 \\
--0_2\\+
 \,\overline{\color{white}{-}\overset{}{1_2}} \,\overline{\color{white}{-}\overset{}{1_2}}
-}&{ +}&{ d) \\ 
-d) \\ +\color{white}                       {-}0_2 \\ 
-\color{white}{-}0_2 \\ +                                     - 1_2 \\
--1_2\\+
 \;\;\overline{\overset{\color{red}{1}}{\color{white}{0}}\overset{}{1_2}} \;\;\overline{\overset{\color{red}{1}}{\color{white}{0}}\overset{}{1_2}}
 }  } 
Zeile 452: Zeile 444:
 The calculation for $d)$ shows the carry, which here has to borrow a bit from the next upper digit. This is similar to the calculation:  The calculation for $d)$ shows the carry, which here has to borrow a bit from the next upper digit. This is similar to the calculation: 
 \begin{align*} \begin{align*}
-\color{white}{-}42_{10} \\ +\color{white}                                    {-}42_{10} \\ 
--23_{10}\\ +                                                  - 23_{10}\\ 
-\;\;\overline{\overset{\color{red}{1}}{1}\overset{}{9_{10}}}+\;\;\overline{\overset{\color{red}{1}}{1}\overset{}{ 9_{10}}}
 \end{align*}  \end{align*} 
  
Zeile 460: Zeile 452:
  
 \begin{align*} \begin{align*}
-\color{white}{-}\,10\boldsymbol{1}0_2 \\ +\color{white}      {-}\,10\boldsymbol{1}0_2 \\ 
--01\boldsymbol{1}1_2\\+                    -   01\boldsymbol{1}1_2\\
 \,\overline{\color{white}{+}\overset{\color{red}{1}}{0}\overset{\color{red}{1}}{0}\overset{\color{red}{\boldsymbol{1}}}{\boldsymbol{1}}\overset{}{1_2}} \,\overline{\color{white}{+}\overset{\color{red}{1}}{0}\overset{\color{red}{1}}{0}\overset{\color{red}{\boldsymbol{1}}}{\boldsymbol{1}}\overset{}{1_2}}
 \end{align*} \end{align*}
Zeile 468: Zeile 460:
  
 \begin{align*} \begin{align*}
-\color{white}{-}\,\boldsymbol{1}_2 \\ +\color{white}       {-}\,\boldsymbol{1}_2 \\ 
--\boldsymbol{1}_2\\+                     -   \boldsymbol{1}_2\\
 \overline{ \overset{\color{red}{1}} {\color{white}{0}} \overset{\color{red}{\boldsymbol{1}}} {\boldsymbol{1}}_2 } \overline{ \overset{\color{red}{1}} {\color{white}{0}} \overset{\color{red}{\boldsymbol{1}}} {\boldsymbol{1}}_2 }
 \end{align*} \end{align*}
  
-The calculation has to be executed as follows: $\boldsymbol{1}_2 - (\boldsymbol{1}_2 + \color{red}{\small{\boldsymbol{1}}} ) = {\boldsymbol{1}}_2 $. Additionally, another carry has to be taken from the next digit.+The calculation has to be executed as follows: $\boldsymbol{1}_2 - (\boldsymbol{1}_2 + \color{red}{\small{\boldsymbol{1}}} ) = {\boldsymbol{1}}_2 $.  
 +Additionally, another carry has to be taken from the next digit.
  
 === In Hexadecimal === === In Hexadecimal ===
  
 The calculation in hexadecimal is conceptually again the same. Some examples, which we will discuss below: The calculation in hexadecimal is conceptually again the same. Some examples, which we will discuss below:
- 
  
 \begin{align*} \begin{align*}
 +\rm 
 \begin{array}{lll} \begin{array}{lll}
-{ +{ a) \\ 
-a) \\ +                       \color{white}{-}15_{16} \\ 
-\color{white}{-}15_{16} \\ +                       -\color{white}{1}3_{16} \\
--\color{white}{1}3_{16}\\+
 \,\overline{\color{white}{-}\overset{}{12_{16}}} \,\overline{\color{white}{-}\overset{}{12_{16}}}
-}&{ +}&{ b) \\ 
-b) \\ +                                           \color{white}{-}23_{16} \\ 
-\color{white}{-}23_{16} \\ +                                       -\color{white}{\rm B}6_{16} \\ 
--\color{white}{B}6_{16}\\ +\overline{\color{white}{-}\overset{\color{red}{1}}{1} {\rm D}_{16}} 
-\overline{\color{white}{-}\overset{\color{red}{1}}{1}D_{16}} +}&{ c) \\ 
-}&{ +                               \color{white}{-}3{\rm F}_{16} \\ 
-c) \\ +                                              -1{\rm A}_{16}\\
-\color{white}{-}3F_{16} \\ +
--1A_{16}\\+
 \color{white}{-}\overline{\color{white}{B}\overset{}{25_{16}}} \color{white}{-}\overline{\color{white}{B}\overset{}{25_{16}}}
-}&{ +}&{ d) \\ 
-d) \\ +                                                   \color{white}{+}\,38_{16} \\ 
-\color{white}{+}\,38_{16} \\ +                                                           +   1{\rm A}_{16}\\ 
-+1A_{16}\\ +\color{white}{+}\overline{\overset{\color{red}{1}}{1}\overset{}{{\rm B}_{16}}}
-\color{white}{+}\overline{\overset{\color{red}{1}}{1}\overset{}{B_{16}}}+
 }  } 
 \end{array} \end{array}
Zeile 507: Zeile 496:
  
 a) This one is (again) simple: looks (again) like a decimal formula.. \\ a) This one is (again) simple: looks (again) like a decimal formula.. \\
-b) Here, both hexadecimal numerals look like decimal numerals, but the result $3_{10} - 6_{10} = 7_{10} + C$ lead to an underflow, where we have to take a carry of $C=-10_{10}$ for a decimal calculation. In hexadecimal, the carry differs. A better way to solve such a subtraction in hexadecimal is to add the carry before: $3_{16} - 6_{16} + \color{red}{10_{16}}$. Then the calculation can be converted to decimal:  $3_{10} - 6_{10} + \color{red}{16_{10}} = 19_{10} - 6_{10} = 13_{10}$. This result has to be converted back to hexadecimal: $13_{10} = D_{16}$. For the next column, the carry has to be considered. \\ +b) Here, both hexadecimal numerals look like decimal numerals, but the result $3_{10} - 6_{10} = 7_{10} + C$ lead to an underflow, where we have to take a carry of $C=-10_{10}$ for a decimal calculation. In hexadecimal, the carry differs. A better way to solve such a subtraction in hexadecimal is to add the carry before: $3_{16} - 6_{16} + \color{red}{10_{16}}$. Then the calculation can be converted to decimal:  $3_{10} - 6_{10} + \color{red}{16_{10}} = 19_{10} - 6_{10} = 13_{10}$. This result has to be converted back to hexadecimal: $\rm 13_{10} = D_{16}$. For the next column, the carry has to be considered. \\ 
-c) For this formula the idea from b) helps, too: for each digit-by-digit subtraction, we will have a look, at whether a transformation to decimal is beneficial. For the rightmost it is: $F_{16}- A_{16} = 15_{10} - 10_{10} = 5_{10}$. The rightmost result is now: $5_{10} = 5_{16}$  \\ +c) For this formula the idea from b) helps, too: for each digit-by-digit subtraction, we will have a look, at whether a transformation to decimal is beneficial. For the rightmost it is: $\rm F_{16}- A_{16} = 15_{10} - 10_{10} = 5_{10}$. The rightmost result is now: $5_{10} = 5_{16}$  \\ 
-d) We can do the same thing here, for $8_{16}-A_{16}$: consider the carry, convert to decimal, calculate, and re-convert to hexadecimal. $8_{10}-10_{10}+ \color{red}{16_{10}} = 24_{10}-10_{10}= 14_{10}$. The result is $E_{16}$.  \\+d) We can do the same thing here, for $8_{16}-A_{16}$: consider the carry, convert to decimal, calculate, and re-convert to hexadecimal. $8_{10}-10_{10}+ \color{red}{16_{10}} = 24_{10}-10_{10}= 14_{10}$. The result is $\rm E_{16}$.  \\
  
 <WRAP column 100%> <panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%> <WRAP column 100%> <panel type="danger" title="Note!"> <WRAP group><WRAP column 7%>{{fa>exclamation?32}}</WRAP><WRAP column 80%>
Zeile 560: Zeile 549:
  
   * [[https://calculator.name/baseconvert/decimal/hexadecimal/|Conversion tool from decimal to hexadecimal]], this tool shows also the steps and can be used vice versa   * [[https://calculator.name/baseconvert/decimal/hexadecimal/|Conversion tool from decimal to hexadecimal]], this tool shows also the steps and can be used vice versa
 +  * The [[https://www.omnicalculator.com/math/binary-fraction|OmniCalculator]] can also calculate binary fractions and comes with another explaination of the manual process. 
   * A short video on the "[[https://www.youtube.com/watch?v=_s5RFgd59ao&ab_channel=Numberphile|Everything Formula]]" which prints all possible output and is directly related to binary encoding.   * A short video on the "[[https://www.youtube.com/watch?v=_s5RFgd59ao&ab_channel=Numberphile|Everything Formula]]" which prints all possible output and is directly related to binary encoding.
  
Zeile 591: Zeile 581:
 Tip: Convert nibble by nibble to binary. Tip: Convert nibble by nibble to binary.
  
-  - $4321H$ +  - $\rm 4321H$ 
-  - $0xCAFE$ +  - $\rm 0xCAFE$ 
-  - $ \$ $$4be79 $ +  - $ \$ $$\rm 4be79 $ 
-  - $ \$ $$398B4C $ +  - $ \$ $$\rm 398B4C $ 
-  - $0x3a4f.4ecd $ +  - $\rm 0x3a4f.4ecd $ 
-  - $\$1.3DAF $+  - $\$\rm 1.3DAF $
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 605: Zeile 595:
 Tip: Use the numerals from Exercise 2.3.1. Tip: Use the numerals from Exercise 2.3.1.
  
-  - $0b10110110$+  - $\rm 0b10110110$
   - $\%10100101$   - $\%10100101$
-  - $1111\, 0111\, 0001\, 0011B$ +  - $\rm 1111\, 0111\, 0001\, 0011B$ 
-  - $0110111.0101b$ +  - $\rm 0110111.0101b$ 
-  - $0.10101b$ +  - $\rm 0.10101b$ 
-  - $0.1001\, 1001\, ... b$+  - $\rm 0.1001\, 1001\, ... b$
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 633: Zeile 623:
 Execute the following operations manually with the given dual numerals. \\ Execute the following operations manually with the given dual numerals. \\
  
-  - $01110111B + 11010101B$ +  - $\rm 01110111B + 11010101B$ 
-  - $01011001B + 10011111B$ +  - $\rm 01011001B + 10011111B$ 
-  - $01100111B - 01001100B$ +  - $\rm 01100111B - 01001100B$ 
-  - $10101110B - 10000111B$ +  - $\rm 10101110B - 10000111B$ 
-  - $0011.111B - 0011.100B$ +  - $\rm 0011.111B - 0011.100B$ 
-  - $001111.011B + 1.010011B$ +  - $\rm 001111.011B + 1.010011B$ 
-  - $01101.100B - 01000.111B$ +  - $\rm 01101.100B - 01000.111B$ 
-  - $100101.11B - 110001.10B$+  - $\rm 100101.11B - 110001.10B$
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 648: Zeile 638:
 Execute the following operations manually with the given dual numerals. \\ Execute the following operations manually with the given dual numerals. \\
  
-  - $1011B \cdot 0101B$ +  - $\rm 1011B \cdot 0101B$ 
-  - $01100101B \cdot 110B$ +  - $\rm 01100101B \cdot 110B$ 
-  - $10101111B : 101B$ +  - $\rm 10101111B : 101B$ 
-  - $11110000B : 1000B$+  - $\rm 11110000B : 1000B$
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
Zeile 658: Zeile 648:
 Execute the following operations manually with the given dual numerals. \\ Execute the following operations manually with the given dual numerals. \\
  
-  - $1334H + 07ABH$ +  - $\rm  1334H +  07ABH$ 
-  - $0DC43H - 0BD19H$ +  - $\rm 0DC43H - 0BD19H$ 
-  - $1F23H + 90E8H$ +  - $\rm  1F23H +  90E8H$ 
-  - $98C5H - 84CAH$ +  - $\rm  98C5H -  84CAH$ 
-  - $234AH + 7EE6H$ +  - $\rm  234AH +  7EE6H$ 
-  - $0F10CH - 0ED43H$+  - $\rm 0F10CH - 0ED43H$
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 674: Zeile 664:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Exercise 2.3.9. One's Complement"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 2.3.9. Two's Complement"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-In the simulation in <imgref pic70> the one's complement of a 4bit value is shown. \\ +In the simulation in <imgref pic70> the two's complement of a 4bit value is shown. \\ 
-On initialization the value is the following:  +On initialization, the value is the following: 
-  * The value for $A$ is $1010_2 = 10_{10}$. The decimal value ''10'' is also shown in the first display from the left in the circuit.  +  * The value for variable $A$ is $1010_2 = 10_{10}$. The decimal value ''10'' is also shown in the first display from the left in the circuit.  
-  * The value for $B$ is $1110_2 = 14_{10}$. The decimal value ''14'' is also shown in the second display from the left in the circuit.  +  * The value for variable $B$ is $1110_2 = 14_{10}$. The decimal value ''14'' is also shown in the second display from the left in the circuit.  
-  * The addition $A+B$ leads to $S = 1000_2 = 8_{10}$. The decimal value ''10'' is also shown in the rightmost display in the circuit.  +  * The addition $A+B$ leads to $S = 1000_2 = 8_{10}$. The decimal value ''10'' is also shown on the rightmost display in the circuit.  
-  * There is another value in the second display from the right. This is called $B'$ and it is equal to $0010_2 = 2_{10}$.+  * There is another value in the second display from the right. This is called $B'$, and it is equal to $0010_2 = 2_{10}$.
    
 <WRAP><well> <WRAP><well>
-<imgcaption pic70|One's Complement></imgcaption> \\ +<imgcaption pic70|Two's Complement></imgcaption> \\ 
-{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&ctz=CQAgzCAMB0l3BOJyWoSALNATAVmwIyQAc28kAbNgOzrbEi6OSMCmAtAQQFBnUjswYCiAIYMAsPlHSJAQUi9I-QcNH1J0gmUwg5PPpLVdsm09pbzsSlUJEIRgrZAnyw3AB4gEppkQkMEgQMAMqeAtgYDI7YauhBDABC4YIYLLYaRLrBeik05gQRuH6u4CwAOgAOedSmGOjseEyyZSBV3BgEDWCmDlDeIoUY3ADuAsQM2JHjk2lQ3ACyIBOMIitzptjQuIsgtat7dSyb27v+B+cbIFs7SxYXOlc33AAyESu4MevH-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-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-CIMJoS2YhPLAZAJRkVz5RXGZLxOYowczScNKc7MRBeSiWzfUvYEmiHyYMTMig7jlLKUad0zwljFP6E8U4Sw4n0zqScW4eSXQiBaTsLw7BcCuBUBgZoYoQwCBSLQTAdIKBXwUS4kAAAKbRXg0aYFMGAIYTBFEAEpuBAA noborder}}+{{url>https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0l3BOJyWoSALNATAVmwIyQAc28kAbNgOzrbEi6OSMCmAtAQQFBnUjswYCiAIYMAsPlHSJAQUi9I-QcNH1J0gmUwg5PPpLVdsm09pbzsSlUJEIRgrZAnyw3AB4gEppkQkMEgQMAJqeAtgYDI7YauhBDABC4YIYLLYaRLrBeik05gQRuH6u4CwAOgAOedSmGOjseEyyZSBV3BgEDWCmDlDeIoUY3ADuAsQM2JHjk2lQ3ACyoi6MgytzptjQuIsgE6t7gSyb27sWB+cbIFs7SwRquINqVzfcADIR+4+fR9eiIAAzACGABsAM6sRjzD6CFbfWGuY79YHgyFMRQwuwHVQiK4sFEQqEYyQ6eE9SxIwoEtHzQyNOZiCSNb4EFqJRR0yIsbQMZmDHQSRIGZQRBnaIr8ywgRLWTlzPp85alRLuGFTCTw9WYSmA0GE9HvUUsTVzPG61FEw2NEjYsi-czm-XQgRdXyOV3av5UvU0xRjRpfGL7YgsP2SOGOMArWrzf3k7HxigSMMBhghowidMprHpnGicSxjPqXlYmPZtQxhEgMujEksJNFhvlkQNvMmQt5hVY77N-NMrGEETZ8Vc8bHOYp-ajwQ6Ue9rVtjS99t52KmbMaad9Oe19jbuZ7kR2jtT6bsdNayfHG2H642q+iHwulYelP+R+mTgrHf+9-TgcTruAEsJ2Q5AZuB4Dj0HYDme8Y-kWg6SCOgH+qO9AgUhUbJru6H7I00zYR2a5lEy6F9Cmo5EfSLAUbu77UVRKyUdB1F9ERlFip+5FgWhGz4aOPIngwjGEcxuEECI1HphxuEMgWNHXFmcm0TEcw9ipICtoJ4qUdM2lzE2mmVseNZ8dyGiKWZRp7Jh8k4b+T65qOylodMmr6Q5EQ2pqNrWZwT7wh6ymdA0o4KsefRDPR35BSssn+hecUSHgvFjtclbbv5U74Re+wpuxKy3vkwlKby+yyaFRZdisUWYB0XSHAM6iqf8wxjPcIgsmorktRcOjWe+LIrEZHU9SwwahrWk1NYyhb7EJ+zWUt5gDeutbvg2+yjU1W0MO2Yb7H0gbzftBZTstkx7cWp2iOKU4aWMU5RE1dGGAtGgxp1ugAMrCvwC0Fl9LQ-RyIoLeKX0CiAP2yuDDB9F96ww+4cabDGF3rU9V0BMcl3XN8MYlYdszRCly149x5NY6KGVUwTaX0gzJLMxu1xnpM0xvkeB6SXTHZ8yVRPZezTKTDWdJCQqc3voK7iS5k4qcAWlzSnDKhCUJytBNDQo2C6mQKTLbKKAAhCAACSAB2ABurAAE4AC4O-0AAaYAADSu9gXsEF7LAAPTux7Qc+0HftByw1t207LvkOUACOXBcOUVvwAQifJ5AqdwEQmdEBnaeQCnSdENnadl5nxeF8XxdV3nFd50nuc58nVdwKnZfZwn1eF2XGc91w5cF93td99Xie18Plc9xPVtDwPufT036ed23s8d1bSi8lLjh9HV0cO879tEv0ihdI46YeuwX1PhA7Us9OguoeM-MNsT3DmwACiCACuYKFBYMHb2vt-YgBCJ7EIPsQh+xCCwH+-9AHwEni3IuddZ4NyXiglORcS5T1bh3QedcK7oOruXMhk826NwHl3TuyDm7EP7vXYetd67jxLkPaehCF5r3QVnNeTdk593XvHLex5zytXYC9OqCCAGn2NOECgQxqyFH4AkNoCdFGFFUCo8M2QGCJxSBgSM2iwAqA0BIKMAgUhgEgiBHovIaD6NyF4RoGQDYRHiKIJIABycIT4ZxKk8c4sIaNmqBMeizVs0EDr0RbP2UwO1AllmrDTQJWtoL+WgrmGJBYwx8xyaYXqnAuruniYWY0Q5UmFj5iyTMU07ilNELiJErwliVP6C8U4SwImRjqK07pLomklM9CcHYrjcCuBUBgZoUorHsBSLQTATIKD3zUd4kAAAKQxXhsC4lMGAIYTB1EAEpwIfi0kEJ8YYGwegbEZJY9yJANhemM7gQA noborder}}
 </well></WRAP> </well></WRAP>
  
   - Why is ''10'' + ''14'' = ''8''? What happens to other values? \\ The input values ''A'' and ''B'' can be changed by clicking on the bit values.    - Why is ''10'' + ''14'' = ''8''? What happens to other values? \\ The input values ''A'' and ''B'' can be changed by clicking on the bit values. 
   - Try to analyze how ''B''' (shown in the brackets) is derived from ''B''.    - Try to analyze how ''B''' (shown in the brackets) is derived from ''B''. 
-  - Have a look at the wiki page of the {{wp>Ones' complement}} to understand how negative numbers are represented in a microcontroller.+  - Have a look at the wiki page of the {{wp>Two's complement}} to understand how negative numbers are represented in a microcontroller. \\ What does the following representation show? \\ {{drawio>introduction_to_digital_systems:SignedIntCircle.svg}}
   - Imagine that you have to rescue data from an old storage device. The interesting bits are given by the boxed area in <imgref picex01>. The bytes are LSB 0 oriented.    - Imagine that you have to rescue data from an old storage device. The interesting bits are given by the boxed area in <imgref picex01>. The bytes are LSB 0 oriented. 
     - What are these values in an unsigned integer?     - What are these values in an unsigned integer?
-    - What are these values in signed integers based on the ones' complement? +    - What are these values in signed integers based on the Two's complement?