ࡱ> tFi%ma;4͏AU*JFIFHH Photoshop 3.08BIM8x7Title: Orange Rust Content Provider: 1996 PhotoDisc, Inc. All rights reserved. Images provided by 1996 Arthur S. Aubry Contact : PhotoDisc, Inc. US Phone 1-800-528-3472 or 206-441-9355 US Fax 206-441-4961 Europe Phone +44 (0) 181 332 2020 Europe Fax +44 (0) 181 332 9034 Images on this disc are governed by the terms of PhotoDisc's End User License Agreement. Images not licensed for use in products for sale, please call 206-441-9355 for information on acquiring the low cost upgrade license required for this use. 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X-(m-1) x-m2#J(Radix Number System Base  2 (binary numbers) 0 1 Base  8 (octal numbers) 0 1 2 3 4 5 6 7 Base  16 (hexadecimal numbers) 0 1 2 3 4 5 6 7 8 9 A B C D E Ft    )Radix OperationsThe same as for decimal numbers+$Conversion From one radix to anotherFrom decimal to binary-$Conversion From one radix to anotherFrom decimal to base-r Separate the number into an integer part and a fraction part For the integer part Divide the number and all successive quotients by r Accumulate the remaindersXROR20Different Bases *$Conversion From one radix to anotherFrom binary to octal Divide into groups of 3 bits Example 11001101001000.1011011 = 31510.554 From octal to binary Replace each octal digit with three bits Example 75643.5704 = 111101110100011.101111000100v%#1*%#  1*,$Conversion From one radix to anotherFrom binary to hexadecimal Divide into groups of 4 bits Example 11001101001000.1011011 = 3348.B6 From hexadecimal to binary Replace each digit with four bits bits Example 7BA3.BC4 = 111101110100011.101111000100v%!/(%!  /(; ComplementsThey are used to simplify the subtraction operation Two types (for each base-r system) Diminishing radix complement (r-1 complement) Radix complement (r complement) 8WOH OM ComplementsAlternative DefinitionN ComplementsAlternative Definition<09 s and 10 s Complementsl9 s complement of 674653 999999-674653 = 325346 9 s complement of 023421 999999-023421 = 976578 10 s complement of 674653 325346+1 = 325347 10 s complement of 023421 976578+1=976579   HExamples$8 s complement of =.1 s and 2 s Complements1 s complement of 10111001 11111111  10111001 = 01000110 Simply replace 1 s and 0 s 1 s complement of 10100010 01011101 2 s complement of 10111001 01000110 + 1 = 01000111 Add 1 to 1 s complement 2 s complement of 10100010 01011101 + 1 = 01011110 : 0: 0  B)Subtraction with Complements of UnsignedM  N Add M to r s complement of N Sum = M+(r^n  N) = M  N+r^n If M > N, Sum will have an end carry r^n, discard it If M<N, Sum will not have an end carry and Sum = r^n  (N  M) (r s complement of N  M) So M  N =  (r s complement of Sum)`S 0   O   &> $>)Subtraction with Complements of Unsigned 65438 - 5623?)Subtraction with Complements of Unsigned 5623 - 65438@)Subtraction with Complements of Unsigned10110010 - 10011111A)Subtraction with Complements of Unsigned10011111 -10110010CSigned Binary NumbersRUnsigned representation can be used for positive integers How about negative integers? Everything must be represented in binary numbers Computers cannot use  or + signs&WSWS1Negative Binary NumbersThree different systems have been used Signed magnitude One s complement Two s complement&'3'32Signed MagnitudeThe leftmost bit is the sign bit (0 is + and 1 is - ) and the remaining bits hold the absolute magnitude of the number Examples -47 = 1 0 1 0 1 1 1 1 47 = 0 0 1 0 1 1 1 1 - f f3 One s complementwReplace each 1 by 0 and each 0 by 1 Example (-6) First represent 6 in binary format (00000110) Then replace (11111001)L2F2$ 4 Two s complementFind one s complement Add 1 Example (-6) First represent 6 in binary format (00000110) One s complement (11111001) Two s complement (11111010) n)g)$DArithmetic AdditionJUsually represented by 2 s complementQ RegistersZThey can hold a groups of binary data Data can be transferred from one register to anotherRProcessor-Memory Registers* S Operations 7Logic Gates - 1 8  ERanges OStudy ProblemsCourse Book Chapter  1 Problems 1  2 1  3 1  7 1  8 1  12 1  16 1  20 1  34 &!5!5PSneak Preview Next time ASSIGNMENT Will be given QUIZ& & . Expect a question from each one of the following Convert decimal to any base Convert between binary, octal, and hexadecimal Binary add, subtract, and multiply Negative numbers v 1 1   Questions /86 ` ̙33` ` ff3333f` 333MMM` f` f` 3` 6xsgţ;z>?" dZ@0?uKZX@ vYwf n?" dd@   @@``PR    @ ` ` p>> K0 phl(  l l 6Ě "xi  NClick to edit Master title  l 6\ "|x  >Click to edit Master text styles Second level Third level adcv!   ? l 0 "  @   @*  l 0d " @  B*  l hA&޽h @ ?computers slidebak 6xsgţ;z Presentation 0 p *(     02 P!   X*   06  !  Z* d  c $ ?*    0D1  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6@ vP   X*   6HE v   Z* H  0޽h/. ? ̙3380___PPT10. 50 P(    0TVY P!  Y D*    0x]Y  ! Y F*    60rQ vP  Y D*    65 v  Y F*  H  0޽h/. ? ̙3380___PPT10.`U 0(  d  Ngֳgֳ ?0ZU 02U 0 Narration  DIGITAL LOGIC DESIGN4 F(0   Hgֳgֳ ? _ by Dr. Fenghui Yao 鶹ý Department of Computer Science Nashville, TN ( n2( n2( n2( n2H( n2( n2[, LH  0޽h ? ̙33y___PPT10Y+D=' = @B +  K0 D*(  r  S Phlxi   x  c $d|lx0  H  0޽h ? 6xsgţ;z___PPT10i. 0s(R+D=' = @B +7/  K0 %'.(  l `p`   `p` ,$D 0n  0"`@`p`    0` SCENTRAL PROCESSING UNIT ( 2ffr  S 4lxi     B \"`p,$D 0 B Control Unit     B\"`p p,$D 0 SArithmetic Logic Unit (ALU) l @Pp $@Pp,$D 0  B\"`@Pp B Registers      <4"`0` :R1    <"`0  :R2    <l"`0@ TRn XB   0DԔ l ` ` '` `,$D 0@ ` ` %` ` @ ` ` #` `n   0"`` `n  0"`` `n  0"`P ` @`n  0"`` `  0x  G Main Memory ( 2 ff  00 P 0 p @Disk ( 2ff  0d P pp DKeyboard ( 2 ff  0p0P 0p CPrinter ( 2ffrl `  "` ,$D 0ZB  s *D> @ `  !` ZB B s *D>@ZB  s *D>  ZB  s *D> ` ZB  s *D> ` ZB  s *D>Z Z` ZB  s *D> ` l  0 & 0,$D  0f  6Z 0   0İp C I/O Devices 2   0  ,$ 0 9Bus 2H  0޽h ? 6xsgţ;z!___PPT10.#Pu`+~^D%' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<* %(D' =-6B'blinds(horizontal)*<3<* D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*$%(D' =-6B'blinds(horizontal)*<3<*$D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*"%(D' =-6B'blinds(horizontal)*<3<*"D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*'%(D' =-6B'blinds(horizontal)*<3<*'D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*&%(D' =-6B'blinds(horizontal)*<3<*&++0+ ++0+ ++0+ +  K0 P$(  r  S lxi   r  S @l|x   H  0޽h ? 6xsgţ;z___PPT10i.$Y+D=' = @B +C  K0 B:((  ( ( B5 lxi  5  ( H5 l|xN 5 X ( 0A ?PPP;H ( 0޽h ? 6xsgţ;z___PPT10i.*` a++D=' = @B +  K0 9$(  r  S  5lxi  5 r  S 5l|x  5 H  0޽h ? 6xsgţ;z___PPT10i.%a+D=' = @B +q  K0 x(  x  c $D#5lxi  5 X  0A?MH  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B +  K0 0(  x  c $(*5lxi  5 x  c $*5lnxT  5 H  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B +q  K0 x(  x  c $15lxi  5 X  0A?M`FH  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B +  K0 @v(  x  c $65lxi  5 x  c $t75lxd 5 X  0A?zFH  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B +  K0  O(  x  c $x?5lxi  5 ~  s *L@5l|x 5 X  0A?J|H  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B +  K0  (  r  S  K5lxi  5 ~  s *K5l|P 5  0  fA  ? ?lS"`?`k    5  0 # lA  ?f ?lS"`? #    5   6A ?f ?S"`? P  H  0޽h ? 6xsgţ;z___PPT10i.&a!+D=' = @B +  K0   l(      B5 lxi  5    BPb5 l|x  5 H   0޽h ? 6xsgţ;z___PPT10i.)b3+D=' = @B +q  K0 ph`d $(  $r $ S R5lxi  5 ~ $ s *h5l| 5   $ 0hA5r  x* 11001011 +10011101 1011010002+ 2   $ 0r5@r  z, 11001011 - 10011101 001011102- 2   $ 0I5ppM  @ 101 * 110 000 1010 +10100 11110JA 2 H $ 0޽h ? 6xsgţ;z___PPT10i.)k+D=' = @B +I  K0 H@ 8(  8 8 Bl5 lx  5  8 N@5 lL` 5 X 8 0A ?p H 8 0޽h ? 6xsgţ;z___PPT10i.,o+D=' = @B +  K0 p <5(  < < H5 lx  5  < N5 lP  5  < 05U  p B 165 23 3 0 2 RB < s *DԔp   < 0ԏ5 0[ J 4 2 3( 2RB  < s *D@  <0 # lA 5?f ?lS"`?4`P  5 5 < 05  @ 0.6875 x 2 = 1 + 0.3750 0.3750 x 2 = 0 + 0.7500 0.7500 x 2 = 1 + 0.5000 0.5000 x 2 = 1 + 0.0000 2RB < s *D`   < 6A 3?f ?S"`?>  3LB < c $D` ` LB < c $D@@H < 0޽h ? 6xsgţ;z___PPT10i., A+D=' = @B +{  K0 zrPD (  Dr D S 5lxi  5 X D 0A ?H D 0޽h ? 6xsgţ;z___PPT10i..Ds+D=' = @B +  K0 `0l(  0 0 B5 lx  5  0 B5 lx  5 H 0 0޽h ? 6xsgţ;z___PPT10i.+ѱ+D=' = @B +  K0 `4l(  4 4 B5 lx  5  4 B 5 lx  5 H 4 0޽h ? 6xsgţ;z___PPT10i.+P+D=' = @B +5  K0 4,P h(  hr h S 5lxi  5 ~ h s *L5l|0  5  h0 ZA &?f ?lS"`?  A  & 5 h 05 p`  nFor n-digit number N> 2 h0 ZA )?f ?lS"`? X ) 5XB  h 0DԔ` `   h 05    Dr-1 complement 2XB  h 0DԔ `   h 0t5H 0 B r complement 2 H h 0޽h ? 6xsgţ;z___PPT10i.^@-30+D=' = @B +u  K0 tlq(  r  S  5lxi  5 ~  s *5l| 5  0 ZA k?f ?lS"`?  k 5H  0޽h ? 6xsgţ;z___PPT10i.m+D=' = @B +  K0  0(    B5 lxi  5   N5 l| 5   6A Y?f ?S"`?0P`B  YH  0޽h ? 6xsgţ;z___PPT10i. '+D=' = @B +  K0  t$(  tr t S 5lxi  5 r t S 5l|x~ 5 H t 0޽h ? 6xsgţ;z___PPT10i._q"+D=' = @B +)  K0 ( @ (  r  S Qlxi  Q x  c $Ql|  Q  0 ZA ??f ?lS"`?`  ? Q 0 ZA [?f ?lS"`? [ Q  6A F?f ?S"`? @ @  F   6A O?f ?S"`?{ B  O   6| Q  Pd   8$7 s complement of H  0޽h ? 6xsgţ;z___PPT10i.+D=' = @B +  K0 4xl(  x x BLQ lxi  Q  x BQ lx* Q H x 0޽h ? 6xsgţ;z___PPT10i._Q+D=' = @B +<  K0 @O$(  r  S #Qlx  Q r  S $Qlx  Q H  0޽h ? 6xsgţ;z80___PPT10.d0h  K0 f^|(  |r | S 3Qlx  Q x | c $4QlNx Q  | 0)Q$  d 65438 10 s complement of 05623 +94377 159815 Discard end carry 10^5 -100000 Answer 59815J 2]^6H | 0޽h ? 6xsgţ;z80___PPT10.`s  K0 RJ6(    B-Q lx  Q   HH Q lNx Q h  0?Q}   05623 10 s complement of 65438 +34562 40185 There is no end carry => -(10 s complement of 40185) -598152 2]zH  0޽h ? 6xsgţ;z80___PPT10.ax  K0 pn(    B@LQ lx  Q   HMQ lNx Q   0\NQ>  " 10110010 2 s complement of 10011111 +01100001 100010011 Discard end carry 2^8 -100000000 Answer 000010011V 2h ` d 9H  0޽h ? 6xsgţ;z80___PPT10.aj  K0 6(    B]Q lx  Q   Hx^Q lNx Q   0_Q$  \ 10011111 2 s complement of 10110010 +01001110 11101101 There is no end carry => -(2 s complement of 11101101) Answer = -00010011V 2K ` Z1H  0޽h ? 6xsgţ;z___PPT10i.b`y+D=' = @B +<  K0 $(  r  S doQlxi  Q r  S RQl|xE  Q H  0޽h ? 6xsgţ;z80___PPT10.c  K0 H:(  Hx H c $uQlxi  Q x H c $dvQl|x  Q  H 0zQ   lNOTE: For negative numbers the sign bit is always 1, and for positive numbers it is 0 in these three systemsm 2mH H 0޽h ? 6xsgţ;zn  K0 L(  Lx L c $Qlxi  Q  L HQ l0  Q : L 0Q   For 8 bits, we can represent the signed integers  128 to +127 How about for N bits?T 2TH L 0޽h ? 6xsgţ;z  K0 P0(  Px P c $Qlxi  Q x P c $DQl|xb Q H P 0޽h ? 6xsgţ;zD  K0 Tl(  T T BPQ lxi  Q  T B$Q l|x  Q H T 0޽h ? 6xsgţ;z  K0 $ (  r  S Qlxi  Q x  c $ģQl| Q   0 Qpb  0+ 5 00000101 +11 00001011 +16 00010000J1 2   0Q0 b  1- 5 11111011 +11 00001011 +6 100000110f2 2  RB  s *D@@0   0Q  =Discard 2  0Q^ p0 4+ 5 00000101 -11 11110101 -6 11111010 J5 2    0Qp 0 B 3- 5 11111011 -11 11110101 -16 111110000 f4 2  LB   c $D`pp RB   s *D p    0Q =Discard 2H  0޽h ? 6xsgţ;z80___PPT10.e ]<  K0 $(  r  S Qlxi  Q r  S pQl"x Q H  0޽h ? 6xsgţ;z80___PPT10.ޱB  K0 *(  r  S |Qlx5  Q X  0A?`P X  0A? X  0A?, `D X  0A ?  H  0޽h ? 6xsgţ;z80___PPT10.%w  K0 pW (    BQ lxi  Q X  0A"?q`X  0A#?qX  0A$?qX  0A%? ` X   0A&? 0 $H  0޽h ? 6xsgţ;z80___PPT10.F  K0 ``.(  ` ` BQ lxi  Q X ` 0A?H ` 0޽h ? 6xsgţ;z80___PPT10.C0s  K0 dy(  d d 6Q xi gLogic Gates - 2  Z0l d 6A?"`H d 0޽h ? 6xsgţ;z80___PPT10.CP`  K0  w(  r  S $Qlxi  Q X  0A?b X  0A?p  RB  s *D p  0Y@ ` DThe gate input 2RB  s *D p   0(Y@ ` EThe gate output 2H  0޽h ? 6xsgţ;z80___PPT10.g;   K0 l(    B<Y lxi  Y   B,Y lx Y H  0޽h ? 6xsgţ;z80___PPT10.ـB  K0 pX*(  r  S  Ylxi  Y x  c $Yl|xo  Y H  0޽h ? 6xsgţ;z80___PPT10.0[s  K0 z(  r  S Ylxi  Y `  c $A: ??. H  0޽h ? 6xsgţ;zy___PPT10Y+D=' = @B + 0 `W(  ^  S  *     c $T  @   M  H  0޽h/. ? ̙33 0 @-X4(  Xd X c $ *   Y X s *QY  @  Y  H X 0޽h/. ? ̙33:xY twsƐo5TmT͡戨BJZ+x55Gyը!R95-wjk[9᜛ݻ [YS suטၻ0=}baO- J32(,xB ("(*(&(. 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WSu3* [4 Z%}XsjxqX)X^-bXq"X/6MJ;"V~!lJGj6 -Ӵ90  *H :>{թy=q#+(ߌe,ZbVkbEƢ5"r-| CMlX 5ʉp*}a qfbB="2"d$006C!FXF%`0  *H 010U VeriSign Trust Network10U VeriSign, Inc.1,0*U #VeriSign Time Stamping Service Root1402U +NO LIABILITY ACCEPTED, (c)97 VeriSign, Inc.0 970512070000Z 991231070000Z01'0%U VeriSign Time Stamping Service10U VeriSign Trust Network1402U +NO LIABILITY ACCEPTED, (c)97 VeriSign, Inc.10U VeriSign, Inc.10UInternet00  *H 0aGܴ^i b1FI:?xZcEO n06- o' Q`5OTOз,B Ѝ^㖴"IBlӘV.:X"e匫xGBh[-}0  *H m`_i{p+b#D Q#tI'\q5+ൖ苧KsH ڋ %VNGWF8V!&;-%2Λ@GmGv0;~*1+ݰUApF-,~00U ?d sE40  *H 0a10UInternet10U VeriSign, Inc.1301U *VeriSign Commercial Software Publishers CA0 990319000000Z 000416235959Z0]10UInternet10U VeriSign, Inc.1301U *VeriSign Commercial Software Publishers CA1F0DU =www.verisign.com/repository/RPA Incorp. by Ref.,LIAB.LTD(c)981>0<U 5Digital ID Class 3 - Microsoft Software Validation v21 0 UUS10U Washington10URedmond10UMicrosoft Corporation10U Microsoft Corporation00  *H 0cGxVVVZP0)to[) G~ЧTGͺt+ԅ"1Y 7[D?\ޭ@`'ZWJs8v69 +T<1 "eGB̎g?lM"rJޒwo0k0 U00 U0U0{Ch8n; c0a10UInternet10U VeriSign, Inc.1301U *VeriSign Commercial Software Publishers CAǏ7ے(<g0!U000  +70 U 0@06 +7 #0)'https://www.verisign.com/repository/CPSThis certificate incorporates by reference, and its use is strictly subject to, the VeriSign Certification Practice Statement (CPS) version 1.0, available in the VeriSign repository at: https://www.verisign.com; by E-mail at CPS-requests@verisign.com; or by mail at VeriSign, Inc., 2593 Coast Ave., Mountain View, CA 94043 USA Copyright (c)1996 VeriSign, Inc. All Rights Reserved. CERTAIN WARRANTIES DISCLAIMED AND LIABILITY LIMITED. WARNING: THE USE OF THIS CERTIFICATE IS STRICTLY SUBJECT TO THE VERISIGN CERTIFICATION PRACTICE STATEMENT. THE ISSUING AUTHORITY DISCLAIMS CERTAIN IMPLIED AND EXPRESS WARRANTIES, INCLUDING WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE, AND WILL NOT BE LIABLE FOR CONSEQUENTIAL, PUNITIVE, AND CERTAIN OTHER DAMAGES. SEE THE CPS FOR DETAILS. Contents of the VeriSign registered nonverifiedSubjectAttributes extension value shall not be considered as accurate information validated by the IA. 64https://www.verisign.com/repository/verisignlogo.gif06U/0-0+)'%http://status.verisign.com/class1.crl0 +700  *H UטTJ0̪qJp\8.FRBmra4fUUм[M?lQ^7ߞ@p<< *.om욛8lKOXؗQ^!ܬzX1d0`0u0a10UInternet10U VeriSign, Inc.1301U *VeriSign Commercial Software Publishers CAU ?d sE40 *H p0 +7 100 *H  1  +70 +7 10  +70 *H  1]0Nɚzq0  *H E]>3 W#;_ةÔWt:v[i~y973* sgV҃?Dʥ1hf_]~ņ[:9 9o#[xo#ή0 *H  100010U VeriSign Trust Network10U VeriSign, Inc.1,0*U #VeriSign Time Stamping Service Root1402U +NO LIABILITY ACCEPTED, (c)97 VeriSign, Inc.C!FXF%`0 *H Y0 *H  1  *H 0 *H  1 991028195229Z0 *H  1:mW0  *H n5jLRI:E{0<U 5Digital ID Class 3 - Microsoft Software Validation v21 0 UUS10U Washington10URedmond10UMicrosoft Corporation10U Microsoft Corporation00  *H 0cGxVVVZP0)to[) G~ЧTGͺt+ԅ"1Y 7[D?\ޭ@`'ZWJs8v69 +T<1 "eGB̎g?lM"rJޒwo0k0 U00 U0U0{Ch8n; c0a10UInternet10U VeriSign, Inc.1301U *VeriSign Commercial Software Publishers CAǏ7ے(<g0!U000  +70 U 0@06 +7 #0)'https://www.verisign.com/repository/CPSThis certificate incorporates by reference, and its use is strictly subject to, the VeriSign Certification Practice Statement (CPS) version 1.0, available in the VeriSign repository at: https://www.verisign.com; by E-mail at CPS-requests@verisign.com; or by mail at VeriSign, Inc., 2593 Coast Ave., Mountain View, CA 94043 USA Copyright (c)1996 VeriSign, Inc. 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ȦHb6 puA6EP4$@7) @@M!@@]AM  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~5Root EntrydO) [@Pictures9Current User,SummaryInformation(`UPowerPoint Document(&_DocumentSummaryInformation8H%Equation.30,Microsoft Equation 3.0&Equation Equation.30,Microsoft Equation 3.05.Equation Equation.30,Microsoft Equation 3.03/Equation Equation.30,Microsoft Equation 3.0&9Equation Equation.30,Microsoft Equation 3.0):Equation Equation.30,Microsoft Equation 3.0F/ 0|DTimes New Roman0z[ 0DArialNew Roman0z[ 0" DWingdingsRoman0z[ 00DArial Narrowan0z[ 0"@DMonotype Sorts0z[ 0 A .  @n?" dd@  @@``   2/T       '   !" l        /?'   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Digital Systems&They manipulate discrete information (A finite number of elements) Example discrete sets 10 decimal digits, the 26 letters of alphabet Information is represented in binary form Examples Digital telephones, digital television, and digital cameras The most commonly used one is DIGITAL COMPUTERSbC/3lC/3  lDigital Computers  Binary SignalsIt means two-states 1 and 0 true and false on and off A single  on/off ,  true/false ,  1/0 is called a bit Example: Toggle switchH"N"3'Byte\Computer memory is organized into groups of eight bits Each eight bit group is called a byte]XWhy Computers Use BinaryThey can be represented with a transistor that is relatively easy to fabricate (in silicon) Millions of them can be put in a tiny chip Unambiguous signal (Either 1 or 0) This provides noise immunityL\+#\+# Analog Signal  Binary SignalDA voltage below the threshold off A voltage above the threshold onr   Binary Signal  Noise on TransmissionIWhen the signal is transferred it will pick up noise from the environment!RecoveryNEven when the noise is present the binary values are transmitted without error$Binary NumbersMA number in a base-r system x = xn-1xn-2 ... x1x0 . x-1 x-2 ... X-(m-1) x-m2#J(Radix Number System Base  2 (binary numbers) 0 1 Base  8 (octal numbers) 0 1 2 3 4 5 6 7 Base  16 (hexadecimal numbers) 0 1 2 3 4 5 6 7 8 9 A B C D E Ft    )Radix OperationsThe same as for decimal numbers+$Conversion From one radix to anotherFrom decimal to binary-$Conversion From one radix to anotherFrom decimal to base-r Separate the number into an integer part and a fraction part For the integer part Divide the number and all successive quotients by r Accumulate the remaindersXROR20Different Bases *$Conversion From one radix to anotherFrom binary to octal Divide into groups of 3 bits Example 11001101001000.1011011 = 31510.554 From octal to binary Replace each octal digit with three bits Example 75643.5704 = 111101110100011.101111000100v%#1*%#  1*,$Conversion From one radix to anotherFrom binary to hexadecimal Divide into groups of 4 bits Example 11001101001000.1011011 = 3348.B6 From hexadecimal to binary Replace each digit with four bits bits Example 7BA3.BC4 = 111101110100011.101111000100v%!/(%!  /(; ComplementsThey are used to simplify the subtraction operation Two types (for each base-r system) Diminishing radix complement (r-1 complement) Radix complement (r complement) 8WOH O<09 s and 10 s Complementsl9 s complement of 674653 999999-674653 = 325346 9 s complement of 023421 999999-023421 = 976578 10 s complement of 674653 325346+1 = 325347 10 s complement of 023421 976578+1=976579   =.1 s and 2 s Complements1 s complement of 10111001 11111111  10111001 = 01000110 Simply replace 1 s and 0 s 1 s complement of 10100010 01011101 2 s complement of 10111001 01000110 + 1 = 01000111 Add 1 to 1 s complement 2 s complement of 10100010 01011101 + 1 = 01011110 : 0: 0  B)Subtraction with Complements of UnsignedM  N Add M to r s complement of N Sum = M+(rn  N) = M  N+ rn If M > N, Sum will have an end carry rn , discard it If M<N, Sum will not have an end carry and Sum = rn  (N  M) (r s complement of N  M) So M  N =  (r s complement of Sum)`RHH H      H J  '? $>)Subtraction with Complements of Unsigned 65438 - 5623?)Subtraction with Complements of Unsigned 5623 - 65438@)Subtraction with Complements of Unsigned10110010 - 10011111A)Subtraction with Complements of Unsigned10011111 -10110010CSigned Binary NumbersRUnsigned representation can be used for positive integers How about negative integers? Everything must be represented in binary numbers Computers cannot use  or + signs&WSWS1Negative Binary NumbersThree different systems have been used Signed magnitude One s complement Two s complement&'3'32Signed MagnitudeThe leftmost bit is the sign bit (0 is + and 1 is - ) and the remaining bits hold the absolute magnitude of the number Examples -47 = 1 0 1 0 1 1 1 1 47 = 0 0 1 0 1 1 1 1 - f f3 One s complementwReplace each 1 by 0 and each 0 by 1 Example (-6) First represent 6 in binary format (00000110) Then replace (11111001)L2F2$ 4 Two s complementFind one s complement Add 1 Example (-6) First represent 6 in binary format (00000110) One s complement (11111001) Two s complement (11111010) n)g)$DArithmetic AdditionJUsually represented by 2 s complementQ RegistersZThey can hold a groups of binary data Data can be transferred from one register to anotherRProcessor-Memory Registers* S Operations 7Logic Gates - 1 8  ERanges OStudy ProblemsCourse Book Chapter  1 Problems 1  2 1  7 1  8 1  20 1  34 1  35 1  36&!.!.PSneak Preview Next time ASSIGNMENT Will be given QUIZ& & . Expect a question from each one of the following Convert decimal to any base Convert between binary, octal, and hexadecimal Binary add, subtract, and multiply Negative numbers v 1 1   Questions /86+  K0 "' (  l `p`   `p` ,$D 0n  0"`@`p`    0}` SCENTRAL PROCESSING UNIT ( 2ffr  S {lxi     B0\"`p,$D 0 B Control Unit     B,\"`p p,$D 0 SArithmetic Logic Unit (ALU) l @Pp $@Pp,$D 0  B\"`@Pp B Registers      <"`0` :R1    <"`0  :R2    <"`0@ TRn XB   0DԔ l ` ` '` `,$D 0@ ` ` %` ` @ ` ` #` `n   0"`` `n  0"`` `n  0"`P ` @`n  0"`` `  0X  G Main Memory ( 2 ff  00 P 0 p @Disk ( 2ff  0 P pp DKeyboard ( 2 ff  0\0P 0p CPrinter ( 2ffXB  0D>`  dB  <D>  XB  0D> ` XB  0D> ` XB  0D> ZZ` XB  0D> ` l  0 & 0,$D 0f  6Z 0   0`p C I/O Devices 2   0l  ,$ 0 9Bus 2H  0޽h ? 6xsgţ;z"___PPT10.#Pu`+dD&' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<* %(D' =-6B'blinds(horizontal)*<3<* D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*$%(D' =-6B'blinds(horizontal)*<3<*$D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+#.<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*'%(D' =-6B'blinds(horizontal)*<3<*'D' =%(D' =%(DG' =4@BBB B%(D' =1:Bvisible*o3>+#.<*&%(D' =-6B'blinds(horizontal)*<3<*&++0+ ++0+ ++0+ +<  K0 @O$(  r  S 'lx   r  S (lx3   H  0޽h ? 6xsgţ;z80___PPT10.d0h.  K0 ~|(  |r | S 5lx   x | c $T6lNx   | 0,$   65438 10 s complement of 05623 +94377 159815 Discard end carry 105 -100000 Answer 59815f 2]PI6H | 0޽h ? 6xsgţ;z80___PPT10.`s  K0 dy(  d d 6 xi gLogic Gates - 2  Z0l d 6A?"`H d 0޽h ? 6xsgţ;z80___PPT10.CP`  K0 l(    B lxi     B lx   H  0޽h ? 6xsgţ;z80___PPT10.ـr0G8,z>tBrO|? QSM(    l*  @ l -.fSiH7PodS$@lu1JCx "d T @   @ [@]̦Ґn . 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Digital Systems&They manipulate discrete information (A finite number of elements) Example discrete sets 10 decimal digits, the 26 letters of alphabet Information is represented in binary form Examples Digital telephones, digital television, and digital cameras The most commonly used one is DIGITAL COMPUTERSbC/3lC/3  lDigital Computers  Binary SignalsIt means two-states 1 and 0 true and false on and off A single  on/off ,  true/false ,  1/0 is called a bit Example: Toggle switchH"N"3'Byte\Computer memory is organized into groups of eight bits Each eight bit group is called a byte]XWhy Computers Use BinaryThey can be represented with a transistor that is relatively easy to fabricate (in silicon) Millions of them can be put in a tiny chip Unambiguous signal (Either 1 or 0) This provides noise immunityL\+#\+# Analog Signal  Binary SignalDA voltage below the threshold off A voltage above the threshold onr   Binary Signal  Noise on TransmissionIWhen the signal is transferred it will pick up noise from the environment!RecoveryNEven when the noise is present the binary values are transmitted without error$Binary NumbersMA number in a base-r system x = xn-1xn-2 ... x1x0 . x-1 x-2 ... X-(m-1) x-m2#J(Radix Number System Base  2 (binary numbers) 0 1 Base  8 (octal numbers) 0 1 2 3 4 5 6 7 Base  16 (hexadecimal numbers) 0 1 2 3 4 5 6 7 8 9 A B C D E Ft    )Radix OperationsThe same as for decimal numbers+$Conversion From one radix to anotherFrom decimal to binary-$Conversion From one radix to anotherFrom decimal to base-r Separate the number into an integer part and a fraction part For the integer part Divide the number and all successive quotients by r Accumulate the remaindersXROR20Different Bases *$Conversion From one radix to anotherFrom binary to octal Divide into groups of 3 bits Example 11001101001000.1011011 = 31510.554 From octal to binary Replace each octal digit with three bits Example 75643.5704 = 111101110100011.101111000100v%#1*%#  1*,$Conversion From one radix to anotherFrom binary to hexadecimal Divide into groups of 4 bits Example 11001101001000.1011011 = 3348.B6 From hexadecimal to binary Replace each digit with four bits bits Example 7BA3.BC4 = 111101110100011.101111000100v%!/(%!  /(; ComplementsThey are used to simplify the subtraction operation Two types (for each base-r system) Diminishing radix complement (r-1 complement) Radix complement (r complement) 8WOH O<09 s and 10 s Complementsl9 s complement of 674653 999999-674653 = 325346 9 s complement of 023421 999999-023421 = 976578 10 s complement of 674653 325346+1 = 325347 10 s complement of 023421 976578+1=976579   =.1 s and 2 s Complements1 s complement of 10111001 11111111  10111001 = 01000110 Simply replace 1 s and 0 s 1 s complement of 10100010 01011101 2 s complement of 10111001 01000110 + 1 = 01000111 Add 1 to 1 s complement 2 s complement of 10100010 01011101 + 1 = 01011110 : 0: 0  B)Subtraction with Complements of UnsignedM  N Add M to r s complement of N Sum = M+(rn  N) = M  N+ rn If M > N, Sum will have an end carry rn , discard it If M<N, Sum will not have an end carry and Sum = rn  (N  M) (r s complement of N  M) So M  N =  (r s complement of Sum)`RHH H      H J  '? $>)Subtraction with Complements of Unsigned 65438 - 5623?)Subtraction with Complements of Unsigned 5623 - 65438@)Subtraction with Complements of Unsigned10110010 - 10011111A)Subtraction with Complements of Unsigned10011111 -10110010CSigned Binary NumbersRUnsigned representation can be used for positive integers How about negative integers? Everything must be represented in binary numbers Computers cannot use  or + signs&WSWS1Negative Binary NumbersThree different systems have been used Signed magnitude One s complement Two s complement&'3'32Signed MagnitudeThe leftmost bit is the sign bit (0 is + and 1 is - ) and the remaining bits hold the absolute magnitude of the number Examples -47 = 1 0 1 0 1 1 1 1 47 = 0 0 1 0 1 1 1 1 - f f3 One s complementwReplace each 1 by 0 and each 0 by 1 Example (-6) First represent 6 in binary format (00000110) Then replace (11111001)L2F2$ 4 Two s complementFind one s complement Add 1 Example (-6) First represent 6 in binary format (00000110) One s complement (11111001) Two s complement (11111010) n)g)$DArithmetic AdditionJUsually represented by 2 s complementQ RegistersZThey can hold a groups of binary data Data can be transferred from one register to anotherRProcessor-Memory Registers* S Operations 7Logic Gates - 1 8ERanges OStudy ProblemsCourse Book Chapter  1 Problems 1  2 1  7 1  8 1  20 1  34 1  35 1  36&!.!.PSneak Preview Next time ASSIGNMENT Will be given QUIZ& & . 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