Atualização das imagens do cap3 #22

a0f179e7 · Eduardo de Santana Medeiros Alexandre · 084c016c · a0f179e7 · a0f179e7
Commit a0f179e7 authored Oct 31, 2014 by Eduardo de Santana Medeiros Alexandre
--- a/livro/capitulos/cap3-sistema-de-numeracao.asc
+++ b/livro/capitulos/cap3-sistema-de-numeracao.asc
@@ -61,7 +61,7 @@ nesse sistema de numeração.

 [[sistema_egipcio]]
 .Sistema de Numeração Egípcio
-image::images/sistema-de-numeracao/sistema_egipicio.png[scaledwidth="70%"]
+image::images/{cap}/sistema_egipicio.eps[scaledwidth="70%"]


 Este sistema adota o princípio aditivo, ou seja, os símbolos possuem seus 
@@ -80,16 +80,16 @@ números, de 7 a 59, nesse sistema de numeração.

 [[fig_sistema_babilonico]]
 .Sistema de Numeração Babilônico
-image::images/sistema-de-numeracao/sistema_babilonico.png[scaledwidth="70%"]
+image::images/{cap}/sistema_babilonico.eps[scaledwidth="70%"]

 [NOTE]
 ==================
 Agora é com você. Qual seria o valor que cada símbolo babilônico, seguindo 
 os exemplos da <<fig_sistema_babilonico>>?

-image:images/sistema-de-numeracao/um_babilonico.png[] = ?
+image:images/{cap}/um_babilonico.eps[] = ?

-image:images/sistema-de-numeracao/dez_babilonico.png[] = ?
+image:images/{cap}/dez_babilonico.eps[] = ?
 ==================

 ==== Sistema de numeração Romano 
@@ -116,7 +116,7 @@ máximo, três vezes.

 [[sistema_romano]]
 .Sistema de Numeração Romano
-image::images/sistema-de-numeracao/sistema_romano.png[scaledwidth="60%"]
+image::images/{cap}/sistema_romano.eps[scaledwidth="60%"]


 ==== Sistema de numeração Indo-Arábico  
@@ -149,7 +149,7 @@ indo-arábicos, ao longo do tempo.  

 [[fig_sistema_indo_arabico]]
 .Sistema de numeração Indo-Arábico
-image::images/sistema-de-numeracao/sistema_arabico.png[scaledwidth="70%"]
+image::images/{cap}/sistema_arabico.eps[scaledwidth="70%"]

 Observe que, inicialmente, os hindus não utilizavam o zero. A criação de um 
 símbolo para o *nada*, ou seja, o zero, foi uma das grandes invenções dos 
@@ -409,7 +409,7 @@ divisão (que resulta em quociente 0) ocupe a posição de mais alta ordem.

 - Conversão do número 19~10~ para a base 2:

-image::images/sistema-de-numeracao/conversao-fazendo-divisao.png[scaledwidth="70%", width="50%"]
+image::images/{cap}/conversao-fazendo-divisao.eps[scaledwidth="70%", width="50%"]


 Logo temos: 
@@ -428,7 +428,7 @@ ____

 Conversão do número 278~10~ para a base 16:

-image::images/sistema-de-numeracao/278_16.png[scaledwidth="50%", width="50%"]
+image::images/{cap}/278_16.eps[scaledwidth="50%", width="50%"]

 Logo temos: 

@@ -465,7 +465,7 @@ NOTE: Soma-se as posições da direita para esquerda, tal como uma soma decimal.

 Solução:

-image::images/sistema-de-numeracao/figura1.png[scaledwidth="40%"]
+image::images/{cap}/figura1.eps[scaledwidth="40%"]

 A tabuada da subtração aritmética binária:

@@ -483,7 +483,7 @@ Por exemplo: 	11100~2~ – 01010~2~ = ?

 Solução:

-image::images/sistema-de-numeracao/figura2.png[scaledwidth="25%"]
+image::images/{cap}/figura2.eps[scaledwidth="25%"]

 NOTE: Não esqueça, subtrai-se as colunas da direita para a esquerda, tal como 
 uma subtração decimal.
@@ -523,7 +523,7 @@ Efetuar: 101~2~ x 110~2~

 Solução:

-image::images/sistema-de-numeracao/figura5.png[scaledwidth="40%"]
+image::images/{cap}/figura5.eps[scaledwidth="40%"]

 No entanto, a multiplicação em computadores é feita, também, por um 
 artifício: para multiplicar A por n somamos A com A (n-1) vezes. 
@@ -579,7 +579,7 @@ Como o próprio nome indica, a representação *sinal* e *amplitude* é dividida
 para representar o sinal, o bit mais à esquerda: *0* para indicar um valor 
 positivo, *1* para indicar um valor negativo. Já o resto dos bits representam seu valor absoluto.

-image::images/sistema-de-numeracao/sinal_magnitude.png[scaledwidth="40%", width="50%"]
+image::images/{cap}/sinal_magnitude.eps[scaledwidth="40%", width="50%"]


 ==== Complemento de 1
@@ -680,7 +680,7 @@ Uma característica do sistema de complemento de dois é que tanto os números
 com sinal quanto os números sem sinal podem ser somados pelo mesmo circuito. 
 Por exemplo, suponha que você deseja somar os números sem sinal 132~10~ e 14~10~. 

-image::images/sistema-de-numeracao/figura3.png[scaledwidth="60%"]
+image::images/{cap}/figura3.eps[scaledwidth="60%"]

 O microprocessador tem um circuito na ULA (Unidade Lógica e 
 Aritmética) que pode somar números binários sem sinal, quando aparece o 
@@ -693,7 +693,7 @@ resposta é: não sabe. A ULA sempre soma como se as entradas fossem números
 binários sem sinal. Sempre produzirá o resultado correto, mesmo se as 
 entradas forem números em complemento de dois.

-image::images/sistema-de-numeracao/figura4.png[scaledwidth="60%"]
+image::images/{cap}/figura4.eps[scaledwidth="60%"]

 Isto comprova um ponto muito importante. O somador na ULA sempre soma padrões 
 de bits como se eles fossem números binários sem sinal. É a nossa 
@@ -720,7 +720,7 @@ Como exemplo temos a subtração de 69~10~ (Minuendo) por 26~10~ (Subtraendo).

 Jogue fora o transporte final:

-image::images/sistema-de-numeracao/transporte-final.png[scaledwidth="100%"]
+image::images/{cap}/transporte-final.eps[scaledwidth="100%"]

 [NOTE]
 ====
@@ -747,7 +747,7 @@ decrementada em 1 a cada casa afastada do ponto.
 Exemplo:

 .Representação de ponto fixo
-image::images/sistema-de-numeracao/ponto-fixo.png[]
+image::images/{cap}/ponto-fixo.eps[]

 ==== Soma e subtração de números fracionários

@@ -757,7 +757,7 @@ subtração binária demonstrada anteriormente.

 Exemplos:

-image::images/sistema-de-numeracao/figura6.png[scaledwidth="50%"]
+image::images/{cap}/figura6.eps[scaledwidth="50%"]

 === Fundamentos da Notação de Ponto Flutuante

@@ -862,7 +862,7 @@ e o campo da mantissa, como mostrado na <<fig_sinal_expoente_mantissa>>.

 [[fig_sinal_expoente_mantissa]]
 .Divisão de 1 byte nos campos da Notação de Ponto Flutuante
-image::images/sistema-de-numeracao/sinal_expoente_mantissa.png[scaledwidth="60%"]
+image::images/{cap}/sinal_expoente_mantissa.eps[scaledwidth="60%"]



@@ -927,7 +927,7 @@ os bits de sinal forem iguais a ‘0’, o maior dos dois valores é aquele que
 apresentar, da esquerda para a direita, um ‘1’ no bit em que os padrões 
 diferem, logo ao comparar:
 
-image::images/sistema-de-numeracao/figura7.png[scaledwidth="40%"]
+image::images/{cap}/figura7.eps[scaledwidth="40%"]

 Conclui-se que o primeiro padrão é maior que o segundo, sem haver a 
 necessidade de decodificar as representações em Ponto Flutuante, tarefa que 
@@ -957,7 +957,7 @@ parcela: 0,125) se perde, como pode ser observado na figura abaixo:

 Logo temos:

-image::images/sistema-de-numeracao/bit_perdido.png[]
+image::images/{cap}/bit_perdido.eps[]

 Ao ignorarmos este problema, e continuarmos a preencher o campo do expoente e 
 do bit do sinal, teremos o padrão de bits 01101010, que representa o valor 2,5 
@@ -1006,7 +1006,7 @@ Processo:

 Equalizando os expoentes, temos:

-image::images/sistema-de-numeracao/figura8.png[]
+image::images/{cap}/figura8.eps[]

 Normalizando:


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