Commit a0f179e7 authored by Eduardo de Santana Medeiros Alexandre's avatar Eduardo de Santana Medeiros Alexandre
Browse files

Atualização das imagens do cap3 #22

parent 084c016c
......@@ -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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