relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application preliminary knowleoblem of key is: according to meaniphing, (1) determine standard volume (units \n in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of comnt and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry prdge (2)--plane gracs review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and arehea combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relatimon units of measuremeonship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和第3章习题参考答案

1、设有一个具有20位地址和32位字长的存储器，问 (1) 该存储器能存储多少字节的信息？

(2) 如果存储器由512K×8位SRAM芯片组成，需要多少片？ (3) 需要多少位地址作芯片选择？ 解：

32?4M字节 (1) 该存储器能存储：220?8220?32220?32?19?8片 (2) 需要

512K?82?8(3) 用512K?8位的芯片构成字长为32位的存储器，则需要每4片为一组进行字

长的位数扩展，然后再由2组进行存储器容量的扩展。所以只需一位最高位地址进行芯片选择。

2、已知某64位机主存采用半导体存储器，其地址码为26位，若使用4M×8位的DRAM芯片组成该机所允许的最大主存空间，并选用内存条结构形式，问； (1) 若每个内存条为16M×64位，共需几个内存条? (2) 每个内存条内共有多少DRAM芯片?

(3) 主存共需多少DRAM芯片? CPU如何选择各内存条? 解：

226?64?4条内存条 (1) 共需

16M?64(2) 每个内存条内共有

16M?64?32个芯片

4M?8226?6464M?64??128个RAM芯片，(3) 主存共需多少 共有4个内存条，故

4M?84M?8CPU选择内存条用最高两位地址A24和A25通过2：4译码器实现；其余的24根

地址线用于内存条内部单元的选择。

3、用16K×8位的DRAM芯片构成64K×32位存储器，要求： (1) 画出该存储器的组成逻辑框图。

(2) 设存储器读/写周期为0.5μS，CPU在1μS内至少要访问一次。试问采用哪种刷新方式比较合理?两次刷新的最大时间间隔是多少?对全部存储单元刷新一遍所需的实际刷新时间是多少? 解：

(1) 用16K×8位的DRAM芯片构成64K×32位存储器，需要用64K?32?4?4?16个芯片，其中每4片为一组构成16K×32位——进行字长位

16K?8数扩展(一组内的4个芯片只有数据信号线不互连——分别接D0?D7、D8?D15、

relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application preliminary knowleoblem of key is: according to meaniphing, (1) determine standard volume (units \orresponds to\n in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of comnt and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry prdge (2)--plane gracs review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and arehea combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relatimon units of measuremeonship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和D16?D23和D24?D31，其余同名引脚互连)，需要低14位地址(A0?A13)作为模块内各个芯片的内部单元地址——分成行、列地址两次由A0?A6引脚输入；然后再由4组进行存储器容量扩展，用高两位地址A14、A15通过2：4译码器实现4组中选择一组。画出逻辑框图如下。

A0?A13 A0?A6 A0?A6 (5) (6) (7) (8) WE RAS D0?7 D8?15 D16?23 D24?31 A0?A6 (9) (10) (11) (12) WE RAS D0?7 D8?15 D16?23 D24?31 A0?A6 (13) (14) (15) (16) WE RAS D0?7 D8?15 D16?23 D24?31 CPU RAS (1) (2) (3) (4) D0?7 D8?15 D16?23 D24?31 WE D0?D31 A14 A15 WE 2-4 译码 RAS0 RAS1 RAS2 RAS3

(2) 设刷新周期为2ms，并设16K?8位的DRAM结构是128?128?8存储阵列，则对所有单元全部刷新一遍需要128次(每次刷新一行，共128行)

若采用集中式刷新，则每2ms中的最后128?0.5?s=64?s为集中刷新时间，不能进行正常读写，即存在64?s的死时间

若采用分散式刷新，则每1?s只能访问一次主存，而题目要求CPU在1μS内至少要访问一次，也就是说访问主存的时间间隔越短越好，故此方法也不是最适合的

比较适合采用异步式刷新：

2ms?15.625?s，可取采用异步刷新方式，则两次刷新操作的最大时间间隔为12815.5?s；对全部存储单元刷新一遍所需的实际刷新时间为：15.5?s?128=1.984ms；采用这种方式，每15.5?s中有0.5?s用于刷新，其余的时间用于访存(大部分时间中1?s可以访问两次内存)。

4、有一个1024K×32位的存储器，由128K×8位的DRAM芯片构成。问： (1) 总共需要多少DRAM芯片? (2) 设计此存储体组成框图。

(3) 采用异步刷新方式，如单元刷新间隔不超过8ms，则刷新信号周期是多少? 解：

1024K?32?8?4?32片，每4片为一组，共需8组 (1) 需要

128K?8relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \ind associate \hen in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和第3章习题参考答案

(2) 设计此存储体组成框图如下所示。

A0?A16 A0?A8 RAS D0?7 D8?15 D16?23 D24?31 D0?7 D8?15 D16?23 D24?31 D0?7 D8?15 D16?23 D24?31 D0?7 D8?15 D16?23 D24?31 CPU (1) (2) (3) (4) (5) (6) (7) (8) RAS1 WE (9) (10) (11) (12) RAS2 WE (13) (14) (15) (16) RAS3 WE WE D0?D31 RAS0 WE (17) D0?7 D8?15 D16?23 D24?31 (21) (22) (23) (24) RAS5 D0?7 D8?15 D16?23 D24?31 (25) (26) (27) (28) RAS6 D0?7 D8?15 D16?23 D24?31 (29) (30) (31) (32) RAS7 D0?7 D8?15 D16?23 D24?31 CPU (18) (19) (20) RAS4 WE WE A0?A16 A17 A18 A19 3-8 译码 RAS0 RAS1 RAS2 RAS3 RAS4 RAS5 RAS6 RAS7

(3) 设该128K?8位的DRAM芯片的存储阵列为512?256?8结构，则如果选择一个行地址进行刷新，刷新地址为A0?A8，那么该行上的2048个存储元同时进行刷新，要求单元刷新间隔不超过8ms，即要在8ms内进行512次刷新操作。采用

8ms?15.625?s进行一次，异步刷新方式时需要每隔可取刷新信号周期为15.5?s。 512 5、要求用256K×l6位SRAM芯片设计1024K×32位的存储器。SRAM芯片有两个控制端：当CS有效时，该片选中。当W/R＝1时执行读操作，当W/R=0时执行写操作。

解： 1024K?32?4?2?8片，共需8片，分为4组，每组2片

256K?16即所设计的存储器单元数为1M，字长为32，故地址长度为20位（A19~A0），所用芯片存储单元数为256K，字长为16位，故占用的地址长度为18位（A17~A0）。由此可用字长位数扩展与字单元数扩展相结合的方法组成组成整个存储器

字长位数扩展：同一组中2个芯片的数据线，一个与数据总线的D15~D0相连，

3