Chapter 10 Computer Peripherals
Chapter 10
Computer Peripherals
10.1(BL1+) Reads and writes can be done in place on disk because the layout of a disk is fixed and unchanging. Therefore, disk data will appear in the same spot each time it is written to disk. The format of a tape is data dependent. Blocks on the tape are of variable length, and locating the exact spot where data is to be written in place can be very difficult, especially because the tape is constantly starting and stopping, and blocks being changed.
10.2. (BL2). Flash memory is non-mechanical and has no motors or other moving parts. This makes flash memory less subject to mechanical shock when the computer is moved (or dropped!). Because there is no moving disk and head that must be moved into position to perform an operation, the flash drive has faster access to the data, especially for READ operations. Compared to flash memory, however, hard disks offer higher capacity, lower cost, and well-developed technology. The characteristics of the physics methods used for the implementation of flash memory also limit device cycle lifetimes, at least at present. Particularly for heavy-usage WRITE operations, the longevity of the hard disk is likely to exceed that of the flash memory. Both hard disk and flash memory offer non-volatility, a major advantage over RAM. Conversely, RAM is cheap and fast, and is easily expanded to meet a user's requirements.
10.3(all BL2-)
a.There are 40 (sectors) 400 (cylinders) 4 (surfaces) = 64,000 blocks on this disk. Thus, each block provides 128 MB/64K = 2 KB of storage. A cluster, therefore, holds 4 2KB = 8 KB.
b.If the disk is rotating at 4800 rpm, the time it takes to make one revolution is 1/4800 minute 60 seconds/minute = 1/80 second. In that time, forty blocks of data, consisting of 2 KB per block is transferred. Therefore the data transfer rate = 80 KB 80 = 6.4 MB/second.
c.The average latency time is the time that the disk takes to make 1/2 revolution. The time for 1/2 revolution = 1/80 sec./2 = approximately 6.67 msec.
10.4a.(BL2) It takes 110 msec. for the disk to make 1/2 revolution, or 220 msec. for a full revolution. Thus, the disk rotates at 1/0.220 revolutions/second or approximately 272 rpm.
b.(BL2) If it takes 220 msec. for a full revolution, and there are 11 sectors on a disk, the time to traverse one sector is 220/11 = 20 msec.
10.5.(BL2) The CAV disk is divided into sectors, and each video frame forms a block in one of those sectors. Since the outer sector blocks are more widely spaced than the inner blocks, it would be possible to place more video frames on the outer tracks by increasing the number of sectors on the outer tracks, and slowing down the disk to provide the correct data transfer rate. Thus, the CLV would have approximately the same number of sectors as the CAV disk on the inner tracks, but many more sectors in the outer tracks. If the number of tracks is the same in each case, the total capacity of the CLV disk will be higher. Since the playing time for each block is the same on either disk (it's defined by the video standard at 30 frames/second), the total playing time of the CLV disk is longer.
10.6(all BL2+)
a.The innermost track has a circumference of d = 3.14 inches. Since the disk is specified at a density of 1630 bytes/inch, the track contains 1630 3.14 = 5120 bytes /10, or 512 bytes per block.
b.The outermost track is of diameter 5 inches, or length five times that of the innermost track. Thus, the outermost track will have 5 10 = 50 blocks.
c.The middle track is of diameter 3 inches. Therefore the capacity of the middle track is approximately 30 blocks, or 15KB. The total capacity of the disk is 15 KB 2000 tracks, or 30 MB.
d.If the transfer rate is 256,000 bytes per second, and the innermost track contains 5120 bytes, the rotation speed is 256,000/5120 = 50 rps, or 3000 rpm. The outermost track contains five times as many bytes, thus, to maintain a steady transfer rate, the speed will decrease to 600 rpm.
10.7 (BL2) Hard disk drives use concentric tracks. The track for a particular block can be easily identified and the head may be rapidly moved to the correct track for read and write operations. CDs and DVDs use a single spiral track. Since the track is spiral, the precise distance of the track from the center of the disk is dependent on the instantaneous angle of the rotating disk. The use of CLV compounds the problem, since the number of blocks per rotation center differs depending on the distance from the center of the disk. Therefore, calculation of the block's distance from the center is only approximate. The result is a considerable amount of search to find the required block.
10.8. (BL2) Since the track is half as wide, and the track length required is also divided by two, the overall capacity is increased by a factor of four, to 18.8 GB.
10.9. (BL2-) There are 40 75 = 3000 characters per page. Each character requires two bytes for its Unicode representation. Therefore, a single page requires 6000 bytes. A 600 MB CD-ROM would hold 600 MB/6 KB = approximately 100,000 pages. An 80 GB flash memory drive would hold 80 GB/6 KB = approximately 13 1/3 million pages.
10.10. (all BL2-)
a. Amount of memory = 640 480 pixels 3 bytes/pixel = 921,600 bytes of video memory.
Amount of memory = 1600 900 pixels 3 bytes/pixel = 4,320,000 bytes of video memory.
Amount of memory = 1400 1080 pixels 3 bytes/pixel = 4,665,600 bytes of video memory.
b. The number of images = 4.7 GB / 921.6 KB, or approximately 5,000 images.
10.11 (all BL2-)
a. Amount of memory = 640 480 pixels x 2 bytes/pixel = 614,400 bytes of video memory.
b. Amount of memory = 640 480 pixels 3 bytes/pixel = 921,600 bytes of video memory.
c. For 30 frames/second video, the required transfer rate is 921,600 30 = approximately 30 MB/second.
10.12(BL2) A 1½" 2" image at 72 dots per inch holds 108 144 = 15552 pixels. At a rate of 30 frames/second, the system must move 466,560 pixels of data every second. At a size of 3" 4", the transfer rate increases to 216 288 30, or 1,866,240 pixels per second.
10.13 (all BL2)
a. Each image requires 640 480 pixels = 307,200 pixels of video. Thirty full images per second (or sixty half images) requires a bandwidth of 9,216,000 pixels/second.
b. Each image requires 1920 1080 pixels = approximately 2 megapixels of video. Sixty full images per second requires a bandwidth of 120 megapixels/second.
10.14 (all BL3) (Note: the resolutions specified indicate that the monitor is "wide screen" 16 9.)
a. The horizontal and vertical sides of a 9 triangle may be calculated from the formula for a right triangle: a2 + b2 = c2. The diagonal for a 9 16 triangle is, therefore, equal to the square root of (81 + 256) = √337 ≈ 18.4. The width of a 14" screen (the vertical ratio will be similar) is
W = (16 x 14) / 18.4 ≈ 13.2"
.
There are 1600 pixels displayed in 13.2". The number of dots/inch = 1600/13.2 ≈ 120 dots/inch.
b. Each pixel requires 1/120 inch for individual pixel display. Converting to metric, 1" = 25.4 mm. or 1/120" = 25.4/120 = 0.21 mm. Therefore a 0.26 mm pixel resolution is nor quite sufficient for this display.
c. For a 1280 720 image, the number of dots/inch = 1280/13.2 ≈ 97 dots/inch. Each pixel requires 25.4/97 ≈ 0.26 mm. The monitor is adequate for this resolution.
10.15(BL2)Twenty-four rows of characters use 480 pixels for display, so each character occupies a block 20 pixels high. Allowing 4 pixels for spacing, the character is 16 pixels high. On a 15" monitor, each pixel requires 15/800", so 16 pixels would generate a text character 0.30" high. On a 600 800 display, 16 pixels would produce a character 15 16/1000 or 0.24" in height. Since each character requires 20 pixels of height including spacing, the screen would display 30 rows.
10.16(BL2)If the gray scale requires a 3 3 matrix, a 600 dot/inch printer is effectively reduced to 200 pixels per inch. In actuality, the gray scale is interpolated, so the resolution is somewhat higher.
10.17(BL1) The primary replaceable in a laser printer is the toner powder. However, the photosensitive drum and corona wire wear out, and must also be replaced from time to time. In an inkjet printer, the cartridges holding the ink must be replaced. In a dot-matrix impact printer, the ribbon must be replaced.
10.18(BL2-)Character mode displays store a standard set of characters that are used to generate the display. One pixel map is stored for each character. The text itself is stored in character code form, typically ASCII or EBCDIC. Thus, storage of the text requires only one byte per character. The image is generated during raster scan. The video generator looks up the appropriate pixel values for each line as it is scanned. This is done repetitively as long as the image is displayed. Character mode display is simple, and requires little storage, however, the display is limited to the character set provided, and to the single font provided by the video card. Such niceties as italics and bold characters are usually not possible.
In graphics mode, each character is stored graphically as part of a pixel image. This makes it possible to present a variety of fonts, character sizes, and the like. The cost is the amount of memory and processing required to store and manipulate graphic images. The image is generated using the usual pixel graphics techniques that are described in the text.
10.19(BL1+) Pixel graphics store an image, one value per pixel. Object graphics store a geometric representation of the image. The number of values depends on the shape of the image. As an example, a circle would require three values: the X and Y coordinates of the circle, and the radius.
In general, the number of values required to store an image in object format is much smaller than the number required for a pixel image. Furthermore, the object image can be rotated, enlarged, or shrunk without loss of resolution. However, the object image must be converted to pixel image form for display. The pixel display will lose resolution if it is altered in size or rotated. (Enlarging a diagonal line represented in pixel form causes “jaggies,” for example.) On the other hand, pixel graphics can be used to display images that can not be represented by a collection of geometric objects, such as photographs or paintings with subtle shadings that vary from pixel to pixel.
10.20(BL1) Formed character printers can only produce the characters that are offered in the formed character set. No graphical images are possible, and there is no flexibility in the fonts. At least, dot-matrix and other types of printers can use the dot matrix to represent graphical images when required. This lack of flexibility has made formed character printers nearly obsolete.