英语翻译TheblockdiagramofasystemisshowninFig.1.Afrequencysettingwordisappliedtoanaccumulator,whichisclockedatthesystemclockfrequencyof,inthiscase,327MHz.Theoriginofthisfrequencyistherequiredchannelspacing(5-kHz

Question

英语翻译
The block diagram of a system is shown in Fig.1.A frequency
setting word is applied to an accumulator,which is clocked at the
system clock frequency of,in this case,327 MHz.The origin of
this frequency is the required channel spacing (5-kHz basic
increment) times the accumulator length (2’9.Hence,for a
5-kHz increment the exact clock frequency is 327.680 MHz.
Obviously different clock frequencies can be chosen for alternative
increments; a particular example is 3.125-kHz increments
from a 204.800-MHz clock.This example would fit the European
12.5-kHz based channel spacings,while 5 kHz would fit
U.S./Japanese 10-kHz channels.The accumulator could be of
arbitrary length,depending on the channel spacing required; the
example constructed had 16 bits.
To avoid the need for storage of a full 360” in the ROM,the
MSB output of the accumulator is used as a sign inverter,which,
with the LSB’s,forms a digitized triangular number sequence.
The ROM contains all the data for 180”,stored in a cosine
sequence in this instance.ROM size is 1 kbit,i.e.,128x8 bits.
With the reflection about zero implicit in the 180’ storage,this is
equivalent to 256 words of data,i.e.,the storage density is equal
to the word length.The minimum ROM would have stored only
90”,but would have entailed both amplitude and phase inversion
of data.The system used here was chosen partly for simplicity,
and partly because a ROM cell of this size already existed in the
library.

Answer 1

框图系统显示的无花果.1.频率设定字是适用于一个蓄电池,这是59分制时钟频率,在这种情况下,327兆赫.起源此频率是需要的频道(5千赫基本增量)倍累加器长度(2'9.因此,为5-千赫增量的精确的时钟频率为327.680兆赫.显然不同时钟频率可以选择替代 加薪; 一 特别的例子是3.125-千赫递增,从204.800-兆赫时钟.这个例子,也符合欧洲12.5千赫的频道间距,wh ile5千赫将配合美国/日本的10千赫频道.蓄电池可任意长度,de 结案频道间距的要求; th e如兴建了16位.为了避免不必要的存储全部360"的光盘,th emsb输出蓄电池是作为一个标志逆变器,它同lsb的,有效值数字三角序号.ROM上包含所有数据为180",圣 ored在一个余弦序列,在这种情况下.ROM大小为1千比特,即12 8x8位元.随着思考为零隐含在180'存储th 就是等于256个字的数据,即th 电子存储密度等于字长.最低光碟会存放只有90",卜 t将给双方振幅和位相反转的数据.该系统用在这里,选择部分作简单 部分因为有ROMcell这种规模已经存在于图书馆.