The Max bandwidth of LTE system is（）。
第1题：
LTE中分两种随机接入过程，分别是『____』的随机接入过程和『____』的随机接入过程。(There are two kinds of random access procedure in LTE system， respectively based on『____』and『____』random access procedure. )
第2题：
整个LTE OMC网管系统包括两大部分：『____』和『____』。(The LTE OMC network management system consists of two parts:『____』and『____』.)
第3题：
阅读以下说明和Java代码，将应填入(n)处的字句写在对应栏内。
[说明]
编写一个字符界面的Java Application 程序，接受用户输入的10个整数，并输出这10个整数的最大值和最小值。
[Java 代码]
import java. io.* ;
public class abc
{
public static void main ((1))
{int i, n=10, max=0, min=0, temp=0;
try {
BufferedReader br = new BufferedReader (
new InputStreamReader ( System.in ) );
max = min =Integer. parselnt ( br. readLine ( ) );
}. (2) ( IOExccption e ) {} ;
for ( i=2 ; i＜=n ; i++ ) {
try {
BufferedReader br = new (3) (
new InputStreamReader ( System. in ) );
temp =(4) ( br. readLine ( ));
if (temp＞max ) max=temp;
if (temp＜min) (5)
} catch (IOException e ) {};
}
System.out.println ( "max="+max+"\nmin="+min );
}
}
第4题：
第5题：
阅读以下说明和Java码，将应填入(n)处的字名写在对应栏内。
[说明] 编写一个字符界面的Java Application 程序，接受用户输入的10个整数，并输出这10个整数的最大值和最小值。
import java. io. * ;
public class abc
{
public static void main(String args [ ] )
{ int i, n = 10 , max = 0 , min = 0 , temp = 0;
try {
BufferedReader br = new BufferedReader(
new InputStreamReader( System. in) );
(1));
} catch ( IOException e ) { } ;
for(i = 2 ;i ＜= n; i ++ ) {
try {
BufferedReader br = new BufferedReader(
new InputStreamReader (System. in) );
temp = Integer. parselnt(br. readLine( ) );
if ( temp ＞ max ) (2)
if (temp ＜ min) (3)
} catch ( IOExeeption e ) { } ;
System. out. println( "max =" + max + "\nmin =" + min);
}
}
第6题：
在2天线、20MHz带宽的配置下，LTE系统同时调度的最大用户数是『____』。(In the 2 antenna, 20MHz bandwidth configuration, the maximum number of users scheduling is『____』in LTE systems. )
第7题：
Although a given waveform. may contain frequencies over a very broad range，as a practical matter any transmission system will be able to accommodate only a limited band of (71) ．This，in turn，limits the data rate that can be carried on the transmission (72) ．Asquare wave has an infinite number of frequency components and hence an infinite (73) ．However, the peak amplitude of the kth frequency component，kf，is only 1/k，so most of the (74) in this wavefonn is in the first few fiequency components．In general，any digitalwaveform. will have (75) bandwidth．If we attempt to transmit this waveform. as a signalover any medium，the transmission system will limit the bandwidth that can be transmitte
（71）
A．frequencies
B．connections
C．diagrams
D．resources
第8题：
LTE中『____』基于主同步信号和辅同步信号进行的。(What carries on basing on the main synchronized signal and the auxiliary synchronized signal in LTE system『____』.)
第9题：
本题定义了一个求两个数的最大值的方法max，并调用该方法计算67和23的最大值。 public class javal{ public static void main(String[]args){ javal temp=new javal； int res=max(67，23)； System．out．println("res="+res)； } static int maX( ){ int maxNum； if(a>b) ； else maxNum=b； ； } }
第10题：
LTE小区参数bandwidth为n50，则该小区带宽为（）MHz。