20 Gb/s PAM-4信号直接调制DBR激光器实现20公里光纤传输

研究论文
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该研究论文探讨了在20公里光纤上传输20 Gb/s四电平脉冲幅度调制（PAM-4）信号的技术，采用了一种直接调制的可调分布式布拉格反射器（DBR）激光器作为关键传输工具。这项工作展示了在长距离通信中的一种创新方法，无需依赖光学放大器和光谱补偿模块，从而降低了系统复杂性和成本。 论文的核心技术是直接调制的DBR激光器，它具有高效率和良好的线性调制性能，能够有效地实现PAM-4信号的高效编码和解码。PAM-4是一种多电平信号编码方式，相比传统的PAM-2，可以提供更高的数据速率，这对于满足未来数据通信网络的需求至关重要。 实验结果表明，通过这种激光器实现的20 Gb/s PAM-4信号能够在20公里的光纤中稳定传输，展现出出色的信号质量。激光器的波长调谐范围达到了惊人的11.5纳米，这意味着它可以在一个宽广的光谱窗口内工作，具有很强的灵活性。更值得一提的是，其3dB带宽超过了10吉赫兹，这确保了在整个波长调谐范围内信号的高速传输。 文章还提到了光学分用技术的无源特性，这减少了额外的信号损失和系统复杂性，对于实现绿色、低成本和高效的光通信网络具有重要意义。此外，论文的作者们还强调了这项工作的潜在应用，包括数据中心间的高速互联、云计算和下一代光纤通信网络的构建。 这篇研究论文为我们提供了关于如何利用直接调制的可调DBR激光器进行长距离、高速率PAM-4信号传输的关键技术细节，为未来的光通信技术发展提供了新的可能性和方向。通过优化激光器的性能和系统设计，有望推动通信行业的进步并促进信息技术的进一步革新。

Decoding of 10-G Optics-Based 50-Gb/s PAM-4 Signal Using Simplified MLSE

2021-02-08 上传
Decoding of 10-G Optics-Based 50-Gb/s PAM-4 Signal Using Simplified MLSE

JSSC 2023.1 all papers

2024-01-28 上传
四、A 100-Gb/s PAM4 Optical Transmitter in a 3-D-Integrated SiPh-CMOS Platform Using Segmented MOSCAP Modulators 本篇论文介绍了一种基于 3D 集成 SiPh-CMOS 平台的高速光学传输器，使用分段 MOSCAP 调制器...

采用45nm SOI CMOS的低噪声高线性度56Gb / s PAM-4光接收器

2021-03-14 上传
PAM-4即4级脉冲幅度调制，是一种高效率的多级调制方案，能够以较低的波特率实现更高的数据传输速率。与传统的非归零(NRZ)编码相比，PAM-4调制方案能够更有效地利用带宽资源，从而在400GbE标准中被多数子标准采用。 ...

逐条注释 ``` if [ -f "/etc/pam.d/password-auth-ac" ];then if grep -q "password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1" /etc/pam.d/password-auth-ac then sed -i '/password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1/s/^#//' /etc/pam.d/password-auth-ac else echo 'password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1' >> /etc/pam.d/password-auth-ac fi if grep -q "password required pam_unix.so remember=5 use_authtok md5 shadow" /etc/pam.d/password-auth-ac then sed -i '/password required pam_unix.so remember=5 use_authtok md5 shadow/s/^#//' /etc/pam.d/password-auth-ac else echo 'password required pam_unix.so remember=5 use_authtok md5 shadow' >> /etc/pam.d/password-auth-ac fi fi ```

2023-07-15 上传
这段脚本的作用是检查/etc/pam.d/password-auth-ac文件是否存在，并对其进行修改或添加配置。具体解释如下： ```shell if [ -f "/etc/pam.d/password-auth-ac" ];then ``` - 检查/etc/pam.d/password-auth-ac文件...

if [ -f "/etc/pam.d/system-auth" ];then if grep -q "password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1" /etc/pam.d/system-auth then sed -i '/password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1/s/^#//' /etc/pam.d/system-auth else echo 'password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1' >> /etc/pam.d/system-auth fi if grep -q "password required pam_unix.so remember=5 use_authtok md5 shadow" /etc/pam.d/system-auth then sed -i '/password required pam_unix.so remember=5 use_authtok md5 shadow/s/^#//' /etc/pam.d/system-auth else echo 'password required pam_unix.so remember=5 use_authtok md5 shadow' >> /etc/pam.d/system-auth fi fi if [ -f "/etc/pam.d/password-auth-ac" ];then if grep -q "password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1" /etc/pam.d/password-auth-ac then sed -i '/password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1/s/^#//' /etc/pam.d/password-auth-ac else echo 'password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1' >> /etc/pam.d/password-auth-ac fi if grep -q "password required pam_unix.so remember=5 use_authtok md5 shadow" /etc/pam.d/password-auth-ac then sed -i '/password required pam_unix.so remember=5 use_authtok md5 shadow/s/^#//' /etc/pam.d/password-auth-ac else echo 'password required pam_unix.so remember=5 use_authtok md5 shadow' >> /etc/pam.d/password-auth-ac fi fi

2023-07-14 上传
subprocess.call("sed -i '/password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1/s/^#//' /etc/pam.d/system-auth", shell=True) else: subprocess....

``` cp /etc/pam.d/su /etc/pam.d/su.bak cp /etc/pam.d/password-auth-ac /etc/pam.d/password-auth-ac.bak cp /etc/pam.d/system-auth /etc/pam.d/system-auth.bak cp /etc/login.defs /etc/login.defs.bak cp /etc/ssh/sshd_config /etc/ssh/sshd_config.bak cp /etc/vsftpd/vsftpd.conf /etc/vsftpd/vsftpd.conf.bak cp /etc/profile /etc/profile.bak cp /etc/motd /etc/motd.bak ```

2023-07-15 上传
1. `cp /etc/pam.d/su /etc/pam.d/su.bak`：备份/etc/pam.d/su文件为/etc/pam.d/su.bak。 2. `cp /etc/pam.d/password-auth-ac /etc/pam.d/password-auth-ac.bak`：备份/etc/pam.d/password-auth-ac文件为/etc/pam.d...

逐条注释 ``` if [ -f "/etc/pam.d/system-auth" ];then if grep -q "password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1" /etc/pam.d/system-auth then sed -i '/password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1/s/^#//' /etc/pam.d/system-auth else echo 'password requisite pam_cracklib.so retry=3 minlen=8 minclass=3 ucredit=-1 lcredit=-1 dcredit=-1 ocredit=-1' >> /etc/pam.d/system-auth fi if grep -q "password required pam_unix.so remember=5 use_authtok md5 shadow" /etc/pam.d/system-auth then sed -i '/password required pam_unix.so remember=5 use_authtok md5 shadow/s/^#//' /etc/pam.d/system-auth else echo 'password required pam_unix.so remember=5 use_authtok md5 shadow' >> /etc/pam.d/system-auth fi fi ```

2023-07-15 上传
这段脚本的作用是检查/etc/pam.d/system-auth文件是否存在，并其进行修改或添加配置。具体解释如下： ```shell if [ -f "/etc/pam.d/system-auth" ];then ``` - 检查/etc/pam.d/system-auth文件是否存在。如果存在...

cat /etc/pam.d/system-auth cat /etc/pam.d/sshd cat /etc/profile cat /etc/passwd 查看密码 cat /etc/shadow cat /etc/group cat /etc/hosts.equiv cat /etc/hosts.rhosts ps aux |grep telnet ps aux |grep rlogin ps aux |grep ftp cat /etc/group cat /etc/ssh/sshd_config umask ls -l /etc/passwd ls -l /etc/group ls -l /etc/shadow ls -l /etc/xinetd.conf ls -l /etc/sudoers ls -l /etc/httpd.conf ls -l /etc/httpd-mpm.conf ls -l /etc/conf/tomcat-users.xml ls -l /etc/conf/web.xml ls -l /etc/conf/server.xml ls -l /etc/my.cnf ls -l /var/mysqllog/logbin.log ps aux | grep syslogd ps aux | grep rsyslogd service syslog status service rsyslog status ps aux | grep audit service auditd status ps aux |grep ssh ls -l /var/log/messages ls -l /var/log/secure ls -l /var/log/audit/audit.log tail -20 /var/log/messages tail -20 /var/log/audit/audit.log tail -n 20 /var/log/messages tail -n 20 /var/log/audit/audit.log cat /etc/logrotate.conf cat /etc/rsyslog.conf auditctl -s auditctl -l lsof -i:21 lsof -i:22 lsof -i:23 这些命令的意思

2023-07-21 上传
1. `cat /etc/pam.d/system-auth`: 显示 `/etc/pam.d/system-auth` 文件的内容，该文件用于管理系统级身份验证和授权。 2. `cat /etc/pam.d/sshd`: 显示 `/etc/pam.d/sshd` 文件的内容，该文件用于管理 SSH 服务的...

linux没有找到/etc/pam.d/common-password

2023-05-25 上传
如果在 Linux 中找不到 `/etc/pam.d/common-password` 文件，可能有以下几种原因： 1. 你的 Linux 发行版中没有使用 PAM（Pluggable Authentication Modules）作为身份验证机制。在这种情况下，你不需要考虑 `/etc/...

逐条注释 ``` if grep -q "auth sufficient pam_rootok.so" /etc/pam.d/su then sed -i '/auth sufficient pam_rootok.so/s/^#//' /etc/pam.d/su else echo 'auth sufficient pam_rootok.so' >> /etc/pam.d/su fi if grep -q "auth required pam_wheel.so group=wheel" /etc/pam.d/su then sed -i '/auth required pam_wheel.so group=wheel/s/^#//' /etc/pam.d/su else echo 'auth required pam_wheel.so group=wheel' >> /etc/pam.d/su fi if grep -q "export TMOUT=300" /etc/profile then sed -i '/export TMOUT=300/s/^#//' /etc/profile else echo 'export TMOUT=300' >> /etc/profile fi source /etc/profile ```

2023-07-15 上传
sed -i '/auth sufficient pam_rootok.so/s/^#//' /etc/pam.d/su else echo 'auth sufficient pam_rootok.so' >> /etc/pam.d/su fi ``` - 检查/etc/pam.d/su文件中是否存在"auth sufficient pam_rootok.so"这一行。 ...
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