For the second time in as many weeks, a Buccaneer has been voted as the NFL's Rookie of the Week. After a standout performance against the Texans in Week 3, linebacker Kwon Alexander has been selected by the fans as the league's top-performing rookie. Last week, quarterback Jameis Winston was named Rookie of the Week. Alexander is the first defensive player to win the award this season.
"Kwon has done a great job for us in the middle of our defense this year and we are excited to see him getting recognized for his level of play," said Buccaneers Head Coach Lovie Smith. "Defensively, we focus on making big plays and forcing takeaways, and that is exactly what Kwon did for us last week in Houston. He is getting more comfortable in his role each week and we look forward to seeing him continue to develop."
Against Houston, Alexander recorded 10 tackles, intercepted a pass and deflected two more passes. Earlier in the week, the rookie said that the interception was the first he's ever recorded at any level. According to Pro Football Focus, Alexander was the league's third-best inside linebacker in pass coverage in Week 3. He's currently the Bucs' second-leading tackler, behind Lavonte David, with 23 tackles.
Alexander outlasted Bills running back Karlos Williams, Eagles linebacker Jordan Hicks, Raiders wide receiver Amari Cooper and Seahawks wide receiver Tyler Lockett to win the award. Cooper had been nominated for the award for two weeks in a row.
