Talk to Sales

Benchmarks

View scores and output across OCR models spanning many document categories.

Want to run these evals on your own documents?

Talk to Sales
Page 1

浙江大学 2016 - 2017 学年 春夏学期

《电磁场与电磁波》课程期中考试试卷

课程号:11120010,开课学院:信电学院

考试形式:一纸开卷,允许带一张 A4 大小手写稿入场

考试日期:2016 年 4 月 26 日,考试时间:120 分钟 (14:00-16:00)

诚信考试,沉着应考,杜绝违纪。

考生姓名:__________ 学号:__________ 所属专业:__________

题序 总分
得分
评卷人

一、选择题(每题 2 分,共 20 分):

1. 传输线特征阻抗为 Z_0 ,负载阻抗为 R_L ,且 Z_0 eq R_L ,若用特性阻抗为 Z_{01} 的 1/4 波长阻抗变换器进行匹配,则 Z_{01} 应满足条件( C )。

A. Z_{01} = Z_0 R_L B. Z_0 = rac{\sqrt{Z_{01} R_L}}{Z_{01}} C. Z_{01} = rac{\sqrt{Z_0 R_L}}{Z_0} D. Z_{01} = R_L

2. 一圆极化波垂直投射于一理想导体平板上(平板和 z 轴垂直,位于 z=b ),入射电场

\mathbf{E} = E_m (\mathbf{x}_0 + j\mathbf{y}_0) e^{-jkz}

则反射波电场为( C )

A. \mathbf{E} = E_m (\mathbf{x}_0 + j\mathbf{y}_0) e^{-jkz} B. \mathbf{E} = -E_m (\mathbf{x}_0 + j\mathbf{y}_0) e^{jk(z-b)}

C. \mathbf{E} = -E_m (\mathbf{x}_0 + j\mathbf{y}_0) e^{jk(z-2b)} D. \mathbf{E} = -E_m (\mathbf{x}_0 + j\mathbf{y}_0) e^{jkz}

3. 一容性负载经过四分之一阻抗变换器后,在导纳圆图上标注在( B )。

A) 上半圆 B) 下半圆 C) 纯电纳圆 D) 纯电导线

4. 右图所示为传输线上电压的驻波分布,判别负载 Z_L 是什么性质的阻抗?

( B )

A. 纯电阻 B. 电阻、电容都有

C. 纯电抗 D. 电阻、电感都有

Graph showing the voltage standing wave distribution on a transmission line. The load is at z=0. The voltage maximum is 8 and the minimum voltage is 2. A voltage minimum occurs at z=0. The z-axis is labeled, and there is a label -0.1λ near the origin.

图示为传输线上电压的驻波分布。负载位于 $z=0$ 处。最大电压 $V_{max}=8$,最小电压 $V_{min}=2$。电压最小值出现在 $z=0$ 处。

Graph showing the voltage standing wave distribution on a transmission line. The load is at z=0. The voltage maximum is 8 and the minimum voltage is 2. A voltage minimum occurs at z=0. The z-axis is labeled, and there is a label -0.1λ near the origin.
Diagram of a transmission line segment with characteristic impedance Z_c = 50Ω terminated by a load impedance Z_L.

传输线段示意图,特征阻抗 $Z_c = 50\Omega$,负载阻抗 $Z_L$。

Diagram of a transmission line segment with characteristic impedance Z_c = 50Ω terminated by a load impedance Z_L.

5. 已知天线的方向性为 1.25,天线效率为 80%,则天线增益为( C )

A. 0 dB B. 1 dB C. 2 dB D. 3 dB

6. 均匀平面波的电场为 \mathbf{E} = \hat{\mathbf{x}}E_0 \sin(\omega t - kz + \pi/6) + \hat{\mathbf{y}}E_0 \cos(\omega t - kz) ,则表明此波是( C )

~ 1 ~