正弦脉宽调制SPWM的基本原理是什么？载波比N、电压调制系数M的定义是什么？改变高频载波电压幅值[tex=1.286x1.286]UQwdF6kT8e/AjfeAHx4Z8w==[/tex]和频率[tex=0.857x1.286]OCXYTmnE7P951YKyy3mCKg==[/tex]为什么能改变逆变器交流输出基波电压[tex=1.0x1.286]FQCty1oDljUrciYr/EgSOA==[/tex]的大小和基波频率[tex=0.857x1.286]wIHxADjGmy3B9mbaEXgSZg==[/tex]

设[tex=0.857x1.0]RIwvPDbuSt4HWDoPsJIiNQ==[/tex]为[tex=0.643x1.286]ZsZs11iKEvfmzDIurZth8g==[/tex]维向量空间的线性变换，[tex=1.0x1.286]FQCty1oDljUrciYr/EgSOA==[/tex]，[tex=1.0x1.286]Y//LTBZVoC62W4jnTxjo3w==[/tex]为[tex=0.857x1.286]ZpwhzmyivskaH5M1X7ozaQ==[/tex]两个线性子空间。证明：如果[tex=6.357x1.214]79Wd/JsaQKi3RBB3vwr8304iWeeyN5oM/4rWSyDZPAAsnzTTv6SnNlFoG43nly5YeoJoNf0NIbIKI72lLYoAAg==[/tex]，则存在线性变换[tex=1.214x1.214]6IZIK8PFueAWRBm6DPIGkwocpMVF+Iu3u3WNxpBxLCg=[/tex]，[tex=1.214x1.214]+9zj08zYkjZ1AMUGplS+zHdldHUYVWJGk++xhKOIn3Q=[/tex]，使[tex=5.214x1.214]lRH8ipEtddswal17Qs6h3CWdC//t2tteBdEVebjRDhqvgLejcab2uVirNE7wMymWTEN3CAxIS/2hslTOepiQUQ==[/tex]，[tex=5.214x1.214]PX4zwKuSAlqAfjlv+Mw9oeF5Q81KUvMV/bB1DtuiaZjo8eqWfBj30uXmyTUr4OJ4CjkpuRaRaMp7Zz2r70eJAg==[/tex]，且[tex=4.714x1.214]CmAnDYdFbTdGenyJbSjUKOdmW4VioZL3X58UpLFCnsGRPoH0Y3sMNxG4hLB53iYq[/tex]。

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4

共有4个站进行码分多址通信。4个站的码片序列为：a：（-1 -1 -1 +1 +1 -1 +1 +1） b：（-1 -1 +1 -1 +1 +1 +1 -1） c：（-1 +1 -1 +1 +1 +1 -1 -1） d：（-1 +1 -1 -1 -1 -1 +1 -1） 现收到这样的码片序列：（-1 +1 -3 +1 -1 -3 +1 +1），则（ ）发送1。

共有四个站进行CDMA通信，现收到这样的码片序列（-3 +1 -1 -1 +1 +1 +1 -3），则下列站点发送了数据1的是（ ） A: （-1 -1 -1 +1 +1 -1 +1 +1） B: （-1 -1 +1 -1 +1 +1 +1 -1） C: （-1 +1 -1 +1 +1 +1 -1 -1） D: （-1 +1 -1 -1 -1 -1 +1 -1）

共有四个站进行码分多址CDMA通信。四个站的码片序列为：A(-1 -1 -1 +1 +1 -1 +1 +1) , B(-1 -1 +1 -1 +1 +1 +1 -1) , C(-1 +1 -1 +1 +1 +1 -1 -1), D(-1 +1 -1 -1 -1 -1 +1 -1) , 现在收到这样的码片序列：（-1 -1 +1 +1 +3 +3 -1 -1 ）