数字信号处理关于奈奎斯特采样频率的题目Let the continuous-time signal is x(t) = 2cos(650πt) – sin(720πt).(b)If x(t) is sampled at the double rate of the Nyquist frequency,what is the frequency of sine wave in the sampled sequenc
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/29 05:52:55
数字信号处理关于奈奎斯特采样频率的题目Letthecontinuous-timesignalisx(t)=2cos(650πt)–sin(720πt).(b)Ifx(t)issampledatthed
数字信号处理关于奈奎斯特采样频率的题目Let the continuous-time signal is x(t) = 2cos(650πt) – sin(720πt).(b)If x(t) is sampled at the double rate of the Nyquist frequency,what is the frequency of sine wave in the sampled sequenc
数字信号处理关于奈奎斯特采样频率的题目
Let the continuous-time signal is x(t) = 2cos(650πt) – sin(720πt).
(b)If x(t) is sampled at the double rate of the Nyquist frequency,what is the frequency of sine wave in the sampled sequence
数字信号处理关于奈奎斯特采样频率的题目Let the continuous-time signal is x(t) = 2cos(650πt) – sin(720πt).(b)If x(t) is sampled at the double rate of the Nyquist frequency,what is the frequency of sine wave in the sampled sequenc
有连续时间信号 x(t) = 2cos(650πt) – sin(720πt).对其抽样,采样频率是奈奎斯特频率的两倍.则输出序列里的正弦波的频率是多少?