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
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![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](/uploads/image/z/9939456-0-6.jpg?t=%E6%95%B0%E5%AD%97%E4%BF%A1%E5%8F%B7%E5%A4%84%E7%90%86%E5%85%B3%E4%BA%8E%E5%A5%88%E5%A5%8E%E6%96%AF%E7%89%B9%E9%87%87%E6%A0%B7%E9%A2%91%E7%8E%87%E7%9A%84%E9%A2%98%E7%9B%AELet+the+continuous-time+signal+is+x%28t%29+%3D+2cos%28650%CF%80t%29+%E2%80%93+sin%28720%CF%80t%29.%28b%29If+x%28t%29+is+sampled+at+the+double+rate+of+the+Nyquist+frequency%2Cwhat+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 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?