对一个最高频率为3kHz的限带连续时间信号抽样构成一个离散时间信号，为了保证从此离散信号中能重建原信号，理论上每秒钟的最少抽样数为多少？如果想用此离散信号重建原信号，则所用理想低通滤波器的截止频率应该是多少？如果所用的是非理想低通滤波器，且过渡带的宽度与通带宽度相同，则每秒的最少抽样数为多少？

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∻Solution by Steps∻ ‖ step 1 ⋮ The Nyquist-Shannon sampling theorem states that to reconstruct a continuous-time signal without aliasing, the sampling frequency must be at least twice the highest frequency of the signal. Given the highest frequency of the signal is 3 kHz, the minimum sampling frequency is: $f_s = 2 imes 3 ext{ kHz} = 6 ext{ kHz}$. ‖ ‖ step 2 ⋮ Therefore, the minimum number of samples per second is: $N = f_s = 6 ext{ kHz} = 6000 ext{ samples/second}$. ‖ ‖ step 3 ⋮ For the ideal low-pass filter used to reconstruct the original signal, the cutoff frequency should be equal to the highest frequency of the original signal, which is: $f_c = 3 ext{ kHz}$. ‖ ‖ step 4 ⋮ If a non-ideal low-pass filter is used, and the transition band width is equal to the passband width, the minimum sampling rate can be calculated as: $N = 2 imes (3 ext{ kHz} + \Delta f)$, where $\Delta f$ is the width of the transition band. Assuming $\Delta f = 3 ext{ kHz}$, we have: $N = 2 imes (3 ext{ kHz} + 3 ext{ kHz}) = 12 ext{ kHz} = 12000 ext{ samples/second}$. ‖ ∻Answer∻ ⚹ Minimum sampling rate: 6000 samples/second; Cutoff frequency: 3 kHz; Non-ideal filter minimum sampling rate: 12000 samples/second. ⚹ ∻Key Concept∻ ⚹ The Nyquist-Shannon sampling theorem is crucial for signal reconstruction. ⚹ ∻Explanation∻ ⚹ The theorem ensures that the sampling rate is sufficient to capture the signal's information without loss, preventing aliasing. ⚹◊What is the Nyquist theorem and how does it relate to the sampling frequency of a continuous-time signal?⍭ Generate me a similar question◊

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