The signal f(t)=10sin(4pit)+sin(48pit)+5sin(89pit) is sampled at the frequency of 5HZ. What frequencies are there in the output ?
时间: 2024-05-27 13:09:31 浏览: 18
When a continuous-time signal is sampled, its spectrum is repeated at multiples of the sampling frequency. This repetition is known as aliasing. To determine the frequencies in the output of the sampled signal, we can apply the Nyquist-Shannon sampling theorem, which states that the sampling frequency must be at least twice the maximum frequency component of the signal. In this case, the maximum frequency component is 89 Hz.
Therefore, the Nyquist frequency is 2*89 = 178 Hz, which means that any frequency component above 178 Hz will be aliased. The sampling frequency is 5 Hz, so the signal spectrum will be repeated at multiples of 5 Hz.
The frequency components in the original signal are 4, 48, and 89 Hz. When these are sampled at 5 Hz, the resulting frequencies are:
- 4 Hz will be sampled as 4 Hz (not aliased)
- 48 Hz will be aliased to 2 Hz (48 - 9*5 = 3, which is closer to 0 than to 5, so it will be aliased to 2)
- 89 Hz will be aliased to 1 Hz (89 - 17*5 = 4, which is closer to 0 than to 5, so it will be aliased to 1)
Therefore, the frequencies in the output are 1 Hz, 2 Hz, and 4 Hz.