01 The eighth assignment

1. Introduction to Exercises

Signal sampling and recovery is a bridge between the continuous time signal and the digital signal world. The sampling theorem requires that the sampling frequency of the spectrum-limited signal be greater than twice the highest frequency of the signal. This sampling frequency is also called the Nyqui of the signal ster frequency. In this way, the continuous-time signal whose highest frequency is less than half the sampling frequency can be uniquely recovered from the sampled data. The sampling spectrum of a single model is relatively easy to determine according to its highest frequency. After the signal is processed, the corresponding highest frequency will change. In the eighth homework, there is an exercise to judge the sampling frequency after the signal operation, which includes the product and convolution between signals. Signal displacement, differentiation, multiplication, scale change, etc. Let us now discuss the solution to this problem.

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2. Problem solving

1. The first sub-question

The first sub-problem, is the product of two signals. The title gives their respective highest frequencies. And assume that omega1 is smaller than omega2. This condition has no effect in the first small question. Their spectrum is distributed within their highest frequency range. According to the Fourier transform frequency domain convolution theorem, the frequency spectrum of the signal product is the convolution of the frequency spectrum of the two signals. The width of the continuous signal convolution is equal to the sum of the widths of the signals involved in the convolution. From this, it can be known that the maximum frequency of the signal product is equal to the sum of the maximum frequencies of the two signals. The corresponding signal sampling frequency is equal to twice the maximum frequency, divided by 2 Pi. From this, the lowest value of the sampling frequency of the first sub-problem is obtained. That is, the Nyquist frequency of the product signal.
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The second sub-problem is the convolution of two signals. There is a Fourier transform time-domain convolution theorem. The spectrum of the convolution result is equal to the product of the spectrum of the two convolution signals. Since they are all spectrum-limited spectrums, their product is also spectrum-limited, and the maximum frequency is determined by the smallest spectrum range in the original signal spectrum. According to the conditions given in the title, the maximum value of the convolution signal spectrum does not exceed omega1. Thus, the maximum frequency of the signal y2 can be obtained, and then its Nyquist frequency can be obtained, and the reciprocal of the sampling frequency is the corresponding sampling time interval.
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▲ Figure 1.2.1 Sampling interval and sampling frequency corresponding to signal 2

▲ 图1.2.1 信号2对应的采样间隔与采样频率

2. The second question

For the second sub-problem, it is to judge the sampling frequency of the result after periodic extension, differentiation, squaring, convolution, modulation and scale transformation of the signal f(t). The Nyquist frequency of the signal f(t) is omega 0. The period extension of the signal, each of which is the translation of the signal, does not affect the change of the highest frequency of the signal, but only affects the phase of the spectrum. After superposition, the highest frequency of the signal is still the same, so the Nyquist frequency of the first problem is still omega 0. The third sub-question, the square of the signal, in the frequency domain, the corresponding spectral convolution, the corresponding frequency bandwidth is doubled. The next small problem is that the signal is sinusoidally modulated. After modulation, the corresponding frequency spectrum shifts left and right, and the highest frequency becomes twice the omega 0.
In the second sub-problem, the signal is differentiated, and the corresponding spectrum is multiplied by j omega. Its highest frequency does not change. Still omega 0. The fourth sub-question is the convolution of the signal itself, and the corresponding spectrum is squared, and the highest frequency is still omega 0. Finally, the signal is scaled, and the corresponding spectrum is scaled inversely. Causes the highest frequency of the frequency to be doubled. This is the Nyquist frequency of the six signals in the second subsection.
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▲ Figure 1.2.2 Nyquist frequency of the signal

▲ 图1.2.2 信号的奈奎斯特频率

※ Summary ※

This article discusses the exercise of judging the Nyquist frequency of the signal in the eighth assignment. Operations in the time domain of the signal can sometimes affect the highest frequency of the result.
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■ Links to related literature:

● Links to related diagrams:

Figure 1.2.1 Sampling interval and sampling frequency corresponding to signal 2
Figure 1.2.2 Nyquist frequency of the signal

Homework Tutorial Video SS2023-HW8: Signal Sampling Frequency