Signal Sampling and Recovery
- Signals and Systems 2023 (Spring) Assignment Requirements - Eighth Assignment
- Signals and Systems 2023 (Spring) Homework Reference Answers - 8th Homework
01 The eighth assignment
1. Introduction to Exercises
In the eighth homework, there are three exercises about signal sampling and recovery. The first sub-problem is to give the expression of the signal f(t), and solve the frequency spectrum and recovery conditions of the sampled signal; the second sub-problem discusses two The sampled spectrum and condition of the multiplied sinc signal. The third exercise discusses the ideally sampled spectrum of a signal after it has been scaled. And determine whether aliasing has occurred. Next, we discuss the solution ideas of these three exercises.
2. Problem solving
1. The first sub-question
The expression of the signal is given in the first sub-problem, including the product of a low-frequency cosine and a high-frequency cosine limit sign, which can be regarded as a signal modulation process. The carrier is a high frequency cosine signal. Using 140Hz to sample the signal, the first small question is to find the frequency spectrum of f(t). The second small question is to find the frequency spectrum of the sampled signal. The third small question, in order to restore f(t) without distortion, what is the bandwidth range of the corresponding low-pass filter. The fourth small question, according to the above analysis, determine the Nyquist frequency of the signal f(t), that is, what is twice its highest frequency. Let us now analyze the solution to this problem.
(1) The first small question
The first small question is to find the spectrum of the signal. Think of it as the modulation process of 15 cosine 16 Pi t signal by cosine 100 Pi t signal, which is convenient for analysis. For 15 cosine 16 Pi t , its spectrum is relatively easy to write, which is the shock spectrum located at plus and minus 16 Pi respectively. Then consider the modulation of the cosine 100Pi t signal in the back, shift the obtained spectrum by 100Pi to the left and right, and double the amplitude, which is the spectrum shifted to the left by 100 Pi. This is the spectrum shifted 100Pi to the right. Simplify the expression to get four shock spectra. Finally, the mathematical expression of the frequency spectrum of f(t) is obtained.
▲ 图1.2.1 f(t) 的频谱
This is to draw the four shock spectra of f(t), and it can be seen that the four shock spectra are distributed in the modulation spectrum plus or minus 100 Pi, with a distance of 16Pi. The impulse spectrum has an intensity of 15 Pi in half. This is the answer to the first question.
▲ 图1.2.2 f(t) 幅度频谱示意图
(2) The second small question
For the second small question, draw the frequency spectrum of the 140Hz ideal sampled signal. According to the principle introduced in the signal sampling course, the signal is discretized in the time domain, and the corresponding frequency spectrum is periodically extended. The sampling frequency determines the previous coefficients. Ts points One of and corresponds to the period of spectrum extension. Therefore, the graph after the frequency spectrum period extension is drawn here. The period of the continuation is 280 Pi, and the strength of the impulse signal is actually divided by the original strength by Ts, corresponding to 1050 Pi, which is the periodic impulse spectrum of the signal after sampling.
▲ 图1.2.3 采样后信号的频谱示意图
(3) The third small question
The third small question is to find the bandwidth of the low-pass filter used when restoring the original low-frequency signal from the sampled signal without distortion. According to the spectrum of the sampled signal drawn just now, the bandwidth of the low-pass filter needs to be able to intercept the low-frequency spectrum of the signal. Therefore, its cut-off frequency should be located between the highest value of the signal spectrum and the lowest value of the extended spectrum of adjacent periods. In this way, the frequency spectrum of the signal can be obtained. According to the plotted signal spectrum, the cutoff frequency range of the low-pass filter can be determined. Between 116 Pi and 164 Pi.
(4) The fourth small question
The fourth small question, about the Nyquist frequency of the signal f(t), according to the spectrum of f(t) obtained from the first small question, it can be known that its highest angular frequency is 116Pi, so the corresponding sampling frequency makes the highest angular frequency double the frequency, and divide by 2. This is the Nyquist frequency of f(t).
2. The second question
The second sub-problem is an exercise in analyzing the signal sampling frequency spectrum, input two signals f1, f2, after multiplying them, the signal f(t) is obtained. These two signals are both sinc functions, so their spectra should be rectangular. Multiply f(t) with the periodic impulse sequence to complete the ideal signal sampling, and the sampling period is T. The sampled signal is fs(t). Find the maximum sampling interval T corresponding to undistorted restoration of f(t) from fs(t). According to the sampling theorem, this maximum sampling time interval should be equal to half of the period corresponding to the maximum frequency of the signal f(t).
▲ 图1.2.4 f1(t),f2(t) 的频谱示意图
Therefore, this topic is to multiply f1 and f2 to obtain the maximum frequency parameter of f(t). According to the functions of f1 and f2, it can be known that their spectrums are holding signals, and the maximum frequencies are 1000Pi and 2000Pi respectively. So after they are multiplied, the corresponding spectrum should be the convolution of their spectrum, the shape is trapezoidal, and the highest frequency is the sum of their highest frequencies, which is equal to 3000Pi. In this way, the maximum value of the corresponding sampling period can be obtained, which is the reciprocal of the highest frequency of the frequency spectrum, and then divided by two.
▲ 图1.2.5 f(t) 的频谱示意图
The second small question is to draw the spectrum of fs(t) after sampling, which should be the spectrum of f(t), and carry out period extension according to the sampling frequency. According to the spectrums of f1 and f2, they are convoluted to obtain the spectrum of f(t). To carry out cycle extension on it, the title requires that the sampling period is exactly equal to twice the highest frequency of the spectrum, so the spectrum after extension is such a continuous trapezoidal signal. The period of the continuation is equal to 6000Pi, and the height is equal to the height of the original signal, multiplied by 6000Pi. This is the spectrum of the signal after sampling. So far, we have got two answers to the second question.
▲ 图1.2.6 采样后信号的频谱示意图
3. The third question
The third sub-problem is to obtain the sampling frequency and spectrum of the signal after scaling operation. For simplicity, the signal spectrum X omega has been given, which is a triangular spectrum. Respectively give x(3t), the sampling period corresponding to x quarter t signal. Use the periodic impulse sequence to sample x(t) and the signals of the two scale operations, the sampling period is Pi/twelfth of Pi, and draw the spectrum of the sampled signal respectively. In fact, this frequency spectrum is the periodic frequency spectrum formed after their respective frequency spectrum period extensions. Finally, determine whether aliasing occurs. That is, whether the period of spectrum extension is greater than twice the highest frequency of the signal. Let us analyze the solution to this problem below.
▲ 图1.2.7 第三小题的问题
For the first small question, perform scaling operations on the signal to find the corresponding sampling period. According to the Fourier transform scaling theorem, the scaling transformation of the signal time domain is just opposite to the scaling transformation of the frequency spectrum. The highest angular frequency of the original signal is 8. After compressing the signal scale three times, the highest frequency becomes 24, and the corresponding period is 2/24 of Pi, which is 12/ of Pi, half of which is the signal Nyquist The sampling period corresponding to the frequency. Equal to 24ths of Pi.
For t/4 of x, after scale transformation, the corresponding highest frequency is 2, and the period is 2/2Pi, half of which corresponds to the sampling period. That is, Pi in half.
For the second sub-question, the signal is ideally sampled, the sampling period is 12th of Pi, and the corresponding spectrum is equal to the periodic extension of the signal spectrum, multiplied by 1/Ts. The period of spectrum extension is equal to 24. This is the periodic spectrum of the discrete signal corresponding to x(t) after sampling. The spectrum is free of aliasing. This is the sampled spectrum of x(3t). After scale transformation, the frequency spectrum width of the signal is also 24, and spectrum aliasing occurs. The corresponding frequency spectrum after period extension is a constant, which in turn means that the signal is a delta(t) function after sampling. This is where x-thirds of the t signal samples have a corresponding frequency spectrum. The spectrum is free of aliasing. This is the answer to the second question.
▲ 图1.2.8 第三小题
※ Summary ※
This article discusses the signal sampling and recovery exercises in the eighth assignment. This is an exercise in the content of the signal sampling theorem.
■ Links to related literature:
- Signals and Systems 2023 (Spring) Assignment Requirements - Eighth Assignment
- Signals and Systems 2023 (Spring) Homework Reference Answers - 8th Homework
● Links to related diagrams:
- Figure 1.2.1 Spectrum of f(t)
- Figure 1.2.2 Schematic diagram of f(t) amplitude spectrum
- Figure 1.2.3 Schematic diagram of the frequency spectrum of the sampled signal
- Figure 1.2.4 Schematic spectrum of f1(t), f2(t)
- Figure 1.2.5 Schematic spectrum of f(t)
- Figure 1.2.6 Schematic diagram of the frequency spectrum of the sampled signal
- Figure 1.2.7 Questions for the third sub-problem
- Figure 1.2.8 The third sub-problem