Basic knowledge about Communication Signal Processing

The following are some posts from the Communication Signal Processing edition of the academic forum.

1. symbol energy problems

I would like to ask you how to solve the problem of the average energy of each symbol modulated by prawns such as 4qam, 16QAM, 32qam, and 64QAM? When I look at other people's programs, the average value of each symbol of 16QAM is 10. I don't know how to find it. I want to know how to add noise to these modulation methods. I have a program as follows:

EAV = 10 * d ^ 2; % energy per symbol

SNR = 10 ^ (snrindb (P)/10); % SNR per bit (given)

Sgma = SQRT (EAV/(8 * SNR); % noise variance

It is calculated based on the mean variance of the noise produced by 16QAM modulation. I don't know what the meanings of these terms mean. Which of the following heroes can help explain them.

A1: Generally, it is considered that all symbols appear. The key is the ratio to the noise power spectral density.

A2:To calculate the average energy of a symbol, you only need to add up the energy of each symbol and divide it by the number of symbols (assuming that each symbol has the same usage probability ). Take 16QAM as an example. In the constellation chart, each quadrant has four points, so you can only calculate the four points in any quadrant (such as the first quadrant. Assume that the minimum distance between a vertex and a vertex is D. Then, the coordinates of these four points are (D/2, D/2), (D/2, 3D/2 ), (3D/2, D/2) and (3D/2, 3D/2 ). Therefore, the energy of the second and third points is (D/2) ^ 2 + (3D/2) ^ 2. You will certainly push the other two points accordingly. Add up all four points of energy, equal to 10 * d ^ 2. Divide by 4 to get the average value of 2.5 * d ^ 2. For convenience, D = 2 is often set. At this time, the average value is equal to 10.

 

2. IFFT complex signal processing

I would like to ask you, in the simulation, the signals after the QAM ing (64 plural) are generated into 64 plural after IFFT. According to the book, at this time, the signal has been modulated into 64 sub-carriers. I would like to ask how these 64 plural numbers are modulated on 64 subcarriers? I understand this. I don't know if it is correct. It is the 64 complex numbers generated. During the conversion of D/, every 0.05us is multiplied by a Rectangular Pulse and modulated to a sub-carrier. In this way, the time for modulation of 64 complex numbers is exactly the same as that of the symbol of 3.2us and an OFDM. The generated OFDM symbol is then modulated to the main carrier. Is that true?

A1: This is the case (this sentence is incorrect)

The 64 plural values of IFFT are the 64 sampling points of an OFDM symbol.

A2:No. After you pass through the QAM map, the 64 plural numbers are the amplitude factors of 64 subcarriers of different frequencies. After the multiplication, the superposition in the time domain is the final OFDM output signal (without considering CP and so on)

Aaa3:On the third floor, the 64 complex numbers before IFFT correspond to 64 subcarriers with different frequencies. After IFFT transformation, the time-domain signals superimposed on these 64 subcarriers are generated, because only Time-domain signals can be transmitted. The D/A rate determines the size of the stator carrier interval, instead of modulated to 64 subcarriers.

 

3. understanding of complex signals

After learning the replay signal in class, I have been confused! Generally, hardware circuits do not support complex operations. The general method is to separate the real and virtual parts in Different Storage regions and perform operations based on the real virtual parts respectively! In this way, I understand that the plural is just a concept in the process of learning. We can use it when making a theory (such as deriving a formula on paper, in practical applications (programming in hardware circuits to implement the theory we have previously deduced), we must replace a plural number with two values () and calculate them according to the algorithm of real numbers, I just need to remember that the result is real, that is virtual? Do you know this is correct? There is another problem, that is, QAM. Orthogonal modulation is better understood by the plural. How is orthogonal modulation implemented in the real number field ??? When a QAM signal is generated, the I and Q signals are orthogonal, or two independent common signals!

Aaa1:I think the landlord's understanding of the complex signal is still in place. As for QAM, the I and Q signals are only orthogonal, because we divide exp signals into sin and cos signals, while sin and cos signals are orthogonal.
If you want to understand the QAM signal, it is easier to use a constellation chart. In a constellation chart, the abscissa and the ordinate are orthogonal. This orthogonal can be understood as the angle offset of 90, which is the angle modulation. Then, QAM has different amplitude. For example, 16QAM ranges from-3 to + 3. -- this is amplitude modulation. This is why we understand QAM as an angle plus amplitude modulation.

 

4. How to move the complex signal from the carrier to the low frequency

For example, I multiply a complex signal a + B * j With the carrier to generate a complex signal S (t) = (a + J * B) * Cos (2 * pI * fc * t) fc = 900 m, So how should we demodulated it? If we use a low-pass filter, we can only obtain its real part in the frequency domain? What's the solution? Which man knows?

A1:The positive and orthogonal Representation Method of the complex signal has two components, called the same phase component and the orthogonal component, which are described in detail in the communication. In radar signals, the baseband complex signal is also called a (coherent) video signal. It has a corresponding acquisition method and is connected to communication in principle. There are many demodulation methods, such as orthogonal mixing low-pass filtering, Hilbert transform frequency shift, and parity separation symbol Transformation (including Hilbert transform, interpolation, and multi-phase filtering ). You can query books and documents.

A2:To modulated the baseband complex signal to a high frequency, the real and virtual parts of the complex signal must be multiplied by the COs and sin waveforms of the high frequency respectively, and then the two products are added up to form a high-frequency real-signal. It is impossible to multiply by one cos. At the receiving end, the high-frequency signal must be multiplied by the COs and sin waveforms of the high-frequency signal to obtain two signals. After each signal is filtered, the real and virtual parts of the baseband complex signal are obtained.

A3:Complex signal processing is not complex is a classic document. You can search and download it on the website. I believe it will be helpful to you. The concept is that the complex signal flow chart is more intuitive and corresponds to the real signal flow chart. You can know the difference between the two.

Modulation and decoding at the level of complex signals is simply multiplied by a Complex Harmonic (carrier or intermediate frequency) component, and the understanding at the level of real signals is cross-coupling multiplication. For details, see this article.

A4:There is no complex signal in the real world, so it is impossible for you to use a single cosine to modulated a complex signal. Only a complete exponent e (J * SITA) can be used to represent the complex number, that is, two orthogonal components. The difference is the cosine wave (or sine wave) of PI/2 ). However, we can simplify the analysis by using the complex signal representation.

A5:From the physical point of view, a complex signal is a signal expressed simultaneously in both the amplitude and phase. First, a nonlinear transformation or time-varying network (Circuit) transformation is required for demodulation, and then a filter for separating signals is used. The ideal feature of this filter is that it has zero attenuation for the entire signal spectrum. 2. Constant Group latency. There is an infinite attenuation of bandwidths other than signals.

A6:Modulated signal: x (t) = Re {S (t) * E ^ (JWT)} = s_r (t) Cos (wt)-s_ I (t) sin (wt );
Demodulated signal: s_r (t) = x (t) * Cos (wt)-> LPF; s_ I (t) = x (t) * sin (wt)-> LPF.

A7: 1. Change to the low frequency multiplied by a sine wave, and the part that shifts above the filter will be fine. However, be careful when calculating the negative frequency of the sine wave.

2. If the frequency is changed to zero, it is impossible to multiply a complex exponential signal. Orthogonal demodulation is generally used to obtain the real and virtual parts of the baseband signal by two orthogonal sine waves. Based on the above communication principles, you can ask the original teacher to pay the tuition fee back.

 

5. How to send the last complex OFDM signal

I have encountered a problem in the simulation of the OFDM system. How can I send the complex data after IFFT after the frequency conversion or the plural data? What is the simulation of the channel, whether to perform operations on the complex number or other operations? Is it possible for the receiving end to restore the real part and the virtual part through downconversion and low-pass at the sending end? I think it would be too easy to consider. If the channel distortion is too serious, the data may not be restored. Thank you!

Aaa1:According to the introduction in some OFDM documents, if, before IFFT, the sequence after IFFT is actually a real sequence after the sequence is bounded and symmetric, then only the real sequence can be sent. (This is very simple) Another method is that prior to IFFT, the sequence is directly transformed by IFFT without being subjected to the symmetric and bounded sequence. In this way, the complex sequence is generated in I, Q is sent on two-way carriers. BTW. There are many documents on the Forum before. For details, refer!

0... N-1. Then, map the digital signal (-1, + 1) to N carriers (that is, the exp term in the formula, that is, IFFT), and combine them, and then send them. For example, if the bandwidth is 1024 MB and the IFFT size is 1024, the bandwidth of MB is divided into small portions. This aims to reduce the bandwidth, in this way, the ISI is reduced. Then the frequency of the first carrier is 100 m/1024, and the frequency of the 2nd carrier is 2*100 m/1024, and so on. In this part, N carriers are orthogonal to each other to eliminate inter-Carrier Interference.

 

6. Zero-padding for OFDM

When I was a beginner in OFDM, I found that there were two ways to complete IFFT. One is to add at the beginning and end to remove unwanted low-frequency components. Another method is to fill in the middle. It seems to be called oversampling.

I think the above two problems are quite dizzy. which method is used in actual application? How many supplements are needed? Please kindly advise!

Aaa1: You are advised to read a book about digital signal processing.

Both of the zero-padding methods you mentioned are available. When performing FFT and IFFT, the highest frequency point is in the middle, which is both the sampling frequency of 1/2, and the second half is the negative frequency image about the sampling frequency. Therefore, the first-end population removes the low frequency component, and the intermediate population is the so-called oversampling. The number of points to be supplemented varies with different situations.

 

7. SNR implementation

If the signal is set to S (T) and the noise is N (t), how can we use MATLAB to achieve the SNR of 10 dB with IFFT ??

Aaa1: You have asked a large number of questions. For the next question, I think this can be calculated as follows: After IFFT is passed, the N-point sequence x (n) is used to calculate the average sum of the square values, obtain the signal power Si and obtain the noise power Ni based on SNR. The mean value is 0, and the positive Random Number of the variance is Ni: N (t) = SQRT (NI) × randn (1, N), which can be used as noise. Different random numbers can be simulated based on different channels, as long as the power Ni is known. According to the passal theorem, the signal has the same energy in the frequency and time fields, so

After IFFT and FFT, the signal power remains unchanged.

 

8. About the QAM

1) after a string of binary sequences are converted by S/P, the I and Q branches are split, and then the upsample and root raised cosine filters are used to implement the D/A conversion. What is the significance of upsample? Is it used for Pulse Shaping?
2) Do the signals output by LPF on the modem end pass through the same root raised cosine filter before sampling? How can I obtain the original input sequence by sampling?

Aaa1: 1)
2) Right, the filter will get the original sequence, but there is a filter delay.

Aaa2: QAM must be 4-digit at least, that is, each symbol has four options. The M-base QAM is often written as M-QAM. 16, 64, and 256 hexadecimal QAM are common. For M-based QAM, SQRT (M)-hexadecimal real number sequence is used for entering I and Q. A number corresponds to a symbol. That is to say, each symbol is represented by a (sampling) Point. However, only one sampling point cannot be used for pulse shaping. Therefore, you must add a sample point before sending it to the root raised cosine filter to increase the sampling rate and use more points to represent each symbol. Upsampling is used to increase the sampling rate. Upsampling generally has two ways, one is to insert (N-1) a zero between each two symbols, the other is to repeat each symbol (N-1) times. The specific method depends on the implementation of the root raised cosine filter. At the receiving end, the signal after the carrier is removed is sampled after the same root raised cosine filter. The result of this sampling is to reduce the sampling rate (downsapling) and return to a point to represent a symbol. Whether the sampling time is correct is very important. The correct sampling time must be calculated carefully, or an appropriate time synchronization circuit must be available. After sampling, the I and Q sequences can be quantified by the decision generator to obtain the original input sequence (if there is no error ). This assumes the ideal situation. In fact, the carrier frequency of the receiver and the transmitter may be different. Therefore, the receiver must have a frequency synchronization circuit. The received signal may have distortion such as inter-code interference. Therefore, the receiver may also have an equalizer. And so on.

Aaa3: inserting (N-1) zeros between every two symbols is a more practical upsampling practice. In this way, each symbol has n samples. Simply use the root raised cosine filter, which is implemented by using FIR. Using FIR to implement the raised cosine filter is to calculate the impulse response of the filter and use the value of the impulse response at the sampling point as the coefficient of the FIR filter. This impulse response is parsed, but is rarely seen in books. However, existing functions in MATLAB can be used directly.

Aaa4: the transition band is (1-r)/(2 * t) in rcsofir in MATLAB)

Aaa5: According to the sampling theorem, to sample a simulated signal without distortion, the sampling frequency must be no less than twice the signal bandwidth. The actual sampling rate is usually twice higher than the signal bandwidth. Oversampling refers to sampling with a sampling rate greater than twice the signal bandwidth. When the rolling down coefficient is r = 0, the bandwidth of the raised cosine filter is | f | 0, and the bandwidth of the raised cosine filter is widened to | f |

 

9. Is the base band signal in OFDM still a base band signal after IFFT?

Excuse me, is the base band signal in OFDM still a base band signal after IFFT? I understand that subcarrier modulation is performed during IFFT, so after IFFT, it should be converted into an intermediate frequency signal. I can see Timothy M. in the article robust frequency and timing synchronization for OFDM, there is also an if lo after IFFT. is it still a baseband signal after the baseband signal passes through IFFT in OFDM ?, Thank you for your explanation.

Aaa1...

Aaa2: only signals after IFFT are OFDM signals and baseband signals. signals before IFFT cannot be called OFDM signals. IFFT can be understood as the completion of the OFDM baseband modulation, even if the common signal occupies a certain amount of Baseband bandwidth, the OFDM baseband modulation is to complete the signal processing on this Baseband bandwidth, it has certain characteristics. After baseband modulation is completed, other processing techniques can be applied, such as intermediate frequency and RF processing.

Aaa3: IFFT is followed by a zero-IF signal. Of course, the carrier needs to move it to the RF.

 

10. Role of shaping filter

Aaa1: The shaping filter reduces inter-code interference and reduces out-of-band radiation. For example, the root raised cosine filter is used for both the sender and receiver, the out-of-band radiation of the system is very low. For details, refer to the analysis and corresponding processing of Inter-code interference in communication principles.

 

11. Should I use a real-time signal or a complex signal?

If it is a real-letter number (in fact, when the theory is mentioned in the book, the PSK modulation signals are all real-letter numbers), in OFDM, according to the original theory, after OFDM modulation, it should be a real-letter pair. Because OFDM is only the sum of multiple PSK modulation signals (assume that each sub-carrier uses the PSK Modulation ). However, in MATLAB simulation, almost all the methods are to adjust the signal as a complex signal (this is also a baseband signal) in the process of PSK modulation, and then perform IFFT on these complex numbers. In this way, if only the real signal after IFFT is obtained, this signal should satisfy the previous conclusion. But why? Is that true? Thank you.

Aaa1:According to the Euler's formula, F1 (t) * Cos (WC * t)-F2 (t) * sin (WC * t) = Re (F1 (t) + J * F2 (t) * exp (J * WC * t) provides a basis for representing real numbers in the complex number in baseband modulation. Orthogonal modulation can be expressed as F1 (t) * Cos (WC * t)-F2 (t) * sin (WC * t) Form (WC is carrier frequency ), therefore, you only need to obtain F1 (t) + J * F2 (T) to obtain the modulation signal. In some books, I/Q two-way signals refer to fi (T) and F2 (t), that is, the actual and virtual parts. Personal Understanding for your reference.

Aaa2:The PSK modulation signal consists of two I Q channels and is modulated to the orthogonal carrier. The author asked another question (why do we use real-letter numbers during OFDM transmission ). You can refer to the signal processing in OFDM. Generally, before IFFT, the signal to be transmitted is collapsed and extended, then IFFT is used to obtain the real part of the transformation (because the Fourier transformation of the real finite sequence is a complex sequence of the bounded symmetry ), in this way, the real sequence is modulated to the carrier for transmission. Modulation in OFDM, including the modulation of transmission symbols (PSK or QAM) and FM.

 

5. Why does the IFFT of the OFDM system only take the real part?

I watched the OFDM for several days. For example, the output of the complex IFFT value at 32 points is also 32 complex numbers. To make the IFFT result contain only real components, a negative frequency part must be constructed (symmetric with the original input ), so it becomes an IFFT of 64 points, and the output result is a 64 real number. The final results of the two methods are 64 messages, but the number of IFFT points increases in the latter method. I don't know why? Hope you can give me some advice. Thank you !!!

Aaa1:After two times of oversampling, you can restore the virtual part from the sent real part. For more information, see "data transmission by frequency-division multiplexing using DFT", S. B. Weinstein, ect

Aaa2:This approach may be possible in a wired OFDM system (in fact, before IFFT, there was a wildcard symmetry. In theory, IFFT is followed by a real-letter number, which is equivalent to halving the transmission rate ). This is because if there is only one line and phase information is not used, only one real signal (real or virtual) can be transmitted simultaneously ). Of course, you can also send real and virtual data in TDM mode, so that the transmission rate will be halved. If the plural signal is meaningful, it must be used when the phase information is used. Two real numbers. If the (phase) Relationship between them is used, these two real numbers should be combined to form a complex number, rather than two independent real numbers.

Aaa3:The OFDM signal can be real or complex. Take the example given by the landlord, take 32 plural numbers, and then splice them with the perfect symmetry. In this case, the real OFDM signal will be obtained after IFFT is implemented. If you want to generate a complex number of OFDM, use 64 complex numbers for IFFT. If baseband transmission is used (that is, carrier modulation is not required), Real-Information signals can only be transmitted, and complex signals cannot be transmitted. On the contrary, if carrier modulation is performed, the complex signal can be transmitted theoretically because the sine and Cosine carriers can be used simultaneously. If only one of them is used, the utilization rate of the frequency band will not reach the maximum, but the requirement for carrier synchronization in the receiver can be reduced without worrying about interference between two signals.

Aaa4:When IFFT is used, in fact, the first number (must be a real number) defines the DC component, the second number (n/2 + 1) defines the highest frequency component, and the last (n/2-1) plural numbers define the negative frequency component. Therefore, the signal frequency band after IFFT is (-FM, FM ). Then, if baseband transmission is used, only the real part can be transmitted, and the bandwidth of the signal is FM. If the communication band (that is, the RF carrier) is used for transmission, one more complex part can be transmitted, however, the signal bandwidth is 2 * FM, so the band efficiency is the same. That article only discusses baseband transmission, so it can only transmit the real part. If you want to transmit another complex part, you must use the communication to bring the transfer, and the signal bandwidth will double.

Aaa5: Now the system is described as follows. If our system used the IFFT at point N and point N are all signal points, then according to the above paper, I only need to transmit the real number after the n-point IFFT to return the original N-point signal (of course, this requires selecting N points from 2n ). The above was initially simulated by me. The problem is: the original OFDM should have two carriers sin and cos at the same time, one is the real part used to load the fundamental frequency and the other is the complex part used to load the fundamental frequency. Because of the above process, I only need to use a single carrier to transmit real signals. Therefore, sin is not used ~ If we put the real number of signals at the next time point to the sin carrier at this time point, isn't this equivalent to doubling the data rate?

Aaa6:From the information theory point of view, the same frequency and efficiency can be achieved in both baseband and band-through transmission. Bandwidth is twice the bandwidth of baseband transmission. If only cosine is used to transmit real-time signals, the band efficiency is only half that of baseband transmission, which means a waste. If cosine is used and sine is used, the band efficiency is doubled compared with that of cosine, but the band efficiency is only the same as that of baseband transmission. This not only applies to OFDM, but also to other systems.

Aaa7:Well... I understand what you want to see as "tables ~

However, if you say ~

The baseband transmission only transmits real-time signals, while the band-pass transmission transmits real-time and virtual signals, with the same bandwidth efficiency. However, one thing above is that the signal is sent from the same IFFT. If the band-pass function only transmits the signal at least half of the bandwidth, why not replace the signal with the real-time signal at the next time? Although the bandwidth efficiency is reduced to half, is the transfer rate doubled?

Aaa8:Passband transmission is generally bilateral, so it is twice the transmission bandwidth of baseband, of course, half the transmission efficiency. If passband transmission of single-band (SSB) is used, the transmission efficiency is the same as that of baseband. When a bilateral band is transmitted, the information transmitted by the two sides of the central frequency is exactly the same (image components of each other). When the receiver is processed in the lower frequency conversion, all information can be obtained by converting any of the two bandwidths to the baseband. However, it is wrong to use COs to transmit half of the data and sin to transmit the other half of the data. Therefore, this does not exist. Use another side to transmit half of the data. It is only possible to use single-band transmission technology. The original two single-band bandwidths transmit different data, but the two single-band bandwidths become irrelevant, but they are two independent bandwidths.

Aaa9:If you review the nature of the Fu's transformation, you will know that only the transformation of the real number function is the Union pair, and the transformation of the complex number function is not symmetric. Therefore, only when the baseband signal is a real signal, what you call "Bilateral band transmission, the information transmitted by the two edges on both sides of the center frequency is exactly the same (mirror components for each other). When the receiver is processed in the lower-frequency conversion, all information can be obtained by converting any of the two bandwidths to the baseband." This sentence is true. If the baseband signal is a complex signal, the two bandwidths of the baseband signal are asymmetrical. The original information cannot be restored only when one side is used. For example, if you have a pure real-number Amplitude Modulation (ASK) baseband signal, the information transmitted by the two bandwidths is exactly the same (mirror component ), the receiver can obtain all information when converting any of the two bandwidths to the baseband. You can transmit such a signal using a single side. However, this is not the case for the QAM signal, because the QAM signal is a complex number, and its two sides are asymmetrical. The original information cannot be restored with only one side. Therefore, you cannot use a single side to transmit the QAM. Finally, you said, "It is wrong to use COs to transmit half of the data and sin to transmit the other half of the data ." This sentence is obviously not correct. The QAM signal uses COs to transmit half of the data and sin to transmit the other half of the data.

Aaa10:1) I would have discussed the real-number baseband signal.

2) there is nothing special about the QAM, and 4qam is the PSK.

3) I wrote a statement about half of the data that I didn't think clearly, mainly for the previous discussion. I want to explain that it is not a side band that transmits half of the data. Cos is the so-called I-path, and sin is the so-called Q-path signal. It seems a little dizzy. The problem is a bit unclear. Is a real baseband signal or a complex baseband signal represented by a baseband signal? We use only one side band signal for demodulation after downconverting the PSK and 16QAM signals transmitted in the frequency band ?!

Aaa11:It seems that everyone is very interested in this question. To facilitate discussion, I will sort out the problem first.

(1) To generate a pure real-number OFDM signal, the common practice is to take n plural numbers from the information data to define the positive frequency (0 ~ FM), and then concatenate them with the bounded symmetry to define the negative frequency (-FM ~ 0 ). Then perform IFFT to obtain the real signal of 2n points. The frequency range is (-FM, FM ). The resulting signal transmits n plural information data. If baseband transmission is used, the bandwidth is FM. The bandwidth is 2fm if it is used for communication.

(2) In order to generate a complex OFDM signal, the information data is directly obtained from the information data 2n complex numbers, and then the IFFT is used directly to obtain the complex numbers of signals, use the cosine and sine carriers to transmit the real and virtual parts respectively. Compared with the process of generating real-number signals, the negative frequency is also used to transmit information data because the product does not need to generate a bounded and symmetric spectrum. In this case, the bandwidth of the RF signal is 2fm, and 2n complex information data is transmitted. Therefore, the band transmission efficiency is the same as that of the baseband transmission.

(3) The lovewa problem stems from the method in an IEEE article. This method is different from the above practice, so it is confusing. The method is to obtain n plural numbers from the information data, and then extract the real part after IFFT. At the receiver end, double sampling to obtain 2N real numbers, the original n pieces of information data are restored. Because only the real part is transmitted and the virtual part is not transmitted, the lovewa problem is: can we use the ability to transmit the virtual part (that is, using the sine carrier) in the communication band to transmit one channel of information, to improve the channel bandwidth utilization.

My concern is: Can I use the (3) method to achieve higher band efficiency than the (2) method?

In my opinion, in method (3), if you use band-based transmission, you can use the sine carrier to transmit another real-part signal, but you cannot obtain the ratio (2) high Band efficiency. You are welcome to express your own opinions.

Aaa12:It is true that 4qam is. However, whether you call it 4qam or PSK, its baseband is a complex signal. Generally, the I path is its real part, and the Q path is its virtual part. The spectrum of each symbol is asymmetric. You cannot use only one side to transfer and restore it.

Aaa13:I think it is feasible to use another route to send another real number ~

Then, the efficiency will be the same as that of (2), 2n data, and 2fm memory.

(3) Is there any possibility of higher efficiency than (2) zookeeper?

I think there is only one more orthogonal sine wave.

This will increase the efficiency to 1.5 times (2)

Aaa14:The commonly used method is mentioned in the summary (2)

As for (3) the lovewa problem stems from the method in an IEEE article. This method is different from the above practice, so it is confusing. The method is to obtain n plural numbers from the information data, and then extract the real part after IFFT. At the receiver end, double sampling to obtain 2N real numbers, the original n pieces of information data are restored. Because only the real part is transmitted and the virtual part is not transmitted, the lovewa problem is: can we use the ability to transmit the virtual part (that is, using the sine carrier) in the communication band to transmit one channel of information, to improve the channel bandwidth utilization.

I haven't read that article yet. Through your discussion, I understand that the real part taken out after IFFT is sampled twice as much as the oversampling, and then the frequency conversion is initiated, right?

There are two problems: (1) Assume that IFFT has n plural points, and the IQ of the transmitting end of the conventional method is

2n points. Use the method in your IEEE article. After sampling twice, the transmitting end is 2n points.

As you said, adding a 2n point to the next one will be added. The transmission information is doubled each time, and the number of Transmit points is doubled, which does not increase the transmission rate.

(2) The method of extracting the real part should consider whether it can be implemented in the RF and whether it can be converted.

I still need to verify whether the methods in this article are correct. However, your ideas are quite meaningful. When we are working on a real project, We Have To Do Sampling before RF, if we can combine oversampling in the article with oversampling in the actual system, theoretically the transmission rate can be improved in the actual system.

Aaa15:I pushed this question carefully.

First, I want to determine that it is not the optimal method to use the ability of transmitting the virtual part through band transmission and then transmit the information one way.

Because you do this, the constellation chart is no longer even, and the anti-interference capability is reduced. The method you use to increase the transmission rate is equivalent to doubling the QAM, however, the anti-interference performance is far inferior to doubling the QAM.

Aaa16:I haven't read that article yet. Through your discussion, I understand that the real part taken out after IFFT is sampled twice as much as the oversampling, and then the frequency conversion is initiated, right?

Ans: Also, it is done when the receiver receives the receiver.

That is to say, on the sending end, There is only enough real data at N points.

There are two problems:

(1) assume that there are n plural points after IFFT, And the IQ of the transmitting end of the conventional method has a total of 2n points. Use the method in your IEEE article. After sampling twice, the transmitting end is 2n points.

As you said, adding a 2n point to the next one will be added. The transmission information is doubled each time, and the number of Transmit points is doubled, which does not increase the transmission rate.

Ans: This question is not true.

(2) The method of extracting the real part should consider whether it can be implemented in the RF and whether it can be converted.

Ans: I don't understand this part, so I can't answer it.

P.s. Let's discuss it again. The conclusion seems to be coming soon ~

Aaa17: Next, let me explain my views on the three methods for generating OFDM signals. In my opinion, method (3) is essentially the same as method (1. Although they are different in practice, their band efficiency is the same when they are transmitted on the channel.

The reasons are as follows.

(1) A real-number time-domain signal, no matter what method is used to generate, its Fu's transformation must be bounded and symmetric. If you have any questions about this, please review the nature of the Fu's transformation.

Therefore, when you extract the real part of a complex time-domain signal, you have changed the Fu's transformation of the acquired signal to a bounded symmetry.

(2) exp (J * 2 * pI * fN * t) is a complex time domain signal. Its Fu's transformation is a line of FN.

(3) exp (J * 2 * pI * fN * t) = cos (2 * pI * fN * t) + J * sin (2 * pI * fN * t ). If exp (J * 2 * pI * fN * t) is obtained, cos (2 * pI * fN * t) is obtained ).

(4) The Fu's transformation of COS (2 * pI * fN * t) is located at the two spectral lines of-FN and FN, rather than one.

(5) The Fu's transformation of Cn * Cos (2 * pI * fN * t + qn) is also located in the two spectral lines of-FN and FN, rather than one. Both CN and Qn are real numbers.

(6) The IFFT calculation process is to multiply n plural numbers with N exp (J * 2 * pI * fN * t) and add them.

(7) Therefore, after obtaining the real part of the IFFT result, the sum of N Cn * Cos (2 * pI * fN * t + qn) is obtained. Each of these items has two spectral lines, and a total of 2n spectral lines are bounded and symmetrical.

(8) The effect of such a processing is the same as that of the concatenation of symmetric Spectral Lines in method (1. The real number signal is sent to the channel. The band width is the same as that of method (1), and N plural numbers are also transmitted. Therefore, the band efficiency of the two is the same.

(9) If cosine is used to transmit such a signal, sine can be used to transmit another signal at the same time, but the bandwidth is doubled compared with baseband transmission. The band efficiency is the same as that of method (2), and the band efficiency is not higher than that of method (2.

I have no intention of arguing with anyone or competing to win or lose. My interest is communication. If you do not agree, stick to your opinion and try to verify it in the experiment. My opinion is for reference only.

Aaa18: Um... start to understand what you mean. It's probably a poor level of information.

I would like to ask another question, which is about the definition of Optimization Efficiency ~

How can we define this concept without sending rate?

This is because I only see how many IFFT points are used and how many snapshots are used.

In this way, can we determine the efficiency of the Dynamic Route directly?

In addition, the ninth point is a bit more efficient than the unknown point. Is this sub-statement as efficient as (2 )?

However, if I take the test to improve the reading rate, will the channel use another channel to send data?

I am more confused ~

Finally, I'm sorry if there is something wrong with others.

I think this kind of discussion is good. I have met such a friend in my research room ~

Your opinion is analyzed from the perspective of data. It is true that you must use data as the basis!

Aaa19:

I read the Signal Analysis book, learned about the concept of asymmetric filter, and also saw that the pulse response of asymmetric filter is an amplitude modulation phase signal, the impulse response of the equivalent low-pass filter of the asymmetric filter is complex.

However, I still don't understand its physical meaning and actual processing methods. I 'd like to explain it again, or if I can give some details about it.

For example, in baseband processing, we still seem to think that the spectrum is symmetric and only focus on the positive frequency domain. In addition, the orthogonal two channels are processed separately according to the real number, filter and ADC.

Aaa20: after reading the paper, it seems that we didn't say that we only use cos for transmission after taking the real part of DFT? It is the same as the general band-pass signal. The real part is the original band-pass signal. It is still composed of COS and sin:

S (t) = a (t) Cos [WT + theta (t)]

= A (t) Cos [theta (t)] coswt-a (t) sin [theta (t)] sinwt

Complex signal representation S (t) = a (t) * E ^ jtheta (t) * E ^ JWT

S (t) = Re {S (t )}

The expression and paper have only one negative number.

In paper, there are n complex signals before the DFT signal. After the DFT signal, the real part is obtained. However, there are still 2N real numbers transmitted, which should be cos and sin. What the receiving end obtains is 2N real numbers, n real numbers not analyzed above.

One confusing sentence in this article is: because only the real part of FT is transmitted, it is necessary to perform two times of sampling. I don't know what is the relationship between 2X sampling and the real part that only transmits ft?

Aaa21 ~

How can we define this concept without sending rate?

This is because I only see how many IFFT points are used and how many snapshots are used.

In this way, can we determine the efficiency of the Dynamic Route directly?

Answer:

The purpose of our discussion is to make a qualitative comparison instead of quantitative calculation of the frequency band efficiency. Therefore, when comparing the frequency band efficiency of several methods, we only mention how many plural symbols are transmitted in the frequency band, we didn't talk about how many BITs each symbol represents. If we know or specify the number of BITs each symbol represents, we will discuss the efficiency of the band by using the number of bits transmitted in the Multi-width band.

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In addition, the ninth point is a bit more efficient than the unknown point. Is this sub-statement as efficient as (2 )?

However, if I take the test to improve the reading rate, will the channel use another channel to send data?

I am more confused ~

Answer:

The original Article only discusses the use of baseband transmission, so there is no need to transfer one path. If only one channel is used, the bandwidth efficiency is half wasted. If another channel is uploaded, this waste can be avoided, and the overall transmission rate is higher than that of only one channel. However, it has long been known that two channels of signals can be transmitted through a pass band, and the band efficiency is no better than the existing technology. In my opinion, it seems difficult to calculate new ideas or achievements.

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Finally, I'm sorry if there is something wrong with others.

Answer:

There is no problem in our discussion. It's just that everyone's time is limited and it's impossible to proceed without limit. The questions discussed are often difficult to understand and understand. Therefore, to a certain extent, we can only seek the same ground while reserving differences. Mutual Respect and reserve their own opinions do not have to compete for a level. Whether you agree or disagree with me, it is appreciated that you raise this question first.

Aaa22:

For example, in baseband processing, we still seem to think that the spectrum is symmetric and only focus on the positive frequency domain;

Answer:

We usually discuss the average power spectrum. The word "average" is very important. The instantaneous spectrum of the psk signal is non-symmetric, but the power spectrum after the average is symmetric.

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In addition, the two orthogonal channels are processed, filtered, and ADC separately according to the real number.

Answer:

A complex number consists of real and virtual parts, such as a + J * B. Both A and B are real numbers. Therefore, the two orthogonal channels are processed according to the actual number.

Negative frequency

In signal processing, the complex signal representation is mainly used for the convenience of mathematical processing, because if the real-time signal representation is used, when processing the signal, a large number of "cross items" will be generated ", this will bring some complexity to the system analysis, and this problem can be mitigated by using the complex signal representation, and because the real and virtual parts of the complex signal are exactly the same as the phase branch in the receiver (I) it is natural to use the complex signal representation in the system because it corresponds to the orthogonal Branch (q. The real signal spectrum is symmetric on both sides, that is to say, there is a negative frequency, but actually the negative frequency does not exist, and the resolution of the complex signal spectrum is just a positive frequency.
To obtain the complex signal corresponding to a real signal, you can multiply the positive frequency spectrum of the real signal and make the negative frequency spectrum equal to zero, the actual engineering implementation of this process is carried out through Hilbert transformation, and such a complex signal is parsed.
For more information about this issue, see Richard L. radar Signal Simulation by Mitchell. artech house, Inc. 1976 or a translation: translated by Chen xunda. radar System Simulation. beijing: National Defense Industry Press, 1982

From the signal and system perspective, I think this is a good understanding:
1. To obtain the system response, the input signal must be convolution with the system;
2. In order to simplify and facilitate numerical processing, people need to find a special basic unit signal. Such special signal has two main characteristics: (1) it can express common signals, (2) The response of such signals is relatively simple;
3. After searching, it is found that exponential signals are suitable for such basic unit signals. Their responses are the product of common values and exponent. In addition, such signals can represent a large number of signals;
4. The key is to express ordinary real-letter numbers as exponential forms, and introduce the concept of virtual numbers (Euler's formula ).
The problem can be understood in many ways. Tomcat's understanding is more about the benefits of expressing signals as plural forms, from my perspective, we need to introduce the concept of virtual numbers to describe the signal in the form of real numbers (of course, in some cases, the signal is originally expressed in the form of real numbers, this is a general form ).
Just a few days after arriving at this website, I feel this website is good and I can learn more from you!

For a real signal, the spectrum is bounded and symmetric, that is, the negative frequency can be completely determined by the positive frequency, which is redundant. For a baseband signal whose highest frequency is FM, if it is modulated to the carrier, the bandwidth of the positive frequency is 2fm. If it is constructed for the baseband signal, it is parsed and then modulated to the carrier, the bandwidth is only FM. In this sense, the resolution signal can reduce the bandwidth by half and reduce the sampling frequency of the band-pass signal.
Of course, from another perspective, after the real-time signal is changed to a complex signal, it is actually changed to a two-way signal, for example, a resolution signal (the real-time signal is the original signal, and the imaginary part is the orthogonal signal ). Therefore, for sampling, from one sampling method to two sampling methods, the actual sampling rate is not reduced.
The complex signal is implemented through two signal channels. If the negative signal is multiplied, it is not only the calculation of the two channels, but also the cross-coupling multiplication. Complex Harmonic x = XR + J * xi = cos (WIT) + J * sin (WIT) = exp (usually it) and multiplication of the plural A + JB:

So what should I do when the complex signal passes through the filter? Is the real and virtual parts handled separately?
How can I understand the filter coefficient? Are they coefficients of the same phase and orthogonal component?

Generally, it is a digital filter of two real numbers, processing the real and virtual parts separately.
However, there are also complex coefficients filters that can directly filter complex signals. In the current radar simulation system, the matched filter sampling in pulse compression is the complex coefficient filter, that is, the input and coefficients of convolution filter and the input are all plural. Sometimes it is convenient to consider the problem and the processing problem from the perspective of the complex signal flow chart, for example, in the direct sampling of the intermediate frequency digital mixing orthogonal transformation.
To promote this function, dual elements have a complex signal (two channels, expressed in units 1, I), and four elements have a supercomplex signal (four channels, expressed in units 1, I, j, and K ), each of these filters has a (Super) coefficient. If you are interested, you can view related documents.