Hello, everybody, and welcome to the series of technical videos on Millimeter Wave Sensing. 大家好，欢迎观看该有关毫米波传感器的技术视频系列 Specifically, we’ll be talking about a sensing technology called FMCW RADARS, which is very popular in both automotive and industrial segments. 具体来说，我们将讨论一项称为FMCW雷达的传感器技术，该技术在汽车和工业领域非常流行 The goal of the series is to give you a short but hopefully fairly in-depth understanding of this class of radars. 该系列的目标是让您简单了解这类雷达，同时又为以后深入了解奠定基础 FMCW stands for Frequency Modulated Continuous Waves. FMCW表示调频连续波 And I’ll explain the reason for this naming a little bit later. 我稍后将解释这样命名的原因 This radar basically measures the range, velocity, and angle of arrival of objects in front of it. 该雷达主要测量其前方物体的距离、速度和到达角 So this series of videos discusses each of these dimensions of sensing in some detail, starting with the range in module one, then moving on to velocity over the next couple of modules, and finally angle estimation in module five. 因此，该视频系列将会比较详细地讨论其中地每个传感维度，从第一个模块开始，然后在接下来地几个模块中继续讨论速度，最后在第五个模块中讨论角度估算 For those of you who are new to millimeter wave sensing, I would recommend that you view all of these videos in sequence. 如果您不熟悉毫米波传感，那么我建议您按顺序观看所有这些视频 The first module is going to start with explaining the basics of FMCW radar operation. 第一个模块将首先介绍FMCW雷达运行地基础知识 And then focus primarily on range estimation using the radar. 然后主要讨论如何使用雷达进行距离估算 So the kind of questions that we will focus on in this module are: you have a radar, you have an object in front of it, how does the radar estimate the range to this object. 在本模块中，我们将重点解答这样的问题，您有一个雷达，并且雷达前方有一个物体，雷达如何估算它与物体之间的距离？ What if there are multiple objects at different ranges from the radar? 如果有多个物体，并且它们与雷达之间的距离是不同的，将会怎样？ How close can two objects get and still hope to be resolved as always two objects? 两个物体能够相距多近而仍然能够有望被解析为两个物体？ What determines the furthest distance that a radar can see? 什么决定雷达可以看到的最远距离 At the heart of an FMCW radar is a signal called a chirp. FMCW雷达的核心是一种称为线性调频脉冲的信号 What is a chirp? 什么是线性调频脉冲？ A chirp is a sinusoid or a sine wave whose frequency increases linearly with time. 线性调频脉冲是频率随时间以线性方式增长的正弦波 So in this amplitude versus time, or A-t plot, the chirp could start as a sine wave with a frequency of, say, fc. 那么，在该振幅-时间图中，或者说A-t图中，线性调频脉冲可能以频率为fc的正弦波开始 And gradually increase its frequency, ending up with a frequency of say, fc plus B, where B is the bandwidth of the chirp. 然后，其频率逐渐增大，最后，假设以fc加B的频率结束，其中B是线性调频脉冲的带宽 Thus The chirp is basically a continuous wave whose frequency is linearly modulated. 因此，线性调频脉冲本质上是一种频率以线性方式进行调制的连续波 Hence the term frequency modulated continuous wave or FMCW for short. 因此，我们使用术语调频连续波，或简称FMCW Now if the same chirp were represented in a frequency versus time plot, or f-t plot, how would it look? 现在，如果在频率-时间图或者说f-t图中显示同一线性调频脉冲，它看起来会是什么样子 Remember that the frequency of the chirp increases linearly with time, linear being the operative word. 请记住，线性调频脉冲的频率随时间以线性方式增大，其中线性是关键词 So in the f-t plot, the chip would be a straight line with a certain slope S. 因此，在f-t图中，线性调频脉冲会是一条具有特定斜率S的直线 And just to put in some typical numbers, this figure for example could represent a chirp which starts at a frequency fc of 77 gigahertz, spans a bandwidth B of 4 gigahertz, thus ending up at a frequency of 81 gigahertz. 只需要放入一些典型的数字，该图就可以表示以77GHz的频率开始的线性调频脉冲，跨越4GHz的带宽B，最终以81GHz的频率结束 The slope S of the chirp defines the rate at which the chirp ramps up. 线性调频脉冲的斜率S定义线性调频脉冲上升的速率 In this example, the chirp is sweeping a bandwidth of 4 gigahertz, with a time period Tc of 40 microseconds, which corresponds to a slope of 100 megahertz per microsecond. 在本示例中，线性调频脉冲跨越4GHz的带宽，具有40微秒的Tc时长，这对应于每微秒100MHz的斜率 As we shall see later, the bandwidth B and the slope S are important parameters which define system performance. 正如我们稍后将看到的，带宽B和斜率S是用于定义系统性能的重要参数 Now that we know what a chirp is, we are ready to understand how an FMCW radar works. 现在我们已经知道什么是线性调频脉冲，那么我们可以了解FMCW雷达的工作原理了 This here is a simplified block diagram of an FMCW radar with a single TX and a single RX antenna. 这是一个简化的FMCW雷达框图，该雷达具有单个TX天线和单个RX天线 The radar operates as follows. 该雷达的工作原理如下 The synthesizer generates a chirp. 合成器生成一个线性调频脉冲 This chirp is transmitted by the TX antenna. TX天线将该线性调频脉冲发射出去 The chirp is reflected off an object. 当遇到物体时，该线性调频脉冲会反射回来 And the reflected chirp is received at the RX antenna. RX天线接收反射的线性调频脉冲 The RX signal and the TX signal are mixed. RX信号和TX信号混合在一起 And the resulting signal is called an IF signal-- IF standing for Intermediate Frequency. 最终生成的信号称为IF信号–IF表示中频 We’ll analyze the IF signal in more detail in the next slide. 我们将在下一张幻灯片中对IF信号进行更详细的分析 But first, let’s spend a little time understanding this component called a mixer. 但是，首先让我们花点时间来了解这个称为混频器的组件 What is a mixer? 什么是混频器？ A mixer has two inputs and one output. 混频器具有两个输入和一个输出 A simple way to understand the mixer is as follows. 下面是了解混频器的一种简单方法 If two sinusoids are input to the two input ports of the mixer, the output of the mixer is a sinusoid with the following two properties. 如果向混频器的两个输入端口输入两个正弦波，那么混频器的输出是具有一下两条性质的正弦波 Property number 1: the instantaneous frequency of the output equals the difference of the instantaneous frequencies of the two input sinusoids. 性质1：输出的瞬时频率等于两个输入正弦波的瞬时频率的差值 So even if these sinusoids, if their frequencies were varying with time, the frequency of the output at any point in time would be equal to the difference of the input frequencies at that point in time. 因此，即使这些正弦波的频率随时间发生变化，任一时刻输出频率也将等于该时刻的输入频率差值 Property number 2: the starting phase of the output sinusoid is equal to the difference of the starting phases of the two input sinusoids. 性质2：输出正弦波的起始相位等于两个输入正弦波的起始相位差 These two properties are illustrated in these equations here, where x1 and x2 are the two inputs, and x_out is the output of the mixer. 这里的方程阐释了这两条性质，其中x1和x2是两个输入，x_out是混频器的输出 So note here that the two inputs have frequencies of omega 1 and omega 2, and starting phases of phi 1 and phi 2, respectively. 那么，在这里请注意，两个输入分别具有频率omega1和omega2以及起始相位phi1和phi2 And the output has a frequency of omega 1 minus omega 2, and a starting phase of phi 1 minus phi 2. 输出具有频率omega1减omega2，并具有起始相位phi1减phi2 Let’s look more closely at the operation of the mixer in the radar. 让我们更加详细地看看雷达中混频器的工作原理 And I think it’s best illustrated using the f-t plot that we talked about earlier. 我想使用我们先前讨论过的f-t图可以对其进行最佳阐释 So the plot here refers to the RF signal. 那么，这里的图表示射频信号 So you have the transmitter chirp here and the received chirp here. 那么，这里是发射器线性调频脉冲，这里是接受到的线性调频脉冲 Note that the received chirp is a time delay replica of the TX chirp. 请注意，接收到的线性调频脉冲是TX线性调频脉冲的副本 And for now, I’m assuming there is only one object in front of the radar. 现在，我假设雷达前方只有一个物体 Hence, only one RX chirp. 因此，只有要给RX线性调频脉冲 Remember from the last slide that the output frequency of the mixer is the difference of the instantaneous frequencies of its two inputs, namely the TX chirp and the RX chirp. 回忆一下上一张幻灯片的内容，混频器的输出频率是其两个输入，即TX线性调频脉冲和RX线性调频脉冲的瞬时频率的差值 So to generate the f-t plot for the IF signal, I just need to subtract this line from this. 那么，为了生成IF信号的f-t图，我需要将这条线从这条线减去 And as you can see, these two lines are at a fixed distance from each other. 正如你所看到的，这两条线相互之间存在固定距离 And that fixed distance is given by the slope of the chirp times the round trip delay. 该固定距离由线性调频脉冲的斜率乘以往返延迟给出 In other words, S-tau. 换句话说，S-tau So a single object in front of the radar produces an IF signal consisting of a single frequency given by S-tau. 因此，雷达前方的单个物体可生成一个包含单个频率的IF信号，该频率由S-tau给出 Now tau, the round trip delay from the radar to the object and back, can also be expressed as twice the distance to the object divided by the speed of light. 现在，tau，即从雷达到物体然后又返回的往返延迟，也可以表示为与物体的距离除以光速，然后乘以2 So this is the fundamental concept to remember. 那么，这是需要记住的基本概念 A single object in front of the radar produces an IF signal with a constant frequency given by S2d/c. 雷达前方的单个物体可生成具有恒定频率的IF信号，该频率由S2d/c给出。 Now, it is important to note that the IF signal is only valid from the time the reflected signal is received at the RX antenna. 现在，应注意，IF信号仅从在RX天线上接收到反射信号开始有效 So if you were to digitize this IF signal using an ADC, you need to make sure that you only pick up samples after this time tau has elapsed, and only up to the time where the TX signal is present. 因此，如果您需要使用ADC对该IF信号进行数字化，那么您需要确保仅在经过时间tau之后再接收样本，并且只能持续到TX信号消失之前 Another point worth noting is that the round trip delay tau is usually a very small fraction of the total chirp time. 另外一点值得注意的是，往返延迟tau通常是总线性调频脉冲时间的很小一部分 So for example, for a radar with a maximum distance of 300 meters and a chirp time of 40 microseconds, this ratio of tau by Tc is only 5%. 例如，对于最大距离为300米并且线性调频脉冲时间为40微秒的雷达，该tau与Tc的比率仅为5% Fourier transforms are at the heart of FMCW radar signal processing. 傅里叶变换是FMCW雷达信号处理的核心 And we will see throughout the series of videos, they are used in range, velocity, and angle estimation. 我们将在整个视频系列中，它们用于距离、速度和角度估算 So we will from time to time take short detours to remind ourselves of relevant properties of Fourier transforms. 因此我们将不时地稍微转移一下话题，回忆一下傅里叶变换地相关性质 A Fourier transform converts a time domain signal into a frequency domain. 傅里叶变换将时域信号转换到频域中 So a single tone in the time domain produces a single tone in the frequency domain. 因此时域中地单个音调会在频域中产生单个音调 Likewise, the two tones in the time domain should result in two peaks in the frequency domain. 类似地，时域中地两个音调应在频域中产生两个峰值 But is that always the case? 但情况是否总是如此呢 So in this example here, within the observation window of T, the red tone completes two cycles while the blue tone completes 2.5 cycles. 那么，在这个示例中，在观测窗口T内，红色音调完成了两个周期，而蓝色音调完成了2.5个周期 And this difference of 0.5 cycles between the red and the blue tone, it seems is not sufficient to resolve the two tones in the frequency spectrum. 红色音调和蓝色音调之间地该0.5个周期差值似乎不足解析频谱中地两个周期 So here you have only a single tone corresponding to the contributions from both these signals. 那么，在这里，您只有单个与这两个信号地贡献对应地音调 Let’s now double the observation window from T to 2T. 现在，让我们将观测窗口加倍，从T增加到2T Doubling the observation window, now results in a difference of one cycle between the red and the blue tones. 现在，将观测窗口加倍，会导致红色音调和蓝色音调之间地差值为一个周期 And as you can see, the two tones are now resolved in the frequency spectrum. 正如您所看到的，现在在频率中解析了这两个音调 So the take away is that longer the observation period, better the resolution. 所以，重点是，观测期越长，解析就越好 And in general, an observation window of T can separate frequency components that are separated by more than 1 by T hertz. 一般而言，观测窗口T可以分隔以高于T分之一赫兹进行分隔的频率分量 This completes our short digression on Fourier transforms. 有关傅里叶变换的简短题外话就谈到这里 So far, we’ve talked about a single object in front of the radar. 到目前为止，我们已经讨论了雷达前方的单个物体 It’s easy to extend this to the case where there are multiple objects in front of the radar. 很容易将它扩展到雷达前方有多个物体的情况 So here, you have a radar transmitting a single chirp, and you get multiple reflected chirps from different objects. 那么，这里有一个雷达，它正在发射单个线性调频脉冲，您获取了多个从不同物体反射的线性调频脉冲 Each delayed by a different amount depending on the distance to that object. 每个脉冲具有不同量的延迟，具体取决于与物体之间的距离 So the IF signal will have tones corresponding to each of these reflections. 因此，IF信号将具有与其中每个反射相对应的音调 And the frequency of these tones, as we learnt, is directly proportional to the range. 正如我们所了解到的，这些音调的频率与距离成正比 So this has the smallest frequency and corresponds to the closest object. 因此，这条具有最小的频率对应于最近的物体 While this corresponds to the farthest. 而这条对应于最远的物体 A Fourier transform on this IF signal well then show up multiple peaks. 有关该IF信号的傅里叶变换会显示多个峰值 And the frequency of these peaks will be directly proportional to the range of the corresponding object. 这些峰值的频率将与对应物体的距离成正比 So again, this corresponds to the closest object, and this one to the farthest. 那么，这还是对应于最近的物体，这对应于最远的物体 The moment we talk about multiple objects, the next natural question is range resolution. 由于我们现在是在讨论多个物体，因此下一个问题自然是距离分辨率 That is, how close can two of these objects be and still be resolved as two peaks in the IF spectrum. 也就是说，其中两个物体能够相距多近而仍然能够在IF频谱中解析为两个峰值 So in this example, we have two reflected chirps from two objects. 那么，在该示例中，我们有两个从两个物体反射的线性调频脉冲 And the corresponding A-t plot of the IF signal shows two sine waves. IF信号的对应A-t图显示了两个正弦波 But the frequencies of these sound waves are so close that they show up as a single peak in the frequency spectrum. 但是这些声波的频率是如此接近，以至于它们在频谱中显示为单个峰值 How do we improve the range resolution of this radar? 我们如何提高该雷达的距离分辨率？ Taking a cue from our recap of Fourier transforms, one option is to extend the observation window of these two sine waves by increasing the length of the IF signal. 可以从我们对傅里叶变换的扼要重述中获得提示，一种选项是通过增大IF信号的长度来扩展这两个正弦波的观测窗口 So that’s what I’ve done here. 那么，我在这里是这么做的 So the chirp is extended which then extends the duration of the IF signal. 那么线性调频脉冲得到了扩展，从而扩展了IF信号的持续时间 And this resolves the two peaks in the frequency domain. 在这个频域中解析了这两个峰值 Note that increasing the duration of the IF signal proportionally increases the bandwidth of the chirp. 请注意，增加IF信号的持续时间能够成正比增加线性调频脉冲的宽度 So that gives us a clue that possibly a larger bandwidth corresponds to a better range resolution. 这就提示我们，更大的带宽可能对应更好的距离分辨率 Now that we have some intuition on how to improve the range resolution of radar, it would be nice to go a step further and actually derive an expression for this range resolution. 现在，我们对如何提高雷达的距离分辨率有一些直观的认识，我们最好更进一步，实际推导该距离分辨率表达式 And as it turns out, it’s not that difficult either. 事实证明，这其实没有那么困难 All you need to know are these two pieces of information, something that we’ve already learned before. 您需要知道的就是这两条信息，我们以前已了解过 So at this point, I would really like to encourage you to pause here and try and derive this expression for the range resolution. 那么，此时，我非常想鼓励您在这里暂停，尝试推导该距离分辨率表达式 So two objects, which are delta d apart in distance, will have the IF frequencies separated by delta f, given by this expression. 那么，两个间距为delta d的物体，其IF频率间隔为delta f，由该表达式给出 For these two frequencies to show up as distinct peaks in the IF frequency spectrum, this frequency separation delta f must be greater than 1 by the duration of the IF signal, which is virtually equal to the duration of the chirp TcC if you discount the small fraction in the beginning, the portion tau arising from the round trip delay. 为了使这两个频率在IF频谱中显示不同的值，该频率间隔delta f必须大于1除以IF信号持续的时间，如果您忽略开始的一小部分，即往返延迟产生的tau部分，那么这基本上等于线性调频脉冲的时间Tc So substituting for using this expression, you know we get this inequality here, which after some rearrangement becomes this. 那么，替换该表达式，您知道我们会得到这里的不等式，在进行一些整理之后变成这样 And note that the slope times the duration of the chirp is actually the bandwidth of the chirp. 请注意，斜率乘以线性调频脉冲的持续时间实际上使线性调频脉冲的带宽 So this expression can be further simplified, and you finally get this expression here, which says that two objects can be separated in the IF frequency spectrum as long as the distance (separation) between them is greater than the ratio of the speed of light to twice the bandwidth of the chirp. 因此，可以进一步简化该表达式，您最终得到这里的表达式，它表示，只要两个物体之间的距离(间隔)大于光速与线性调频脉冲带宽的两倍之比，就可以在IF频谱中分离它们 So the takeaway here is that the range resolution depends only on the bandwidth swept by the chirp, and is given by this expression over here-- speed of light divided by twice the bandwidth. 那么，这里的重点是，距离分辨率仅取决于线性调频脉冲覆盖的带宽，由这里的表达式给出–光速除以带宽的两倍 Time for a question now. 现在有两个问题 So you have two chirps here-- Chirp A and Chirp B. 那么，这里有两个线性调频脉冲–线性调频脉冲A和B Chirp A has twice the duration of Chirp B. A的持续时间是B的两倍 But both of them have the same bandwidth. 但它们具有相同的带宽 Which of these two chirps gives you a better range resolution? 这两个线性调频脉冲中的哪一个可以提供更好的距离分辨率？ So if you think about it, both of these chirps have the same bandwidth, B. 那么，如果您对其进行考虑，这两个线性调频脉冲具有相同的带宽B So from the formula that we just derived, c by 2B, both of them should have the same range resolution. 因此，根据我们刚刚推导的公式c除以2B，它们应具有相同的距离分辨率 But then, Chirp A has a longer duration and hence a longer observation window of the IF signal. 但是，线性调频脉冲A具有更长的持续时间，因此具有更长的IF信号观测窗口 So intuitively, if you take into consideration the properties of Fourier transforms, this chirp, Chirp A, should have a better resolution than just Chirp B. 因此，凭直觉，如果您考虑傅里叶变换的性质，线性调频脉冲A的分辨率应好于线性调频脉冲B How do we resolve this contradiction? 我们如何解决该矛盾 Something for you to think about. 请考虑一下这个问题 So we’ve talked about the IF signal, and that the frequency of tones in the IF signal is directly proportional to the range of objects. 那么，我们已经讨论了IF信号，IF信号中音调频率与物体的距离成正比 In most radars, what happens is that the IF signal is digitized for subsequent processing. 在大多数雷达中，发生的情况是，会对IF信号进行数字化，以供后续处理 So it’s first low pass filtered, and then digitized by an ADC, and sent to a suitable processor such as a DSP. 那么，它首先会进行低通滤波，然后由ADC进行数字化，接着被发送到合适的处理器，如DSP The DSP could begin by doing a Fourier transform to estimate the range of objects, and subsequently do other kinds of processing to estimate the velocity and angle of arrival of these objects. DSP可能首先执行傅里叶变换，以估算物体的距离，随后执行其他种类的处理，以估算这些物体的速度和到达角 And this is something we’ll get to in subsequent modules. 这是我们将在后续模块中讨论的内容 When ever we are digitizing a signal. 每当我们要对信号进行数字化时 we need to know what is the bandwidth of interest so that the low pass filter and the ADC sampling rate can be appropriately set. 我们就需要知道目标带宽，以便可以适当地设置低通滤波器和ADC采样率 So let’s say we are interested in objects from zero to a maximum distance of, say, dmax. 那么，假设我们对零到最大距离dmax之间的物体感兴趣 The maximum IF signal. 最大的IF信号 The maximum frequency of the IF signal is then going to be S2dmax/c. IF信号的最大频率将为S2dmax/c And correspondingly, the bandwidth of interest is going to be from zero to this maximum IF frequency which means that the low pass filter should have a cut-off frequency, which is beyond this IF max. 相应地，目标带宽将从零到该最大IF频率，这意味着低通滤波器的截止频率应高于该IF_max And also the ADC should have a sampling rate which is greater than the same value. 此外，ADC具有高于该同一值的采样率 So you can see here that the maximum sampling rate of the ADC can limit the maximum distance that the radar can see. 因此，您可以在这里看到，ADC的最大采样率可能会限制雷达可以看到的最大距离 Note that the maximum IF bandwidth depends on the product of the slope and the maximum distance. 请注意，最大IF带宽取决于斜率与最大距离的乘积 So if the ADC sampling rate and hence the IF bandwidth is a bottleneck in the sensor, you can always trade off the slope and the maximum distance. 因此，如果ADC采样率和IF带宽是传感器的瓶颈，那么您始终可以对斜率和最大距离进行折衷。 And typically, radar’s tend to use smaller slopes for larger dmax. 通常，雷达倾向于针对较大的dmax使用较小的斜率 Time for another question. 现在有另一个问题 Re-visiting our earlier example, what more can we say about these two chirps? 重新查看我们先前的示例，对于这两个线性调频脉冲，我们还可以讨论什么？ Chirp A and B have the same bandwidth. 线性调频脉冲A和B具有相同的带宽 But Chirp A takes twice as long as Chirp B. 但线性调频脉冲A的长度是B的两倍 A good time pause the video, and try to answer this question. 现在您应该暂停一下视频，尝试回答该问题 Since both A and B have the same bandwidth, they of course, have the same range resolution. 由于A和B具有相同的带宽，因此它们当然具有相同的距离分辨率 But note that Chirp A has half the slope of Chirp B. 但是，请注意，线性调频脉冲A的斜率是B的一半 So for the same maximum range requirement, or for the same dmax, Chirp A would require only half the IF bandwidth, which translates to an ADC with a smaller sampling rate. 因此，对于相同的最大距离要求，或对于相同的dmax，线性调频脉冲A仅需要一半的IF带宽，这意味着ADC具有较小的采样率 So while Chirp A has the advantage of a more relaxed ADC requirement, Chirp B of course has the advantage of requiring only half the measurement time. 因此，线性调频脉冲A具有ADC要求更宽松的优势，而线性调频脉冲B当然也具有仅需要一半测量时间的优势 So that is the trade going on here. 那么，这就是需要进行折衷的地方 So this slide summarizes all that we have discussed so far. 那么，该幻灯片总结了到目前为止我们已经讨论的所有内容 This is a block diagram of an FMCW radar with a single transmit and a single receive antenna. 这是具有单个发射天线和单个接收天线的雷达方框图 Let’s go with the sequence of events involved in estimating the range of an object. 让我们来查看估算物体的距离所涉及的一系列事件 So first, the synthesizer or synth generates a chirp. 那么，首先，合成器生成一个线性调频脉冲 This chirp is transmitted over the TX antenna. 该线性调频脉冲通过TX天线进行发射 It is reflected off multiple objects in front of the radar. 到达雷达前方的多个物体后反射回来 And the receiver sees delayed versions of this chirp. 接收器看到该线性调频脉冲的延迟版本 The received signal and the transmitted signal are mixed to create an IF signal. 接收到的信号和发射的信号进行混合，从而产生IF信号 This IF signal consists of multiple tones and the frequency of each of these tones is proportional to the range of the corresponding object. 该IF信号包含多个音调，其中每个音调的频率与对应物体的距离成正比 The IF signal is then low pass filtered and digitized. 然后，IF信号进行低通滤波并数字化 And note that the sampling rate of the ADC must be commensurate with the maximum distance that we wish to see. 请注意，ADC的采样率必须与我们希望看到的最大距离相称 The digitized data is then processed. 然后对数字化数据进行处理 An FFT is performed on this data. 对该数据执行FFT And location of the peaks in the frequency spectrum directly correspond to the range of objects. 频谱中的位置直接对应于物体的距离 Note that here I’ve plotted the FFT with range on the x-axis rather than the IF frequency, which is OK. 请注意，在这里，我绘制FFT时x轴上显示的是距离，而不是IF频率，这是可以的 Because as we’ve learned, the IF frequency is directly proportional to the range. 因为正如我们所了解到的，IF频率与距离成正比 This FFT is called range FFT because it resolve objects in range. 该FFT称为距离FFT，因为它在距离方面对物体进行解析 And this term range FFT is something that you will see a lot in FMCW literature. 对于距离FFT这一术语，您将在FMCW文献中经常看到它 This slide over here summarizes some of the key concepts and formulas that we see in this module. 这里幻灯片总结了我们在该模块中看到的部分关键概念和公式 First, an object at a distance of D produces an IF frequency of S2d/c. 首先，距离为d的物体会生成IF频率S2d/c Range resolution depends only on the bandwidth spanned by the chirp and is given by the speed of light divided by twice the bandwidth. 距离分辨率仅取决于线性调频脉冲跨越的带宽，由光速除以带宽的两倍给出 The ADC sampling rate Fs, limits the maximum range dmax that the radar can see. ADC采样率Fs限制雷达可以看见的最大距离dmax The other thing: when we talk about bandwidth and FMCW radars, there are usually two bandwidths that are important. 其他要点：当我们讨论带宽和FMCW雷达时，通常有两种重要的带宽 The RF bandwidth and the IF bandwidth. 射频带宽和IF带宽 And it’s important to clearly distinguish between both of these. 应清晰地区分这两者，这一点很重要 So the RF bandwidth is the bandwidth spanned by the chirp. 那么，射频带宽是线性调频脉冲跨越的带宽 A larger RF bandwidth directly translates to a better range resolution. 较大的射频带宽可直接转换为较好的距离分辨率 RF bandwidths are typically in the range of a few hundred of megahertz to several gigahertz. 射频带宽的范围通常为几百MHz至几GHz An RF bandwidth of 4 gigahertz, for example, translates to a range resolution of 4 centimeters. 例如，4GHz的射频带宽可转换为4厘米的距离分辨率 An RF bandwidth of 400 megahertz translates to a range of resolution of about 30 centimeters. 400MHz的射频带宽可转换为大约30厘米的距离分辨率 The other bandwidth is the IF bandwidth. 另一种带宽是IF带宽 A larger IF bandwidth primarily enables the radar to see a larger maximum distance. 较大的IF带宽主要可以使雷达看到较大的最大距离 Also enables faster chirps. 还可以实现较快的线性调频脉冲 By faster chirps, I mean chirps with higher slopes. 我说较快的线性调频脉冲，是指具有较高斜率的线性调频脉冲 The IF bandwidth of typical radars is in the low megahertz region. 典型雷达的IF带宽处于低MHz范围内 So that’s one of the things about FMCW radars, that you can have an RF signal spanning a large bandwidth of, say, 4 gigahertz, but yet your ADC would only need to sample a signal of a few megahertz. 那么，这是有关FMCW雷达的要点之一，您可以具有跨越较大带宽的射频信号，比如4GHz，但您的ADC仅需要对几MHz的信号进行采样 This concludes the first module in the series. 该系列的第一个模块到此结束 We focused primarily on range estimation using the FMCW radar. 我们主要讨论了如何使用FMCW雷达进行距离估算 Here’s a question to motivate the subsequent modules. 这里有一个问题，用以引出后续模块 So there are two objects equidistant from the radar. 有两个与雷达的距离相同的物体 How will the range FFT look like? 距离FFT看起来会使什么样的？ Now since these objects are equidistant, the range FFT will have a single peak corresponding to this range d and incorporating the effects of both these objects. 现在，由于这些物体的距离是相同的，因此距离FFT将具有与该距离d相对应并受到这两个物体影响的单个峰值 So how then do we separate out these two objects. 那么，我们如何分离这两个物体呢 It turns out that if these two objects have different velocities relating to the radar, then they can be separated out by further signal processing. 事实证明，如果这两个物体具有不同的相对于雷达的速度，那么可以通过进一步的信号处理分离它们 And to understand that, we need to really look at the phase of the IF signal which is something we will be doing in the next module. 要理解这一点，我们需要实际看看IF信号的相位，这是我们将在下一个模块中介绍
