FMCW Radars - Module 5 : Angle Estimation(介绍mmWave传感器：FMCW雷达-模块5：角度估算)

Welcome to this fifth module in this introductory series on FMCW radars.
欢迎观看有关FMCW雷达的本介绍系列的第五个模块
The past four modules have focused on sensing along two dimensions, range and velocity.
前面的四个模块重点介绍沿距离和速度两个维度进行感应
This module is going to focus on sensing along the third dimension, namely angle.
该模块将重点介绍第三个维度，即角度，进行感应
So the kind of questions we are going to answer in this module are, you have a radar, you have an object in front of it, how does the radar estimate the angle of arrival of this object?
因此，我们要在本模块中回答的问题就是，您有一个雷达，它前面有一个物体，雷达如何估算该物体的到达角
What if there are multiple objects at different angles but possibly have the same range and same relative velocity?
如果在不同的角度存在多个物体，但可能具有相同的距离和相同的相对速度，会怎么样？
What determines the maximum angular field of view of the radar?
雷达的最大角度视场有哪些决定因素
What does the angular resolution of the radar depend on?
雷达的角度分辨率取决于什么
Recall from our modules that the phase of the IF signal is very sensitive to small changes in the distance of the object, specifically a small change delta d in the distance of the object results in a phase change omega given by 4 pi delta d by lambda.
回忆一下我们之前的模块内容，您应该记得，IF信号的相对物体距离的微小变化非常敏感，具体而言，物体距离的微小变化deltad会导致相位变化omega，其值为4pideltad除以lambda
Angle estimation exploits a similar concept.
角度估算利用了类似的概念
Angle estimation requires at least 2 RX antennas.
角度估算需要至少两个RX天线
What is exploited here is the differential distance of the object to each of these antennas.
这里利用的是物体相对每个天线的差分距离
So the transmit antenna transmits a signal that is a chirp.
那么，发射天线发射一个线性调频脉冲信号
It is reflected off the object, and you can imagine one ray going from the object to the first RX antenna and another ray going from the object to the second RX antenna.
它在物体上进行反射，您可以想象一束射线从物体到达第一个RX天线，另一束设线从物体到达第二个RX天线
And in this example, the ray to the second RX antenna has to travel a little longer, an additional distance of delta d, to get there.
在该示例中，到达第二个RX天线的射线必须传播稍远一点的距离，即额外的距离deltad，才能到达那里
This additional distance results in an additional phase of omega equal to 2 pi delta d by lambda.
该额外的距离会导致额外的相位omega它等于2pideltad除以lambda
So this is the phase difference between the signal at this antenna and the signal at this antenna.
这便是该天线处的信号和该天线处的信号之间的相位差
So we can see here that these two expressions are very similar, and actually they’re almost the same except for a factor of 2.
因此，我们可以在这里看到，这两个表达式非常类似，实际上，除了因数2之外，它几乎是相同的
So something for you to think about.
接下来，请你思考下面的问题
Why are these two expressions off by a factor of two.
这两个表达式为什么具有因数2的差异
So this figure here explains the dependence of this additional distance on the angle of arrival.
这个图说明了该额外距离与到达角的相关性
The object is assumed to be far enough compared to this distance d between the two antennas, that the rays from the object to the RX antennas are assumed to be parallel.
假设与两个天线之间的距离d相比，物体足够远，从而可以假设从物体到达RX天线的射线是平行的
So here I apologize for this abuse of notation.
抱歉，我滥用了符号
Henceforth in this module, d will actually refer to the distance between two consecutive antennas.
在该模块的后续部分，d实际上指两个连续天线之间的距离
So here d is the distance between the two RX antennas and theta is the angle of arrival of the object with respect to the radar.
因此，这里的d是两个天线之间的距离，theta是物体相对于雷达的达到角度
And this is the additional distance of the ray to the second antenna compared to the first antenna.
这是与第一个天线相比，设线到第二个天线的额外距离
And using basic geometry, you can see that this is a right-angle triangle here with the hypotenuse being d and this angle theta.
使用基本的几何学知识，您可以发现这是一个直角三角形，其中斜边为d，以及这个角theta
So this additional distance that the ray has to travel turns out to be d sine theta.
因此，设线必须传播的该额外距离结果为dsintheta
So the transmitter antenna transmits a frame a chirps, and the data is received at each of these antennas.
那么，发射器天线发射一个线性调频脉冲帧，每个天线会接收该数据
And each of these antennas processes the data to create a 2D-FFT matrix with a peak corresponding to the range and velocity of the object.
每个天线会处理该数据，以创建一个2D-FFT矩阵，其中包含与物体的距离和速度相对应的峰值
So here you have the 2D-FFT peak corresponding to this receiver and another 2D-FFT matrix corresponding to this receiver.
那么，这里是与该接收器相对应的2D-FFT峰值，以及与该接收器相对应的另一个2D-FFT矩阵
Note that the location of the peak is going to be virtually identical for both of these 2D-FFTs.
请注意，峰值的位置将几乎与这两个2D-FFT相同
We have discussed earlier that the peak location is very insensitive to small changes small changes in distance between the radar and the object.
我们先前已讨论过，峰值对应位置对雷达与物体之间距离的微小变化非常敏感
However, the phase difference between these two peaks is going to be given by 2 pi d sine theta, d sine theta being the additional distance, divided by lambda.
不过，这两个峰值的相位差将为2pidsintheta，dsintheta是额外的距离，然后除以lambda
And once you have measured this phase difference by comparing these two signals, the signals at these two peaks, you can just invert this equation to calculate the angle of arrival.
您通过比较这两个到达信号，这两个峰值处的信号来测量该相位差之后，您可以求该公式的反函数，以计算到达角
If you examine this expression omega equals 2 pi d sine theta by lambda, the relationship between the quantity that we want to estimate, namely theta, and the measured phase difference omega is a nonlinear one because of sine theta.
如果您看看该表达式，omega等于2pidsintheta除以lambda，您会发现，由于sintheta，我们要估算的量，即theta，与所测量的相位差之间的关系是非线性的
So this is the first time in the series we’re encountering such a situation.
那么，这是在该系列中我们第一次遇到这种情况
So if you would recall, the expression from velocity estimation from the second module is given by omega equals 4 pi V Tc by lambda.
那么，您可以回想一下，第二个模块中速度估算中的表达式为omega等于4pivTc除以lambda
And you can see that the estimated phase difference depends linearly on the velocity.
您可以看到，估算的相位差以线性方式依赖于速度
Likewise, in the first module, the relationship between the IF frequency and the range is linear.
类似地，在第一个模块中，IF频率与距离之间具有线性关系
So if you look at this plot of sine theta versus theta, close to theta equal to 0, since theta is very sensitive to theta.
因此，如果您看看该sintheta与theta图，当theta接近于0时，sintheta对theta非常敏感
So small changes in theta produce comparable changes in sine theta.
因此，theta的微小变化会导致sintheta产生同等大小的变化
But this sensitivity decreases as theta increases.
但该敏感性会随着theta的增大而降低
And close to theta equal to 90, sine theta becomes quite insensitive to theta.
当theta接近于90度时，sintheta变得对theta非常不敏感
Hence, estimation of theta is more error prone as theta increases.
因此，随着theta逐渐增大，theta估算越来越容易产生误差
That is shown here in this figure over here.
这里的图对此进行了显示
Angle estimation is best when the object is directly in front of the radar.
当物体位于雷达正前方时，角度估算会达到最佳值
That is theta equal to 0.
这意味着theta等于0
And as theta increases and approaches 90 degrees, the estimation accuracy decreases, of course, in the presence of noise.
随着theta逐渐增大并接近于90，当然，由于存在噪声，估算准确性会下降
In the earlier module, when we were talking about velocity estimation, we saw that there was an upper limit on the velocity that can be measured unambiguously by the radar.
在前面的模块中，当我们讨论速度估算时，我们看到雷达能够不模糊测量的速度具有上限
We called this the maximum velocity of the radar.
我们将此称为雷达的最大速度
Is there a similar limit on the maximum unambiguous angle that can be measured by the radar?
雷达可以测量的最大不模糊角度是否存在类似的限制
And that is what we are going to discuss in this slide.
这正是我们要在该幻灯片中讨论的内容
So if you imagine this representing the phasor corresponding to the peak of the 2D-FFT for an object to the left of the radar, this phasor moves anticlockwise when you go from the first RX antenna to the second RX antenna.
那么，如果您将这想象成表示与雷达左侧物体的2D-FFT峰值相对应的相量，当您从第一个RX天线转到第二个RX天线时，该相量沿逆时针方向移动
Likewise, for an object to the right of the radar, the phasor moves clockwise, as shown here.
类似地，对于雷达右侧地物体，相量将沿顺时钟方向移动，如此处所示
So the measurement of this movement is unambiguous, only as long as the movement either in the clockwise or in the anticlockwise direction is less than 180 degrees or pi radians.
因此，只要沿顺时针或逆时针方向方向的移动小于180度或pi弧度，对该移动的测量就是不模糊的
If that were not the case, as illustrated here, that this phasor has moved from here to here, one cannot say if this movement is because of movement of a degrees in the anticlockwise direction or a movement of b degrees in the clockwise direction.
如果情况不是这样，如此处所示，该相量从这里移动到这里，那么您就无法分辨该移动是沿逆时针方向移动了响应的度数，还是沿顺时针方向移动了b度数
So basically, unambiguous measurement of velocity requires that the phase change between the two antennas be less than 180 degrees.
因此，基本而言，速度的不模糊测量要求两个天线之间的相位变化小于180度
And plugging in the value of omega from the previous slides and rearranging this a little bit, we see that the maximum angle that can be measured by the radar has to be less than this expression over here.
然后，插入前面的幻灯片中的omega值，并稍微对其进行重新排列，我们看到雷达可以测量的最大角度必须小于这里的表达式
Sine inverse of lambda by 2d, while d is the spacing between the two antennas.
lambda除以2d的反正弦，其中d是两个天线之间的距离
So basically the takeaway here, the maximum field of view that can be serviced by two antennas space d apart is given by theta max equals sine inverse of lambda by 2d.
所以，此处的重点是，两个天线之间的距离d可以服务的最大视场为thetamax除以2d的反正弦
Note that a spacing d of lambda by 2 between the two antennas results in the largest possible field of view of plus or minus 90 degrees.
请注意，在两个天线之间，lambda除以2再除以距离d可导致最大的可能视场正90度或负90度
So far, we’ve talked about a single object in front of the radar and estimating its angle of arrival by measuring the phase difference across two RX antennas.
到目前为止，我们已经讨论了雷达前方的单个物体，以及通过测量两个RX天线之间的相位差来估算其到达角
Now consider two objects in front of the radar at the same range and the same relative velocity with respect to the radar, such that both these objects fall in the same range velocity bin in the 2D-FFT.
现在请考虑雷达前方的两个物体，它们相对于雷达具有相同的距离和相同的速度，这样这两个物体将处于2D-FFT中相同的距离速度单元
So this is the 2D-FFT from the first antenna.
那么，这是来自第一个天线的2D-FFT
And this is the 2D-FFT from the second antenna.
这是来自第二个天线的2D-FFT
So there is a single peak in this 2D-FFT.
因此该2D-FFT中有单个峰值
But the signal at the peak will have contributions from phasors corresponding to both these objects, as illustrated here for this antenna and here for the second antenna.
但峰值处的信号将具有来自对应于这两个物体的相量贡献，这里对这个天线进行了说明，这里对第二个天线进行了说明
So the simple phase comparison technique that we talked about earlier will not work.
因此，我们先前讨论的简单相位比较技术将不再适用
What is the solution?
解决方案是什么
It’s going to be very analogous to what we did earlier with the module on velocity.
它将非常类似于我们先前对速度模块采取的解决方案
We increase the number of RX antenna from two and create an area of N antennas N receive antennas and the two-dimensional FFT at all these antennas is going to have a peak at the same location.
我们将RX天线的数量从两个增加到N个，并创建N个接收天线的区域，所有这些天线处的二维FFT将在同一个位置具有一个峰值
And the signal at this series of peaks is going to create a discrete sequence consisting of two rotating phases, as shown here.
该系列峰值处的信号将创建一个包含两个旋转相位的离散序列，如此图所示
And FFT on the sequence will show up as two peaks at omega 1 and omega 2, where omega 1 and omega 2 are the rates of rotation in radians per sample for the two objects.
该序列上的FFT将显示为omega1和omega2处的两个峰值，其中omega1和omega2是两个物体的旋转速度，以每个样本的弧度数为单位
So you read the location of these two peaks from the FFT, and then just back-calculate the angle of arrivals for the two objects, as shown here.
那么，您通过FFT读取这两个峰值的出现位置，然后反演计算两个物体的到达角，如此图所示
We call this particular FFT here, which is performed across RX antennas, as the angle FFT
我们将此处在RX天线上执行的特定FFT称为角度FFT
Now that we have introduced the angle FFT, the next natural question is, what is the resolution of this FFT?
现在我们已经介绍了角度FFT，那么下一个问题自然是，该FFT的分辨率是多少
That is for two objects with angles of arrival of theta and theta plus delta theta relative to the radar.
该问题针对两个物体，它们相对于雷达分别具有theta和theta加dletatheta的到达角
How small can delta theta get with the two objects showing up as distinct peaks in the angle FFT?
当两个物体在角度FFT中显示为不同的峰值时，deltatheta可以达到多小的值
This is something that we’ve done for range resolution in module 1, velocity resolution in module 2.
我们已经在模块1中针对距离分辨率以及在模块2中针对速度分辨率解决了该问题
And we’ll do something very similar here for angle resolution.
在这里，我们将针对角度分辨率执行非常类似的操作
To derive this expression for angle resolution, all you need to remember are the following: an object with an angle of arrival of theta has a discrete frequency of omega given by 2 pi d sine theta by lambda , d being the separation between the antennas.
要针对角度分辨率推导该表达式，只需记住以下两点：到达角为theta的物体具有离散频率omega，它为2pidsintheta除以lambda，d为两个天线之间的距离
And the criterion for separation in the frequency domain is that the separation of angular frequencies delta omega must be greater than 2 pi by N, N being the number of samples in the FFT.
频域中距离的标准是，角频率的距离deltaomega必须大于2pi除以N，N为FFT中的样本数
So at this point, you can perhaps pause the video and try to derive this question of this expression yourself.
那么，此时，您或许可以暂停视频，尝试自己推导处该表达式
Two objects with angles of arrival of theta and theta plus delta theta will have their discrete frequencies separated by this expression over here.
到达角分别为theta和theta加deltatheta的两个物体将通过这里的表达式区分其离散频率
Now to simplify this expression, remember from calculus that the derivative of sine theta is cos theta, which means that the ratio of change in sine theta for a corresponding small change delta theta is equal to cos theta.
现在，为了简化该表达式，您应该记得在微积分中，sintheta的导数是costheta，这意味着对应的微小变化deltatheta的sintheta变化了等于costheta
So I can replace this expression here with cos theta times delta theta, which is what I do here.
因此，我可以将这里的表达式替换为costheta乘以deltatheta，就像这样
Also recall from the properties of discrete Fourier transforms that we discussed in module 2, that for our two peaks to be separated in the frequency domain, the separation of their angular frequencies should be greater than 2 pi by N, N being the number of samples input to the FFT, which in this case is also the number of antennas in the array.
此外，请回忆一下我们在模块2中讨论过的离散傅里叶变换的特性，对于我们要在频域中分隔的两个峰值，其角频率的距离应大于2pi除以N，N为FFT的样本输入数量，在本例中也是阵列中天线的数量
So substituting this expression over here, I get this inequality and then rearranging this a little bit, I get this inequality which is actually the expression for the angle of resolution of the radar.
那么，替换此处的表达式，我得到这个不等式，然后稍微对其进行重新排列，我得到这个不等式，它实际上是表示雷达分辨率角度的表达式
Also note that since N is the number of antennas in the array, and d is the space in between consecutive antennas, N times d is actually the length of the antenna array.
还要注意，由于N是阵列中天线的数量，d是两个连续天线之间的距离，因此N乘以d实际上是天线阵列的长度
So you can say that the angular resolution increases as the length of the receiver antenna array increases.
因此，您可以说角分辨率随着接收器天线阵列的长度的增大而增大
Now, resolution is often quoted assuming that the interval antenna spacing is lambda by 2 and that theta is 0.
现在，当我们说分辨率时，通常假设两个天线之间的距离为lambda除以2，并且theta为0
And substituting that in this expression you get this expression for the angle of resolution, which is actually independent of theta and the interval antenna spacing.
在该表达式中替换它，您将得到该表达式实际上以弧度为单位，它实际上取决于theta和天线之间的距离
One thing to remember is that this expression here is actually in units of radians.
需要记住的一点就是，这里的表达式实际上以弧度为单位
So you would need to multiply this by 180 divided by pi to convert it to degrees.
因此，您需要将其乘以180度再除以pi，以将其转换为度数
If you look at this expression, one interesting thing is that the angle of resolution depends on theta.
如果您看看该表达式，有一点很有意思，即分辨率角度取决于theta
This is something that we did not see in the context of range or velocity.
这在距离或速度背景中未出现过
So both the range and velocity resolution were independent of actual values of range and velocity.
因此距离和速度分辨率都独立于实际的距离和速度值
The reason for this is, again, the nonunion nature of sine theta.
其原因还是在于sintheta的不一致特性
So for two objects separated by delta theta, their angular frequencies in the angle FFT are actually further apart at theta equal to 0 and come closer to each other as theta increases, even though the separation delta theta is the same in both cases.
因此，对于由deltatheta分隔的两个物体，其在角度FFT中的角频率在theta等于0实际上离得更远，并且随着theta的增大彼此越来越靠近，即使在两种情况下距离deltatheta相同也是如此
As you might have probably figured out by now, angle estimation and velocity estimation in FMCW radar depend on very similar concepts.
正如您现在可能已经了解的那样，FMCW雷达中的角度估算取决于非常类似的概念
Actually, the mathematical underpinnings are almost identical.
实际上，数学基础几乎是相同的
So I thought it would be useful to have a slide comparing the two.
因此我认为通过一个幻灯片来比较这两者应该有所帮助
Angle estimation exploits the phase change across chirps which are separated in space.
角度估算利用线性调频脉冲之间的相位变化，这些脉冲在空间上是分离的
Velocity estimation also exploits phase change, but it’s a across chirps, which are separated in time.
速度估算也利用相位变化，但它是针对在时间上分离的线性调频脉冲
Angle resolution depends on the antenna array length.
因此，角度分辨率取决于天线的阵列长度
Note that Nd actually is the total length of the antenna array.
请注意，Nd实际上是天线阵列的总长度
So angle resolution is inversely proportional to the length of the antenna array.
因此，角度分辨率与天线阵列的长度成反比
Velocity estimation is inversely proportional to the length of the frame.
速度估算与帧地长度成反比
The maximum angle that can be unambiguously measured depends on the distance, d, between two consecutive antennas.
可以不模糊测量的最大角度取决于这两个连续天线之间的距离d
Smaller this distance, the larger the theta max.
该距离越小，thetamax就越大
The maximum velocity that can be unambiguously measured depends on the time Tc between consecutive chirps.
可以不模糊测量的最大速度取决于连续线性调频脉冲之间的时间Tc
Again, smaller the Tc, the larger Vmax.
类似地，Tc越小，Vmax就越大
This slide summarize the angle estimation process and places it in context within the FMCW radar system.
此幻灯片总结了角度估算过程，并将其置于FMCW雷达系统的背景中
So you have synthesizer synth, also known as the LO or local oscillator, which generates a chirp.
那么，您具有合成器synth，也称为LO或本机振荡器，它可以生成线性调频脉冲
The chirp is transmitted by the transmit antenna.
线性调频脉冲由发射天线进行发射
It is reflected off objects in front of the radar.
它从雷达前方的物体上进行反射
And the reflective signal is received at each of these receive antennas.
在每个接收天线处接收反射信号
The synth signal is routed to each of these RX antennas, and the received signal is mixed with the LO signal to create an IF signal.
synth信号路由到其中的每个RX天线，接收到的信号与LO信号混合，以创建IF信号
The IF signal is low pass filter and samples by an ADC.
IF信号经过低通滤波器，并由ADC进行采样
This ADC data is processed across an entire frame to create a 2D-FFT grid, one such 2D-FFT grid for each of these RX antennas.
该ADC数据在真个帧中进行处理，以创建一个2D-FFT网络，针对每个RX天线创建一个此类2D-FFT网格
Peaks in this 2D-FFT grid correspond to the range and velocity of objects.
该2D-FFT网格中的峰值于物体的距离和速度相对应
An FFT across the corresponding peaks in this series of TX antennas is called an angle FFT, and peaks in this angle FFT directly correspond to the angle of arrival of objects.
该系列TX天线中对应峰值上的FFT称为角度FFT，该角度FFT中的峰值直接对应于物体的到达角
I’d like to close this module with a quick question.
我要通过一个简短的问题来结束该模块
So you have two stationary objects at the same range from the radar.
那么您有两个静止的物体，它们到达雷达的距离相等
And this radar has one transmit and two receive antennas.
该雷达具有要给发射天线和两个接收天线
Is it possible to estimate the angle of arrival of both these objects using this radar?
是否可以使用该雷达估算这两个物体的到达角
Something for you to think about.
请考虑一下这个问题
Anyway, this brings us to the end of this module.
总之，该模块就到此结束了
I hope that modules 1 through 5 have given you a reasonable understanding of millimeter wave sensing using FMCW radars and hopefully you enjoyed it too.
我希望通过对模块1至5，您已经对使用FMCW雷达进行毫米波传感有了一定的了解，并且希望您喜欢它
We do hope to add more modules in the future to build on what you’ve seen so far.
我们希望将来以您到目前为止学习到的内容为基础添加更多的模块
So until then, goodbye.
在那之前，再见

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