2.9 DOWN-RANGE AND CROSS-RANGE IMAGING – SIGNAL DEFINITION
We now want to extend our previous work to both down-range and cross-range imaging. We will also extend the problem to include a more general case of squinted SAR. Squinted SAR is normally associated with strip-map SAR but the development here also applies to spot-light SAR. As before, we will start by defining the signal that the SAR processor must work with since this will give insight into how to process the signal.
2.9.1 Signal Definition
The geometry of interest is a modification of the geometry of Figure 13 and is contained in Figure 24. The main difference between Figure 13 and Figure 24 is that in Figure 13 the center of the imaged area lies on the x-axis of the coordinate system while in Figure 24 it does not. This offset of the imaged area center will result in additional Doppler considerations plus a phenomena termed range cell migration (RCM), both of which complicate SAR processing. Another, minor, difference is that the coordinates of the scatterer are relative to the center of the imaged area. We did this as a convenience.
Since we are considering both down-range and cross-range imaging the transmit waveform will be pulsed instead of CW. In practical SAR, the pulses are phase coded, usually with LFM, to achieve the dual requirements of large bandwidth to achieve fine range resolution and long duration to achieve sufficient energy. In this development we will use narrow, uncoded (unmodulated) pulses to avoid complicating the development with pulse coding and the associated matched filter or stretch processing. The extension to coded pulses is relatively straightforward.
Given the above we write the transmit signal as
(68)
where
(69)
and is the pulse width. The sum notation means a sum over all and is used to indicate that the waveform is, in theory, infinite duration. We will later make it finite duration.
The signal from a single scatterer at (see Figure 24) is
(70)
where
. (71)
2.9.1.1 Removal of the Carrier and Gross Doppler
As before, the first operation we will perform is removal of the carrier. However, in addition, we will also remove what we term gross Doppler. Removal of gross Doppler is necessary in some applications in that this Doppler is large relative to the PRF and has the potential of causing problems with aliasing and Doppler ambiguities.
To determine the gross Doppler we examine the phase of the returned (RF) signal. From Equation (70) this phase is
. (72)
We can find the frequency from
. (73)
The first term is the carrier frequency and the second is the Doppler frequency. We define the gross Doppler as the Doppler frequency at , and . This yields
. (74)
In Equation (74), is the squint angle. For the unsquinted SAR we considered in the CW development, was zero because was zero.
Given the above, we remove and from the received signal by multiplying by the heterodyne signal,
. (75)
We also “normalize away” as before to leave the baseband signal
. (76)
2.9.1.2 Single-pulse Matched Filter
The next step in processing is to send through a matched filter matched to the transmit pulse. The (normalized) output of the matched filter is
(77)
where
. (78)
2.9.1.3 Generation of the Sampled Signal
Recall that for the CW case we sampled at intervals of . We will do the same for the pulsed case. However, for each pulse (each ) we will also subsample at intervals of , the pulse width. We will start sampling, relative to each transmit pulse, at some
(79)
and continue sampling to some
(80)
where
. (81)
Between and we obtain approximately
(82)
range samples. We will do this times to form samples, which we will collect into a by element array for further processing.
Mathematically, we sample at
(83)
to yield
. (84)
2.9.2 Preliminary Processing Considerations
If we were to directly extend our CW processing methodology we would, for each , remove a quadratic phase term and then perform a DFT across . Unfortunately, the situation is complicated by the range sampling so that this straightforward approach is not directly applicable. We will need to perform an interim step first.
2.9.2.1 Range Cell Migration Correction
Figure 25 contains a plot of for the parameters of EXAMPLE 1 (Section 2.8, Table 1) with the added condition that we are using a pulse with a pulse width of 0.5 m or . We are also considering a single target at and . With this resolution we will have 50/0.5 = 100 down-range cells. We force this to an odd number of 101 and index the range cells from -50 to 50. Thus goes from -50 to 50 and goes from -120 to 120. (We already determined that we had cross range samples.)
Since we are considering a single scatterer in the center of the imaged area we expect, at first blush, the return to be at range sample 0. However, this is not the case because the location of the target return depends upon range to the scatterer, not the x location of the target. Since the range is given is given by
(85)
it will vary as a function of . This is why we see a curved line in Figure 25 instead of the straight line we would like to see.
The problem with the curved line comes when we try to apply the quadratic phase correction and take the Fourier transform to form the image. Specifically, we want to perform both of these operations across for each range cell, i.e., each . In fact, we should apply the quadratic phase correction and Fourier transform along the curved line. We get around this problem by “warping” the plot of Figure 25 so that the curved line becomes a straight line. The method we use is interpolation. A straight forward method of performing this interpolation is via the Fourier transform method. Specifically, compute the Fourier transform (using the FFT) for each , apply the appropriate linear phase shift and take the inverse Fourier transform. The amount of linear phase shift depends upon the distance, in time, we want to move the samples. With some thought, it will be obvious that all range samples at the particular will be moved by the same amount.[6]
The algorithm we use is as follows: From Equation (85), we note that the minimum value of , when , occurs when and is equal to . We decide that we want this range to correspond to a down-range value of . For each we compute
. (86)
This then becomes the range correction based on the assumption that when . We use this with the Fourier transform method to move the samples in range. As an implementation note, for the 101 range cells of this example, I used a 128 point FFT. After the inverse FFT I simply discarded the last 27 elements of the array.
The result of applying the above methodology to the plot of Figure 25 is shown in Figure 26. As can be seen, the curved line of Figure 25 is now a straight line, albeit somewhat distorted. The distortion is due to the fact that we sample in range at multiples of the pulse width, . It turns out that this distortion causes problems in the image. We will address this later.
The methodology discussed above will adjust the returns from all scatterers by the same amount for each value of . Furthermore, the amount that the returns are adjusted is determined by the properties of a hypothetical scatterer located at the center of the imaged area. As an example, we consider three scatterers that are located at the same but at values of -23, 0 and 23 meters (range sample numbers of -46, 0 and 46). The resulting uncorrected plot of is shown in Figure 27. It will be noted that this figure uncovers a problem: The curved trace of the scatterer at is cutoff. To correct this problem we find that we need to sample over a larger number of range cells. From Figure 25 we note that the trace spans about 12 or 13 range samples. Thus, to be sure we get the complete trace for a scatterer at the far-range end of the imaged area we need to extend the range samples to an upper limit of 50+12. We will round this to 65. The result of this is shown in Figure 28. The resulting RCM corrected image is shown in Figure 29. It will be noted that there are three straight lines located at , as they should.
It will be noted that the curved lines of Figure 27 are the same, as are the three straight lines of Figure 28. As another example we place the three scatterers at . That is, at diagonal corners and the center of the imaged area. The resulting uncorrected and corrected plots of are shown in Figures 30 and 31. Careful examination of Figure 30 shows that the three curved lines are not exactly the same. Also, the top and bottom straight lines of Figure 31 are not exactly horizontal. It turns out that, in some applications, this can cause problems and an interim processing step must be used to eliminate the problem. This interim step is discussed in Section 2.12.
2.9.3 Quadratic Phase Removal and Image Formation
Now that we have an algorithm that performs RCMC we need to develop an algorithm for removing the quadratic phase. We will want to remove the quadratic phase from the RCMCed signal. The information we need is in the phase of (Equation (84)) at the peak of the tri( ) function (i.e. along the curved ridge before RCMC).
If we refer to of Equation (77) we find we want to examine the information in the phase of at
. (87)
The problem with this equation is that appears on both sides of the equation and is embedded in a square root on the right side. As a result, solving for will involve the solution of a rather complex quadratic equation. To avoid this we seek a simpler approach. Specifically, we ask the question: Does the phase of vary slowly enough to allow the use of an approximate value of ?
We can write the phase of , from Equation (77), as
. (88)
From calculus we know that we can relate variations in to variations of by
. (89)
If we perform the partial derivative we get
. (90)
We are interested in the variation of over the times we are taking measurements. Specifically, from to . Thus, we use . Further, we let . With this we have
. (91)
Figure 32 contains a plot of vs. as the top plot. For reference, the bottom curve is a plot of pulse-to-pulse phase change vs. . As can be seen the pulse-to-pulse phase change ranges between about -1000 and +1000 degrees while the phase variation, or phase error, over is between -6×10-3 and +6×10-3 degrees. This says that varies slowly over , and thus, that it will be reasonable to compute at , or even rather than via the more accurate form of Equation (88).
Given this we can now examine to formulate a quadratic phase correction scheme. We can write
(92)
We recognize the first term of the last equality of Equation (92) as a constant phase that we do nothing about. The second term is zero since, by Equation (74), . Finally, the third term is the quadratic phase that we want to eliminate. It will be noted that this quadratic phase term is exactly the same as the quadratic phase term in the CW problem. Thus, to perform the quadratic phase correction we multiply each row of the RCMCed signal space array by
. (93)
We are now in a position to formulate an algorithm for creating a cross/down-range image.
2.10 ALGORITHM FOR CREATING A CROSS- & DOWN-RANGE IMAGE
· We assume we have a sampled base-band signal of the form given by Equation (84). Note: this signal has had the gross Doppler, , removed.
· Perform RCMC using the FFT methodology with the corrections given in Equation (86). The RCMC is applied to all range cells for each pulse (each ).
· Perform the quadratic phase correction by multiplying the returns for each range cell by the of Equation (93).
· Take the FFT across pulses, for each range cell.
· Transform the frequency and range delay axes of the output of the FFTs to cross-range and down-range, and plot the image.
2.11 EXAMPLE 2
As an example, we extend EXAMPLE 1, Section 2.8. Table 2 is a repeat of Table 1 with additions and modifications consistent with the cross- & down- range methodology.
Table 2 – Parameters Used in SAR EXAMPLE 2
Parameter / ValueWidth of image area, / 50 m
Depth of image area, / 50 m
SAR wavelength, / 0.03 m
Aircraft velocity, / 50 m/s
Synthetic array length, / 600 m
Number of scatterers, / 3
Waveform PRI, / 50 ms
Down-range resolution, / 0.5 m
Center of Imaged Area (m) / (20000,200)
Scatterer locations, (m) / (-23,0), (0,0), (23,0)
Scatterer powers, (w) / 1, 1, 1
As with EXAMPLE 1, we have so that goes from -120 to 120 and we transmit 241 pulses over a time period of -6 to 6 seconds. Given the down-range resolution of 0.5 m we compute a pulse width of
. (94)
From EXAMPLE 1, we recall that the cross-range resolution is also 0.5 m. From the previous examples, we recall that we need to extend by 6 to 8 m to account for the RCM of scatterers near the far down-range of the imaged area. These extra range cells need to be trimmed before we make the image. We also recall that, since our is smaller than the minimum dictated by the width of the imaged area our SAR image will need to be trimmed in cross-range before we make the image.
When we formed the image for EXAMPLE 1, we used an FFT length that was longer than the number of samples because we wanted a smooth linear plot. Since we are only forming an image for this example, we can limit the FFT length to the nearest power of two greater than . Since is 241, a 256 point FFT will suffice.
When I implemented the aforementioned algorithm, with the additional steps indicated in the previous paragraphs, the image of Figure 33 was the result.