SYNTHETIC APERTURE RADAR - UAH



2.5 SAR SIGNAL CHARACTERIZATION

In order to formulate a SAR processor we first need to characterize the signal that the SAR processor will operate upon. Although our previous discussions treated SAR cross-range imaging as an antenna problem, for the rest of the development we will cast the problem in the Doppler domain. Past experience in this course indicates that the Doppler formulation is easier to understand than the antenna formulation.

2.5.1 Derivation of the SAR Signal

As we have done so far, we will initially consider only the cross-range problem. We will later extend the discussions to the down-range and cross-range problem. Since we are considering the cross-range problem we start by considering a CW transmit signal of the form

[pic]. (24)

If we consider the geometry of Figure 13 the signal returned from the “nth” scatterer located at [pic] is

[pic] (25)

where

[pic][1] (26)

and [pic] is the y position of the aircraft at some time [pic]. If we assume that the aircraft is flying at a constant velocity of [pic] and [pic] occurs at [pic] of Figure 13 we get

[pic]. (27)

We assume that the total time for the aircraft to travel a distance of [pic] is [pic] and that the aircraft starts at [pic] when [pic]. With this we get

[pic]. (28)

We note that the area to be imaged has a cross-range width of [pic] and a down range length of [pic]. The region is centered in cross-range at [pic] and in down-range at [pic].

In Equation (25), [pic] is the normalized signal power associated with the nth scatterer. It is determined from the radar range equation, without the [pic] term (since it is incorporated into Equation (25)).

If we assume that [pic], [pic] and [pic] are small relative to [pic] we can replace [pic] in the denominator of Equation (25) by [pic] and write

[pic] (29)

where [pic] is the wavelength of the transmit signal.

Since the information needed to form the image is in the second exponential term we will eliminate the first exponential term by heterodyning (which is done in the actual radar). We also eliminate the [pic] term through normalization to yield the baseband signal

[pic]. (30)

If we have [pic] scatterers in the image region the resulting baseband signal would be

[pic] (31)

2.5.2 Examination of the Phase of the SAR Signal

Since the information we seek is in the phase of [pic] we want to examine it. To proceed we need to examine [pic]. We can write

[pic]. (32)

We note that [pic], [pic] and [pic]. This means that the second and third terms of the last square root are small relative to 1. This, in turn, allows us to write

[pic]. (33)

If we substitute this into Equation (30 we get

[pic]. (34)

2.5.2.1 Linear Phase, or Constant Frequency, Term

The first exponential is a phase caused by range delay to the scatterer that we must live with. The second term is a linear phase term or a term that we associate with frequency. In fact, the frequency represented by this term is

[pic]. (35)

This tells us that [pic] has a constant frequency term that depends upon the scatterer cross range position, [pic]. [pic] also depends upon the aircraft velocity, [pic], and the radar wavelength, [pic]. However both of these are known (and fixed). Finally, [pic] also depends upon [pic]. If we assume that all of the scatterers are at the same [pic] (which we can do since we are only concerned with the cross-range problem) and note that [pic] we get the previous assertion that

[pic]. (36)

This discussion tells us that, if we can determine [pic], we can determine [pic] from

[pic]. (37)

2.5.2.2 Quadratic Phase, or LFM, Term

The third exponential of Equation (34) is a quadratic phase, or linear frequency modulation, term and causes problems. We can write the quadratic phase as

[pic]. (38)

With the previous assumption that [pic], [pic] is approximately the same for all scatterers. This means that we can remove it by a mixing or heterodyning process.[2] If we do this we will be left with only the magnitude, constant phase term and the [pic]-dependent frequency term. This is what we want.

2.5.3 Extracting the Cross-Range Information

Once we remove the quadrtic phase we have

[pic] (39)

for a single scatterer. For the more general case of [pic] scatterers we have

[pic]. (40)

The forms of Equations (39) and (40) tell us that we can extract the information we want by taking the Fourier transform of [pic] or, more generally, [pic]. From our experience with Fourier transforms, this will give us a response that peaks at the frequencies [pic]. The heights of the peaks will be proportional to [pic]. If we use Equation (37) to plot this as intensity ([pic]) vs. [pic] we have a one dimensional image.[3]

To verify the above, and begin to understand SAR resolution properties, we will find the Fourier transform of [pic]. Indeed

[pic]. (41)

To perform the integration we need to realize that [pic] and thus [pic] is measured only over [pic]. Thus, we assume that [pic] is zero outside of these limits and write

[pic] (42)

where

[pic]. (43)

Figure 14 contains a plot [pic] vs. [pic]. It will be noted that the response has a peak at [pic], or at [pic] and that the peak has a height of [pic]. The width of the peak is [pic] which means that the SAR image will have a resolution of

[pic]. (44)

If we change the horizontal axis to [pic] using the relation (see Equation (37))

[pic] (45)

we get the plot of [pic] vs. [pic] of Figure 15. This plot has a peak at [pic] with a height of [pic]. From Equation (44), if the resolution of Figure 14 is [pic] the resolution of Figure 15 is

[pic] (46)

since [pic]. This is the same as the resolution we obtained from the linear array approach (see Equation (18)).

2.6 PRACTICAL IMPLEMENTATION

In the previous section we found that the processing methodology we must use to form an image is to form a Fourier transform of [pic] (or [pic]). However, this approach makes the tacit assumption that [pic] is a continuous function of time. Thinking ahead to when we will consider both cross-range and down-range imaging we realize that the SAR will transmit a pulsed signal rather than a CW signal. Because of this, we recognize that [pic] will not be a continuous-time signal but a discrete-time signal with samples spaced by the radar PRI, [pic]. If we also recognize that we will perform the processing with digital signal processors (DSPs) we reinforce the realization that [pic] is, in fact, a discrete-time signal, [pic] or [pic].

2.6.1 A Discrete-Time Model

For a single scatterer (i.e. [pic]) we get (from Equation (34))

[pic]. (47)

After we (digitally) remove the quadratic phase term, the signal we process to form the image is

[pic]. (48)

Since [pic] (and [pic] for multiple scatterers) is a discrete-time signal we use the discrete-time Fourier transform (DFT) to find its Fourier transform. Specifically we find

[pic]. (49)

As with the continuous-time Fourier transform, we limit the sum by considering that we gather data only from [pic] to [pic] or for [pic]. If we use [pic] the limits on [pic] become

[pic] (50)

where it is understood that we round, or truncate, [pic] to the nearest integer.

If we use this in Equation (49), with Equation (48) and Equation (35), we get

[pic]. (51)

We note that this is similar to Equation (42).

Figure 16 contains a plot of [pic] vs.[pic]. As can be seen, it has a peak at [pic] as did Figure 14. However, it also has peaks at [pic]. In fact, if we recall the theory associated with discrete-time signals and the DFT we recognize that [pic] will have peaks at [pic] where [pic] is an integer. All peaks except the one corresponding to [pic] are ambiguities and are undesirable. In terms of SAR, they result in what are termed ghost images. We want to be sure we choose [pic], and the characteristics of the SAR antenna, to avoid these ghosts since they can result in misleading SAR images. The SAR antenna was mentioned because it acts as a spatial anti-aliasing filter.

2.6.2 Other Considerations

As we did before, we want to change the horizontal axis of Figure 16 to cross-range distance rather than frequency. To do so we use Equation (45). This results in the plot of [pic] shown in Figure 17. The ambiguities are shown in this figure and are located at

[pic]. (52)

The above tells us that we want to choose the PRI such that all scatterers lie within ±1/2 ambiguity. That is we want to choose the PRI such that all [pic] satisfy

[pic]. (53)

Since all scatterers of interest lie within the imaged area this says that we want to choose the PRI such that

[pic]. (54)

In fact we usually choose the PRI such that

[pic] (55)

so as to be sure that the SAR antenna beam adequately attenuates targets outside of the imaged region. After we form the larger image, because we will be using a larger effective [pic], we will truncate it to include only the region of interest.

We can turn Equation (54) around and use it to find an upper bound on PRI. Specifically we solve Equation (54) for [pic] to yield

[pic] (56)

or from Equation (55)

[pic]. (57)

If we consider an earlier example where [pic], [pic] and [pic], and consider an aircraft velocity of [pic] we get

[pic], (58)

which is an easy constraint to satisfy. When we consider down-range imaging we will impose a lower limit on [pic] so as to satisfy unambiguous range operation. However, that lower limit is generally well below the upper limit of Equation (58).

We now want to summarize the above as an algorithm that we can implement to form a cross-range image.

2.7 AN ALGORITHM FOR CREATING A CROSS-RANGE IMAGE

• We assume we have a base-band, CW signal (see Equation (30) and Equation (31))

• We sample this signal at intervals of [pic] and generate [pic] samples where

o [pic] (Equation (57))

o [pic] (Equation (50))

o [pic] (Equation (28))

o [pic] (Equation (46))

In these equations [pic], [pic], [pic], [pic], and [pic] are known. The samples are taken for [pic] between [pic] and [pic] or for [pic] between [pic] and [pic].

• Remove the quadratic phase by multiplying the sampled signal by

[pic] (59)

This gives [pic].

• Compute the DFT of [pic]. This is most easily done using an FFT. The minimum FFT length is [pic] although we usually choose the FFT length to be a power of 2 greater than [pic]. In “real” applications we often adjust various SAR parameters so that [pic] is close to a power of 2. For purposes of this class we choose an FFT length much greater (4 to 16 times) than [pic] so that the resulting frequency plot is smooth.

• If [pic] is the length of the FFT, the frequency spacing between output FFT taps is

[pic] (60)

After the front and rear halves of the FFT outputs are swapped (think fftshift) the frequencies of the taps are

[pic] (61)

• Transform the frequency scale to cross-range using

[pic] (Equation (45))

and plot the magnitude of the FFT output vs. [pic]. This doesn’t produce an image but a linear plot as shown in Figure 16.

• To generate a pseudo image, create an array of zeros where the number of columns is equal to the number of samples needed to cover the width, [pic], of the image area and a comparable number of rows. If there are [pic] rows, replace row [pic] with the appropriate FFT outputs. Use the resulting array to create the pseudo image.

2.8 EXAMPLE 1

To illustrate the above, we consider a specific example. The parameters of this example are given in Table 1.

Table 1 – Parameters Used in SAR Example 1

|Parameter |Value |

|Width of image area, [pic] |50 m |

|Depth of image area, [pic] |50 m |

|Range to image area center, [pic] |20 Km |

|SAR wavelength, [pic] |0.03 m |

|Aircraft velocity, [pic] |50 m/s |

|Synthetic array length, [pic] |600 m |

|Number of scatterers, [pic] |3 |

|Scatterer locations, [pic] (m) |([pic],0), ([pic],20), ([pic],-15) |

|Scatterer powers, [pic] (w) |1, 0.25, 0.09 |

Given these we can compute some of the SAR parameters indicated in the algorithm description. Specifically:

[pic] and (62)

[pic]. (63)

We will choose a PRI of 50 ms, that is we choose

[pic]. (64)

This gives

[pic]. (65)

With this the SAR starts sampling at [pic] and samples until [pic]. The samples are taken every [pic] and a total of [pic] samples are used. This means that we need, as a minimum, a 256 point FFT. In order to produce a smooth plot we will use a 2048 point FFT.

We note that, since we chose [pic] the actual width of the area included in the image is

[pic]. (66)

Once we form the image we will need to discard the FFT outputs outside of the range of ±25 m (after the conversion from frequency to y position)

The resolution of the SAR image is

[pic]. (67)

This means that we ought to be able to distinguish scatterers separated by about 1 m or greater, and maybe down to 0.5 m separation if their relative powers and phases allow this.

Before processing the SAR signal using the previously discussed algorithm we need to generate the SAR signal. To do so we use Equation (31) with [pic]. We generate 241 samples of [pic] starting at [pic] and ending at [pic] in steps of [pic].

In my code I generated [pic], [pic], using Equation (26) and Equation (27). I then used these [pic], along with the [pic] values in Equation (30) to compute the three [pic]. Finally, I summed the three [pic] to form [pic].

Figure 18 is a linear plot of the [pic] for [pic] and Figure 19 is a pseudo image. To form the pseudo image I first created an array of zeros that had 101 rows (which is [pic]) and a number of columns equal to the number of [pic] values in the linear plot. I next loaded the [pic] values from the linear plot in the 51st row of the array. Next I used the commands

imagesc(x, y, max(max([pic]))-[pic]);colormap gray

to generate the image. The use of max(max([pic]))-[pic]creates an image with a white background, which is really a negative image. I did this to save on printer ink.

In examining Figure 18 we note that [pic] has three peaks at the y positions of the scatterers. Further, the heights of the peaks are [pic]. The image (Figure 19) shows three dots at the given scatterer positions. It will be noted that the dots are different shades of gray, indicating different amplitudes.

To check the aforementioned resolution statement I moved the y positions of the scatterers to -1, 0 and 1 m and reformed the linear plot and image. These are shown in Figure 20 and Figure 21. The linear plot clearly shows three peaks but the relative amplitudes are somewhat different than those of Figure 18. This is due to the sidelobes of the Fourier transform response function. The presence of the three scatterers can also be seen in the image of Figure 21.

As another interesting experiment I eliminated the quadratic phase removal step of the SAR processing algorithm. The results are shown in Figure 22 and Figure 23 (I reverted to the original locations of the scatterers). As can be seen, the peaks are spread and the image is blurred in the y direction. In SAR lingo, we say that the image is not focused. In fact, the process of removing the quadratic phase is sometimes termed focusing of the SAR image.

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[1] For now we are assuming that the aircraft (platform) is at an altitude of zero. The extension to non-zero altitude is straightforward.

[2] Note that this is similar to stretch processing wherein we remove the quadratic phase in the mixer.

[3] Again, note the similarity to stretch processing

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[pic]

Figure 13 – Geometry Used to Develop Signal Representation – Cross-range Imaging

[pic]

Figure 23 – Image of [pic] - Three Scatterers at -15, 0, 20 m – Without Quadratic Phase Removal

[pic]

Figure 22 – Linear plot of [pic] - Three Scatterers at -15, 0, 20 m – Without Quadratic Phase Removal

[pic]

Figure 21 – Image of [pic] - Three Scatterers at -1, 0, 1 m

[pic]

Figure 20 – Linear plot of [pic] - Three Scatterers at -1, 0, 1 m

[pic]

Figure 19 – Image of [pic] - Three Scatterers at -15, 0, 20 m

[pic]

Figure 18 – Linear plot of [pic] - Three Scatterers at -15, 0, 20 m

[pic]

Figure 17 – Plot of [pic] vs. [pic] Using a Discrete-time signal with [pic]

[pic]

Figure 16 – Plot of [pic] vs. [pic] Using a Discrete-time signal with [pic]

[pic]

Figure 15 – Plot of [pic] vs. [pic]

[pic]

Figure 14 – Plot of [pic] vs. [pic]

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