My question is related to a theoretical part of the book: Digital Processing of Synthetic Aperture Radar Data, Algorithms and Implementation by I.G. Cumming et al. [Chapter 4. Equation 4.25 (for those who know the book)].

Scene geometry as shown in the book

Basically what I undertand that is said there, is that if I have my transmit pulse waveform $s_{pul}(\tau)$ and some ground reflectivity function $g_r(\tau)$, "the reflected energy at any illumination instant is a convolution ..." as: $$ s_r(\tau) = g_r(\tau) \ast s_{pul}(\tau) $$

An example is given for a point target at a distance $R_a$ from the radar having a magnitude $A_0$ accounting for the RCS. The assumed reflectivity function is given as: $$ g_r(\tau) = A_0 \delta(\tau - \frac{2R_a}{c}) $$ which would yield the backscattered signal as: $$ s_r(\tau) = A_0 s_{pul}(\tau - 2\frac{2R_a}{c}) $$

So far so good, for this point target case. If I think in terms of the Fourier Transforms, the dirac-delta would correspond to a constant (and a phase slope for the shift), which is multiplied by the spectrum of my transmit signal. However, what if I assume a different ground reflectivity, lets say a bumpy surface modeled by a sinusoid function. Its FT would be more of a "peaky" signal in the frequency domain. Multiplying this with the FT of my transmit signal would be some sort of filtering, no?

I have the feeling that I am missing an important part here, or that the convolution equation may be a simplification, leaving out certain aspects that skew my understanding. E.g. what maybe confuses me is the dependency in time via $\tau$, but there is no spatial dependency. I know from electromagnetics courses that for scattered fields there is typically also some green's function involved, but there is always some dependency on the location afaik. Maybe a key phrase from the book is "... reflected signal at any illumination instant ...".

I would be happy for some links, reference or any hints to better understand this convolution equation.


2 Answers 2


The FT of the ground reflectivity is indeed multiplied by the FT of the transmit waveform. If the ground's FT were "peaky", then this would be reflected in the product of the two. (The spectrum of the waveform could itself be "peaky"). In reality, however, the observed spectrum of the ground will be well-modeled by the central limit theorem, since it will be the sum of the spectrum of thousands/millions/a whole lot of small scattering elements. Regardless of the individual spectrums, their sum will be approximately Gaussian.

Even if it were "peaky" (and it often is for individual targets), so what? The goal of the radar is to measure/estimate the ground/target reflectivity. If it is peaky, then that is what we want to know.

The convolution model is not an approximation. Rather, it is exactly what you get back. In the simplest case, for each target you get back a scaled copy of the waveform that is delayed according to the target's distance. The receiver measures the sum of each of these delayed and scaled copies of the waveform. The sum of scaled and delayed copies of a waveform is, by definition, the convolution of that waveform with another signal that is defined by the scaling/delay pairings.

For example, let $w(t)$ be the transmit waveform. Suppose there is a target $\tau_1$ seconds away whose reflection scales the waveform by $\alpha_1 \in \mathbb{C}$. Now let there be a second target with delay $\tau_2$ and scalar $\alpha_2$. What the receiver sees is the following:

$ r(t) = \left( w(t) \ast \alpha_1 \delta (t-\tau_1) \right) + \left( w(t) \ast \alpha_2 \delta (t-\tau_2) \right). $

By adding the two target profiles (what I am calling their delay/scaling pair), we get the profile for the entire hypothetical scene, $s(t)$. When sensed by the radar, we see:

$ s(t) = \alpha_1 \delta( t - \tau_1) + \alpha_2 \delta( t - \tau_2 ) \Rightarrow r(t) = w(t) \ast s(t). $

For cases where a target's reflectivity is a function of frequency, the convolution model is still accurate. What changes in this case is that the individual profiles are no longer the simple $\alpha \delta( t - \tau)$, but rather a more complicated function that reflects its response at each frequency. Nonetheless, the profile for the scene, $s(t)$, is still the sum of the individual profiles and the waveform is convolved with this sum.


The spatial dependency is implicit in the relationship $\tau -\frac{2R_a}{c}$. It is easier to form the convolution using the time of flight rather than position on the ground since the radar really detects the time of arrival of the reflection. Consider what happens when there is a sudden change in elevation in the imaging area - the timing of various returns to the radar becomes highly dependent on the the geometry of the of the radar and the ground terrain.

Additionally, if you consider a specific the reflectivity at a specific delay, then this reflectivity value at that particular delay is an aggregate value that is generated by multiple returns from the ground along a conic section in azimuth (assuming a simplified flat earth model).

The use of convolution is indeed a simplification in several ways:

  1. Assumes reflectivity is linear. It neglects multipath between reflectors. This is commonly seen in SAR images of bridges & buildings
  2. Same as item 1 but neglects when there a large number of targets within a SAR image cell which leads to speckle. This is commonly seen in SAR images of grassy areas.
  3. Neglects motion of objects e.g. trees blowing in the wind
  4. It also assumes that the target reflectivity doesn't change much with either the depression angle or azimuth angle.
  5. It neglects the idea of the obfuscation of one target by another during the creation of the synthetic aperture.

It is still a good approximation but it does have its shortcomings.

  • $\begingroup$ Thanks for the reply, I understand what you are saying ... I guess... but still, viewed from the Fourier Transform perspective, why does the ground reflectivity function not act as a sort of "filter"? If I multiply in the FT domain my transmit spectrum with a ground reflectivity spectrum from a bumpy terrain (e.g. sine wave), the return spectrum is only a peak at the frequency of the sinusoid. I am missing the understanding at this part, where does my line of thinking go wrong? $\endgroup$
    – Lucas
    Commented Jun 21, 2023 at 8:58
  • $\begingroup$ If it can be written as a convolution, you can definitely view it as a filter. In this case the, frequencies of the signal pulse are only exciting those parts of the ground reflectivity response. Actually, that's another approximation - the ground response is not the same over all frequencies, but for most SAR bandwidths and frequencies it is a decent approximation. $\endgroup$
    – David
    Commented Jun 21, 2023 at 14:03

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