We discuss in this segment the simplest restoration filter one can derive, the so-called inverse filter. The inverse filter has traditionally been developed for an LSI system, and it can therefore be implemented in the discrete frequency domain. In this case, you just invert the degradation operator, at least for the frequencies it is invertible. We present actually the Inverse filter as the solution to a least squares problem which results in the generalized Inverse filter even when that the degradation is not linear especially invariant. We demonstrate the performance of the filter in restoring an image which is blurred due to motion. We're going to use the same example for the material presented during this and next week so as to be able to make direct comparisons among the various restoration feeders. The main drove of Inverse filter is that it amplifies the noise present in the data. So even if the noise is not visible in the observed image, it is greatly amplified in the restored image. Due to this main drawback, an improved version of the filter will be presented next. The simplest possible restoration filter one can devise is the Inverse filter. Here's the degradation equation we're using, y the observed image. H, the known degradation, and f, the original image we try to find, and the additive noise. With a model like this, the only constraint during possible degradation, is that it is linear. [SOUND] So given the degradation model, the simplest possible solution approach is to minimize this squared error between the observation and hf. A necessary condition for this functional to have a minimum, is that it's gradient is equal to 0. And by the way, this is the same gradient that I denoted like this earlier on. So we do know that the norm can be expanded as follows. [BLANK_AUDIO] [SOUND] So this term is independent on F. It can go away, and then as I showed earlier, the gradient of this term is minus 2, H transposed Y. Y's the gradient of the quadratic term is 2 times H transpose H f. So this should be equal to 0, and therefore we obtain the so-called normal equations. So the solution f is equal to the generalized inverse. This is this plus here. So generalized Inverse. If H transpose H is invertible, then that's the Regular Inverse. So this solution, again, is for just any matrix with no particular structure. Now, if the degradation system is linearly special in variant, we then know that H is a block circulant matrix and we furthermore know that they can take the solution. Equation to the discrete frequency domain, and if we do so we end up with this solution. So according to this the restored image had frequency discreet frequencies UV is equal to 0 when the frequency parts of the system as those frequencies is equal to 0. Otherwise, it's given by this expression here. This expression actually can be rather simplified because I have H transposed, I'm sorry, H cong, conjugated UV, YUV, and then the magnitude in the denominator, is H complex conjugate, times H. So these guys cancel out and this is just the observation of the frequency domain over the frequency response of the system in the frequency domain. And clearly you can obtain this expression by just look at the convolution equation, and taking it to the discrete frequency domain, and dividing by this since convolution becomes multiplication. Clearly the main disadvantage of the inverse filter is the fact that the noise is ignored as mentioned, and if we look at this expression here right, y is equal to h u v f u v divide by eight uv. So this will give me the through solution plus the noise, then, divided by Huv. Now, the noises are assumed to be broadband and, therefore for those frequencies that Huv is small. I have, course divided by something very small. So, this gives me a large number that's the noise of litigation. I can try to control the noise amplification by introducing a threshold here. So I'm not looking for exact 0s of the frequency response here, but for values that are smaller than a threshold t, otherwise for values greater than t I use the other expression. So let's see how this inverse filter performs through some examples. So here's an example would be using, kind of, throughout this presentation to compare different restoration approaches. There's the original image, cameraman. It's blurred by a system H here which introduces motion blur over eight pixels in the horizontal direction. And then noise is added so that the blurred signal to noise ratio is 20 dB, so this is clearly blurred. The noise is hardly visible, but through the restoration, when it's amplified it will be really visible and bothersome. So with this setting given this image here this is our observation again y with no H. This is the description of H and we try to get an estimate of the original limit. Well let's at the impulse frequency response of the degradation system. It's introducing one dimensional motion, blur along the horizontal direction. So, this is a point spread function, the impulse response. It's normalized therefore the height of each of these samples is one over eight. So in in two dimensions if from left to right h n one and two it's equal to h n one which is this one if I rename it times delta at m two. So it take if I take the fully transformed of this signal here we do know that the sink function. And so this is the discriptfully transform. Let's call this the u domain. So I show the magnitude of H u,v discrete frequencies for any v. So this is independent of v so it will be the same for every v. So it centered. So this a 56 point DFT. This is therefore 128. The height is 1, since this is normalized. And then it was chosen carefully, this motion blur over eight pixels because the 0s are at integer multiples of 256 over 8. Let's say times l. So this is equal to 32l.L So the space in between the zero is 32, therefore this is at 128 plus 32, so this is at 160. The next one is at 192, and the last one is a 224. Right and similar in the other side this is zero at 96, 0 at 64 and a 0 at 32. So there exact 0s at these discrete frequencies. So if I look at the two dimensional frequency response, the magnitude squared. It looks like this, it's this sync function so this is U, this is V, axis. So, the shape changes along U but it does not change along V. It's independent of V, alright? So this is the, the frequency response of the system that is introducing the 1 dimensional motion blur and this is what we want to invert. That this is all about, right? The restoration. I want to, to invert in the frequency domain this shape and this will give me the inverse filter. We see here two results. That are obtained by this thresholded inverse filter for the case under consideration. Motion blur over eight pixels plus 20 dB blur signals [UNKNOWN] ration. So, two different thresholds are utilized here 10 to the minus 16 and .01. The main observation is that both restorations, they look very similar actually, and they're both buried in noise. So this is not an acceptable result. We show here also one slice of the frequency response of the restoration filter, which is equal to one over the frequency response of the degradation system. We see that the at the locations, frequency locations of exact 0, the generalized inverse is also equal to 0. Without do, using this generalized in this idea, one of the 0s should blow up. They should be infinity values. At 0 the frequency is one. Right, it's normalized while another one is one. And then at higher frequencies, because the frequency response becomes, of the degradation system is smaller. When I invert it, I see larger values of the restoration filter at high, higher frequencies. Let's see now how we can control the result by modifying the threshold using a higher threshold. We show here two additional restorations, by this thresholder inverse filterer for the case on the course of iteration. So the thresholds are increased to this value and this value. And we see now that there is some distant control of the noise amplification. As a mat, as a matter of fact, this might be a. Acceptable restoration result. We also show the frequency response of the corresponding restoration filters. Since there are small, smaller values in the frequency response of the gradation system at higher frequencies these are the ones that are fresh [UNKNOWN] when I look at the inverse filter down here. The main characteristic is that the high frequencies are tapered off. The values are, are increase, decreased, with high frequencies in both these, restorations. I've kept the vertical axis intentionally at, the same kind of length here, so that it's clear it's easy to compare the values. Restoration filters, so while at low frequencies the inverse filter is, is faithful. At high frequencies the Inverse filter is, is altered due to this threshold, but this is a desirable effect, since the noise simplification is controlled. That way. We increase the values of the threshold even further to these values. And we see that as T increases, the noise is further controlled, so we see less noise at the output of the restored image. The frequency response of the filter is out, out of considerably. I feel for this threshold here that you see it's only at low frequency and these mid frequencies that the that there restoration filter has some values, and everything else is 0. We see some of the so called rinding artifacts here that we'll talk a little bit about later. So these are artifacts, undesirable effects in the image. So maybe this is also an acceptable restoration. So we do see here that the inverse filter by itself is not doing an acceptable job because primarily it amplifies noise. Through the introduction of this threshold, we're able to control the noise amplification in an ad hoc fashion. We don't know. What the proper value of T is. However, one might experiment around, as we did here, and choose a restoration that is suitable for their purposes.