EE299 Midterm Winter 2007 Solutions

Transcription

1 EE299 Midterm Winter 2007 Solutions 1. (25 points) You have an audio signal with a 20kHz sampling rate. (a) (7 points)what is the time between samples? T s = 1 = 1 =.00005sec =.05ms F s (b) (10 points) The person who recorded the signal forgot to use an anti-aliasing filter before sampling, and the frequencies above 8kHz are corrupted by aliasing. You don t have the original signal, so cannot resample. Describe the filter specifications you would need to get rid of the bad frequency region in the digital signal. You need a low pass filter to remove frequencies above 8kHz. In the analog domain, the cut-off frequency would be 8kHz. Since you are working in the digital domain, the cut-off is (8kHz/10kHz)=0.8, which is what you would use in a MATLAB filter design function. (An answer of 8kHz was acceptable.) (c) (8 points) Now that you ve filtered the data, you could get away with a lower sampling frequency and save bits by resampling. What new sampling rate would you use? Using Shannon s sampling theorem with a signal bandwidth of B = 8kHz: F s > 2B = 2(8kHz) = 16kHz 1

2 2. (15 points) The two problems below deal with frequency content. You need only do one of these two circle the one you wish to have graded. (a) The plots below show the frequency content of two sections from the guitar and castanets sound file. Which corresponds to the guitar and which to the castanets? Explain your answer. The castanets have more energy at high frequencies than the guitar, as we discovered from the fact that we could not hear the guitar after we ran the combined signal through a high-pass filter. Noticing that energy in the 5kHz-20kHz range is higher in the left figure than the right, you can conclude that the left figure is the castanets and the right is the guitar. You might also notice harmonics in the right figure, which would be characteristic of the guitar note, but not of the castanets. Several people interpreted these figures as time plots, which led to incorrect answers. It is important to understand the difference between time and frequency plots. The time plots that correspond to the above frequency plots are given below. 2

3 (b) The two figures below correspond to the frequency content of the original and a filtered version of the rose image, left and right respectively. What kind of filter was used? How does the filtered image look compared to the original? Justify your answers. In the two figures above, low frequencies correspond to the left and top sides of the image. The picture on the right has less energy (more blue) at higher frequencies, so it corresponds to an image that has been processed with a low pass filter (LPF). The filtered image will look blurry compared to the original, since an LPF has a smoothing effect. The image that corresponds to the frequency content in the figure on the right is given below. 3. (30 points) Explain how the following operations change how the signal sounds or looks. (a) (10 points) For an audio signal, you use an 8kHz sampling rate to play out a signal, not remembering that the true sampling rate is 16kHz. When we change the sampling rate, we are changing the time between samples, which in this case is doubled. So the signal will sound slower and lower pitch, and it will take twice as long to play. Note that since we are changing the sampling rate at the time of playing out the signal and not at the time of sampling, there is no aliasing involved. Aliasing only comes into play when sampling or resampling to a lower rate. 3

4 (b) (10 points) For a square grayscale image, you multiply all rows of the image by a triangle function (starts at zero and ramps up to one, then ramps back down to zero). Then you multiply all columns by the same function. Multiplying the image by 1 leaves it as is, and multiplying it by 0 turns it to black. So multiplying a row by a triangle will make it like the original in the middle and gradually darkening to black at the edges. Since this is a two dimensional triangle, then the image darkens in moving towards all four edges. The figure below illustrates the multiplying function, an example image, and the scaled version. (c) (10 points) For a color image, you add a constant to every pixel in the red plane, but cap it at the maximum, e.g. Y=X; for i=1:n, for j=1:m Y(i,j,1)=min(X(i,j,1)+30,255) end and you completely zero out the green plane. Eliminating the green means that where ever there was pure green, there will be black, and that there can be no white in the image. The new image will be a mix of red, blue and purple. Adding a constant to the red plane means that the reds will be brighter and there will be less contrast among the brightest regions. It also means that there will be no black, but in this case the constant is so small that you can t really see this. The effect of adding a constant to red without any blue and with an 0.75 level of blue is illustrated in the figure below that shows the full range of red-only values at the top of the image. The parrot image sequence shows the effect of removing all green (middle image) from the original (left image) and then adding the constant of 30 to the red plane (right image). 4

5 4. (15 points) You want to make a photo-quality (300 PPI, pixels per inch) print of a picture that you took with your cell phone over the weekend. The cell phone has an image size of 1200x900 (about 1 mega pixel), what size will the print be? What could you do if you wanted the print to be larger? The pixels per inch is given as 300 PPI (pixels per inch) and the image size is 1200x900, therefore the printed size will be, 1200 pixels 300 pixels inch 900 pixels 300 pixels inch = 4 3. There are at least three possible things you could do to make the print larger: Resample the image to have enough pixels to match your print size. The quality of the image will decrease rapidly as the scale is increased and the result will be a blocky image. Lower the PPI. Decreasing the PPI will cause the image to be printed larger, however the pixels will start to become noticeable (and distracting) at lower PPI s. A PPI of 50 would mean a print size of 24 x18, but will look grainy, or pixelated. Retake the picture with a higher resolution camera. 5. (15 points) A bank wants to keep a database of signature scans. The images are binary (pixels are either black or white). Propose a method for compressing the images so that they take up less space when they are stored. Your method can be either lossy or lossless, but keep in mind that you don t want to corrupt the images so much that the bank can t tell them apart. The idea is to take advantage of the fact that the vast majority of the scan is white, so it should be possible to save bits by coding large runs or areas of white pixels using fewer bits than the number of pixels in those areas. A lossless method talked about in the book is run-length coding, where the image is viewed as a stream of bits representing pixel values, and instead of storing one bit for each pixel you store the number of pixels in a row that have the same value. The image shown here is pixels, so runs could be coded with 19 bits, since that is the number needed to specify a length between 0 and = For this image, there are 4432 runs, so the total number of bits in the compressed image is 4432 runs 19 bits/run = 84, 208 pixels (or 84, 209 including the value of the first bit). This gives a compression ratio of /84209 = 4.4. Another lossless compression idea for images containing very few black pixels is to simply store a list of all the black pixel locations in the image. In order for this to work (meaning, actually save bits), it requires that the ratio of white pixels to black pixels be greater than or equal to the number of bits needed to specify a pixel location. For example, in this image it takes 19 bits to specify a pixel location, so we would need to have (on average, for these types of images) at least 19 white bits for every black bit in the image (this white to black ratio would give a compression ratio of 20:19). In this particular signature scan, the ratio is only about 18 white bits per black bit, so this compression scheme might actually cost more bits, on average, for signature images. Lossy compression is probably not the ideal way to tackle this problem, for two reasons. One is that the images are to be used to compare signatures, which is a task possibly carried out by computer. The lossy compression methods we talked about in class take advantage of the human brain and visual sensitivities, but information in the image which seems insignificant to human viewers might be very important in the computerized algorithm. A second reason lossy compression is not ideal is that the image is binary, and so it s not possible to take advantage of lossy compression methods that change some pixel values slightly (the only way you can change a black pixel is to make it white.) Trying to do something like JPEG which requires coding DCT transform coefficients might actually cost bits in this case, since presumably the transform coefficients would require more than one bit each to store. 5