Digital Image processing Gonzalez - Solution Manual (3rd by Gonzalez

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Assuming a grid spacing of 1 unit, the minimum distance between any two points of different clusters then must greater than 2(n − 1). In other words, these points must be separated by at least the distance spanned by n − 1 cells along the mask diagonal. 19 (a) Numerically sort the n 2 values. The median is ζ = [(n 2 + 1)/2]-th largest value. 38 CHAPTER 3. PROBLEM SOLUTIONS (b) Once the values have been sorted one time, we simply delete the values in the trailing edge of the neighborhood and insert the values in the leading edge in the appropriate locations in the sorted array.

4(c), which is the sum of two sine waves in this case. For some values of sampling, the sum of the two sines combine to form a single sine wave and a plot of the samples would appear as in Fig. 8 of the book. Other values would result in functions whose samples can describe any shape obtainable by sampling the sum of two sines. 5 Starting from Eq. 2-20), ∞ f (t ) + g (t ) = f (τ)g (t − τ)d τ. −∞ The Fourier transform of this expression is ∞ ℑ f (t ) + g (t ) = −∞ ∞ = ⎡ ⎣ ⎤ ∞ f (τ)g (t − τ)d τ⎦ e −j 2πμt d t −∞ ⎡ ⎤ ∞ f (τ) ⎣ g (t − τ)e −∞ −j 2πμt d t ⎦ d τ.

Because, by the convolution theorem, the Fourier transform of the spatial convolution of two functions is the product their transforms, it follows that the Fourier transform of a tent function is a sinc function squared. 8 (a) We solve this problem by direct substitution using orthogonality. Substituting Eq. 4-4) yields M −1 Fm = ⎡ 1 ⎣ M n =0 = 1 M = Fm M −1 r =0 ⎤ M −1 Fr e r =0 −j 2πr n /M ⎦ −j 2πm n /M e ⎡ ⎤ Fr ⎣ e −j 2πr n /M e −j 2πm n /M ⎦ M −1 n =0 where the last step follows from the orthogonality condition given in the problem statement.

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