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Solutions Manual for Medical Imaging Signals and Systems 2nd Edition by Prince IBSN 9780132145183

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Solutions Manual for Medical Imaging Signals and Systems 2nd Edition by Prince IBSN 9780132145183 Full download: https://goo.gl/QKZ5Ni medical imaging signals and systems 2nd edition pdf medical imaging signals and systems jerry l. prince pdf medical imaging signals and systems pdf medical imaging signals and systems pdf download medical imaging signals and systems prince pdf medical imaging signals and systems + solutions manual download medical imaging signals and systems solution manual medical imaging signals and systems solutions manual pdf
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  =   = −∞   = −∞   Solutions Manual for Medical Imaging Signals and Systems 2nd Edition by Prince IBSN 9780132145183 Full download: http://downloadlink.org/p/solutions-manual-for-medical-imaging-signals-and-systems-2nd-edition-by-prince-ibsn-9780132145183/  2   Signals and S ys t e ms   S IGNALS AND THEIR PROPERTIES Solution 2.1(a) δ s   (x, y) = P ∞   −∞   separable signal. P ∞   n = −∞   δ (x −   m, y −   n) = P ∞   δ (x −   m) ·   P ∞   δ (y −   n), therefore it is a   (b) δ l   (x, y) is separable if sin(2 θ ) = 0. In this case, either sin θ   = 0 or cos θ   = 0, δ l   (x, y) is a product of a constant function in one axis and a 1-D delta function in another. But in general, δ l   (x, y) is not separable. (c) e(x, y) = exp[  j 2 π ( u 0   x + v 0   y)] = exp(  j 2 π u 0   x ) · exp(  j 2 π v 0   y) = e 1D (x; u 0 ) · e 1D (y; v 0   ), where e 1D (t; ω ) =   exp(  j 2 π ω t ) .  Therefore, e(x, y) is a separable signal.  (d) s(x, y) is a separable signal when u 0 v 0 = 0. For example, if u 0 = 0, s(x, y) = sin(2 π v 0 y) is the product of a constant signal in x and a 1-D sinusoidal signal in y. But in general, when both u 0 and v 0 are nonzero, s(x, y) is not separable. Solution 2.2 (a) Not periodic. δ (x, y) is non-zero only when x = y = 0. (b) Periodic. By definition ∞   comb(x, y) = X   ∞   X δ (x −   m, y −   n) .   For arbitrary integers M and N , we have m= −∞   n = −∞   comb(x + M, y +  N ) =   ∞ ∞   X X δ (x −   m + M, y −   n +  N ) m= −∞   n = −∞ ∞ ∞   = X X   δ (x −    p, y −   q) [let  p = m −   M, q = n −    N ]  p = −∞   q = −∞    = comb(x, y) .   2  3 3 CHAPTER 2 : SIGNALS AND SYSTEMS x   x   s   So the smallest period is 1 in both x and y directions. (c) Periodic. Let f   (x + T x   , y) = f   (x, y), we have sin(2 π x) cos(4 πy ) = sin(2 π (x + T x   )) cos(4 πy ) .  Solving the above equation, we have 2 π T x = 2k  π  for arbitrary integer k. So the smallest period for x is T x0 = 1. Similarly, we find that the smallest period for y is T y0 = 1 /  2.  (d) Periodic. Let f   (x + T x   , y) = f   (x, y), we have sin(2 π (x + y)) = sin(2 π (x + T x + y )) .  So the smallest period for x is T x0 = 1 and the smallest period for y is T y0 = 1. (e) Not periodic. We can see this by contradiction. Suppose f   (x, y) = sin(2 π (x 2 + y 2   )) is periodic; then there exists some T x such that f   (x + T x   , y) = f   (x, y), and sin(2 π (x 2 + y 2   )) = sin(2 π ((x + T x   ) 2 + y 2   ))   = sin(2 π (x 2 + y 2 + 2xT x + T   2   )) .  In order for the above equation to hold, we must have that 2xT x + T   2 = k for some integer k. The solution for T x depends on x. So f   (x, y) = sin(2 π (x 2 + y 2   )) is not periodic. (f) Periodic. Let f  d   (m + M, n) = f  d   (m, n). Then sin π  m cos π  n  = sin π  (m + M ) cos π  n  .  5 5 5 5 Solving for M , we find that M = 10k for any integer k. The smallest period for both m and n is therefore 10. (g) Not periodic. Following the same strategy as in (f), we let f  d   (m + M, n) = f  d   (m, n), and then   sin   1   m   5  1 cos n  5  = sin   1  (m + M )   5   1  cos n .   5   The solution for M is M = 10k  π . Since f  d   (m, n) is a discrete signal, its period must be an integer if it is to be periodic. There is no integer k that solves the equality for M = 10k  π  for some M . So, f  d   (m, n) = sin 1 m cos 1 n is not periodic. 5 5   Solution 2.3 (a) We have E ∞   ( δ s   ) =  Z ∞   Z ∞   δ 2   (x, y) dx dy −∞   = lim  −∞   lim Z X Z Y ∞   ∞   X δ (x −   m, y −   n) dx dy X   →∞   Y →∞ − X   − Y   m= −∞   n = −∞ = lim lim (2 b X   c + 1)(2 b Y c + 1) X   →∞   Y →∞   = ∞   ,  4 4 CHAPTER 2 : SIGNALS AND SYSTEMS s   2   where b X   c is the greatest integer that is smaller than or equal to X . We also have P ∞   ( δ s   ) = lim  lim 1 Z X Z Y  δ 2   (x, y) dx dy X   →∞   Y →∞   4X Y   − X − Y   = lim lim 1 Z X Z Y ∞   ∞   X δ (x −   m, y −   n) dx dy X   →∞   Y →∞   4X Y   − X − Y   m= −∞   n = −∞   = lim lim  (2 b X   c   + 1)(2 b Y   c   + 1) X   →∞   Y →∞   4X Y   4 b X   cb Y   c  2 b X   c   + 2 b Y   c  1   = lim lim + + X   →∞   Y →∞   = 1 .  4X Y  4X Y  4X Y   (b) We have E ∞   ( δ l   ) =  Z ∞   Z ∞   | δ (x cos θ   + y sin θ   −   l ) | 2   dx dy −∞   −∞ Z ∞   Z ∞   =   δ (x cos θ   + y sin θ   −   l) dx dy −∞   −∞     Z ∞   1 dx, sin θ   = 0  | sin θ | 1 −∞   =   Z   ∞   1   E ∞   ( δ l   ) = ∞   .   | cos θ | dy, cos θ   = 0  −∞   Equality 1 comes from the scaling property of the point impulse. The 1-D version of Eq. (2.8) in the text is δ (ax) = 1 δ (x). Suppose cos θ   = 0. Then  | a |   Therefore, δ (x cos θ   + y sin θ   −   l) =  Z ∞   1   δ   | cos θ |   x + y sin θ  cos θ   1   l   −   cos θ .  We also have δ (x cos θ   + y sin θ   −   l)dx =   −∞   .   | cos θ |   P ∞   ( δ l   ) = lim  lim 1 Z X Z Y  | δ (x cos θ   + y sin θ   −   l) | dx dy X   →∞   Y →∞   4X Y   − X − Y   = lim lim 1 Z X Z Y  δ (x cos θ   + y sin θ   −   l)dx dy . X   →∞   Y →∞   4X Y   − X − Y   Without loss of generality, assume θ   = 0 and l = 0, so that we have sin θ   = 0 and cos θ   = 1. Then it follows
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