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Solution Manual for Medical Imaging Signals and Systems, 2nd Edition

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    Medical Imaging Signals and Systems (2e): Solutions Manual Version 1.4 (December 21, 2016) Jerry L. Prince Electrical and Computer Engineering Whiting School of Engineering Johns Hopkins University Jonathan M. Links Environmental Health Sciences Bloomberg School of Public Health Johns Hopkins University

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    Contents Preface 1 Signals and Systems 2 Image Quality 37 Physics of Radiography 61 Projection Radiography 70 Computed Tomography 96 The Physics of Nuclear Medicine 134 Planar Scintigraphy 141 Emission Computed Tomography 162 The Physics of Ultrasound 176 Ultrasound Imaging Systems 188 Physics of Magnetic Resonance 208 Magnetic Resonance Imaging 216 iii

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    2 Signals and Systems S IGNALS AND THEIR PROPERTIES Solution 2.1 (a) δ s ( x, y ) = ∑ ∞ m = −∞ ∑ ∞ n = −∞ δ ( x − m, y − n ) = ∑ ∞ m = −∞ δ ( x − m ) · ∑ ∞ n = −∞ δ ( y − n ) , therefore it is a separable signal. (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 πu 0 x ) · exp( j 2 πv 0 y ) = e 1D ( x ; u 0 ) · e 1D ( y ; v 0 ) , where e 1D ( t ; ω ) = exp( j 2 πωt ) . 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 πv 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 ) = ∞ ∑ m = −∞ ∞ ∑ n = −∞ δ ( x − m, y − n ) . For arbitrary integers M and N , we have comb ( x + M, y + N ) = ∞ ∑ m = −∞ ∞ ∑ n = −∞ δ ( x − m + M, y − n + N ) = ∞ ∑ p = −∞ ∞ ∑ q = −∞ δ ( x − p, y − q ) [ let p = m − M, q = n − N ] = comb ( x, y ) . 2

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    3 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 πx ) cos(4 πy ) = sin(2 π ( x + T x )) cos(4 πy ) . Solving the above equation, we have 2 πT x = 2 kπ for arbitrary integer k . So the smallest period for x is T x 0 = 1 . Similarly, we find that the smallest period for y is T y 0 = 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 x 0 = 1 and the smallest period for y is T y 0 = 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 + 2 xT x + T 2 x )) . In order for the above equation to hold, we must have that 2 xT x + T 2 x = 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 5 m ) cos ( π 5 n ) = sin ( π 5 ( m + M ) ) cos ( π 5 n ) . Solving for M , we find that M = 10 k 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 5 m ) cos ( 1 5 n ) = sin ( 1 5 ( m + M ) ) cos ( 1 5 n ) . The solution for M is M = 10 kπ . 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 = 10 kπ for some M . So, f d ( m, n ) = sin ( 1 5 m ) cos ( 1 5 n ) is not periodic. Solution 2.3 (a) We have E ∞ ( δ s ) = ∫ ∞ −∞ ∫ ∞ −∞ δ 2 s ( x, y ) dx dy = lim X →∞ lim Y →∞ ∫ X − X ∫ Y − Y ∞ ∑ m = −∞ ∞ ∑ n = −∞ δ ( x − m, y − n ) dx dy = lim X →∞ lim Y →∞ (2 b X c + 1)(2 b Y c + 1) = ∞ ,

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    4 CHAPTER 2 : SIGNALS AND SYSTEMS where b X c is the greatest integer that is smaller than or equal to X . We also have P ∞ ( δ s ) = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y δ 2 s ( x, y ) dx dy = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y ∞ ∑ m = −∞ ∞ ∑ n = −∞ δ ( x − m, y − n ) dx dy = lim X →∞ lim Y →∞ (2 b X c + 1)(2 b Y c + 1) 4 XY = lim X →∞ lim Y →∞ { 4 b X cb Y c 4 XY + 2 b X c + 2 b Y c 4 XY + 1 4 XY } = 1 . (b) We have E ∞ ( δ l ) = ∫ ∞ −∞ ∫ ∞ −∞ | δ ( x cos θ + y sin θ − l ) | 2 dx dy = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x cos θ + y sin θ − l ) dx dy © 1 =            ∫ ∞ −∞ 1 | sin θ | dx, sin θ 6 = 0 ∫ ∞ −∞ 1 | cos θ | dy, cos θ 6 = 0 E ∞ ( δ l ) = ∞ . 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 | a | δ ( x ) . Suppose cos θ 6 = 0 . Then δ ( x cos θ + y sin θ − l ) = 1 | cos θ | δ ( x + y sin θ cos θ − l cos θ ) . Therefore, ∫ ∞ −∞ δ ( x cos θ + y sin θ − l ) dx = 1 | cos θ | . We also have P ∞ ( δ l ) = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y | δ ( x cos θ + y sin θ − l ) | 2 dx dy = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y δ ( x cos θ + y sin θ − l ) dx dy . 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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    5 that P ∞ ( δ l ) = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y δ ( x ) dx dy = lim X →∞ lim Y →∞ 1 4 XY ∫ Y − Y { ∫ X − X δ ( x ) dx } dy = lim X →∞ lim Y →∞ 1 4 XY ∫ Y − Y 1 dx = lim X →∞ lim Y →∞ 2 Y 4 XY = lim X →∞ 1 2 X = 0 . (c) We have E ∞ ( e ) = ∫ ∞ −∞ ∫ ∞ −∞ | exp [ j 2 π ( u 0 x + v 0 y )] | 2 dx dy = ∫ ∞ −∞ ∫ ∞ −∞ 1 dx dy = ∞ . And also P ∞ ( e ) = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y | exp[ j 2 π ( u 0 x + v 0 y )] | 2 dx dy = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y 1 dx dy = 1 . (d) We have E ∞ ( s ) = ∫ ∞ −∞ ∫ ∞ −∞ sin 2 [2 π ( u 0 x + v 0 y )] dx dy © 2 = ∫ ∞ −∞ ∫ ∞ −∞ 1 − cos[4 π ( u 0 x + v 0 y )] 2 dx dy = ∫ ∞ −∞ ∫ ∞ −∞ 1 2 dx dy − ∫ ∞ −∞ ∫ ∞ −∞ cos[4 π ( u 0 x + v 0 y )] 2 dx dy © 3 = ∞ . Equality © 2 comes from the trigonometric identity cos(2 θ ) = 1 − 2 sin 2 ( θ ) . Equality © 3 holds because the first integral goes to infinity. The absolute value of the second integral is bounded, although it does not

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    6 CHAPTER 2 : SIGNALS AND SYSTEMS converge as X and Y go to infinity. We also have P ∞ ( s ) = lim X →∞ lim Y →∞ 1 4 XY ∫ X − X ∫ Y − Y sin 2 [2 π ( u 0 x + v 0 y )] dx dy = lim X →∞ lim Y →∞ 1 4 XY ∫ Y − Y { ∫ X − X 1 − cos[4 π ( u 0 x + v 0 y )] 2 dx } dy = lim X →∞ lim Y →∞ 1 4 XY ∫ Y − Y [ X + sin[4 π ( u 0 X + v 0 y )] − sin[4 π ( − u 0 X + v 0 y )] 8 πu 0 ] dy © 4 = lim X →∞ lim Y →∞ 1 4 XY ∫ Y − Y [ X − sin(4 πu 0 X ) cos(4 πv 0 y ) 4 πu 0 ] dy = lim X →∞ lim Y →∞ 1 4 XY ( 2 XY − 2 sin(4 πu 0 X ) sin(4 πv 0 Y ) (4 π ) 2 u 0 v 0 ) = 1 2 . In order to get © 4 , we have used the trigonometric identity sin( α + β ) = sin α cos β + cos α sin β . The rest of the steps are straightforward. Since s ( x, y ) is a periodic signal with periods X 0 = 1 /u 0 and Y 0 = 1 /v 0 , we have an alternative way to compute P ∞ by considering only one period in each dimension. Accordingly, P ∞ ( s ) = 1 4 X 0 Y 0 ∫ X 0 − X 0 ∫ Y 0 − Y 0 sin 2 [2 π ( u 0 x + v 0 y )] dx dy = 1 4 X 0 Y 0 ( 2 X 0 Y 0 − 2 sin(4 πu 0 X 0 ) sin(4 πv 0 Y 0 ) (4 π ) 2 u 0 v 0 ) = 1 4 X 0 Y 0 ( 2 X 0 Y 0 − 2 sin(4 π ) sin(4 π ) (4 π ) 2 u 0 v 0 ) = 1 2 . S YSTEMS AND THEIR PROPERTIES Solution 2.4 Suppose two LSI systems S 1 and S 2 are connected in cascade. For any two input signals f 1 ( x, y ) , f 2 ( x, y ) , and two constants a 1 and a 2 , we have the following: S 2 [ S 1 [ a 1 f 1 ( x, y ) + a 2 f 2 ( x, y )]] = S 2 [ a 1 S 1 [ f 1 ( x, y )] + a 2 S 1 [ f 2 ( x, y )]] = a 1 S 2 [ S 1 [ f 1 ( x, y )]] + a 2 S 2 [ S 1 [ f 2 ( x, y )]] . So the cascade of two LSI systems is also linear. Now suppose for a given signal f ( x, y ) we have S 1 [ f ( x, y )] = g ( x, y ) , and S 2 [ g ( x, y )] = h ( x, y ) . By using the shift-invariance of the systems, we can prove that the cascade of two LSI systems is also shift invariant: S 2 [ S 1 [ f ( x − ξ, y − η )]] = S 2 [ g ( x − ξ, y − η )] = h ( x − ξ, y − η ) .

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    7 This proves that two LSI systems in cascade is an LSI system To prove Eq. (2.46) we carry out the following: g ( x, y ) = h 2 ( x, y ) ∗ [ h 1 ( x, y ) ∗ f ( x, y )] = h 2 ( x, y ) ∗ ∫ ∞ −∞ ∫ ∞ −∞ h 1 ( ξ, η ) f ( x − ξ, y − η ) dξ dη = ∫ ∞ −∞ ∫ ∞ −∞ h 2 ( u, v ) [∫ ∞ −∞ ∫ ∞ −∞ h 1 ( ξ, η ) f ( x − u − ξ, y − v − η ) dξ dη ] du dv = ∫ ∞ −∞ ∫ ∞ −∞ ∫ ∞ −∞ ∫ ∞ −∞ h 2 ( u, v ) h 1 ( ξ, η ) f ( x − u − ξ, y − v − η ) dξ dηdu dv = ∫ ∞ −∞ ∫ ∞ −∞ h 1 ( ξ, η ) [∫ ∞ −∞ ∫ ∞ −∞ h 2 ( u, v ) f ( x − ξ − u, y − η − v ) du dv ] dξ dη = h 1 ( x, y ) ∗ [ h 2 ( x, y ) ∗ f ( x, y )] . This proves the second equality in (2.46). By letting α = u + ξ , and β = v + η , we have g ( x, y ) = ∫ ∞ −∞ ∫ ∞ −∞ ∫ ∞ −∞ ∫ ∞ −∞ h 2 ( u, v ) h 1 ( ξ, η ) f ( x − u − ξ, y − v − η ) dξ dηdu dv = ∫ ∞ −∞ ∫ ∞ −∞ [∫ ∞ −∞ ∫ ∞ −∞ h 2 ( α − ξ, β − η ) h 1 ( ξ, η ) dξ dη ] f ( x − α, y − β ) dα dβ = [ h 1 ( x, y ) ∗ h 2 ( x, y )] ∗ f ( x, y ) , which proves the second equality in (2.46). To prove (2.47) we start with the definition of convolution g ( x, y ) = ∫ ∞ −∞ ∫ ∞ −∞ h 2 ( ξ, η ) h 1 ( x − ξ, y − η ) dξ dη = h 1 ( x, y ) ∗ h 2 ( x, y ) . We then make the substitution α = x − ξ and β = y − η and manipulate the result g ( x, y ) = ∫ −∞ + ∞ ∫ −∞ + ∞ h 2 ( x − α, y − β ) h 1 ( α, β )( − dα ) ( − dβ ) = ∫ + ∞ −∞ ∫ + ∞ −∞ h 1 ( α, β ) h 2 ( x − α, y − β ) dα dβ = ∫ + ∞ −∞ ∫ + ∞ −∞ h 1 ( ξ, η ) h 2 ( x − ξ, y − η ) dξ dη = h 2 ( x, y ) ∗ h 1 ( x, y ) , where the next to last equality follows since α and β are just dummy variables in the integral.

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    8 CHAPTER 2 : SIGNALS AND SYSTEMS Solution 2.5 1. Suppose the PSF of an LSI system is absolutely integrable. ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy ≤ C < ∞ (S2.1) where C is a finite constant. For a bounded input signal f ( x, y ) | f ( x, y ) | ≤ B < ∞ , for every ( x, y ) , (S2.2) for some finite B , we have | g ( x, y ) | = | h ( x, y ) ∗ f ( x, y ) | = ∣ ∣ ∣ ∣ ∫ ∞ −∞ ∫ ∞ −∞ h ( x − ξ, y − η ) f ( ξ, η ) dξdη ∣ ∣ ∣ ∣ ≤ ∫ ∞ −∞ ∫ ∞ −∞ | h ( x − ξ, y − η ) | · | f ( ξ, η ) | dξdη ≤ B ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy ≤ BC < ∞ , for every ( x, y ) (S2.3) So g ( x, y ) is also bounded. The system is BIBO stable. 2. We use contradiction to show that if the LSI system is BIBO stable, its PSF must be absolutely integrable. Suppose the PSF of a BIBO stable LSI system is h ( x, y ) , which is not absolutely integrable, that is, ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy is not bounded. Then for a bounded input signal f ( x, y ) = 1 , the output is | g ( x, y ) | = | h ( x, y ) ∗ f ( x, y ) | = ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy, which is also not bounded. So the system can not be BIBO stable. This shows that if the LSI system is BIBO stable, its PSF must be absolutely integrable. Solution 2.6 (a) If g ′ ( x, y ) is the response of the system to input ∑ K k =1 w k f k ( x, y ) , then g ′ ( x, y ) = K ∑ k =1 w k f k ( x, − 1) + K ∑ k =1 w k f k (0 , y ) = K ∑ k =1 w k [ f k ( x, − 1) + f k (0 , y )] = K ∑ k =1 w k g k ( x, y )

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    9 where g k ( x, y ) is the response of the system to input f k ( x, y ) . Therefore, the system is linear. (b) If g ′ ( x, y ) is the response of the system to input f ( x − x 0 , y − y 0 ) , then g ′ ( x, y ) = f ( x − x 0 , − 1 − y 0 ) + f ( − x 0 , y − y 0 ); while g ( x − x 0 , y − y 0 ) = f ( x − x 0 , − 1) + f (0 , y − y 0 ) . Since g ′ ( x, y ) 6 = g ( x − x 0 , y − y 0 ) , the system is not shift-invariant. Solution 2.7 (a) If g ′ ( x, y ) is the response of the system to input ∑ K k =1 w k f k ( x, y ) , then g ′ ( x, y ) = ( K ∑ k =1 w k f k ( x, y ) ) ( K ∑ k =1 w k f k ( x − x 0 , y − y 0 ) ) = K ∑ i =1 K ∑ j =1 w i w j f i ( x, y ) f j ( x − x 0 , y − y 0 ) , while K ∑ k =1 w k g k ( x, y ) = K ∑ k =1 w k f k ( x, y ) f k ( x − x 0 , y − y 0 ) . Since g ′ ( x, y ) 6 = ∑ K k =1 g k ( x, y ) , the system is nonlinear. On the other hand, if g ′ ( x, y ) is the response of the system to input f ( x − a, y − b ) , then g ′ ( x, y ) = f ( x − a, y − b ) f ( x − a − x 0 , y − b − y 0 ) = g ( x − a, y − b ) and the system is thus shift-invariant. (b) If g ′ ( x, y ) is the response of the system to input ∑ K k =1 w k f k ( x, y ) , then g ′ ( x, y ) = ∫ ∞ −∞ K ∑ k =1 w k f k ( x, η ) dη = K ∑ k =1 w k (∫ ∞ −∞ f k ( x, η ) dη ) = K ∑ k =1 w k g k ( x, y ) , where g k ( x, y ) is the response of the system to input f k ( x, y ) . Therefore, the system is linear.

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    10 CHAPTER 2 : SIGNALS AND SYSTEMS On the other hand, if g ′ ( x, y ) is the response of the system to input f ( x − x 0 , y − y 0 ) , then g ′ ( x, y ) = ∫ ∞ −∞ f ( x − x 0 , η − y 0 ) dη = ∫ ∞ −∞ f ( x − x 0 , η − y 0 ) d ( η − y 0 ) = ∫ ∞ −∞ f ( x − x 0 , η ) dη. Since g ( x − x 0 , y − y 0 ) = ∫ ∞ −∞ f ( x − x 0 , η ) dη , the system is shift-invariant. Solution 2.8 From the results in Problem 2.5, we know that an LSI system is BIBO stable if and only if its PSF is absolutely integrable. (a) Not stable. The PSF h ( x, y ) goes to infinite when x and/or y go to infinity. ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy = ∫ ∞ −∞ ∫ ∞ −∞ ( x 2 + y 2 ) dx dy = ∫ ∞ −∞ [ ∫ ∞ −∞ x 2 dx ] dy + ∫ ∞ −∞ [ ∫ ∞ −∞ y 2 dy ] dx . Since ∫ ∞ −∞ x 2 dx = ∫ ∞ −∞ y 2 dy is not bounded, then ∫ ∞ −∞ ∫ ∞ −∞ ( x 2 + y 2 ) dx dy is not bounded. (b) Stable. ∫ ∞ −∞ ∫ ∞ −∞ | h ( x, y ) | dx dy = ∫ ∞ −∞ ∫ ∞ −∞ (exp {− ( x 2 + y 2 ) } ) dx dy = [ ∫ ∞ −∞ e − x 2 dx ] 2 = π , which is bounded. So the system is stable. (c) Not stable. The absolute integral ∫ ∞ −∞ ∫ ∞ −∞ x 2 e − y 2 dx dy = ∫ ∞ −∞ x 2 [ ∫ ∞ −∞ e − y 2 dy ] dx = ∫ ∞ −∞ √ πx 2 dx is unbounded. So the system is not stable. Solution 2.9 (a) g ( x ) = ∫ ∞ −∞ f ( x − t ) f ( t ) dt. (b) Given an input as af 1 ( x ) + bf 2 ( x ) , where a, b are some constant, the output is g ′ ( x ) = [ af 1 ( x ) + bf 2 ( x )] ∗ [ af 1 ( x ) + bf 2 ( x )] = a 2 f 1 ( x ) ∗ f 1 ( x ) + 2 abf 1 ( x ) ∗ f 2 ( x ) + b 2 f 2 ( x ) ∗ f 2 ( x ) 6 = ag 1 ( x ) + bg 2 ( x ) , where g 1 ( x ) and g 2 ( x ) are the output corresponding to an input of f 1 ( x ) and f 2 ( x ) respectively. Hence, the system is nonlinear. (c) Given a shifted input f 1 ( x ) = f ( x − x 0 ) , the corresponding output is g 1 ( x ) = f 1 ( x ) ∗ f 1 ( x ) = ∫ ∞ −∞ f 1 ( x − t ) f 1 ( t ) dt = ∫ ∞ −∞ f ( x − t − x 0 ) f 1 ( t − x 0 ) dt.

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    11 Changing variable t ′ = t − x 0 in the above integration, we get g 1 ( x ) = ∫ ∞ −∞ f ( x − 2 x 0 − t ′ ) f 1 ( t ′ ) dt ′ = g ( x − 2 x 0 ) . Thus, if the input is shifted by x 0 , the output is shifted by 2 x 0 . Hence, the system is not shift-invariant. C ONVOLUTION OF SIGNALS Solution 2.10 (a) f ( x, y ) δ ( x − 1 , y − 2) = f (1 , 2) δ ( x − 1 , y − 2) = (1 + 2 2 ) δ ( x − 1 , y − 2) = 5 δ ( x − 1 , y − 2) (b) f ( x, y ) ∗ δ ( x − 1 , y − 2) = ∫ ∞ −∞ ∫ ∞ −∞ f ( ξ, η ) δ ( x − ξ − 1 , y − η − 2) dξ dη = ∫ ∞ −∞ ∫ ∞ −∞ f ( x − 1 , y − 2) δ ( x − ξ − 1 , y − η − 2) dξ dη = f ( x − 1 , y − 2) ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − ξ − 1 , y − η − 2) dξ dη = f ( x − 1 , y − 2) = ( x − 1) + ( y − 2) 2 (c) ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − 1 , y − 2) f ( x, 3) dx dy © 1 = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − 1 , y − 2) f (1 , 3) dx dy = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − 1 , y − 2)(1 + 3 2 ) dx dy = 10 ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − 1 , y − 2) dx dy © 2 = 10 Equality © 1 comes from the Eq. (2.7) in the text. Equality © 2 comes from the fact: ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − 1 , y − 2) dx dy = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x, y ) dx dy = 1 .

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    12 CHAPTER 2 : SIGNALS AND SYSTEMS (d) δ ( x − 1 , y − 2) ∗ f ( x + 1 , y + 2) © 3 = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − ξ − 1 , y − η − 2) f ( ξ + 1 , η + 2) dξ dη © 4 = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − ξ − 1 , y − η − 2) f (( x − 1) + 1 , ( y − 2) + 2) dξ dη = ∫ ∞ −∞ ∫ ∞ −∞ δ ( x − ξ − 1 , y − η − 2) f ( x, y ) dξ dη © 5 = f ( x, y ) = x + y 2 © 3 comes from the definition of convolution; © 4 comes from the Eq. (2.7) in text; © 5 is the same as © 2 in part (c). Alternatively, by using the sifting property of δ ( x, y ) and defining g ( x, y ) = f ( x + 1 , y + 2) , we have δ ( x − 1 , y − 2) ∗ g ( x, y ) = g ( x − 1 , y − 2) = f ( x − 1 + 1 , y − 2 + 2) = f ( x, y ) = x + y 2 . Solution 2.11 (a) f ( x, y ) ∗ g ( x, y ) = ∫ ∞ −∞ ∫ ∞ −∞ f ( ξ, η ) g ( x − ξ, y − η ) dξ dη = ∫ ∞ −∞ ∫ ∞ −∞ f 1 ( ξ ) f 2 ( η ) g 1 ( x − ξ ) g 2 ( y − η ) dξ dη f ( x, y ) ∗ g ( x, y ) = (∫ ∞ −∞ f 1 ( ξ ) g 1 ( x − ξ ) dξ ) (∫ ∞ −∞ f 2 ( η ) g 2 ( y − η ) dη ) . Hence, their convolution is also separable. (b) f ( x, y ) ∗ g ( x, y ) = ( f 1 ( x ) ∗ g 1 ( x )) ( f 2 ( y ) ∗ g 2 ( y )) .

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    13 Solution 2.12 g ( x, y ) = f ( x, y ) ∗ h ( x, y ) = ∫ ∞ −∞ ∫ ∞ −∞ f ( x − ξ, y − η ) h ( ξ, η ) dξdη = ∫ ∞ −∞ ∫ ∞ −∞ ( x − ξ + y − η )exp {− ( ξ 2 + η 2 ) } dξdη = ( x + y ) ∫ ∞ −∞ ∫ ∞ −∞ e − ξ 2 − η 2 dξdη − ∫ ∞ −∞ ∫ ∞ −∞ ξe − ξ 2 − η 2 dξdη − ∫ ∞ −∞ ∫ ∞ −∞ ηe − ξ 2 − η 2 dξdη = ( x + y ) [∫ ∞ −∞ e − ξ 2 dξ ] 2 − ∫ ∞ −∞ e − η 2 [∫ ∞ −∞ ξe − ξ 2 dξ ] dη − ∫ ∞ −∞ e − ξ 2 [∫ ∞ −∞ ηe − η 2 dη ] dξ = π ( x + y ) (S2.4) We get (S2.4) by noticing that since ξ is an odd function and e − ξ 2 is an even function, we must have ∫ ∞ −∞ ξe − ξ 2 dξ = 0 . Also, ∫ ∞ −∞ e − ξ 2 dξ = √ π . F OURIER TRANSFORMS AND THEIR PROPERTIES Solution 2.13 (a) See the solution to part (b) below. The Fourier transform is F 2 { δ s ( x, y ) } = δ s ( u, v ) (b) F 2 { δ s ( x, y ; ∆ x, ∆ y ) } = ∫ ∞ −∞ ∫ ∞ −∞ δ s ( x, y ; ∆ x, ∆ y ) e − j 2 π ( ux + vy ) dx dy δ s ( x, y ; ∆ x, ∆ y ) is a periodic signal with periods ∆ x and ∆ y in x and y axes. Therefore it can be written as a Fourier series expansion. (Please review Oppenheim, Willsky, and Nawad, Signals and Systems for the definition of Fourier series expansion of periodic signals.) δ s ( x, y ; ∆ x, ∆ y ) = ∞ ∑ m = −∞ ∞ ∑ n = −∞ C mn e j 2 π ( mx ∆ x + ny ∆ y ) ,

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    14 CHAPTER 2 : SIGNALS AND SYSTEMS where C mn = 1 ∆ x ∆ y ∫ ∆ x 2 − ∆ x 2 ∫ ∆ y 2 − ∆ y 2 δ s ( x, y ; ∆ x, ∆ y ) e − j 2 π ( mx ∆ x + ny ∆ y ) dx dy = 1 ∆ x ∆ y ∫ ∆ x 2 − ∆ x 2 ∫ ∆ y 2 − ∆ y 2 ∞ ∑ m = −∞ ∞ ∑ n = −∞ δ ( x − m ∆ x, y − n ∆ y ) e − j 2 π ( mx ∆ x + ny ∆ y ) dx dy. In the integration region − ∆ x 2 < x < ∆ x 2 and − ∆ y 2 < y < ∆ y 2 there is only one impulse corresponding to m = 0 , n = 0 . Therefore, we have C mn = 1 ∆ x ∆ y ∫ ∆ x 2 − ∆ x 2 ∫ ∆ y 2 − ∆ y 2 δ ( x, y ) e − j 2 π ( 0 · x ∆ x + 0 · y ∆ y ) dx dy = 1 ∆ x ∆ y . We have: δ s ( x, y ; ∆ x, ∆ y ) = 1 ∆ x ∆ y ∞ ∑ m = −∞ ∞ ∑ n = −∞ e j 2 π ( mx ∆ x + ny ∆ y ) . Therefore, F 2 { δ s } = ∫ ∞ −∞ ∫ ∞ −∞ δ s ( x, y ; ∆ x, ∆ y ) e − j 2 π ( ux + vy ) dx dy = ∫ ∞ −∞ ∫ ∞ −∞ 1 ∆ x ∆ y ∞ ∑ m = −∞ ∞ ∑ n = −∞ e j 2 π ( mx ∆ x + ny ∆ y ) e − j 2 π ( ux + vy ) dx dy = ∞ ∑ m = −∞ ∞ ∑ n = −∞ 1 ∆ x ∆ y ∫ ∞ −∞ ∫ ∞ −∞ e j 2 π ( mx ∆ x + ny ∆ y ) e − j 2 π ( ux + vy ) dx dy = ∞ ∑ m = −∞ ∞ ∑ n = −∞ 1 ∆ x ∆ y F 2 { e j 2 π ( mx ∆ x + ny ∆ y ) } = ∞ ∑ m = −∞ ∞ ∑ n = −∞ 1 ∆ x ∆ y δ ( u − m ∆ x , v − n ∆ y ) © 5 = ∞ ∑ m = −∞ ∞ ∑ n = −∞ 1 ∆ x ∆ y · ∆ x ∆ yδ ( u ∆ x − m, v ∆ y − n ) F 2 { δ s } = δ s ( u ∆ x, v ∆ y ) Equality © 5 comes from the property δ ( ax ) = 1 | a | δ ( x ) .

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