Solution manual for medical imaging signals and systems 2nd edition by prince

Since fdm, n is a discrete signal, its period must be an integer if it is to be periodic... We use contradiction to show that if the LSI system is BIBO stable, its PSF must be absolutely

Signals and Systems

SIGNALS AND THEIR PROPERTIES

Solution 2.1

(a) δs(x, y) =P∞

m=−∞

P∞ n=−∞δ(x − m, y − n) =P∞

m=−∞δ(x − m) ·P∞

n=−∞δ(y − n), therefore it is aseparable signal

(b) δl(x, y) is separable if sin(2θ) = 0 In this case, either sin θ = 0 or cos θ = 0, δl(x, y) is a product of aconstant function in one axis and a 1-D delta function in another But in general, δl(x, y) is not separable

(c) e(x, y) = exp[j2π(u0x+v0y)] = exp(j2πu0x)·exp(j2πv0y) = e1D(x; u0)·e1D(y; v0), where e1D(t; ω) =exp(j2πωt) Therefore, e(x, y) is a separable signal

(d) s(x, y) is a separable signal when u0v0 = 0 For example, if u0 = 0, s(x, y) = sin(2πv0y) is the product

of a constant signal in x and a 1-D sinusoidal signal in y But in general, when both u0and v0are nonzero,s(x, y) is not separable

m=−∞

∞X

n=−∞

δ(x − m, y − n) For arbitrary integers M and N , we have

comb(x + M, y + N ) =

∞X

m=−∞

∞X

n=−∞

δ(x − m + M, y − n + N )

=

∞X

p=−∞

∞X

q=−∞

δ(x − p, y − q) [let p = m − M, q = n − N ]

= comb(x, y)

So the smallest period is 1 in both x and y directions.

(c) Periodic Let f (x + Tx, y) = f (x, y), we have

sin(2πx) cos(4πy) = sin(2π(x + Tx)) cos(4πy) Solving the above equation, we have 2πTx = 2kπ for arbitrary integer k So the smallest period for x is

Tx0= 1 Similarly, we find that the smallest period for y is Ty0= 1/2

(d) Periodic Let f (x + Tx, y) = f (x, y), we have

sin(2π(x + y)) = sin(2π(x + Tx+ y))

So the smallest period for x is Tx0= 1 and the smallest period for y is Ty0= 1

(e) Not periodic We can see this by contradiction Suppose f (x, y) = sin(2π(x2+ y2)) is periodic; then thereexists some Txsuch that f (x + Tx, y) = f (x, y), and

5n

.Solving for M , we find that M = 10k for any integer k The smallest period for both m and n is therefore10

(g) Not periodic Following the same strategy as in (f), we let fd(m + M, n) = fd(m, n), and then

sin 1

5m

cos 1

5n



The solution for M is M = 10kπ Since fd(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 = 10kπ for some M So, fd(m, n) =sin 15m
cos 1

m=−∞

∞X

where bXc is the greatest integer that is smaller than or equal to X We also have

P∞(δs) = lim

X→∞ lim

Y →∞

14XY

m=−∞

∞X

cos θ



Z X

−X

Z Y

−Yδ(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

Z X

−X

Z Y

−Yδ(x) dx dy

X→∞ lim

Y →∞

14XY

X→∞ lim

Y →∞

14XY

Z Y

−Y1dx

X→∞ lim

Y →∞

2Y4XY

X→∞

12X

P∞(e) = lim

X→∞ lim

Y →∞

14XY

converge as X and Y go to infinity We also have

P∞(s) = lim

X→∞ lim

Y →∞

14XY

Z X

−X

Z Y

−Ysin2[2π(u0x + v0y)] dx dy

X→∞ lim

Y →∞

14XY

X→∞ lim

Y →∞

14XY

X→∞ lim

Y →∞

14XY

2XY −2 sin(4πu0X) sin(4πv0Y )

(4π)2u0v0



2.

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 X0 = 1/u0and Y0 = 1/v0, we have an alternative way tocompute P∞by considering only one period in each dimension Accordingly,

4X0Y0

2X0Y0−2 sin(4πu0X0) sin(4πv0Y0)

So the cascade of two LSI systems is also linear Now suppose for a given signal f (x, y) we have S1[f (x, y)] =

g(x, y), and S2[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:

S2[S1[f (x − ξ, y − η)]] = S2[g(x − ξ, y − η)] = h(x − ξ, y − η)

This proves that two LSI systems in cascade is an LSI system

To prove Eq (2.46) we carry out the following:

To prove (2.47) we start with the definition of convolution

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)|

=

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,

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

k=1

wkfk(x, −1) +

KX

k=1

wkfk(0, y)

=

KX

k=1

wk[fk(x, −1) + fk(0, y)]

=

KX

k=1

wkgk(x, y)

where gk(x, y) is the response of the system to input fk(x, y) Therefore, the system is linear.

(b) If g0(x, y) is the response of the system to input f (x − x0, y − y0), then

k=1

wkfk(x, y)

! KX

i=1

KX

j=1

wiwjfi(x, y)fj(x − x0, y − y0),

while

KX

k=1

wkgk(x, y) =

KX

k=1

wkfk(x, y)fk(x − x0, y − y0)

Since g0(x, y) 6=PK

k=1gk(x, y), the system is nonlinear

On the other hand, if g0(x, y) is the response of the system to input f (x − a, y − b), then

g0(x, y) = f (x − a, y − b)f (x − a − x0, y − b − y0)

= g(x − a, y − b)and the system is thus shift-invariant

(b) If g0(x, y) is the response of the system to inputPK

k=1

wkfk(x, η) dη

=

KX

k=1

wkgk(x, y),

where gk(x, y) is the response of the system to input fk(x, y) Therefore, the system is linear

On the other hand, if g0(x, y) is the response of the system to input f (x − x0, y − y0), then

(c) Not stable The absolute integralR∞

where g1(x) and g2(x) are the output corresponding to an input of f1(x) and f2(x) respectively

Hence, the system is nonlinear

(c) Given a shifted input f1(x) = f (x − x0), the corresponding output is

Changing variable t0= t − x0in the above integration, we get

(d)δ(x − 1, y − 2) ∗ f (x + 1, y + 2) =3

Hence, their convolution is also separable

(b)

f (x, y) ∗ g(x, y) = (f1(x) ∗ g1(x)) (f2(y) ∗ g2(y))

δs(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 thedefinition of Fourier series expansion of periodic signals.)

δs(x, y; ∆x, ∆y) =

∞X

m=−∞

∞X

n=−∞

Cmnej2π(mx∆x +ny∆y) ,

Z ∆y2

− ∆y 2

∞X

m=−∞

∞X

n=−∞

δ(x − m∆x, y − n∆y)e−j2π(mx∆x +ny∆y) dx dy

In the integration region −∆x2 < x < ∆x

2 and −∆y2 < y < ∆y2 there is only one impulse corresponding to

Z ∆y2

−∆y2δ(x, y)e−j2π(0·x+0·y) dx dy

m=−∞

∞X

m=−∞

∞X

n=−∞

ej2π(mx∆x +ny∆y)e−j2π(ux+vy)dx dy

=

∞X

m=−∞

∞X

m=−∞

∞X

m=−∞

∞X

m=−∞

∞X

 Z ∞

−∞

1

√2πσ2e−(y2+j4πσ2vy)/2σ2dy

(b) Similarly, assuming f (x) is real and f (x) = −f (−x),

(b) Scaling property: F2(fab)(u, v) = |ab|1 F2(f ) ua,vb
.



(c) Convolution property: F2(f ∗ g)(u, v) = F2(g)(u, v) · F2(f )(u, v)

(d) Product property: F2(f g)(u, v) = F (u, v) ∗ G(u, v).

Z 1/2

−1/2sin(2πux)dx, ejθ= cos θ + j sin θ

=

Z 1/2

−1/2cos(2πux)dx

= sin(πu)πu

= sinc(u) Therefore, we have F {sinc(x)} = rect(u) Using Parseval’s Theorem, we have

For the sinc function, P∞= 0, because E∞is finite.

Solution 2.18

Since the signal is separable, we have

F [f (x, y)] = F1D[sin(2πax)]F1D[cos(2πby)] ,

F1D[sin(2πax)] = 1

2j[δ (u − a) − δ (u + a)] ,

F1D[cos(2πby)] = 1

2[δ (v − b) + δ (v + b)] So,

F [f (x, y)] = 1

4j[δ(u − a)δ(v − b) − δ(u + a)δ(v − b) + δ(u − a)δ(v + b) − δ(u + a)δ(v + b)] Now we need to show that δ(u)δ(v) = δ(u, v) (in a generalized way):

δ(u)δ(v) = 0, for u 6= 0, or v 6= 0Therefore,

Letting u = q cos φ and v = q sin φ, the above equation becomes:

F (u, v) =

Z ∞ 0

Z 2π

0sin(2πqr sin θ)1

= 2

Z π

0cos(2πqr sin θ)dθ

= 2πJ0(2πqr)

1 holds because cos(−θ) = cos(θ), and sin(θ) = − sin(θ)

Based on the above derivation, we have proven (2.108)

Solution 2.20

The unit disk is expressed as f (r) = rect(r) and its Hankel transform is

F (q) = 2π

Z ∞ 0

f (r)J0(2πqr)r dr

= 2π

Z ∞ 0rect(r)J0(2πqr)r dr

= 2π

Z 1/2

0

J0(2πqr)r dr Now apply the following change of variables

From mathematical tables, we note that

Z x

0

J0()d = xJ1(x) Therefore,

−3

−2

−1 0 1 2 3 0 0.2 0.4 0.6 0.8 1

x y

(b) The transfer function of the function is the Fourier transform of the impulse response function:

H(u, v) = F {h(x, y)}

= F {e−πx2}F {e−πy2/4}, since h(x, y) is separable

= 4e−π(u2+4v2)

αw2



At the point halfway between two adjacent bars, we have

α

hsinαw2



− sinαw −π

2

i

(b) From the line spread function alone, we cannot tell whether the system is isotropic The line spread function

is a “projection” of the PSF During the projection, the information along the y direction is lost

(c) Since the system is separable with h(x, y) = h1D(x)h1D(y), we know that

2

hsincπ

α(u − α/2π)

+ sincπ

α(u + α/2π)

i.Therefore, the transfer function is

H(u, v) = π

2

hsincπ

α(u − α/2π)

+ sincπ

α(u + α/2π)

i

hsincπ

α(v − α/2π)

+ sincπ

α(v + α/2π)

i

APPLICATIONS, EXTENSIONS AND ADVANCED TOPICS

Solution 2.23

(a) The system is separable because h(x, y) = e−(|x|+|y|)= e−|x|e−|y|.(b) The system is not isotropic since h(x, y) is not a function of r =px2+ y2.Additional comments: An easy check is to plug in x = 1, y = 1 and x = 0, y = √

2 into h(x, y) Bynoticing that h(1, 1) 6= h(0,√

2), we can conclude that h(x, y) is not rotationally invariant, and hence notisotropic

Isotropy is rotational symmetry around the origin, not just symmetry about a few axes, for example the and y-axes h(x, y) = e−(|x|+|y|)is symmetric about a few lines, but it is not rotationally invariant

x-When we studied the properties of Fourier transform, we learned that if a signal is isotropic then its Fouriertransform has a certain symmetry Note that the symmetry of the Fourier transform is only a necessary, butnot sufficient, condition for the signal to be isotropic

e−ηdη



= 2e−|x|.(d) The response is

-( )x-y

h<0 +h<0

x-y h>0x-y+h<0 h>0x-y+h>0

Prob-lem 2.23(d)

I η ∈ (−∞, 0) In this interval, x − y + η < η < 0 |η| = −η, |x − y + η| = −(x − y + η);

II η ∈ [0, −(x − y)) In this interval, x − y + η < 0 ≤ η |η| = η, |x − y + η| = −(x − y + η);

III η ∈ [−(x − y), ∞) In this interval, 0 ≤ x − y + η < η |η| = η, |x − y + η| = x − y + η

Based on the above analysis, we have:

x-y h<0x-y+h>0 h>0x-y+h>0

Prob-lem 2.23(d)

I η ∈ (−∞, −(x − y)) In this interval, η < x − y + η < 0 |η| = −η, |x − y + η| = −(x − y + η);

II η ∈ [−(x − y), 0) In this interval, η < 0 ≤ x − y + η |η| = −η, |x − y + η| = x − y + η;

III η ∈ [0, ∞) In this interval, 0 ≤ η < x − y + η |η| = η, |x − y + η| = x − y + η

Based on the above analysis, we have:

(a) Yes, it is shift invariant because its impulse response depends on x − ξ

(b) By linearity, the output is

.Using the linearity of the Fourier transform and the Fourier transform pairs

only high frequency components Formal proof:

= f (t) +

Z −∞

t2U0sinc(2U0(y))dy

1 − 1

2−

Z t

02U0sinc(2U0(y))dy t > 0

1

2 −

Z t

02U0sinc(2U0(y))dy t > 0

−1/T, T /2 < t < T

0, otherwise

The impulse response is plotted in Fig S2.4

The absolute integral of h(t) isR∞

−∞|h(t)|2dt = 2/T So The system is stable when T > 0 The system isnot causal, since h(t) 6= 0 for −T < t < 0

(b) The response of the system to a constant signal f (t) = c is

−t/T + 1, T /2 < t < T

The response of the system to the unit step signal is plotted in Figure S2.5

Figure S2.5 The response of the system to the unit step signal See Problem 2.26(c).

(d) The Fourier transform of a rect function is a sinc function (see Problem 2.17) By using the properties of theFourier transform (scaling, shifting, and linearity), we have

H(u) = F {h(t)}

= −0.5e−j2πu(−0.75T )sinc(0.5uT ) + sinc(uT ) − 0.5e−j2πu(0.75T )sinc(0.5uT )

= sinc(uT ) − cos(1.5πuT ) sinc(0.5uT ) (e) The magnitude spectrum of h(t) is plotted in Figure S2.6

−200 −15 −10 −5 0 5 10 15 20 0.2

0.4 0.6 0.8 1 1.2 1.4

(f) From the calculation in part (d) and the plot in part (c), it can be seen that |H(0)| = 0 So the output of thesystem does not have a DC component The system is not a low pass filter The system is not a high-passfilter since it also filters out high frequency components As T → 0, the pass band of the system moves tohigher frequencies, and the system tends toward a high-pass filter

Solution 2.27

(a) The inverse Fourier transform of ˆH(%) is

ˆh(r) = F−1{ ˆH(%)}

=

Z ∞

−∞

ˆH(%)ej2πr%d%

G(%) = 0 Therefore, the responses of a ramp filter to a constant function is g(r) = 0 ii) The Fourier transform of asinusoid function f (r) = sin(ωr) is

F (%) = 1

2j

hδ(% − ω2π) − δ(% +

ω2π)

i.Hence,

h

δ% − ω2π

Therefore, the response of a ramp filter to a sinusoid function is



= rectu

a,

vb

G(u, v) = rectu

a,

vb

+1

2[δ(u − A, v − B) + δ(u + A, v + B)] ,

which is plotted in Figure S2.7 In order for an ideal low pass filter to recover f (x, y), the cutoff frequencies of the

filter must satisfy

|a|/2 < U < A and |b|/2 < V < B The Fourier transform of h(x, y) is rect 2Uu ,2Vv
; therefore, the impulse response is

h(x, y) = F−1nrect u

2U,

v2V

A sinc function, sinc(x), is shown in Figure 2.4(b)

(b) If the sampling period is ∆x1= 1/2, we have

g1(m) = g(m/2) =



1, −2 ≤ m ≤ 2

0, otherwise .Its DTFT is

G1(ω) = FDTFT{g1(m)}

= ej2ω+ ejω+ 1ej0ω+ e−jω+ 2e−j2ω

= 1 + 2 cos(ω) + 2 cos(2ω) The DTFT of g1(m) is shown in Figure S2.8

Figure S2.8 The DTFT g1(m) See Problem 2.30(b).

(c) If the sampling period is ∆x2= 1, we have

g2(m) = g(m) =



1, −1 ≤ m ≤ 1

0, otherwise .Its DTFT is

G2(ω) = FDTFT{g2(m)}

= ejω+ 1ej0ω+ e−jω

= 1 + 2 cos(ω) The DTFT of g2(m) is shown in Figure S2.9

(d) The discrete version of signal g(x) can be written as

g1(m) = g(x − m∆x1), m = −∞, · · · , −1, 0, 1, · · · , +∞ The DTFT of g1(m) is

is the continuous Fourier transform of the product of g(x) and δs(x; ∆x1) evaluated as u = ω/(2π∆x1)

Using the product property of the continuous Fourier transform, we have:

G1(ω) = F {g(x)} ∗ F {δs(x; ∆x1)}|u=ω/(2π∆x

1 )

= G(u) ∗ comb(u∆x1)|u=ω/(2π∆x

1 ).The convolution of G(u) and comb(u∆x1) is to replicate G(u) to u = k/∆x1 Since u = ω/(2π∆x1),

G1(ω) is periodic with period Ω = 2π

(e) The proof is similar to that for the continuous Fourier transform:

m=−∞

e−jωm

∞X

n=−∞

x(m − n)y(n)

=

∞X

n=−∞

" ∞X

m=−∞

e−jωmx(m − n)

#y(n)

=

∞X

n=−∞

e−jωn

" ∞X

k=−∞

e−jωkx(k)

#y(n)(let k = m − n)

=

∞X

n=−∞

e−jωnFDTFT{x(m)}y(n)

= FDTFT{x(m)}FDTFT{y(m)}

(f) First we evaluate the convolution of g1(m) with g2(m):

Then by direct computation, we have

FDTFT{g1(m) ∗ g2(m)} = 3 + 3 × 2 cos(ω) + 2 × 2 cos(2ω) + 2 cos(3ω)

= 3 + 6 cos(ω) + 4 cos(2ω) + 2 cos(3ω)

On the other hand, we have

FDTFT{g1(m)} = 1 + 2 cos(ω) + 2 cos(2ω)and

FDTFT{g2(m)} = 1 + 2 cos(ω)

So, the product of the DTFT’s of g1(m) and g2(m) is

FDTFT{g1(m)}FDTFT{g2(m)} = [1 + 2 cos(ω)][1 + 2 cos(ω) + 2 cos(2ω)]

= 1 + 4 cos(ω) + 2 cos(2ω)+4 cos2(ω) + 4 cos(ω) cos(2ω)

= 1 + 4 cos(ω) + 2 cos(2ω)+41 + cos(2ω)

as a Fourier series expansion (Please review Oppenheim, Willsky, and Nawad, Signals and Systems for thedefinition of... cos(θ), and sin(θ) = − sin(θ)

Based on the above derivation, we have proven (2.108)

Solution 2.20

The unit disk is expressed as f (r) = rect(r) and its Hankel transform is... impulse response depends on x − ξ

(b) By linearity, the output is

.Using the linearity of the Fourier transform and the Fourier transform pairs