SOLUTIONS MANUAL FOR BIOMEDICAL SIGNAL AND IMAGE ... · solutions manual for biomedical signal and...

SOLUTIONS MANUAL FOR BIOMEDICAL SIGNAL AND IMAGE PROCESSING 2ND EDITION NAJARIAN

SOLUTIONS


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Chapter 2

Problem 2

We have )()(1 tgtg α= , where α>1

Let tu α= . This implies that α/ut = where α is the scaling factor with 1>α

Then

∫ ∫∞

∞−

∞

∞−

−−

====

αααα α fGudeugdteatgtgFTtgFT jfujft 1)/()()()}({)}({ /

1

Problem 4

)()(

)()()(

)()(

)()()}()({

21

)(

21

vGuG

dtdetyex

ddtetyx

dtedtyxtgtgFT

tvu

jft

jft

⋅=

−−=

−=

−=∗

∫∫

∫ ∫

∫ ∫

−−−

−

−

ττττ

τττ

τττ

ττ

As F(u) = g(t), we have

Problem 6

∫== − dfeuGuGFtg jft)(21)}({)( 1

π

Let ,' tt −= we get

∫ −=− dfejfGtg jft '

)(21)( '

π

Interchanging 't and f we get

∫

∫

==−

==−

−

−

)}({)()(

)}({)()(2

'2'

''

'

'

tGFdtetGfg

tGFdtetGfg

ftj

jft

π

π



Problem 8:

Use the same code for problem 1-2 d to find the fft of the fignal

Order=9;

Problem 10:

Wn=[200]/(samplerate/2); [bl,al]=butter(n,Wn); Filteredsig=filter(bl,al,SIG);



Chapter 3

I=imread('p_3_2.jpg');

Problem 2

I=rgb2gray(I);

figure;imshow(I);

S=size(I);

I=double(I);

I=dec2base(I,2);

newS=size(I)

J=zeros(S(1),S(2));

for i=1:newS(1)

k=char(I(i,:));

k(1) = '0';

k(2) = '0';

k(3) ='0';

// k(4) = '0'; //Uncomment this line to solve section b

k=base2dec(k,2);

a=fix(i/S(1))+1;

b=mod(i,S(1));

if b == 0

b=S(1);

a=a-1;

end

J(b,a)=k;

end

figure;imshow(J,[0 255]);

Image before elimination of any bit(file: p_3_2.jpg)



Courtesy of CRC Press/Taylor & Francis Group

t

δΔ(t)

Δ/2

1/Δ

–Δ/2

Figure 2.1 δΔ(t) function.

002x001.eps




t

δ(t)

Figure 2.2 Visual representation of an impulse function.

002x002.eps




tt0

δ(t– t0)

Figure 2.3 Shifted impulse function.

002x003.eps




tT

PΠ(t)

A

Figure 2.4 Unit pulse function.

002x004.eps




3

2

1

0

–1

–2

–30 1 2 3 4 5

Time t

x(t)

6 7 8 9 10

Figure 2.5 A given time signal.

002x005.eps




1

0.5

|X( f )|

f–0.5 –0.25 0.50.25

Figure 2.6 FT of the time signal in Figure 2.5.

002x006.eps




g(n)

g(n– 1)

g(n– 2)

SHIFT1 g(n)

SHIFT2 g(n)

Figure 2.7 Circular shift of a signal g(n).

002x007.eps




g(n)STRETCH2 g(n)

Figure 2.8 Stretch of a signal.

002x008.eps




g(n)REPEAT2 g(n)

Figure 2.9 Repeat of a signal.

002x009.eps




8

6

4

2

0

–2

–4

–60 10 20 30 40 50

Time

Tim

e sig

nal

60 70 80 90 100

Figure 2.10 Signal x defined in the time domain.

002x010.eps




140

120

100

80

60

40

20

00 5 10 15 20 25

Frequency

Mag

nitu

de o

f DFT

30 35 40 45 50

Figure 2.11 Magnitude of DFT of signal x.

002x011.eps




50

100

150

200

250

300

50 100 150 200x

y

250 300 350

Figure 2.12 Image g(x, y). (Courtesy of Andre D’Avila, MD, Heart Institute (InCor), University of Sao Paulo, Medical School, Sao Paulo, Brazil.)

002x012.eps




50

100

150

200

250

300

50 100 150 200u

v

250 300 350

Figure 2.13 Magnitude of 2-D DFT of image g(x, y).

002x013.eps




P( f ) Q( f )H( f )

Figure 2.14 Filtering signals and images using filter H( f ).

002x014.eps




1

|H|

fD0

Figure 2.15 Ideal low-pass filter.

002x015.eps




1

|H|

fD0

Figure 2.16 Ideal high-pass filter.

002x016.eps




1

|H|

fD0 D1

Figure 2.17 Ideal band-pass filter.

002x017.eps




|H|1

D0(a) f

1——––√2

|H|1

D0(b) f

1——––√2

Figure 2.18 (a) Low-pass Butterworth filter and (b) high-pass Butterworth filter.

002x018.eps




0

–100

–200

–3000 50 100 150 200 250

Frequency (Hz)

Mag

nitu

de (d

B)

300 350 400 450 500

0

–400–200

–600

–1000–800

0 50 100 150 200 250Frequency (Hz)

Phas

e (de

gree

s)

300 350 400 450 500

Figure 2.19 Frequency response of Butterworth filter designed in Example 2.3.

002x019.eps



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