Biosignal and Biomedical Image Processing phần 1 pps

... NY 10 016 , U.S.A. tel: 212 -696-9000; fax: 212 -685-4540 Distribution and Customer Service Marcel Dekker, Inc., Cimarron Road, Monticello, New York 12 7 01, U.S.A. tel: 800-228 -11 60; fax: 845-796 -17 72 Eastern ... constant value over frequency. TLFeBOOK 12 Chapter 1 −3 db means that the ratio is 1/ 1. 414 , and −20 db means the signal is 1/ 10 of the noise in RMS units. Figure...

Ngày tải lên: 06/08/2014, 00:20

... box and record the following output: Frequency (Hz): 2 10 20 60 10 0 12 5 15 0 200 300 400 V out volts rms: .15 10 −7 0 .1 10 −3 0.002 0.2 1. 5 3.28 4.47 4.97 4.99 5.0 What is the cutoff frequency and ... the anti-aliasing filter, and the system bandwidth is explored in a problem at the end of this chapter. Example 1. 1. An ECG signal of 1 volt peak-to-peak has a bandwidth of...

Ngày tải lên: 06/08/2014, 00:20

... 10 for j = 1: M % Generate M (10 0) images % Generate sine; scale between 0 and 1 phase = 10 *pi*j/M; % Shift phase 18 0 (pi) over 12 images I_sin (1, :) = .5 * sin(2*pi*x ؉ phase) ؉ .5’; for i = 1: N ... *sin(w *12 ) ؉ .1* randn (1, N); % Double sin, a sawtooth s2 = sawtooth(w*5,.5)؉ .1* randn (1, N); % and a periodic % function s3 = pulstran((0:999),(0:5)’ *18 0,kaiser (10 0,3)...

Ngày tải lên: 06/08/2014, 00:20

... 223 Yule-Walker, 11 8invertible, 18 8 Laplacian, 312 , 353, 369 Filter design, 94, 311 – 313 three-stage FIR, 11 1 11 7linear, 308– 310 log, 369 three-stage IIR, 11 9 12 3 two-stage FIR, 10 9 11 1lowpass, 13 14 optimal, ... evoked response, 34 10 0, 10 2, 10 3, 10 6, 12 5, 15 2– 15 3, 15 6 16 2, 16 3, 16 5, 17 7, Wavelet analysis, 17 7– 211 17 8, 18 1, 18 5, 19 3, 292...

Ngày tải lên: 06/08/2014, 00:21

... 0 .1, and an upper passband between 500 and 10 00 Hz (assuming f s /2 = 10 00) with the same ripple as the lower passband. The f , a , and dev vectors would be: f = [10 0 300 400 500] ; a = [10 1] ; ... of these ranges and is the same length as a . For example, assume you desire a bandstop filter that has a passband between 0 and 10 0 with a ripple of 0. 01, a stopband between 30...

Ngày tải lên: 06/08/2014, 00:20

... 6 .1 tn = (1: N/4)/fs; % Time vector to create % sinusoids b = (1: N)/fs; % Time vector for plotting x = [zeros(N/4 ,1) ; sin(2*pi*f1*tn) ’; sin(2*pi*f2*tn)’; zeros(N/4 ,1) ]; ti = ( (1: N1/2)/fs) *10 ; ... the Eqs. ( 21) and (22): G 0 (z) = H 1 (z) = z −N H 0 (z 1 )( 21) G 1 (z) =−H 0 (−z) = z −N H 1 (z 1 )(22) where the second equality comes from the relationship expressed in...

Ngày tải lên: 06/08/2014, 00:20

... 0:(lf -1) h1(i 1) = ( -1) vi * h0(lf-i); end g1 = h1(lf: -1: 1); % Highpass filter coeffi- % cients using order % flip, Eq. (24) % Calculate filter outputs for all levels for i = 1: L lpx = y (1: lseg); ... range: 0 <∆< 1 10LP x (12 ) where L is the length of the FIR filter and P x is the power in the input signal. P X can be approximated by: P x Ϸ 1 N − 1 ∑ N n =1 x 2 (n) (...

Ngày tải lên: 06/08/2014, 00:20

... .25*N; U = [1 1; 1 M; N M; N 1] ; % Input quadrilateral TLFeBOOK 304 Chapter 11 F IGURE 11 .1 The influence of aliasing due to undersampling on two images with high spatial frequency. The aliased images ... Example 11 .3 and Figure 11 .5A and B % Linear filtering. Load the blood cell image % Apply a 32nd order lowpass filter having a bandwidth of .12 5 % fs/2, and a highpass...

Ngày tải lên: 06/08/2014, 00:20

... Figure 11 .13 . The original reference image is shown on the left, and the input image TLFeBOOK Image Segmentation 357 F IGURE 12 .11 Textural pattern used in Example 12 .3. The horizontal and vertical patterns ... Alternatively, the islice routine could have been used as in Example 12 .2. % Example 12 .3 and Figures 12 .11 , 12 .12 , and 12 .13 % Analysis of texture pa...

Ngày tải lên: 06/08/2014, 00:21