carotid plaque texture analysis using 3 dimensional volume ultrasonic imaging
Float Analysis - Powerful Technical Indicators Using Price and Volume
... Information found in Daily Graphs, April 30 , 1999: 50 32 Float Analysis Findings Twenty-five stocks produced a total of 37 formations: 24 bottoms and 13 tops Twenty out of 25 stocks, or 80%, ... 131 Summary 132 Appendix A—A Compendium of Float Analysis Formations 135 Appendix B—The Dilemma of Differing ... has been for me — Steve Woods Float Analysis Powerful Technical Indicators Using Price and Volume CHAPTER ONE Watching the Float: The New Look of Price and Volume Charts and Their Relationship...
- 251
- 704
- 0
practical packet analysis using wireshark to solve real world network problems phần 3 ppt
... with their IP addresses Choose the interface you wish to use, and click Capture (Figure 3- 3) Figure 3- 3: Selecting an interface on which to perform your packet capture Your packet capture should ... WinPcap is checked, and click Install (Figure 3- 2) The installation process should begin Figure 3- 2: Selecting the option to install the WinPcap driver 30 C pt er About halfway through the Wireshark ... well as the duration of the current capture (Figure 3- 4) Figure 3- 4: The Capture window shows a brief summary of the packets being captured 32 C pt er Wait about a minute or so, and when you are...
- 18
- 527
- 0
Accurate and efficient three dimensional electrostatics analysis using singular boundary elements and fast fourier transform on multipole (FFTM)
... 16 32 64 128 256 Linear Extrapolation 256 (c.f [25]) Coefficients, Qs 1.12172 1.1 231 3 1.1 234 8 1.1 236 3 1.1 237 4 1.1250 1.1280 31 Chapter Two -Dimensional Singular Elements Next, the effect of using ... 28 4 .3. 1 Coaxial conductor example 28 4 .3. 2 Parallel conductor example 33 4 .3. 3 Biased element distribution effect for M = 36 4.4 Conclusion for Two -Dimensional ... squares sense [29], by hybrid-combined methods [30 , 31 ], penalty-combined methods [32 ] and also their combinations [33 ] 3. 4 Comments on the Three Approaches 3. 4.1 Mesh refinement techniques The mesh...
- 168
- 159
- 0
Báo cáo y học: "Segment-orientated analysis of two-dimensional strain and strain rate as assessed by velocity vector imaging in patients with acute myocardial infarction"
... -2.22 ± 1.40 1.85 ± 1.17 LE 1- 50% -11. 73 ±4.28 -0.69 ± 0.26 0.58 ± 0 .34 2 .30 ± 1.25 -1.60 ± 1. 03 -1 .30 ± 0. 93 LE ≥ 51% -10 .34 ± 4.76 -0.62 ± 0.25 0. 53 ± 0 .32 1.60 ± 1.26 1.10 ± 0.94 -0.99 ± 0.84 ... (cm/s) infarcted -10 .37 ± 4.75 -0.62 ± 0.25 0. 53 ± 0 .32 1.61 ± 1.27 -1.10 ± 0.94 -1.00 ± 0.84 adjacent -11.45 ± 4.55 -0.68 ± 0.25 0.60 ± 0 .34 2. 43 ± 1 .33 -1.65 ± 1.08 -1 .36 ± 0.96 non-infarcted ... Echocardiogr 2000; 1 (3) : 154-170 Mirsky I, Parmley WW Assessment of passive elastic stiffness for isolated heart muscle and the intact heart Circ Res 19 73; 33 (2): 233 -2 43 Pavlopoulos H, Nihoyannopoulos...
- 8
- 683
- 0
Network Traffic Analysis Using tcpdump
... >dnsquery.net.10 83: 3| 17/8/0 (488) (DF) 4500 0204 6af9 4000 fe11 1 439 2222 2222 33 33 333 3 0 035 043b 01f0 b 034 00 03 838 0 0001 0011 0008 0000 08 63 6f75 6e74 6572 32 0a 736 5 7874 ‘port 53 and udp[10] ... 10 11 a002 7d78 92b4 0000 0204 05b4 0402 080a 12 13 14 15 16 17 18 19 20 21 22 23 0076 3b6c 0000 0000 01 03 030 0 28 29 30 31 32 33 34 35 36 37 38 39 24 25 26 27 TCP header length = 10 * = 40 bytes ... sparky.com ftp: S 106 739 3 832 :106 739 3 832 (0) win 32 120 (DF) 4500 003c 0f5a 4000 4006 731 7 xxxx xxxx (16) yyyy yyyy 0457 0015 3f9f 232 8 0000 0000 (32 ) TCP header...
- 76
- 392
- 0
Electronics and Circuit Analysis Using MATLAB P1
... statement D1 = A1./B1 gives the result D1 = 0 .33 33 4.0000 1.7500 3. 0000 2.0000 2.5000 0.5714 0 .33 33 0.5000 0.4000 and the statement E1 = A1.\B1 gives E1 = 3. 0000 0.2500 The array operation of raising ... 3 + j + j3 w= then we can represent it in MATLAB as w = [1+j 2-2*j; 3+ 2*j 4 +3* j] which will produce the result w= 1.0000 + 1.0000i 2.0000 - 2.0000i 3. 0000 + 2.0000i 4.0000 + 3. 0000i If the ... for matrix t3, t3(2 ,3) denotes the entry in the second row and third column Using the colon as one of the subscripts denotes all of the corresponding row or column For example, t3(:,4) is the...
- 23
- 467
- 1
Electronics and Circuit Analysis Using MATLAB
... statement D1 = A1./B1 gives the result D1 = 0 .33 33 4.0000 1.7500 3. 0000 2.0000 2.5000 0.5714 0 .33 33 0.5000 0.4000 and the statement E1 = A1.\B1 gives E1 = 3. 0000 0.2500 The array operation of raising ... (dB) 34 34 34 32 30 Solution MATLAB Script % Bode plot for capacitively coupled amplifier f = [20 40 80 100 120 2000 5000 8000 10000 12000 15000 20000]; g = [ 10 30 32 34 34 34 34 32 30 10 5]; semilogx(f, ... frequency using a logarithmic scale for the frequency and a linear scale for the gain Frequency (Hz) 20 40 80 100 120 Gain (dB) 10 30 32 34 Frequency (Hz) 2000 5000 8000 10000 12000 Gain (dB) 34 34 34 ...
- 400
- 511
- 2
Electronics and Circuit Analysis Using MATLAB P3
... 6.0 Solution MATLAB Script diary ex3_2.dat % y1 = bitatd _3( -1.25) y2 = bitatd _3( 2.57) y3 = bitatd _3( 6.0) diary function Y_dig = bitatd _3( X_analog) % % bitatd _3 is a function program for obtaining ... string in quotes For example, the MATLAB statement disp( 3- by -3 identity matrix’) will result in 3- by -3 identity matrix and disp(eye (3, 3)) will result in 0 0 Echo The echo command can be used ... 1999 CRC Press LLC 3. 3 The Fibonacci sequence is given as 1 13 21 34 … Write a MATLAB program to generate the Fibonacci sequence up to the twelfth term Print out the results 3. 4 The table below...
- 17
- 440
- 0
Electronics and Circuit Analysis Using MATLAB P2
... (dB) 34 34 34 32 30 Solution MATLAB Script % Bode plot for capacitively coupled amplifier f = [20 40 80 100 120 2000 5000 8000 10000 12000 15000 20000]; g = [ 10 30 32 34 34 34 34 32 30 10 5]; semilogx(f, ... frequency using a logarithmic scale for the frequency and a linear scale for the gain Frequency (Hz) 20 40 80 100 120 Gain (dB) 10 30 32 34 Frequency (Hz) 2000 5000 8000 10000 12000 Gain (dB) 34 34 34 ... of ten students in a class STUDENT 10 EXAM #1 81 75 95 65 72 79 93 69 83 87 EXAM #2 78 77 90 69 73 84 97 72 80 81 EXAM #3 83 80 93 72 71 86 94 67 82 77 (a) Plot the results of each examination...
- 16
- 453
- 1
Electronics and Circuit Analysis Using MATLAB P6
... known, the voltage V3 can be obtained using circuit analysis tools Suppose V3 is If the values of V3 = Vm ∠ 3 , then the time domain voltage V3 (t) is v (t ) = Vm3 cos( wt + 3 ) The following ... voltages v3_abs = abs(V (3) ); v3_ang = angle(V (3) )*180/pi; fprintf('voltage V3, magnitude: %f \n voltage V3, angle in degree: %f', v3_abs, v3_ang) diary The following results are obtained: voltage V3, ... C I3 Figure 6.7 Unbalanced Three-phase System Simplifying Equations (6 .31 ), (6 .32 ) and (6 .33 ), we have 110∠0 = (6 + j 13) I (6 .34 ) 110∠ − 120 = (4 + j 2) I (6 .35 ) 110∠120 = (6 − j12.5) I (6 .36 )...
- 39
- 561
- 0
Electronics and Circuit Analysis Using MATLAB P5
... 10e-6; r1 = 1e3; © 1999 CRC Press LLC tau1 = c*r1; t = 0:0.002:0.05; v1 = 10*(1-exp(-t/tau1)); r2 = 10e3; tau2 = c*r2; v2 = 10*(1-exp(-t/tau2)); r3 = 1e3; tau3 = c*r3; v3 = 10*(1-exp(-t/tau3)); plot(t,v1,'+',t,v2,'o', ... =0 dt R1 R2 (5 .32 ) C2 dv (t ) V2 − V1 + + i1 = dt R2 (5 .33 ) Using loop analysis V2 = i1 R3 + L The output di1 ( t ) dt (5 .34 ) y (t ) is given as y ( t ) = v1 ( t ) − v ( t ) (5 .35 ) Simplifying ... Equations (5 .32 ) to (5 .34 ), we get dv1 ( t ) V V 1 )V1 + + s = −( + dt C1 R1 C1 R2 C1 R2 C1 R1 (5 .36 ) dv ( t ) V V i = − − dt C2 R2 C2 R2 C2 (5 .37 ) di1 ( t ) V2 R3 = − i1 dt L L (5 .38 ) Expressing...
- 33
- 714
- 0
Electronics and Circuit Analysis Using MATLAB P4
... 21 Z = Z31 Z n1 Z12 Z22 Z32 Zn Z 13 Z 23 Z 33 Zn3 Z1n Z2 n Z3 n Z nn I1 I 2 I = I3 . In and ∑ V1 ∑ V2 V = ∑ V3 ... inspection, Vb = V1 − V4 (4 .31 ) Using Ohm’s Law Ia = V4 − V3 (4 .32 ) Using KCL at node 1, and supernode 1-2, we get V1 V1 − V4 V − V3 + − 5Vb + =0 10 Using Equation (4 .31 ), Equation (4 .33 ) simplifies to ... node 3, we get V3 − V4 V3 − V2 + −5= which simplifies to − 0125V2 + 0 .32 5V3 − 0.2V4 = (4 .36 ) Using KVL for loop 1, we have − 10 I a + Vb + 5I a + 8( I a + 5) = (4 .37 ) Using Equations (4 .31 ) and...
- 27
- 471
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P12 pdf
... using equation (12.65) a4 = 1; a3 = wl1 + wl2 + wl3 + wh; a2 = wl1*wl2 + wl1*wl3 + wl2*wl3 + wl1*wh + wl2*wh + wl3*wh; a1 = wl1*wl2*wl3 +wl1*wl2*wh + wl1*wl3*wh + wl2*wl3*wh; a0 = wl1*wl2*wl3*wh; ... obtained using Equations (12 .32 ) and (12 .33 ), respectively The change in I C for each temperature is calculated using Equation (12 .36 ) MATLAB Script: % Bias stability % rb1=50e3; rb2=10e3; re=1.2e3; ... MATLAB Script %Frequency response of CE Amplifier rc=4e3; rb1=60e3; rb2=40e3; rs=100; rce=60e3; re=1.5e3; rl=2e3; beta=150; vcc=10; vt=26e -3; vbe =0.7; cc1=2e-6; cc2=4e-6; ce=150e-6;, rx=10; cpi=100e-12;...
- 52
- 559
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P11 doc
... have (11 .31 ) (11 .32 ) Also VO = V1 − I R R2 (11 .33 ) Using Equations (11 .30 ), (11 .31 ) and (11 .32 ), Equation (11 .33 ) becomes VO = − VO R2 − (V + VO A) A R1 IN (11 .34 ) Simplifying Equation (11 .34 ), we ... w); d3 = 1/(c*r2 (3) ); den3 = [1 d3]; h3 = freqs(num1,den3,w); semilogx(f,abs(h1),'w',f,abs(h2),'w',f,abs(h3),'w') xlabel('Frequency, Hz') ylabel('Gain') axis([1.0e-2,1.0e6,0,260]) text(5.0e-2 ,35 ,'R2 ... 8e5*pi; p3 = 1.6e6*pi; p = [p1 p2 p3]; % zeros z = [0]; const = 2.62e21; % convert to poles and zeros and % find the frequency response a3 = 1; a2 = p1 + p2 + p3; a1 = p1*p2 + p1*p3 + p2*p3; a0...
- 35
- 526
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P10 ppt
... ln = q q ni (10 .34 ) and Using Equations (10 .31 ) to (10 .34 ), we have VC = kT N A N D ln q ni2 (10 .35 ) It should be noted that Equations (10 .30 ) and (10 .35 ) are identical Typically, ... junction with at 30 0 oK is 2 .39 *10 13 , N A = 1018 cm -3 , N D = 1015 cm -3 and ni (a) Calculate the contact potential (b) Plot the junction potential for source voltages of -1.0V to 0.3V 10.5 For the ... Junction potential versus source voltage % using equation(10 .36 ) contact potential is t = 30 0; na = 1.0e17; nd = 1.0e14; nisq = 1.04e20; q = 1.602e-19; k = 1 .38 e- 23; % calculate contact potential vc...
- 43
- 483
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P9 docx
... 0. 133 e-12 0.2 1.79e-12 0 .3 24.02e-12 0.4 0 .32 1e-9 0.5 4 .31 e-9 0.6 57.69e-9 0.7 7.726e-7 Solution diary ex9_1.dat % Diode parameters vt = 25.67e -3; v = [0.1 0.2 0 .3 0.4 0.5 0.6 0.7]; i = [0. 133 e-12 ... MATLAB program diary ex9_8.dat % Zener diode voltage regulator vs1 = -30 ; vs2 = -35 ; rl =10e3; rs = 2e3; i = -50e -3: 5e -3 :0; vz = -20 + 0.05*i; m = length(i); i(m+1) = 0; vz(m+1) = -10; i(m+2) ... V') From the MATLAB program, we have idiode = 9 .30 37e-004 vdiode = 0.69 63 Thus I D = 0. 930 4 mA and V D = voltage during the iteration process 0.69 63 V Figure 9.7 shows the diode Figure 9.7 Diode...
- 41
- 550
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P8 ppt
... N W30 = W31 = e − j 2π − j 4π W32 = e = −0.5 − j 0.866 = −0.5 + j 0.866 W 33 = W30 = W34 = W31 Using Equation (8.41), we have G[0] = ∑ g[n]W30 = + + = n =0 G[1] = ∑ g[n]W3n = g[0]W30 + g[1]W31 ... a g (at ) ⇔ (8 .30 ) Duality G (t ) ⇔ g ( − f ) (8 .31 ) Time shifting g (t − t ) ⇔ G ( f ) exp( − j 2πft ) (8 .32 ) Frequency Shifting exp( j f C t ) g (t ) ⇔ G ( f − f C ) (8 .33 ) Definition in ... G[1] = ∑ g[n]W3n = g[0]W30 + g[1]W31 + g[2]W32 n=0 = + 2(−05 − j0866) + (−05 + j0866) = −05 − j0866 G[2] = ∑ g[n]W32n = g[0]W30 + g[1]W32 + g[2]W34 n =0 = + 2(−05 + j0866) + (−05 −...
- 27
- 433
- 0
Tài liệu Electronics and Circuit Analysis Using MATLAB P7 pdf
... + Z3 Z3 [Z ] = Z + Z3 Z3 We can convert the z-parameters to y-parameters [refs and 6] and we get Z + Z3 Z1Z2 + Z1Z3 + Z Z3 − Z3 y12 = Z1 Z2 + Z1Z3 + Z2 Z3 − Z3 y 21 = Z1Z2 + Z1Z3 + ... (7. 53) Using Equation (7.50), the equivalent y-parameters of the bridge-T network are Z + Z3 + Z4 Z1 Z2 + Z1 Z3 + Z2 Z3 Z3 =− − Z4 Z1 Z2 + Z1Z3 + Z Z3 Z3 =− − Z4 Z1Z2 + Z1Z3 + Z2 Z3 Z1 + Z3 = ... a1 = [3 2; 1]; a2 = [3 4; 0.5 1]; a3 = [3 8; 0.25 1]; a4 = [3 16; 0.125 1]; % equivalent transmission parameters a = a1*(a2*(a3*a4)) diary The value of matrix a is a= 112.2500 39 .37 50 7 .3 630 .0000...
- 29
- 482
- 0
Tài liệu Electronics and Circuit Analysis using MATLAB pptx
... statement D1 = A1./B1 gives the result D1 = 0 .33 33 4.0000 1.7500 3. 0000 2.0000 2.5000 0.5714 0 .33 33 0.5000 0.4000 and the statement E1 = A1.\B1 gives E1 = 3. 0000 0.2500 The array operation of raising ... (dB) 34 34 34 32 30 Solution MATLAB Script % Bode plot for capacitively coupled amplifier f = [20 40 80 100 120 2000 5000 8000 10000 12000 15000 20000]; g = [ 10 30 32 34 34 34 34 32 30 10 5]; semilogx(f, ... frequency using a logarithmic scale for the frequency and a linear scale for the gain Frequency (Hz) 20 40 80 100 120 Gain (dB) 10 30 32 34 Frequency (Hz) 2000 5000 8000 10000 12000 Gain (dB) 34 34 34 ...
- 386
- 341
- 0
Tìm thêm: hệ việt nam nhật bản và sức hấp dẫn của tiếng nhật tại việt nam xác định các nguyên tắc biên soạn khảo sát các chuẩn giảng dạy tiếng nhật từ góc độ lí thuyết và thực tiễn khảo sát chương trình đào tạo gắn với các giáo trình cụ thể xác định thời lượng học về mặt lí thuyết và thực tế tiến hành xây dựng chương trình đào tạo dành cho đối tượng không chuyên ngữ tại việt nam điều tra đối với đối tượng giảng viên và đối tượng quản lí nội dung cụ thể cho từng kĩ năng ở từng cấp độ phát huy những thành tựu công nghệ mới nhất được áp dụng vào công tác dạy và học ngoại ngữ mở máy động cơ lồng sóc các đặc tính của động cơ điện không đồng bộ hệ số công suất cosp fi p2 đặc tuyến hiệu suất h fi p2 đặc tuyến mômen quay m fi p2 đặc tuyến tốc độ rôto n fi p2 động cơ điện không đồng bộ một pha sự cần thiết phải đầu tư xây dựng nhà máy thông tin liên lạc và các dịch vụ phần 3 giới thiệu nguyên liệu chỉ tiêu chất lượng theo chất lượng phẩm chất sản phẩm khô từ gạo của bộ y tế năm 2008