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Nghiên cứu tác động của các yếu tố và sự thỏa mãn thương hiệu đến lòng trung thành thương hiệu của khách hàng sử dụng máy ảnh gương lật kỹ thuật số DSLR (DSLR – Digital Single-lens Reflex).

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"1. Kinh nghiệm thương hiệu: Kinh nghiệm thương hiệu là một yếu tố quan trọng trong việc xây dựng lòng trung thành thương hiệu của khách hàng. Nghiên cứu này sẽ phân tích các yếu tố liên quan đến kinh nghiệm thương hiệu của khách hàng với máy ảnh gương lật kỹ thuật số DSLR ở khu vực TP. HCM. Để đạt được mục tiêu này, nghiên cứu sẽ phân tích các yếu tố như: thời gian sử dụng máy ảnh, sự hài lòng về chất lượng sản phẩm, sự hài lòng về dịch vụ hỗ trợ, sự hài lòng về giá cả, sự hài lòng về thiết kế và các yếu tố khác liên quan đến kinh nghiệm th"

BÀI TẬP LỚN MƠ HÌNH HĨA HỆ THỐNG Sinh viên: Nguyễn Đức Dũng Lớp: DTD53-DH2 MSV: 45880 Đề 07: Dùng máy tính khảo sát q trình q độ hệ thống điều khiển tự động Cho hệ thống điều khiển tự động có sơ đồ cấu trúc sau: U(t) K1 + T1.s _ K2 (1 + T2 s)(1 + T3 s) K3 K1=50; K2=5; K3=0,5; T1=0,1; T2=0,01; T3=0,2 Nhiệm vụ sinh viên: 1) Viết phương trình sai phân * Tính W(p) K1 K2 g + T1s (1 + T2 s)(1 + T3 s) W(s) = K 1K K 1+ (1 + T1.s)(1 + T2 s)(1 + T3 s) K 1K W(p) = T T T s3 + T T s + (T + T + T T + T T )s + + K K K 2 3 Thay s= z −1 × T z +1 y(t) K1 K W(z) = T T T  2(z − 1)  + T T  2(z − 1)  + (T + T + T T + T T )  2(z − 1)  + + K K K 3 2 3  ÷ ÷ ÷  T(z + 1)   T(z + 1)   T(z + 1)  Nhân tử mẫu với T3(z+1)3 Ta được: K K T z + 1) W(z) = 31 2 ( Az + B z + C z + D Với: A = 8T1T2T3+4(T1T2+T2T3+T1T3)T+2(T1+T2+T3)T2+(1+K1K2K3)T3 B = -24T1T2T3+4(-T1T2-T2T3-T1T3)T+2(T1+T2+T3)T2+(3+3K1K2K3)T3 C = 24T1T2T3+4(-T1T2-T2T3-T1T3)T+2(-T1-T2-T3)T2+(3+3K1K2K3)T3 D = -8T1T2T3+4(T1T2+T2T3+T1T3)T+2(-T1-T2-T3)T2+(1+K1K2K3)T3 Ta có hàm sai phân sau: K1K 2T ( z + 1) Y(z) = U(u) Az + B z + C z + D Y(z)(Az3 + Bz2 + Cz+D) = K1K2T3(z+1)3U(z) Az3 Y(z) + Bz2 Y(z) + CzY(z)+DY(z) = K1K2T3(z3U(z) + 3z2U(z) + 3zU(z) + U(z)) AY(k+3) + BY(k+2) + CY(k+1) + DY(k) = K 1K2T3[U(k+3) + 3u(k+2) + 3U(k+1) + U(k)] u(t) = 1(t) nên u(k+3) = u (k+2) = u(k+1) = u(k) = Vậy ta có: AY(k+3) + BY(k+2) + CY(k+1) + D(k) = 8K1K2T3 Ta có phương trình sai phân:  B C D 8K1K 2T  Y(k + 3) =  − Y(k + 2) − Y(k + 1) − Y(k) + ÷ A A A  A  Viết phương trình mơ Private Sub close_Click() MsgBox ("ket thuc chuong trinh") End End Sub Private Sub ve_Click() Dim k1 As Double Dim k2 As Double Dim k3 As Double Dim t As Double Dim Y(1000) As Double Dim u(1000) As Double k1 = txtK1.Text k2 = txtK2.Text k3 = txtK3.Text t1 = txtT1.Text t2 = txtT3.Text t3 = txtT3.Text t = txtT.Text A = * t1 * t2 * t3 + * t1 * t2 * t + * t ^ * t1 + * t ^ * t2 + * t ^ * t3 + * t ^ * t1 * t3 + * t ^ * t2 * t3 + t ^ + k1 * k2 * k3 * t3 B = -24 * t1 * t2 * t3 - * t1 * t2 * t + * t ^ * t1 + * t ^ * t2 + * t ^ * t3 + * t ^ * t1 * t3 + * t ^ * t2 * t3 + * t ^ + * k1 * k2 * k3 * t3 C = 24 * t1 * t2 * t3 - * t1 * t2 * t - * t ^ * t1 - * t ^ * t2 - * t ^ * t3 - * t ^ * t1 * t3 - * t ^ * t2 * t3 + * t ^ + * k1 * k2 * k3 * t3 D = -8 * t1 * t2 * t3 - * t1 * t2 * t - * t ^ * t1 - * t ^ * t2 - * t ^ * t3 - * t ^ * t1 * t3 - * t ^ * t2 * t3 + t ^ + k1 * k2 * k3 * t3 E = * k1 * k2 * t ^ For k = To 997 Y(k + 2) = (-B / A) * Y(k + 2) + (-C / A) * Y(k + 1) + (-D / A) * Y(k) + E / A Next k For i = To 100 If i Mod 10 = Then txty.Text = txty + " " + "y[" + CStr(i) + "] = " + Format(CStr(Y(i)), "#0.0000000") + vbNewLine End If Next i ' cac chi tieu chat luong 'dau on dinh yod yod = Y(1000) ykod.Text = yod 'ymax ymax = Y(1) For k = To 1000 If Y(k) > ymax Then ymax = Y(k) End If Next k ykmax.Text = ymax 'do qua dieu chinh omax omax = 100 * (ymax - yod) / yod qdc.Text = omax ' Thoi gian qua Tqd = 1000 Do While (Abs(Y(Tqd) - yod) < 0.05) Tqd = Tqd - Loop Tqd = Tqd * t txtTod.Text = Tqd ' Tmax tmax = For k = To 1000 If Y(k) = ymax Then tmax = k * t End If Next k tmax.Text = tmax ' ve thi dothi1_Click dothi1.ForeColor = vbBlue k=0 w=0 dothi1.CurrentX = X dothi1.CurrentY = w For k = To 1000 X = k * 0.01 w = Y(k) dothi1.Line -(X, w) Next k End Sub Private Sub dothi1_Click() Dim i As Single dothi1.Scale (-0.1, 9)-(4, -0.5) ' tao khung toa dothi1.Line (-0.1, 0)-(4, 0), vbBlack ' Draw X axis For i = To Step 0.5 dothi1.Line (i, -0.1)-(i, 0), vbBlack ' Ðánh dâ'u If i * 10 Mod = Then dothi1.Print i Next i dothi1.Line (0, -0.1)-(0, 9), vbBlack ' Draw Y axis For i = To Step 0.5 dothi1.Line (-0.1, i)-(0, i), vbBlack ' ve duong thang If i * 10 Mod = Then dothi1.Print i ' danh so Next i End Sub Chọn số bước tính Chọn k = 1000 để hệ ổn định In kết In 100 giá trị y[k] – cách 10 giá trị in số Kết sau: y[0] = 0,00000 y[10] = 3,68448 y[20] = 32,03244 y[30] = -4461,06121 y[40] = -284526,13747 y[50] = -6793502,71019 y[60] = 149107193,53138 y[70] = 19529955477,02810 y[80] = 741082475430,86900 y[90] = 3860719448918,80000 y[100] = -1098015348216610,00000 y[110] = -63078218070728100,00000 y[120] = -1315507434488950000,00000 y[130] = 42832916022369100000,00000 y[140] = 4497917775195740000000,00000 y[150] = 155825809834508000000000,00000 y[160] = 127280430092793000000000,00000 y[170] = -265991219719729000000000000,00000 y[180] = -13867487266353400000000000000,00000 y[190] = -246191638351017000000000000000,00000 y[200] = 11623352234149500000000000000000,00000 y[210] = 1026817236717670000000000000000000,00000 y[220] = 32346110477408500000000000000000000,00000 y[230] = -136479612541711000000000000000000000,00000 y[240] = -63570601045505100000000000000000000000,00000 y[250] = -3022375478643470000000000000000000000000,00000 y[260] = -43865342165956500000000000000000000000000,00000 y[270] = 3035552854226180000000000000000000000000000,00000 y[280] = 232433782971078000000000000000000000000000000,00000 y[290] = 6614636150777710000000000000000000000000000000,00000 y[300] = -67179174483246700000000000000000000000000000000,00000 y[310] = -15014373592445400000000000000000000000000000000000,00000 y[320] = -652723131454455000000000000000000000000000000000000,00000 y[330] = -7225244336363390000000000000000000000000000000000000,00000 y[340] = 770891244214737000000000000000000000000000000000000000,00000 y[350] = 52181464618815200000000000000000000000000000000000000000,00000 y[360] = 1328623316235240000000000000000000000000000000000000000000,00000 y[370] = -23102990778674900000000000000000000000000000000000000000000,00000 y[380] = -3508744848681370000000000000000000000000000000000000000000000,0000 y[390] = -139588202562866000000000000000000000000000000000000000000000000,00 000 y[400] = -1024263033259720000000000000000000000000000000000000000000000000,0 0000 y[410] = 191579954404095000000000000000000000000000000000000000000000000000, 00000 y[420] = 1161903217850810000000000000000000000000000000000000000000000000000 0,00000 y[430] = 2609860827561510000000000000000000000000000000000000000000000000000 00,00000 y[440] = -693214697099962000000000000000000000000000000000000000000000000000 0000,00000 y[450] = -812040219266446000000000000000000000000000000000000000000000000000 000000,00000 y[460] = -295327715563375000000000000000000000000000000000000000000000000000 00000000,00000 y[470] = -948199842531811000000000000000000000000000000000000000000000000000 00000000,00000 y[480] = 4678514272059970000000000000000000000000000000000000000000000000000 0000000000,00000 y[490] = 2565825524804100000000000000000000000000000000000000000000000000000 000000000000,00000 y[500] = 4980085519063640000000000000000000000000000000000000000000000000000 0000000000000,00000 y[510] = -193262068538534000000000000000000000000000000000000000000000000000 0000000000000000,00000 y[520] = -186235765702272000000000000000000000000000000000000000000000000000 000000000000000000,00000 y[530] = -617365272559801000000000000000000000000000000000000000000000000000 0000000000000000000,00000 y[540] = 9006118994411800000000000000000000000000000000000000000000000000000 000000000000000000,00000 y[550] = 1125896988254690000000000000000000000000000000000000000000000000000 0000000000000000000000,00000 y[560] = 5618150557039900000000000000000000000000000000000000000000000000000 00000000000000000000000,00000 y[570] = 9129564054332940000000000000000000000000000000000000000000000000000 000000000000000000000000,00000 10 y[580] = -514215714633533000000000000000000000000000000000000000000000000000 000000000000000000000000000,00000 y[590] = -423448789554699000000000000000000000000000000000000000000000000000 00000000000000000000000000000,00000 y[600] = -127291793866740000000000000000000000000000000000000000000000000000 0000000000000000000000000000000,00000 y[610] = 8757582197602720000000000000000000000000000000000000000000000000000 000000000000000000000000000000,00000 y[620] = 2675428252744640000000000000000000000000000000000000000000000000000 000000000000000000000000000000000,00000 y[630] = 1219288872126040000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000,00000 y[640] = 1575740610102050000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000,00000 y[650] = -132406103547415000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000,00000 y[660] = -954798359685769000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000,00000 11 y[670] = -258232779810392000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000,00000 y[680] = 3451373287451230000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000,00000 y[690] = 6286673601144060000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000,00000 y[700] = 2621351323910450000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000,00000 y[710] = 2452447189041830000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000,00000 y[720] = -332636330415992000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000,00000 y[730] = -213527013976251000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000,00000 y[740] = -513615933802611000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000,00000 y[750] = 1099568249072160000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000,00000 12 y[760] = 1462311597952060000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000,00000 y[770] = 5578393588733760000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000,00000 y[780] = 3042133247478480000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000,00000 y[790] = -819537366269937000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000,00000 y[800] = -473611713724388000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000,000 00 y[810] = -996274434641621000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000,00 000 y[820] = 3172180357548620000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000,0 0000 y[830] = 3369637050195580000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 ,00000 y[840] = 1173789526168500000000000000000000000000000000000000000000000000000 13 0000000000000000000000000000000000000000000000000000000000000000000 00,00000 y[850] = 1282496572691670000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00,00000 y[860] = -198702599083455000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 000000,00000 y[870] = -104173925514398000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000,00000 y[880] = -186887406018150000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 000000000,00000 y[890] = 8631758536470610000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 000000000,00000 y[900] = 7696377316951530000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000000,00000 y[910] = 2438518236965230000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 0000000000000,00000 14 y[920] = -950856118203114000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000000000,00000 y[930] = -475244605153684000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000000000000,00000 y[940] = -227162934032092000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000,00000 y[950] = -334161227584675000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000,00000 y[960] = 2258615573520880000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000,00000 y[970] = 1743038248668050000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000,00000 y[980] = 4991439200562990000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000,00000 y[990] = -487841059811595000000000000000000000000000000000000000000000000000 15 0000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000,00000 Vẽ đường cong trình độ Kết chạy chương trình sau: Hình Kết mơ VisualBas Các tiêu chất lượng hệ Ymax = 53.5823 Yod = 53.5823 Sigma = 1025.23(%) Tmax = 1(s) Tod = 36.8 Dùng Matlab – Simulink mô lại hệ thống K1 = 50; K2 = 5; K3 = 0.5; T1 = 0.1; T2 = 0.01; T3 = 0.2; num=[K1*K2]; den=[T1*T2*T3 T1*T2+T1*T3+T2*T3 T1+T2+T3 1+K1*K2*K3]; step(num,den); 10 title('Dac tinh qua cua he DKTD'); 11 ylabel('y(t)'); 12 xlabel('t,sec'); 13 grid on 16 Hình 1.2 Kết mơ Matlab Kết luận: Hệ cho ổn định theo thời gian Kết khảo sát Matlab hoàn toàn trùng với kết thu VB, chứng tỏ tính đắn phương pháp dùng để mơ hình hóa hệ thống 8.Nhận dạng hệ thống Từ đường cong độ thu nhờ phần mềm mô ta nhận thấy: - Đường cong xuất phát từ gốc tọa độ cho thấy hàm truyền kín hệ bậc tử số nhỏ bậc mẫu số - Giá trị xác lập ổn định số khác nên dễ thấy hàm độ đơn điệu tăng độ điều chỉnh nên kết luận số thời gian tử nhỏ số thời gian mẫu So với hàm truyền cho: W(s)= 100 0,02s3 + 0,3s + s + 100,1 +Bậc tử (=0 ) < bậc mẫu (=3) nên hàm số xác lập giá trị khác 17 +Hằng số thời gian tử (=0) < số số thời gian mẫu nên hàm độ đơn điệu tăng Kết hoàn toàn phù hợp Thiết kế giao diện 10.Tài liệu tham khảo [1] Matlab Simulink dành cho kỹ sư điều khiển tự động NGUYỄN PHÙNG QUANG-NXB KHOA HỌC VÀ KỸ THUẬT-2006 [2] Bài giảng Visual Basic Khoa Công Nghệ Thông Tin-Trường Đại Học Hàng Hải-2008 18 19 ... 8T1T2T3+4(T1T2+T2T3+T1T3)T+2(T1+T2+T3)T2+(1+K1K2K3)T3 B = -2 4T1T2T3+4(-T1T2-T2T3-T1T3)T+2(T1+T2+T3)T2+(3+3K1K2K3)T3 C = 24T1T2T3+4(-T1T2-T2T3-T1T3)T+2(-T1-T2-T3)T2+(3+3K1K2K3)T3 D = -8 T1T2T3+4(T1T2+T2T3+T1T3)T+2(-T1-T2-T3)T2+(1+K1K2K3)T3... t1 - * t ^ * t2 - * t ^ * t3 - * t ^ * t1 * t3 - * t ^ * t2 * t3 + * t ^ + * k1 * k2 * k3 * t3 D = -8 * t1 * t2 * t3 - * t1 * t2 * t - * t ^ * t1 - * t ^ * t2 - * t ^ * t3 - * t ^ * t1 * t3 -. .. * 0.01 w = Y(k) dothi1.Line -( X, w) Next k End Sub Private Sub dothi1_Click() Dim i As Single dothi1.Scale (-0 .1, 9 )-( 4, -0 .5) ' tao khung toa dothi1.Line (-0 .1, 0 )-( 4, 0), vbBlack ' Draw X axis

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