Masinamehaanika Kodutöö nr. 2 Üliõpilane: Matriklinumber: Rühm: MAHB41 Kuupäev: 08.05.2012 Õppejõud: Merle Randrüüt Ülesanne 1 r = OA = 250mm = AC = 900mm ja a) Punkti A koordinaadid , sõltuvus funktsiooni pöördenurgast b) Määrata punkti C koordinaadid xC , yC funktsioonina pöördenurgast c) Matlab-i kood r = 0.25; l = 0.9; xB = 0.4 yB = 0.3; phi = linspace (0, 2*pi, 361); xC = zeros(1, 361); yC = zeros(1, 361); %Tsükkel for k=1:361 gamma = atan((xBr*cos(phi(k)))/(yBr*sin(phi(k)))); xA = r*cos(phi(k)); yA = r*sin(phi(k)); xC(k) = xA+l*sin(gamma); yC(k) = yA+l*cos(gamma); end figure(1) hold off plot(xC, yC, 'linewidth', 2) title('Punkti C trajektoor') xlabel('x [m]') ylabel('y [m]') Ülesanne 2 a) Vedru-massi süsteemi omavõnkesagedus Liikumisvõrrand: Omavõnkesagedus: b) Vabavõnkumiste periood c) Ülekandefunktsioon d) Matlab-i programm
tektseditor mate LaTeX r = 0.1; l = 0.2; rpm = 2000; omega = rpm*pi/30; phi = linspace(0, 2*pi, 361); v_B = zeros(1, 361); om_AB = zeros (1, 361); for ind = 1:361, r_sin_phi = r*sin(phi(ind)); r_cos_phi = r*cos(phi(ind)); sq = sqrt(l^2-r_sin_phi^2); v_B = -r_sin_phi*(1+r_cos_phi/sq)*omega; om_AB = -r_cos_phi/sq*omega; end figure (1) hold off plot(phi, v_B, 'linewidth', 2) hold on plot([0 2*pi], [0 0], 'color', 'black') title ('liuguri kiirus') xlabel('varphi [rad]') ylabel('v_B [m/s]')
PI_yreg %käsufail pidevaja regulaatori sünteesiks PI_yreg.m käsufaili sisu: % PI järgivsüsteemi süntees % Integraalne TS väljundi järgi (järgimiseks) + TS oleku järgi % Laiendatud olekuvektoriga süsteem % ~ % X = [ X ; Z ] % . % Z = R - Y = R - CX % % ! ~ ~ ~ % U = +K*X +Ki*Z = -K*X , K = [-K -Ki] % % P - soovutud suletud süsteemi pooluste paigutus n + dim(Y) tükki. nnn=size(A,1); rrr=size(B,2); % olekumuutujate arv ja sisendite arv if exist('C'), y_r=size(C,1); A2=[A zeros(nnn,y_r); -C zeros(y_r,y_r)]; B2=[B;zeros(y_r,rrr)]; r2=rank(ctrb(A2,B2)) %juhitavuse maatriksi astak else disp('C maatriks puudub!') end 5. Regulaatorite süntees diskreetajas. Q2=diag([1/(0.2*0.2) 0 1/(0.7*0.7) 0 10]) R2=5/(100*M*M) K2=lqr(A2, B2, Q2, R2) % Q2 ja K2 on laiendatud olekuvektoriga mudeli kaalumaatriksid, R2 on laiendatud olekuvektoriga mudeli juhitavuse maatriksi astak. pi_yregd %käsufail diskreetaja regulaatori sünteesiks PI_yreg.m käsufaili sisu:
2 Pöördpendli modelleerimine ja juhtimine. Rain Jõearu 040737 IASB Tallinn 2008 1. Mudeli lähteandmed X0 = [-0.1; 0; 0; 0] - algolek Xs = [0; 0; 0,7; 0] seadesuurus X(t) - olek A = 0 1 0 0; 17.64 0 0 0; 0 0 0 1; -0.784 0 0 0 ] B = [0; -0.3333; 0; 0.2] C=eyes(4) D=zeros(4,2) G = [0; 0; 0; 0] - olekuhäiringu sisendmaatriks M= 5 - mass X1 pendli nurk rad X2 - pendli nurga muutumise kiirus X3 - pendli asend X4 - pendli asendi muutmise kiirus U(t) - Jõud N, 2. Vormistatud eksperimendi lühiselgitus Ülesandeks oli pendli hoidmine püsti asendis nii, et juhtimine toimuks võimalikult kiiresti ja parameetrid oleksid ettenähtud piiride. Samuti pidime kontrollima võimalikke suurimaid olekutaastaja vigu, mille puhul süsteem on veel antud piirides
m x¨ =ku-kx m x¨ +kx=ku 20 m=k m x¨ + 20 mx = 20 mu x¨ + 20 x= 02 u s 2 X ( s)+ 20 X ( s )= 20 U ( s ) X ( s) 20 T ( s)= = U ( s ) s 2+ 02 d) Sisend u(t) on ajavahemikus [0, 2t1] siinusfunkt- sioon amplituudiga u1 = 10mm, mille nurksagedus on sama, mis vabavõnkuva süsteemi omavõnkesagedus 0. Väljaspool seda vahemikku on sisendi väärtus null. syms nyy t1 k u1 k = 5; t1 = 1.9; m = 0.5 u = zeros (1, 401); u(1:201) = linspace (0, sin(a), 201); nyy = k/m TF = tf (nyy,[1 0 nyy]); figure (1) t = linspace (0, 4*t1, 401); lsim (TF, u, t) u1 = 10; title(`Võnkumise graafik')
- g 0 0 0 M M Lineaarne mudel on saadud lineariseerimisega: sin(0+) ja cos(0+) 1 kui || 0.2rad mudeli viga jääb < 1%. Mudeli kirjeldamine Matlabis: % parameetrite initsialiseerimine g = 9.8; M = 2; m = 0.2; l = 0.6; % olekumudeli maatriksite sisestamine A = [0 1 0 0; g*(M + m)/(M*l) 0 0 0; 0 0 0 1; -m/M*g 0 0 0] B = [0; -1/(M*l); 0; 1/M] C = eye(4) % ühikmaatriksi loomine et Y = X D = zeros(4,1) % nullise maatriksi loomine sys=ss(A,B,C,D) % olekumudeli maatriksite koondamine ühe nime alla 2. Määrake süsteemi omaväärtused (poolused) ja stabiilsus. Kontrollige, et süsteem oleks täielikult juhitav olekutagasiside kasutamiseks. ov=eig(sys.a) % omaväärtused ja stabiilsuse määramine rank(ctrb(sys)) % juhitavuse kontroll 3. Teisendage olekumudeli diskreetaega (sobiva diskreetimis-sammuga): td = 0.1 % diskreetimissamm (-takt)
In program they test cars and they make one quick lap with that car. Their racing driver is called The Stig. They make some challanges with old and cheap cars and they show which car is best in that category. They have guests in show. Their guest drive fast lap with Kia Cee´d which is normal price car and not very fast. Then they take the lap time and but it on the board. They use the wall, which called "Cool Wall". There is many cars which are cools, un cools, seriously un cools or sub zeros, like they think. Every episode they change car positions. The latest episode was very funny. UK Top Gear competed with Australian Top Gear team. UK won competition overall because they cheated with last challenge. They should use James May in WRC racing but they knew May isn´t fast driver and they used the Stig, like they used to beat Germans in racing. The presenters are journalists. They like to drive with a car and drink beer at the bar. All of them
· more than one ski / skis · more than one Barrymore / Barrymores Words that end in -ch, x, s or s-like sounds, however, will require an -es for the plural: · more than one witch / witches · more than one box / boxes · more than one gas / gases · more than one bus / buses · more than one kiss / kisses · more than one Jones / Joneses BUT! Photos, kilos, pianos, zeros, studios, radios. When the word ends in the letter -y: · country / countries · family / families · story / stories Nouns ending in -f: (should be learned by heart) · calf / calves · elf / elves · half / halves · knife / knives · life / lives · leaf / leaves · loaf / loaves · scarf / scarves (s) · self / selves · sheaf / sheaves
3A916 + 24D16 = 5F616 3A9 + 24D --------- 5F6 9 + D = 16 A+4+1=F 3+2=5 6. Variable X contains the number of bytes to be read from an external device. Using a binary shift, write one line of pseudocode to calculate the number of bits to be read and store the result in Y. 1 byte = 8 bits Shift left = X ∗ 2𝑛 , where n stands for how big the shift is. 23 = 8, so we have to shift left by 3. Y = X shl 3 // shifts left by 3 bits, filling the empty bits on the right with zeros 7. You have two Boolean variables A and B. Write down a logic expression using only the operators AND, OR and NOT that will evaluate to true if A and B are equal and to false otherwise. (A AND B) OR ((NOT A) AND (NOT B)) 8. You have two strings stored in variables S1 and S2. Both strings are stored in ASCII format and contain 8 bits per character and 20 characters per string. Write pseudocode that will
a) 121 – 185 b) -70 – 88 == 4. Convert back to verify answer == 5. Calculate the following without converting the number base. Show calculations. a) 3A916 + 24D16 == 5. 5F616 == 6. Variable X contains the number of bytes to be read from an external device. Using a binary shift, write one line of pseudocode to calculate the number of bits to be read and store the result in Y. == 6. Y = X shl 3 // shifts left by 3 bits, filling the emply bits on the right with zeros (3 bits) == 7. You have two Boolean variables A and B. Write down a logic expression using only the operators AND, OR and NOT that will evaluate to true if A and B are equal and to false otherwise. == 7. (A and B) or ((not A) and (not B)) == 8. You have two strings stored in variables S1 and S2. Both strings are stored in ASCII format and contain 8 bits per character and 20 characters per string. Write pseudocode that will
koodiga M = 4;%RS koodi pikkuse määramise koefitsient teooria N_rs = 2^M - 1; %kodeeritud sõna pikkus plokkides K_rs=N_rs-2*LL; %infosõna plokkide arv L=M*K_rs; %infobittide arv kodeeritud sõnas-üleminek plokkidest bittide peale %T = floor((N - K)/2);%Vea parandamise võimekus %x2 = x2(:); t2ide = mod(L - mod(length(x2(:)),L),L);%kahe järgmise reaga moodustatakse liiasus, et kokku sobitada koodide pikkused msg_t2ide = cat(2, x2(:)', zeros(1, t2ide)); rs_encode = rsenco(msg_t2ide,N_rs,K_rs);%RS kodeerimine x = rs_encode(:);%muutujale "x" määratakse muutuja "rs_encode" väärtused jaotutud üheks veeruks või reaks % Tsükli loomine for S_N = SNR_VEC % tsükli alustamine %____________________________________________________________ sim('skeem'); % üleminek Simulinki keskonda y = y(1:length(x)); % lõigatakse liigsed arvud maha %____________________________________________________________
There could potentially be thousands of 1's or 0's in sequence. If the encoded data contains long 'runs' of logic 1's or 0's, this does not result in any bit transitions. The lack of transitions prevents the receiver DPLL from reliably regenerating the clock making it impossible to detect the boundaries of the received bits at the receiver. This is the reason why Manchester coding is used in Ethernet LANs. A binary encoding and transmission scheme in which ones and zeros are represented by opposite and alternating high and low voltages, and in which there is no return to a reference (zero) voltage between encoded bits. I2C-kood Siin on välja eraldatud Manchesteri kood. Eesmärk on ka lahti saada taktsignaalist. Andmesiini testitakse taktsignaali kõrgetel nivoodel. Kui üks pool muudab taktsignaali, siis teine pool testib ja seab oma taktsignaali paika. Kui taktsignaal on madal, siis ülekannet ei toimu. Ülekandesignaalina võib kasutada ka voolu muutust
flamingoes a hero heroes m m ar H a Gr elp With some nouns that end in -o, you can add either -s or -es to form the plural. Singular Plural Plural a mango mangoes mangos a mosquito mosquitoes mosquitos a zero zeroes zeros a buffalo buffaloes buffalos 28 Some plural nouns don't follow the -s rule. They don't end in -s, -es, -ies or -ves. Instead, the word changes form. mouse mice goose geese Word File Singular Plural child children
The xComputer Applet The xComputer applet is divided into three sections. The right-hand third of the applet represents the main memory of the simulated computer. This section of the applet shows the 1024 locations in xComputer's memory. These locations are numbered from 0 to 1023. Each line in the memory shows a location number (in blue) and the value stored in that location. When the program first starts up, the memory contains only zeros. The scroll bar can be used to view any part of memory. The "Control" section of the applet is used to interact with and control the xComputer. For example, you can use text-input boxes and buttons in the Control section to enter programs and data into the xComputer's memory. Once a program has been loaded into memory, you can tell the xComputer to execute it. You'll learn about controlling the xComputer as you work through the lab.
. . . Jefferson went on to say that if the cylinder be six inches long ("which probably will be a convenient length, as it may be spanned between the middle finger & thumb of the left hand, while in use") the number of wheels would total 36, and the number of ways in which they can be strung on the spindle to form different ciphers for different correspondents would come to 36 factorial, or 1 X 2 X 3 X ... X 35 X 36, which Jefferson calculated almost exactly as "372 with 39 cyphers [zeros] added to it." In fact, 36 factorial is 371,993,326,789,901,217,467,999,448,150,835, 200,000,000. Had the President recommended his own system to Secretary of State James Madison, he would have endowed his country with a method of secret communication that would almost certainly have withstood any cryptanalytic attack of those days. Instead he appears to have filed and forgotten it. It was not rediscovered among his papers in the Library of Congress until 1922, coincidentally the year the U
The falling edge of -CS starts a conversion, which takes a maximum of 7.5 ms. When -CS goes low, the MAX1242 drives its data output pin low. After the conversion is complete, the MAX1242 drives the data output pin high. The processor can then read the data a bit at a time by toggling the clock line and monitoring the MAX1242 data output pin. After the 10 bits are read, the MAX1242 provides two sub-bits, S1 and S0. If further clock transitions occur after the 13 clocks, the MAX1242 outputs zeros. Figure 2.15 shows how a MAX1242 would be connected to a microcon- troller with an on-chip SPI/Microwire interface. The SCLK signal goes to the SPI SCLK signal on the microcontroller, and the MAX1242 DOUT signal con- nects to the SPI data input pin on the microcontroller. One of the micro- controller port bits generates the -CS signal to the MAX1242. Note that the -CS signal starts the conversion and must remain low until the conversion is complete
book blurbs. This is a dazzling and highly useful work." --A. J. Jacobs, editor-at-large of Esquire magazine and author of The Know-It-All "Tim is Indiana Jones for the digital age. I've already used his advice to go spear shing on remote islands and ski the best hidden slopes of Argentina. Simply put, do what he says and you can live like a millionaire." --Albert Pope, derivatives specialist at UBS World Headquarters "Reading this book is like putting a few zeros on your income. Tim brings lifestyle to a new level--listen to him!" --Michael D. Kerlin, McKinsey & Company consultant to Bush-Clinton Katrina Fund and a J. William Fulbright Scholar "Part scientist and part adventure hunter, Tim Ferriss has created a road map for an entirely new world. I devoured this book in one sitting --I have seen nothing like it." --Charles L. Brock, chairman and CEO of Brock Capital Group; former CFO, COO, and general counsel of Scholastic, Inc
annaabi.ee 
