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Over the weekend I came across a pi approximation using durations of years and weeks (image below, Wolfram, eq. 89), accurate to 6 digits using the average Gregorian year (365.2425 days).
Here it is in MATLAB. I divided by 1 week at the end rather than multiplying by its reciprocal because you can’t divide a numeric by a duration in MATLAB (1/week).
weeks = @(n)n*days(7);
piApprox = ((years(13)-weeks(6))/years(13) + weeks(3)) / weeks(1)
% piApprox = 3.141593493469302
Here’s a breakdown
- The first argument becomes 12.885 yrs / 13 yrs or 0.99115
- Add three weeks: 0.99115 + 3 weeks = 21.991 days
- The reduced fraction becomes 21.991 days / 7 days
Now it looks a lot closer to the more familiar approximation for pi 22/7 but with greater precision!
Need help about FPGA Based VSC HVDC Real Time Simulation Model.
This person used computer version to build a keyboard input, and used standard flag semaphore for the positions.
Flag semaphore is used mostly by sailors to be able to communicate optically over a distance; it does not need anything more than make-shift flags (but binoculars or telescopes can help.) Trained users can go faster than you might guess.
Chen, Rena, and I are at a community management event. It's great to be with others talking about relationships, trust, and co-creation.
I'm in a community conference in Boston today and see what snacks we get! The organizer said it's a coincidence, but it's definitly a good idea to have them in our MathWorks community meetings.
A research team found a way to trick a number of AI systems by injecting carefully placed nonsense -- for example being able able to beat DeepMind's Go game.
This video discusses the "Cody" bridge, which is a pedestrian bridge over a canal that has been designed to move up and out of the way when ships need to travel through. The mathematics of the bridge movement are discussed and diagrammed. It is unique and educational.
Recently developed: a "microscope" based on touch and stereo vision.
Using touch removes the possibility of optical confusion -- for example, black on touch is only due to shape, not due to the possibility that the object has a black patch.
Sorry, you might need a Facebook account to watch the video.
(Sorry - it should be 2023b by now.)
spy
I'm curious how the community uses the hold command when creating charts and graphics in MATLAB. In short, hold on sets up the axes to add new objects to the axes while hold off sets up the axes to reset when new objects are added.
When you use hold on do you always follow up with hold off? What's your reasoning on this decision?
Can't wait to discuss this here! I'd love to hear from newbies and experts alike!
Calling all students! New to MATLAB or need helpful resources? Check out our MATLAB GitHub for Students repository! Find MATLAB examples, videos, cheat sheets, and more!
Visit the repository here: MATLAB GitHub for Students
Imagine x is a large vector and you want the smallest 10 elements. How might you do it?
The way we've solved ODEs in MATLAB has been relatively unchanged at the user-level for decades. Indeed, I consider ode45 to be as iconic as backslash! There have been a few new solvers in recent years -- ode78 and ode89 for example -- and various things have gotten much faster but if you learned how to solve ODEs in MATLAB in 1997 then your knowledge is still applicable today.
In R2023b, there's a completely new framework for solving ODEs and I love it! You might argue that I'm contractually obliged to love it since I'm a MathWorker but I can assure you this is the real thing!
I wrote it up in a tutorial style on The MATLAB Blog https://blogs.mathworks.com/matlab/2023/10/03/the-new-solution-framework-for-ordinary-differential-equations-odes-in-matlab-r2023b/
The new interface makes a lot of things a much easier to do. Its also setting us up for a future where we'll be able to do some very cool algorithmic stuff behind the scenes.
Let me know what you think of the new functionality and what you think MathWorks should be doing next in the area of ODEs.
To solve the puzzle, first unscramble each of the words on the left. Then rearrange the letters in the yellow shaded boxes to complete the sentence on the right.
If you enjoyed this puzzle let me know with a like or in the comments below and I'll post more of them. Please don't post your answer, or any hints, and spoil it for those who come across this puzzle after you!! If you want to check your answer, you can messge me your guess through the link on my profile card (click on my name, Rena Berman, above and then on the envelope icon in the top right corner of the profile card that appears).
I am trying to make a simulink model to use a MPC to reduce power consumption of HVAC system in an electric vehicle during cool down from ambient temperature to a set point temperature. Any help regarding this would be appreciated
Hi Everyone,
I have a high-order filter and I need to determinate its R,L,C values. In the simulation, the filter is unstable for some values. But I want to find values for stability of the filter. For second order system has some formulation between Q and phase margin so that it can be obtain the stability ciriteria to choose components.
My transfer function denominator is 5th order polynomial, and it is not possible to find roots in terms of R,L,C, Lm. I need to factor it like one 1th and two second order as shown in below.
D(s) = (1+s/wp) (1+s/wo1Q1 + (s/wo1)²) (1+s/wo2Q2 + (s/wo2)²)
Is there any methods to reduce order of transfer function?(Maybe, it can be possible for only in certain frequency range, can be renounced high-frequency )
Kindly help me correct this code to function properly. I am just learning MATLAB. i cannot get the output in abc frame. This is the code:
%----------- Define input and state parameters-----------------------------
clc
v_dc = 350; % DC input voltage in V
m = 0.841; % modulation index
C = 4000e-6; % DC buss capacitance in uf
L_1 = 2.5e-3; % Inverter side inductance in mH
L_2 = 2.5e-3; % Load side inductance in mH
L = 0; % load inductance
C_f = 10e-6; % filter capacitance in uf
R_f = 0.7; % damping resistance in ohms
R_L = 20; % load resistance in ohms
f_s = 10e3; % switching frequency
f = 60; % System frequency
R_s = 0.01; % Capacitance of the DC circuit
I_d = 8.594; % steady state current
w = 2*pi*f; % System angular Frequency
% Define initial steady state values
v_c = 349.4; i_d = 8.594; i_q = -0.213; v_df = 285; v_qf = -120; i_Ld = 8.594; i_Lq = 0.85;
%------------------S V P W M Generator-------------------------------------
% Define reference vector Uref
U_mag = m*v_dc/2; % Magnitude of Uref
% Define switching vectors
U1 = [v_dc/2;0]; % Vector Q1
U2 = [v_dc/4;sqrt(3)*v_dc/4]; % Vector Q2
U3 = [-v_dc/4;sqrt(3)*v_dc/4]; % Vector Q3
U4 = [-v_dc/2;0]; % Vector Q4
U5 = [-v_dc/4;-sqrt(3)*v_dc/4]; % Vector Q5
U6 = [v_dc/4;-sqrt(3)*v_dc/4]; % Vector Q6
% Define sector angles
theta1 = pi/6;
theta2 = pi/2;
theta3 = 5*pi/6;
theta4 = 7*pi/6;
theta5 = 3*pi/2;
theta6 = 11*pi/6;
% Define duty cycles for each switch using a for loop
for t=0:1/f_s:1/f % Time variable from 0 to one cycle of system frequency with steps of switching frequency
U_phase = w*t; % Phase of Uref (t is time variable)
U_alpha = U_mag*cos(U_phase); % Alpha component of Uref
U_beta = U_mag*sin(U_phase); % Beta component of Uref
if (0 <= U_phase) && (U_phase < theta1) % Sector 1
T1 = (sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T2 = (-sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T0 = 1 - T1 - T2;
d_a(round(t)+1) = T1 + T0/2;
d_b(round(t)+1) = T2 + T0/2;
d_c(round(t)+1) = T0/2;
elseif (theta1 <= U_phase) && (U_phase < theta2) % Sector 2
T3 = (sqrt(3)*U_beta - U_alpha)/(2*v_dc);
T2 = (sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T0 = 1 - T3 - T2;
d_a(round(t)+1) = T0/2;
d_b(round(t)+1) = T2 + T0/2;
d_c(round(t)+1) = T3 + T0/2;
elseif (theta2 <= U_phase) && (U_phase < theta3) % Sector 3
T3 = (sqrt(3)*U_beta - U_alpha)/(2*v_dc);
T4 = (-sqrt(3)*U_beta - U_alpha)/(2*v_dc);
T0 = 1 - T3 - T4;
d_a(round(t)+1) = T0/2;
d_b(round(t)+1) = T0/2;
d_c(round(t)+1) = T3 + T0/2;
elseif (theta3 <= U_phase) && (U_phase < theta4) % Sector 4
T5 = (-sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T4 = (-sqrt(3)*U_beta - U_alpha)/(2*v_dc);
T0 = 1 - T5 - T4;
d_a(round(t)+1) = T5 + T0/2;
d_b(round(t)+1) = T0/2;
d_c(round(t)+1) = T4 + T0/2;
elseif (theta4 <= U_phase) && (U_phase < theta5) % Sector 5
T5 = (-sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T6 = (sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T0 = 1 - T5 - T6;
d_a(round(t)+1) = T5 + T0/2;
d_b(round(t)+1) = T6 + T0/2;
d_c(round(t)+1) = T0/2;
elseif (theta5 <= U_phase) && (U_phase < theta6) % Sector 6
T1 = (sqrt(3)*U_beta + U_alpha)/(2*v_dc);
T6 = (sqrt(3)*U_beta - U_alpha)/(2*v_dc);
T0 = 1 - T1 - T6;
d_a(round(t)+1) = T1 + T0/2;
d_b(round(t)+1) = T0/2;
d_c(round(t)+1) = T6 + T0/2;
end
end
%-------------------------Define system matrices---------------------------
% Create Three-phase SVPWM VSI Inverter
% System matrix Nx-by-Nx matrix
A = [-1/(C*R_s),-sqrt(3)*m/(2*C),0,0,0,0,0;
sqrt(3)*m/(3*L_1),-R_f/(3*L_1),w,-1/(2*L_1),-sqrt(3)/(6*L_1),-R_f/(3*L_1),0;
0,-w,-R_f/(3*L_1),-sqrt(3)/(6*L_1),-1/(2*L_1),0,R_f/(3*L_1);
0,1/(2*C_f),-sqrt(3)/(6*C_f),0,w,-1/(2*C_f),sqrt(3)/(6*C_f);
0,sqrt(3)/(6*C_f),1/(2*C_f),-w,0,-sqrt(3)/(6*C_f),-1/(2*C_f);
0,R_f/(3*(L_2+L)),0,1/(2*(L_2+L)),sqrt(3)/(6*(L_2+L)),((-3*R_L-R_f)/(3*(L_2+L))),w;
0, 0, R_f/(3*(L_2+L)), -sqrt(3)/(6*(L_2+L)), 1/(2*(L_2+L)), -w, ((-3*R_L-R_f)/(3*(L_2+L)))];
% Define input matrix
B = [1/(C*R_s),-sqrt(3)*i_d/(2*C);d_a*v_dc,(sqrt(3)*v_c)/L_1;d_b*v_dc,0;d_c*v_dc,0;0,0;0,0;0,0]; % Nx-by-Nu input matrix
% Define output matrix
C = [0 1 0 0 0 0 0; % Ny-by-Nx matrix
0 0 1 0 0 0 0;
0 0 0 1 0 0 0;
0 0 0 0 1 0 0;
0 0 0 0 0 1 0;
0 0 0 0 0 0 1];
% Feedthrough matrix
D = zeros(6, 2); % Ny-by-Nu matrix
% create state-space model object
sys = ss(A,B,C,D);
% Define initial conditions and input
x0 = [v_c; i_d; i_q; v_df; v_qf; i_Ld; i_Lq]; % Initial state vector
t = 0:1e-6:0.5; % Time vector for simulation
u = repmat([v_dc;m],1,length(t)); % repeat u for each time step
% Simulate the system
[y, ~, x] = lsim(sys, u, t, x0);
% Extract the states
v_c_sim = x(:, 1);
i_d_sim = x(:, 2);
i_q_sim = x(:, 3);
v_df_sim = x(:, 4);
v_qf_sim = x(:, 5);
i_Ld_sim = x(:, 6);
i_Lq_sim = x(:, 7);
% Extract the outputs
v_abc_sim = y(:, 1:3);
i_abc_sim = y(:, 4:6);
v_dq_sim = y(:, 4:5);
i_dq_sim = y(:, 2:3);
% Plot the variables
figure;
subplot(4, 2, 1);
plot(t, v_c_sim);
xlabel('Time');
ylabel('v_c');
title('Capacitor Voltage');
subplot(4, 2, 2);
plot(t, i_d_sim);
xlabel('Time');
ylabel('i_d');
title('d-Axis Current');
subplot(4, 2, 3);
plot(t, i_q_sim);
xlabel('Time');
ylabel('i_q');
title('q-Axis Current');
subplot(4, 2, 4);
plot(t, v_df_sim);
xlabel('Time');
ylabel('v_df');
title('d-Component Filter Voltage');
subplot(4, 2, 5);
plot(t, v_qf_sim);
xlabel('Time');
ylabel('v_qf');
title('q-Component Filter Voltage');
subplot(4, 2, 6);
plot(t, i_Ld_sim);
xlabel('Time');
ylabel('i_Ld');
title('d-Axis Load Current');
subplot(4, 2, 7);
plot(t, i_Lq_sim);
xlabel('Time');
ylabel('i_Lq');
title('q-Axis Load Current');
% Perform coordinate transformation from dq frame to abc frame for currents
i_a_sim = cos(w*t)*i_d_sim - sin(w*t)*i_q_sim;
i_b_sim = cos(w*t - 2*pi/3)*i_d_sim - sin(w*t - 2*pi/3)*i_q_sim;
i_c_sim = cos(w*t + 2*pi/3)*i_d_sim - sin(w*t + 2*pi/3)*i_q_sim;
% Perform coordinate transformation from dq frame to abc frame for voltages
v_a_sim = cos(w*t)*v_df_sim - sin(w*t)*v_qf_sim;
v_b_sim = cos(w*t - 2*pi/3)*v_df_sim - sin(w*t - 2*pi/3)*v_qf_sim;
v_c_sim = cos(w*t + 2*pi/3)*v_df_sim - sin(w*t + 2*pi/3)*v_qf_sim;
Many thanks
The MATLAB Answers community is an invaluable resource for all MATLAB users, providing selfless assistance and support. However, with the emergence of AI-based chatbots, like chatGPT, there may be concerns about the future relevance and utility of the MATLAB Answer community. What are your thoughts?
Hello, Recently I have started working on a thermal management project at my Institute. I am taking some hints from en example of EV thermal management provided by MATLAB. I am having an issue in deciding how the battery current and powertrain heat load is calculated as a function of vehicle speed. It would be helpful if someone could provide some links or references regarding this. (Also, attaching a screenshot of the example for better understanding)
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