Hauptinhalt
Ergebnisse für
Hi everyone,
I've recently joined a forest protection team in Greece, where we use drones for various tasks. This has sparked my interest in drone programming, and I'd like to learn more about it. Can anyone recommend any beginner-friendly courses or programs that teach drone programming?
I'm particularly interested in courses that focus on practical applications and might align with the work we do in forest protection. Any suggestions or guidance would be greatly appreciated!
Thank you!
I have picked the title but don't know which direction to take it. Looking for any and all inspiration. I took the project as it sounded interesting when reading into it, but I'm a satellite novice, and my degree is in electronics.
"What are your favorite features or functionalities in MATLAB, and how have they positively impacted your projects or research? Any tips or tricks to share?
function ans = your_fcn_name(n)
n;
j=sum(1:n);
a=zeros(1,j);
for i=1:n
a(1,((sum(1:(i-1))+1)):(sum(1:(i-1))+i))=i.*ones(1,i);
end
disp
Gabriel's horn is a shape with the paradoxical property that it has infinite surface area, but a finite volume.
Gabriel’s horn is formed by taking the graph of 
with the domain 
and rotating it in three dimensions about the 
axis.
with the domain and rotating it in three dimensions about the axis.There is a standard formula for calculating the volume of this shape, for a general function 
.Wwe will just state that the volume of the 
solid between a and b is:
.Wwe will just state that the volume of the solid between a and b is:
The surface area of the solid is given by:
One other thing we need to consider is that we are trying to find the value of these integrals between 1 and ∞. An integral with a limit of infinity is called an improper integral and we can't evaluate it simply by plugging the value infinity into the normal equation for a definite integral. Instead, we must first calculate the definite integral up to some finite limit b and then calculate the limit of the result as b tends to ∞:
Volume
We can calculate the horn's volume using the volume integral above, so
The total volume of this infinitely long trumpet isπ.
Surface Area
To determine the surface area, we first need the function’s derivative:
Now plug it into the surface area formula and we have:
This is an improper integral and it's hard to evaluate, but since in our interval 
So, we have :
Now,we evaluate this last integral
So the surface are is infinite.
% Define the function for Gabriel's Horn
gabriels_horn = @(x) 1 ./ x;
% Create a range of x values
x = linspace(1, 40, 4000); % Increase the number of points for better accuracy
y = gabriels_horn(x);
% Create the meshgrid
theta = linspace(0, 2 * pi, 6000); % Increase theta points for a smoother surface
[X, T] = meshgrid(x, theta);
Y = gabriels_horn(X) .* cos(T);
Z = gabriels_horn(X) .* sin(T);
% Plot the surface of Gabriel's Horn
figure('Position', [200, 100, 1200, 900]);
surf(X, Y, Z, 'EdgeColor', 'none', 'FaceAlpha', 0.9);
hold on;
% Plot the central axis
plot3(x, zeros(size(x)), zeros(size(x)), 'r', 'LineWidth', 2);
% Set labels
xlabel('x');
ylabel('y');
zlabel('z');
% Adjust colormap and axis properties
colormap('gray');
shading interp; % Smooth shading
% Adjust the view
view(3);
axis tight;
grid on;
% Add formulas as text annotations
dim1 = [0.4 0.7 0.3 0.2];
annotation('textbox',dim1,'String',{'$$V = \pi \int_{1}^{a} \left( \frac{1}{x} \right)^2 dx = \pi \left( 1 - \frac{1}{a} \right)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} V = \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
dim2 = [0.4 0.5 0.3 0.2];
annotation('textbox',dim2,'String',{'$$A = 2\pi \int_{1}^{a} \frac{1}{x} \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} dx > 2\pi \int_{1}^{a} \frac{dx}{x} = 2\pi \ln(a)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} A \geq \lim_{a \to \infty} 2\pi \ln(a) = \infty$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
% Add Gabriel's Horn label
dim3 = [0.3 0.9 0.3 0.1];
annotation('textbox',dim3,'String','Gabriel''s Horn', ...
'Interpreter','latex','FontSize',14, 'EdgeColor','none', 'HorizontalAlignment', 'center');
hold off
daspect([3.5 1 1]) % daspect([x y z])
view(-27, 15)
lightangle(-50,0)
lighting('gouraud')
The properties of this figure were first studied by Italian physicist and mathematician Evangelista Torricelli in the 17th century.
Acknowledgment
I would like to express my sincere gratitude to all those who have supported and inspired me throughout this project.
First and foremost, I would like to thank the mathematician and my esteemed colleague, Stavros Tsalapatis, for inspiring me with the fascinating subject of Gabriel's Horn.
I am also deeply thankful to Mr. @Star Strider for his invaluable assistance in completing the final code.
References:
When it comes to MOS tube burnout, it is usually because it is not working in the SOA workspace, and there is also a case where the MOS tube is overcurrent.
For example, the maximum allowable current of the PMOS transistor in this circuit is 50A, and the maximum current reaches 80+ at the moment when the MOS transistor is turned on, then the current is very large.
At this time, the PMOS is over-specified, and we can see on the SOA curve that it is not working in the SOA range, which will cause the PMOS to be damaged.
So what if you choose a higher current PMOS? Of course you can, but the cost will be higher.
We can choose to adjust the peripheral resistance or capacitor to make the PMOS turn on more slowly, so that the current can be lowered.
For example, when adjusting R1, R2, and the jumper capacitance between gs, when Cgs is adjusted to 1uF, The Ids are only 40A max, which is fine in terms of current, and meets the 80% derating.
(50 amps * 0.8 = 40 amps).
Next, let’s look at the power, from the SOA curve, the opening time of the MOS tube is about 1ms, and the maximum power at this time is 280W.
The normal thermal resistance of the chip is 50°C/W, and the maximum junction temperature can be 302°F.
Assuming the ambient temperature is 77°F, then the instantaneous power that 1ms can withstand is about 357W.
The actual power of PMOS here is 280W, which does not exceed the limit, which means that it works normally in the SOA area.
Therefore, when the current impact of the MOS transistor is large at the moment of turning, the Cgs capacitance can be adjusted appropriately to make the PMOS Working in the SOA area, you can avoid the problem of MOS corruption.
I am trying to earn my Intro to MATLAB badge in Cody, but I cannot click the Roll the Dice! problem. It simply is not letting me click it, therefore I cannot earn my badge. Does anyone know who I should contact or what to do?
Hi everyone,
I need deep orientation to make calculation of speed and Angle for the absolute encoder RM22SC with signal (data+, Data-, Clock +, Clock -) using Launchpad F28379D and Simulink.
I did interface the absolute encoder with IC DS26LS32CN and I did get signal Data and Clock. I did use the GPIO20 for Data and GPIO21 for Clock and connect both to the Matlab Function block to get as output the position. See the code on attached. The output of the Matlab function times 2*pi/8192 to get the angle. However, I don't get anything as value.
Matlab Fuction Block code
function position = decodeSSI(data, clock)
%#codegen
persistent bitCounter shiftRegister prevClock
if isempty(bitCounter)
bitCounter = uint32(0);
shiftRegister = uint32(0);
prevClock = uint32(0);
end
% Parameters
numBits = 13; % Number of bits in the SSI word
% Rising edge detection for clock
clock = uint32(clock); % Ensure clock is of type integer
clockRisingEdge = (clock == 1) && (prevClock == 0);
prevClock = clock;
if clockRisingEdge
bitCounter = bitCounter + 1;
% Shift in the data bit
shiftRegister = bitor(bitshift(shiftRegister, 1), uint32(data));
% Check if we have received the full word
if bitCounter == numBits
position = shiftRegister;
% Reset for the next word
bitCounter = uint32(0);
shiftRegister = uint32(0);
else
position = uint32(0); % or NaN to indicate incomplete data
end
else
position = uint32(0); % or retain the last valid position
end
end
Hello everyone, i hope you all are in good health. i need to ask you about the help about where i should start to get indulge in matlab. I am an electrical engineer but having experience of construction field. I am new here. Please do help me. I shall be waiting forward to hear from you. I shall be grateful to you. Need recommendations and suggestions from experienced members. Thank you.
I recently wrote up a document which addresses the solution of ordinary and partial differential equations in Matlab (with some Python examples thrown in for those who are interested). For ODEs, both initial and boundary value problems are addressed. For PDEs, it addresses parabolic and elliptic equations. The emphasis is on finite difference approaches and built-in functions are discussed when available. Theory is kept to a minimum. I also provide a discussion of strategies for checking the results, because I think many students are too quick to trust their solutions. For anyone interested, the document can be found at https://blanchard.neep.wisc.edu/SolvingDifferentialEquationsWithMatlab.pdf
hello i'm working on simulation using simulink which is my title is ENHANCING BATTERYENERGY STORAGE SYSTEMSTHROUGH MODULAR MULTILEVEL CONVERTER WITH STATE-OF-CHARGE BALANCING CONTROL. i already build 9 level mmc. but i dont have any idea for state of charge balancing control.please any suggestion and explain.
Kindly link me to the Channel Modeling Group.
I read and compreheneded a paper on channel modeling "An Adaptive Geometry-Based Stochastic Model for Non-Isotropic MIMO Mobile-to-Mobile Channels" except the graphical results obtained from the MATLAB codes. I have tried to replicate the same graphs but to no avail from my codes. And I am really interested in the topic, i have even written to the authors of the paper but as usual, there is no reply from them. Kindly assist if possible.
Hi, I'm looking for sites where I can find coding & algorithms problems and their solutions. I'm doing this workshop in college and I'll need some problems to go over with the students and explain how Matlab works by solving the problems with them and then reviewing and going over different solution options. Does anyone know a website like that? I've tried looking in the Matlab Cody By Mathworks, but didn't exactly find what I'm looking for. Thanks in advance.
Any one have deep learning reinforcement based speed control of induction motor?
错误使用 ipqpdense
The interior convex algorithm requires all objective and constraint values to be finite.
出错 quadprog
ipqpdense(full(H), f, A, B, Aeq, Beq, lb, ub, X0, flags, ...
出错 MPC_maikenamulun
[X, fval,exitflag]=quadprog(H,f,A_cons,B_cons,[],[],lb,ub,[],options);
Twitch built an entire business around letting you watch over someone's shoulder while they play video games. I feel like we should be able to make at least a few videos where we get to watch over someone's shoulder while they solve Cody problems. I would pay good money for a front-row seat to watch some of my favorite solvers at work. Like, I want to know, did Alfonso Nieto-Castonon just sit down and bang out some of those answers, or did he have to think about it for a while? What was he thinking about while he solved it? What resources was he drawing on? There's nothing like watching a master craftsman at work.
I can imagine a whole category of Cody videos called "How I Solved It". I tried making one of these myself a while back, but as far as I could tell, nobody else made one.
Here's the direct link to the video: https://www.youtube.com/watch?v=hoSmO1XklAQ
I hereby challenge you to make a "How I Solved It" video and post it here. If you make one, I'll make another one.
Let’s Code! Make a Cody Video
Ever seen a
The Ans Hack is a dubious way to shave a few points off your solution score. Instead of a standard answer like this
function y = times_two(x)
y = 2*x;
end
you would do this
function ans = times_two(x)
2*x;
end
The ans variable is automatically created when there is no left-hand side to an evaluated expression. But it makes for an ugly function. I don't think anyone actually defends it as a good practice. The question I would ask is: is it so offensive that it should be specifically disallowed by the rules? Or is it just one of many little hacks that you see in Cody, inelegant but tolerable in the context of the surrounding game?
Incidentally, I wrote about the Ans Hack long ago on the Community Blog. Dealing with user-unfriendly code is also one of the reasons we created the Head-to-Head voting feature. Some techniques are good for your score, and some are good for your code readability. You get to decide with you care about.
The study of the dynamics of the discrete Klein - Gordon equation (DKG) with friction is given by the equation :
In the above equation, W describes the potential function:
to which every coupled unit 
adheres. In Eq. (1), the variable $
$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and 
is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient 
, while
is the coefficient of the nonlinear cubic term.
adheres. In Eq. (1), the variable $$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient , whileis the coefficient of the nonlinear cubic term.For the DKG chain (1), we will consider the problem of initial-boundary values, with initial conditions
and Dirichlet boundary conditions at the boundary points 
and 
, that is,
and , that is,
Therefore, when necessary, we will use the short notation 
for the one-dimensional discrete Laplacian
for the one-dimensional discrete Laplacian
Now we want to investigate numerically the dynamics of the system (1)-(2)-(3). Our first aim is to conduct a numerical study of the property of Dynamic Stability of the system, which directly depends on the existence and linear stability of the branches of equilibrium points.
For the discussion of numerical results, it is also important to emphasize the role of the parameter 
. By changing the time variable 
, we rewrite Eq. (1) in the form
. By changing the time variable , we rewrite Eq. (1) in the form
. We consider spatially extended initial conditions of the form:
where 
is the distance of the grid and 
is the amplitude of the initial condition We also assume zero initial velocity:
the following graphs for 
and 
and % Parameters
L = 200; % Length of the system
K = 99; % Number of spatial points
j = 2; % Mode number
omega_d = 1; % Characteristic frequency
beta = 1; % Nonlinearity parameter
delta = 0.05; % Damping coefficient
% Spatial grid
h = L / (K + 1);
n = linspace(-L/2, L/2, K+2); % Spatial points
N = length(n);
omegaDScaled = h * omega_d;
deltaScaled = h * delta;
% Time parameters
dt = 1; % Time step
tmax = 3000; % Maximum time
tspan = 0:dt:tmax; % Time vector
% Values of amplitude 'a' to iterate over
a_values = [2, 1.95, 1.9, 1.85, 1.82]; % Modify this array as needed
% Differential equation solver function
function dYdt = odefun(~, Y, N, h, omegaDScaled, deltaScaled, beta)
U = Y(1:N);
Udot = Y(N+1:end);
Uddot = zeros(size(U));
% Laplacian (discrete second derivative)
for k = 2:N-1
Uddot(k) = (U(k+1) - 2 * U(k) + U(k-1)) ;
end
% System of equations
dUdt = Udot;
dUdotdt = Uddot - deltaScaled * Udot + omegaDScaled^2 * (U - beta * U.^3);
% Pack derivatives
dYdt = [dUdt; dUdotdt];
end
% Create a figure for subplots
figure;
% Initial plot
a_init = 2; % Example initial amplitude for the initial condition plot
U0_init = a_init * sin((j * pi * h * n) / L); % Initial displacement
U0_init(1) = 0; % Boundary condition at n = 0
U0_init(end) = 0; % Boundary condition at n = K+1
subplot(3, 2, 1);
plot(n, U0_init, 'r.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title('$t=0$', 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-3 3]);
grid on;
% Loop through each value of 'a' and generate the plot
for i = 1:length(a_values)
a = a_values(i);
% Initial conditions
U0 = a * sin((j * pi * h * n) / L); % Initial displacement
U0(1) = 0; % Boundary condition at n = 0
U0(end) = 0; % Boundary condition at n = K+1
Udot0 = zeros(size(U0)); % Initial velocity
% Pack initial conditions
Y0 = [U0, Udot0];
% Solve ODE
opts = odeset('RelTol', 1e-5, 'AbsTol', 1e-6);
[t, Y] = ode45(@(t, Y) odefun(t, Y, N, h, omegaDScaled, deltaScaled, beta), tspan, Y0, opts);
% Extract solutions
U = Y(:, 1:N);
Udot = Y(:, N+1:end);
% Plot final displacement profile
subplot(3, 2, i+1);
plot(n, U(end,:), 'b.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title(['$t=3000$, $a=', num2str(a), '$'], 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-2 2]);
grid on;
end
% Adjust layout
set(gcf, 'Position', [100, 100, 1200, 900]); % Adjust figure size as needed
Dynamics for the initial condition , 
, for 
, for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from 
and the appearance of values 
for which the solution converges to a non-linear equilibrium point 
Parameters: 
, for , for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from and the appearance of values for which the solution converges to a non-linear equilibrium point Parameters: 
Detection of a stability threshold 
: For 
, the initial condition , 
, converges to a non-linear equilibrium point
.
: For , the initial condition , , converges to a non-linear equilibrium point.Characteristics for 
, with corresponding norm 
where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch 
This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch 
, with corresponding norm where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch By further reducing the amplitude, distinct values of 
are discerned: 1.9, 1.85, 1.81 for which the initial condition 
with norms 
respectively, converges to a non-linear equilibrium point of branch 
This equilibrium point has norm 
and energy 
. The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition 
converges to geometrically different non-linear states of branch 
as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes 
and 
respectively.
are discerned: 1.9, 1.85, 1.81 for which the initial condition with norms respectively, converges to a non-linear equilibrium point of branch This equilibrium point has norm and energy . The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition converges to geometrically different non-linear states of branch as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes and respectively. Refference:
There are a host of problems on Cody that require manipulation of the digits of a number. Examples include summing the digits of a number, separating the number into its powers, and adding very large numbers together.
If you haven't come across this trick yet, you might want to write it down (or save it electronically):
digits = num2str(4207) - '0'
That code results in the following:
digits =
4 2 0 7
Now, summing the digits of the number is easy:
sum(digits)
ans =
13
Hello and a warm welcome to everyone! We're excited to have you in the Cody Discussion Channel. To ensure the best possible experience for everyone, it's important to understand the types of content that are most suitable for this channel.
Content that belongs in the Cody Discussion Channel:
- Tips & tricks: Discuss strategies for solving Cody problems that you've found effective.
- Ideas or suggestions for improvement: Have thoughts on how to make Cody better? We'd love to hear them.
- Issues: Encountering difficulties or bugs with Cody? Let us know so we can address them.
- Requests for guidance: Stuck on a Cody problem? Ask for advice or hints, but make sure to show your efforts in attempting to solve the problem first.
- General discussions: Anything else related to Cody that doesn't fit into the above categories.
Content that does not belong in the Cody Discussion Channel:
- Comments on specific Cody problems: Examples include unclear problem descriptions or incorrect testing suites.
- Comments on specific Cody solutions: For example, you find a solution creative or helpful.
Please direct such comments to the Comments section on the problem or solution page itself.
We hope the Cody discussion channel becomes a vibrant space for sharing expertise, learning new skills, and connecting with others.
Website auswählen
Wählen Sie eine Website aus, um übersetzte Inhalte (sofern verfügbar) sowie lokale Veranstaltungen und Angebote anzuzeigen. Auf der Grundlage Ihres Standorts empfehlen wir Ihnen die folgende Auswahl: United States.
Sie können auch eine Website aus der folgenden Liste auswählen:
Amerika
- América Latina (Español)
- Canada (English)
- United States (English)
Europa
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom(English)
Asien-Pazifik
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)