how to convert this java program into matlab can you suggest me an ideaI don't know of any automatic translators that you can use. So, just start writing. None of use are going to spend all that time to do it for you, though it shouldn't be long since MATLAB ...
Open in MATLAB Online Hello Ryan, There are 2 ways to go about it. One possible way is: ThemeCopy 7 * sqrt(2) / 2 Another possible and more general way to compute any x to the power of y is ThemeCopy 7 * 2^0.5 / 2 Hope this helps. Thanks, Aparajith 1 Comment Ryan on ...
답변:Abhishek Gupta2020년 12월 16일 태그 "powers" 웹사이트 선택 번역된 콘텐츠를 보고 지역별 이벤트와 혜택을 살펴보려면 웹사이트를 선택하십시오. 현재 계신 지역에 따라 다음 웹사이...
You have drawn the lens in matlab in 3D, how is it done? I used the lens equation y=sqrt((n^2-1)*((x-f)^2))+((2)*(f)*(n-1)*(x-f))); to draw a hyperbolic lens. where n is the refractive index of the lens, f is the focal length of the lens. The above equation...
= sqrt(sum_err_sqrd / numpoints); fprintf('%-16s%14.12f %14.12f %9.6 %4.1f %4.1f\n', , alpha, beta, avg_err, decibels(rms_err, decibels(peak_err)); end It took 2 minutesto convert is from Cto Matlab. 1 Comment EROL EREN GÖZLÜon 6 May 2019 Thank you very...
Possibly your intent was to multiply that by 3, and not raise 3 to that power. We don't know, at least not until we can get the mind reading toolbox in working order. It is just such a buggy thing. ;-) 3*sqrt((16*T)/(tau*pi*(...
As in MATLAB, if the step is omitted, it defaults to 1. Notice that you had to pass the stop value 7 so that the array stopped at 6. However, the size of the resulting array is 7 - 1 = 6 elements long. Next, you should see how to change the step size: Python In [5]: ...
[x,y]=meshgrid(-18:1:18);r=sqrt(x.^2+y.^2)/2;z=cos(r)./r;mesh(x,y,z)colorbar Output: We can use themeshc()function to draw counters below the surface plot instead of themesh()function. We can also use themeshz()function to draw curtains below the surface plot instead of...
The variables q1, q2, p1, and p2 are declared to represent the coordinates of two points: (q1, p1) and (q2, p2). These variables are then assigned specific values, which correspond to the coordinates of the two points in Euclidean space. Here, we use the Math.sqrt function to compute...
r= sqrt(y(1)^2 + y(2)^2); K1=500; K2 = 500; F=interp1(ti,xx,t); %osc1 dz(1)= (mu - r^2)*y(1) - y(3)*y(2) +K1.*F; dz(2) = (mu - r^2)*y(2) + y(3)*y(1); dz(3) = (-K1.*F) * (y(2)/sqrt(y(1)^2 + y(2)^2)); ...