How can I take an integer like n=1:1:50 and get the output for each value?

7 次查看（过去 30 天）
Tasin Nusrat
Tasin Nusrat 2021-10-30
回答： Sulaymon Eshkabilov 2021-10-30
I want to plot my output for each input and the input is from 1 to 50, but I am not sure why my code is not working.
n=1:50; %my range, I want to check for each n values
f=n*175*10^6; % want to get frequency for each n
c=(sin(n*pi*d))/(n*pi*d) %d,tr,T these are my fixed values
d1=(sin((n*pi*tr)/T))/((n*pi*tr)/T);
In=2*d*Ipp*c*d1;
E=(In*f*L)/(0.8*r); %I am not getting E for each n values

请先登录，再回答此问题。

回答（2 个）

Walter Roberson
Walter Roberson 2021-10-30
Ran in:
d = rand() %since it was not defined
d = 0.0687
T = rand() %since it was not defined
T = 0.3380
tr = rand() %since it was not defined
tr = 0.7222
L = rand() %since it was not defined
L = 0.8296
r = rand() %since it was not defined
r = 0.7237
Ipp = randi(255); %since it was not defined
n=1:50; %my range, I want to check for each n values
f=n*175*10^6; % want to get frequency for each n
c=(sin(n*pi*d))./(n*pi*d) %d,tr,T these are my fixed values
c = 1×50
0.9923 0.9692 0.9316 0.8803 0.8169 0.7431 0.6608 0.5722 0.4797 0.3857 0.2925 0.2024 0.1175 0.0397 -0.0295 -0.0887 -0.1371 -0.1742 -0.1996 -0.2137 -0.2171 -0.2105 -0.1951 -0.1724 -0.1438 -0.1110 -0.0756 -0.0394 -0.0039 0.0293
d1=(sin((n*pi*tr)/T))./((n*pi*tr)/T);
In=2*d*Ipp*c.*d1;
E=(In.*f*L)/(0.8*r);
plot(n, E)
Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021-10-30
Without knowing the size of your variables and just presuming some variables to be scalar and others to be a vector, here is the corrected code that is based on the color operator (:) :
n=1:50; %my range, I want to check for each n values
f=n*175*10^6; % want to get frequency for each n
c=(sin(n(:).*pi*d))./(n(:).*pi*d) %d,tr,T these are my fixed values
d1=(sin((n(:).*pi*tr)./T))./((n(:)*pi*tr)./T);
In=2*d(:).*Ipp*c*d1;
E=(In*f(:).*L)./(0.8*r); %I am not getting E for each n values
Presumably, you have here d and L or r to be a vector with a certain length.

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by