I have a problem about calculating with Delta Function. I am trying write matlab code for these function.
I wrote the following code for this function.
n = -5:1:7;
x = delta(n+1) - delta(n) + unit(n+1) - unit(n-2);
stem(n,x,'fill');
axis([-6 8 -1.5 1.5])
xlabel('n')
ylabel('x[n]')
grid
But I am getting the following error.
I don't now how can ı write the delta function in this way ,
function y = unit(x);
y = zeros(size(x));
y(x>0) = 1;
end
Can you help me write the delta function as the 'unit' function you see above? Thank you for your helping.

 采纳的回答

Star Strider
Star Strider 2021-3-24

2 个投票

If you have the Symbolic Math Toolbox, use the dirac function. It can be used with non-symbolic arguments as well.

4 个评论

dirac() is also available for double() and single() without symbolic toolbox.
When I write the code like this,
n = -5:1:7;
x = dirac(n+1) - dirac(n) + unit(n+1) - unit(n-2);
subplot(2,2,3);stem(n,x,'fill');
axis([-6 8 -1.5 1.5])
xlabel('n')
ylabel('x[n]')
grid
I get a graph like this,
But my main goal is to get this graph, what can I do about it?
That is not a true Dirac δ function.
delta = @(x) x==0;
unit = @(x) x>=0;
n = -5:1:7;
x = delta(n+1) - delta(n) + unit(n+1) - unit(n-2);
stem(n,x,'fill');
axis([-6 8 -1.5 2])
xlabel('n')
ylabel('x[n]')
grid
Thank you for your help Mr. Roberson. I got the result I wanted.

请先登录,再进行评论。

更多回答(1 个)

The examples above refer to the Kronecker delta function for discrete signals. If you want to apply the Dirac delta function in simulation to continuous-time systems, the following code is enough:
function y = delta_dirac(u)
[n,m] = size(u);
if max(n,m) ==1
dt = 1e-6; % Define a small time increment for the delta function
else
dt = u(2) - u(1);
end
y = zeros(n,m);
for i=1:max(m,n)
if u(i) == 0
y(i) = 1/dt;
else
y(i) = 0;
end
end

类别

帮助中心File Exchange 中查找有关 Graph and Network Algorithms 的更多信息

产品

Community Treasure Hunt

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

Start Hunting!

Translated by