how replace dirac to diracz when need ?

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rodrigo figueiredo
rodrigo figueiredo 2021 年 8 月 20 日
回答済み: Star Strider 2021 年 8 月 20 日
Q=f(dirac(x - 10.5)*sin(x)*(10+i))
Q=diracz(x - 10.5)*sin(x)*(10+i)
function p=diractz(x)
if x ==0
p=1;
else
p=0;
end
end

回答 (1 件)

Star Strider
Star Strider 2021 年 8 月 20 日
I am not certain what you want to do.
Since the independent variable may never exactly be 10.5, I drew that separately (‘i’ is not defined other than by default being the imaginary operator):
diracz = @(t) t==0;
x = linspace(0, 20, 1E+4);
Q = @(x) diracz(x-10.5) .* sin(x) * 10+1i;
figure
plot(x, real(Q(x)), '-b')
hold on
plot([10.49 10.5 10.51], real(Q([10.49 10.5 10.51])), '-r')
hold off
grid
ylim([-10 10])
Note that the Dirac delta function is only defined on real values.
.

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