draw a function in the matlab

How to draw abs(g(k)) in the matlab when a=0,k=+-2pi or a=0,k=+-6pi

Answer 1

[Chunru]

a=0;
k = (-10:1:10)*2*pi;
g = -2*pi./(4*pi^2+(a-k*1i).^2) +2*pi./(4*pi^2+(a+k*1i).^2) -(a-k*1i)./(36*pi^2+(a-k*1i).^2) +...
    + (a+k*1i)./(36*pi^2+(a+ k*1i).^2);
stem(k/(2*pi), abs(g))
xlabel('k/(2\pi)')

[Chris Lin]

May I ask what is the k = (-10:1:10)*2*pi means?

[Chunru]

k is a vector having the values of -20pi, -18pi, .., 0, ..., 20pi.

[Chris Lin]

If I have the original code as follows, how can I continue to draw what lg(x)l looks like when a=0,k=+-2pi or a=0,k=+-6pi?Thanks.

syms x a k real

syms f(x)

assume(a>0)

f(x) = sin(2*pi*x) + cos(6*pi*x)

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

g(k) = int(inner, x, -inf, inf)

[Chunru]

It seems that your integration over infity is not converging (as long as the symbolic math thinks).

syms x a k real
syms f(x)
assume(a>0)
a = 0

a = 0

f(x) = sin(2*pi*x) + cos(6*pi*x)

f(x) = 

inner = f(x)*exp(-a*abs(x))*exp(-1i*k*x)

inner = 

g(k) = int(inner, x, -inf, inf)

g(k) = 

kval = (-5:5)*2*pi;
gval = g(kval)

gval = 

Answer 2

[Walter Roberson]

When a = 0 then there is no damping being done on the exp(-i*k*x). We also know that f(x) does not have any damping. Therefore the integral can only converge if exp(-i*k*x) converges to 0, but in order for that to happen, k would have to be a purely imaginary number with negative coefficient, which is not the case.

Therefore you cannot plot this when a is 0.