%% *Example 2.4 *
% We already looked at the result $\mathbf y$ of sampling $f\left(x\right)=2sin\left(12\pi 
% x\right)+\ldotp 5sin\left(36\pi x\right)$ at 48 points. Here it is again. 

f = 2*sin(12*pi*1/48*(0:47))+0.5*sin(36*pi*1/48*(0:47));
plot(1/48*(0:47),f,'o')
hold on
for j = 1:48
    plot([1/48*(j-1),1/48*(j-1)],[0,2*sin(12*pi*1/48*(j-1))+0.5*sin(36*pi*1/48*(j-1))],'g')
end
hold off
%% 
% Here's the plot of imaginary part of the discrete Fourier transform $\mathbf 
% Y$ of $\mathbf y$.

y = -48*[zeros(1,6) 1 zeros(1,11) .25 zeros(1,11) -.25 zeros(1,11) -1 zeros(1,5)];
plot(1/48*(0:47),y,'o')
hold on
for j = 1:48 
    plot([1/48*(j-1),1/48*(j-1)],[0,y(j)],'g')
end
hold off
%% 
% Notice that this is skew symmetric about the *Nyquist point,* the midpoint 
% of the range.
% 
% **