why the fft results of these functions are not what I expect?

I have three functions; a sinusoid, square and sawtooth. The fft of sine is what we expect in fourier domain. But I expect to see a sinc function for the square. But why its not what I expect?

close all;
clear all;
fs=1e5;
a=-0.01;
b=0.01;
t = a:1/fs:b;
x = sin(1000*2*pi*t); 
y = sawtooth(500*2*pi*t);
z = square(2000*2*pi*t);
N=length(x);
figure, subplot(1,3,1), plot(t,x);xlim([a/10 b/10])
subplot(1,3,2), plot(t,y);xlim([a/10 b/10])
subplot(1,3,3), plot(t,z);xlim([a/10 b/10])

A = 2*fftshift(abs(fft(x))/N);
b = 2*fftshift(abs(fft(y)))/N;
c = 2*fftshift(abs(fft(z)))/N;
if mod(N,2) == 0 % N is even
   f = ( (-N/2) : ((N-2)/2) )/N*fs;
 else            % N is odd
   f = ( (-(N-1)/2) : ((N-1)/2) )/N*fs;
end
figure
plot(f, c); xlim([-fs fs])

figure
plot(f, b);xlim([-fs fs])

figure
plot(f, A);xlim([-fs fs])

[Paul]

Hi 2NOR_Kh,

Why is the DFT of z expected to be a sinc (or approximately a sinc) function?

[2NOR_Kh]

Because it is a square wave in time domain.

[Paul]

Yes, but why would this square wave in the time domain have a Fourier transform that is a sinc? The Fourier transform of a rectangular pulse is a sinc, but this is a square wave, not a rectangular pulse.

[2NOR_Kh]

You are right but what about the energy of the signals?

in time domain the area under the curve of the all these three function in time domain is zero which can be represented as their energy and I expect the overal energy in the other domain be zero but I cant see that. Or maybe I am missing something.

[Paul]

The energy of a signal is not the area under the curve. link to reference

[2NOR_Kh]

thanks alot. It was a great help.