Getting the magnitude of FFT of a sine wave

I am new to fft and a bit confused about the magnitude. This is what I have

A =1;

t = 0:0.01:1-0.01;

x = A*cos(2*pi*10*t);

xdft = fft(x,1024);

When I plot abs(xdft) I get a magnitude of the length of the wave * amplitude /2. I am confused because I though the magnitude would be 1024*A/2. What am I missing? also since I am trying to learn fft and understand the output it spits out very well, can someone direct me to a good source to start with? Thank you.

Answer 1

[Wayne King]

You should not zero pad here because as is your frequency of 10 Hz falls directly on a DFT bin.

 A =1;
 t = 0:0.01:1-0.01;
 x = A*cos(2*pi*10*t);
 xdft = fft(x);
 plot(abs(xdft))

Your data, x, is 100 samples in length. So you get 100*A/2 at -10 Hz and +10 Hz. That is equal to 50.

The zero pad length does not affect the amplitude, just the length of the data vector.

You can of course do:

xdft = fft(x)/length(x);
plot(abs(xdft))

[Nina]

Thank you Wayne, sorry but I am not sure what you mean by padding by 0, I may have padded without realizing it, can you show me how not to pad?Thank you so much again.

[Wayne King]

I did in my example above, you wrote fft(x,1024) that is padding out to 1024. I wrote just fft(x)

[Nina]

Oh I see, thank you :)