Finding the dominant frequency of a time series data using fft

% Hi can anyone explain me why using this two procedures I get two diferent dominant frequencies for the same signal? % Note: the data can be downloaded from here: http://www.filedropper.com/data_21

s = load('data.mat');                                                      % load the data
Signal= s.data;
Fs = 60;
t = 0: 1/Fs:length(Signal)/Fs-(1/Fs);
% plot data
subplot(3,1,1); plot(t,Signal,'m');
xlabel('Time','FontWeight','bold','FontSize',10)
ylabel('Amplitude','FontWeight','bold','FontSize',10)

%%Procedure 1 
N = length(Signal);
Signal = Signal - mean(Signal);                                            % to remove the frequency at 0 (or D-C offset)
sigFFT = (abs(fft(Signal)).^2);                  
bin_vals = 0 : N-1;
fax_Hz = bin_vals*Fs/N;
N_2 = ceil(N/2);
power = sigFFT(1:N_2);                                                     % get magnitude 
freq = fax_Hz(1:N_2);                                                      % get freq in Hz
subplot(3,1,2); plot(freq,power,'k')
xlabel('Frequency (Hz)','FontWeight','bold','FontSize',10)
ylabel('Power','FontWeight','bold','FontSize',10)
xlim([0 35]); title('Periodogram')

%%Procedure 2 
%(obtained from here: http://www.mathworks.com/matlabcentral/answers/183642-finding-the-dominant-frequency-of-a-time-series-data-using-fft-matlab)

x = Signal - mean(Signal);
nfft = 512;                                                                % next larger power of 2
y = fft(x,nfft);                                                           % Fast Fourier Transform
y = abs(y.^2);                                                             % raw power spectrum density
y = y(1:1+nfft/2);                                                         % half-spectrum
[v,k] = max(y);                                                            % find maximum
f_scale = (0:nfft/2)* Fs/nfft;                                             % frequency scale
subplot(3,1,3);
plot(f_scale, y),axis('tight'),grid('on'),title('Dominant Frequency')
xlim([0 35]);

But insted of this if I use this signals below I get the same results between procedure 1 and 2

Fs =500;                  % sample frequency
f1 = 3; f2 = 13; f3 = 30; % frequency
theta =0;                 % phase shift
A0 =0;                    % offset
A = 1;                    % amplitude
t = 0: 1/Fs:1 -(1*1/Fs);  % time

Sig1 = A0 + A*sin(2*pi*f1*t + theta);
Sig2 = A0 + A*cos(2*pi*f2*t + theta) ;
Sig3 = A0 + A*sin(2*pi*f3*t + theta);
SignalSum  = Sig1 + Sig2 + Sig3;

[Luis Omar]

fft works better when the total number of the N-points is multiple of power of 2^n like 1024, 512, 2048 etc