MATLAB: How to execute fft and ifft

fftifft

I tried to execute this code but for some reason the graph of the ifft is not a perfect sine wave. Can someone please instruct me as to what I did wrong? I got the fft code from the internet and I want to get the ifft
Main Code
fo = 4; %frequency of the sinewave
Fs = 100; %sampling rate 
Ts = 1/Fs; %sampling time interval
t = 0:Ts:1-Ts; 
n = length(t); %number of samples 
y = sin(2*pi*fo*t);
figure(1)
plot(t,y)
grid on
[YfreqD,freqRng] = positiveFFT(y,Fs);
figure(2)
stem(freqRng,abs(YfreqD));
B = ifft(YfreqD)*length(y);
t1 = (0:(1-Ts)/(length(B)-1):1-Ts);
figure(3)
plot(t1,B)
grid on
Positive fft code
function[X,freq] = positiveFFT(x,Fs)
N = length(x);
k = 0:N-1;
T = N/Fs;
freq = k/T; %create the frequency range
X = fft(x)/N; %normalize the data 
cutOff = ceil(N/2);
X = X(1:cutOff);
freq = freq(1:cutOff);

Best Answer

Hi thereb, the sine wave is right there, but the problem is that after FFTing the your y data, YfreqD contains half the samples than the original. When calculating the iFFT, that length is kept and plot(t1,B) plots half the original data.
You can try plot like this to see the points you are plotting in each case:
figure(1)
plot(t,y,t,y,'r*')
figure(3)
plot(t1,B,t1,B,'r*')
You can add a higher sampling rate at the beginning of your code and your figure(3) will appear like a less "straight" sine wave.
fo = 4; %frequency of the sinewave
Fs = 200; %sampling rate 
Ts = 1/Fs; %sampling time interval
t = 0:Ts:1-Ts; 
n = length(t); %number of samples 
y = sin(2*pi*fo*t);
figure(1)
plot(t,y,t,y,'r*')
grid on
[YfreqD,freqRng] = positiveFFT(y,Fs);
figure(2)
stem(freqRng,abs(YfreqD));
B = ifft(YfreqD)*length(y);
t1 = (0:(1-Ts)/(length(B)-1):1-Ts);
figure(3)
plot(t1,B,t1,B,'r*')
grid on
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