Example 1: FFT of a Simple Sinusoid

GUIDE: Mathematics of the Discrete Fourier Transform (DFT). Example 1: FFT of a Simple Sinusoid

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Example 1: FFT of a Simple Sinusoid

Our first example is an FFT of the simple sinusoid


where we choose $\ (frequency $f_s/4$) and $T=1$(sampling rate set to 1). Since we’re using the FFT, the signal length $N$ must be a power of $2$. Here is the Matlab code:
echo on; hold off; diary off;
% !/bin/rm -f examples.dia; diary examples.dia    % For session log
% !mkdirs eps                                     % For figures

% Example 1: FFT of a DFT sinusoid

% Parameters: N = 64; % Must be a power of two T = 1; % Set sampling rate to 1 f = 0.25; % Sinusoidal frequency in cycles per sample A = 1; % Sinusoidal amplitude phi = 0; % Sinusoidal phase n = [0:N-1]; x = cos(2*pi*n*fT); % Signal to analyze X = fft(x); % Spectrum

Let’s plot the time data and magnitude spectrum:

% Plot time data
ni = [0:.1:N-1]; % Interpolated time axis hold on; plot(ni,cos(2*pi*ni*f*T),‘-’); title(‘Sinusoid Sampled at 14 the Sampling Rate’); xlabel(‘Time (samples)’); ylabel(‘Amplitude’); text(-8,1,‘a)’);

% Plot spectral magnitude magX = abs(X); fn = [0:1.0/N:1-1.0/N]; % Normalized frequency axis subplot(3,1,2); stem(fn,magX) xlabel(‘Normalized Frequency (cycles per sample))’); ylabel(‘Magnitude (Linear)’); text(-.11,40,‘b)’);

% Same thing on a dB scale spec = 20*log10(magX); % Spectral magnitude in dB subplot(3,1,3); plot(fn,spec); axis([0 1 -350 50]); xlabel(‘Normalized Frequency (cycles per sample))’); ylabel(‘Magnitude (dB)’); text(-.11,50,‘c)’); print -deps eps/example1.eps; hold off;

Figure 9.1:Sampled sinusoid at $f=f_s/4$. a) Time waveform. b) Magnitude spectrum. c) DB magnitude spectrum.

The results are shown in Fig. 9.1. The time-domain signal is shown in Fig. 9.1a, both in pseudo-continuous and sampled form. In Fig. 9.1b, we see two peaks in the magnitude spectrum, each at magnitude $32$ on a linear scale, located at normalized frequencies $f=
0.25$ and $f= 0.75 = -0.25$. Since the DFT length is $N=64$, a spectral peak amplitude of $32 = (1/2) 64$ is what we expect, since


when $\. This happens at bin numbers $k = (0.25/f_s)N
= 16$ and $k = (0.75/f_s)N = 48$ for $N=64$. However, recall that Matlab requires indexing from $1$, so that these peaks will really show up at index $17$ and $49$ in the magX array.

The spectrum should be exactly zero at the other bin numbers. How accurately this happens can be seen by looking on a dB scale, as shown in Fig. 9.1c. We see that the spectral magnitude in the other bins is on the order of $300$ dB lower, which is close enough to zero for audio work.

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