Frequencies of complex exponentials in Discrete Fourier Transform

Question

What I understand is that using DFT, we are representing a given discrete signal using a basis of complex exponentials of different harmonic frequencies. If I am taking 16 point DFT of a signal sampled at 1 KHz, k=0 corresponds to 0 Hz, k=1 corresponds to 62.5Hz etc. Is this correct?

In the figure, a cosine of 187.5Hz is shown and it is sampled at 1KHz. The sampled signal is actually the cosine part of the complex exponential of k=3 in a 16 point DFT. What confuses me is that, although the continuous time signal is periodic with frequency 187.5Hz, the sampled signal is not having the same period.

If the sampled signal is not of frequency 187.5Hz, how can we say that basis vectors of k=3 picks up the signal component of 187.5Hz?

Answer 1

It's not necessary for a discrete-time signal to have the same period as the continuous-time signal that it represents. As long as the continuous-time signal is band-limited and as long as the sampling frequency is greater than twice the bandwidth of the continuous-time signal, the discrete-time signal can perfectly represent the continuous-time signal.

In your example, the continuous-time signal is

$$x(t)=\cos(2\pi f_0t)\tag{1}$$

and the discrete-time signal is

$$x_d[n]=x(nT)=\cos(2\pi f_0Tn)\tag{2}$$

$x_d[n]$ perfectly represents $x(t)$ if $f_0T<\frac12$ is satisfied. In general, $x_d[n]$ is not even a periodic sequence. It is only periodic if $f_0T$ is rational.

With the choice $f_0T=3/16$ we make sure that $x_d[n]$ perfectly represents $x(t)$, and we also make sure that $x_d[n]$ is periodic with period $16$. Note that there can't be $3$ periods of the discrete-time signal in the DFT frame, simply because $16/3$ is not an integer.

Since $x_d[n]$ has period $16$, taking a length $16$ DFT results in a single frequency component. With $f_0T=3/16$, that component occurs at frequency index $k=\pm 3$, which exactly represents a continuous-time sinusoid at frequency $3/16\cdot f_s$ (with $f_s=1/T)$.

Answer 2

"187.5Hz is shown and it is sampled at 1KHz."

Sampling is better represented with units of samples per second.

$$ \frac{ 187.5 \frac{cycles}{second} }{ 1000 \frac{samples}{second} } \cdot 16 \frac{samples}{frame} = 3\frac{cycles}{frame} $$

Since your frequency is a whole number, there will be no leakage in the DFT.

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While we're at it, have a unit based look at the exponential term of the DFT definition:

$$\frac{2\pi}{N}kn$$

$$ \frac{ \frac{radians}{cycles} }{ \frac{samples}{frame} } \cdot \frac{cycles}{frame} \cdot samples = radians $$

One of the most important things to know about the DFT is that the bin index ($k$) represents the frequency in terms of cycles per frame. (Except in MATLAB where it is off by one.)