DFT sample point k < N has negative frequency

(From: Schaum's DSP outline, 2nd edition, page 254, problem 6.35)

A signal $$x_a(t)$$ that is bandlimited to 10 kHz is sampled with a sampling frequency of $$f_s = 20$$ kHz. The DFT of N=1000 samples of x[n] is then computed, that is:

$$X[k]=\sum_{n=0}^{N-1} x[n] e^{-j(2\pi / N)nk}$$

with N=1000.

(a) to what analog frequency does the index … k=150, and k=800 correspond?

A few equations:

$$\omega_k = \frac{2 \pi k}{N}$$

$$f_k = \frac{\omega_k f_s}{2 \pi}$$

For k=150:

$$\omega_k = \frac{2 \pi}{1000} 150 = \frac{3}{10}\pi$$

$$f_k= \frac{3\pi 20000}{2 \pi 10} = 3000\ \ hz$$

Ok no problem so far:

For k=300:

$$\omega_k = \frac{2 \pi}{1000} 800 = \frac{4}{5}\pi$$

Here's where i hit a problem:

book says, for k=800, we need to be careful. Because $$X(e^{j\omega})$$ is periodic:

$$X(e^{j\omega}) = X(e^{j\omega + 2\pi})$$

Therefore, k=800 corresponds to the frequency:

$$\omega_k = \frac{2\pi}{N}k = \frac{2\pi}{N}(k - N) = -200\frac{2\pi}{N}$$

My question is this:

what's the reason we need to use a negative frequency here? Since k=800 < N=1000? Wouldn't it just be:

$$\omega_k = \frac{2 \pi 800}{1000} = \frac{8\pi}{5} = 5.02...$$

Is this just the book being strange or am i missing some understanding of DFT? I would have thought you wouldn't need to worry about $$2\pi$$ unless k > 1000?

The DTFT frequency $$\omega_k$$ corresponding to the $$N$$-point DFT index $$k$$ is given by (frequency sampling relation)

$$\omega_k = \frac{2 \pi}{N} k ~~~,~~~ k = 0,1,...,N-1.$$

The confusion might arise with the interpretation that the, on the continuous-time domain these frequencies correspond to:

$$f_k = \frac{f_s}{N} k ~~~,~~~ k = 0,1,...,N-1 ?$$

But no. Because, after about $$k = N/2$$ which corresponds to $$\omega = \pi$$ and $$f = f_s/2$$, you will be sampling the negative portion of the shifted spectrum: $$X(e^{j (\omega-2\pi)})$$, which corresponds to negative discrete-time frequencies of $$X(e^{j \omega})$$ due to the periodicity of DTFT $$X(e^{j\omega})$$.

A more relevant (eventually the same) interpretation of the CTFT, DTFT, DFT frequency relation can be given by the following.

Let $$x_c(t)$$ be bandlimited to $$-B \leq f and sampled at $$F_s = 2B$$ samples per second. The base period of the spectrum of $$x[n]$$ is between $$-\pi \leq \omega < \pi$$. Note that only the left member is included and the right member is discluded.

Now assume an $$N$$-point DFT $$X[k]$$ of $$x[n]$$ is computed. According to the frequency sampling interpretation of DFT, samples of $$X[k]$$ are given by :

$$X[k] = X(e^{j \frac{2\pi}{N} k}) ~~~~ , k = ...-3,-2,-1,0,1,2,3...$$

Note very carefully, that eventhough most of the time we restrict our attention of the DFT index $$k$$ into the range of $$0 \leq k < N$$, indeed, the relation is valid for all $$k$$ which is a manifestation of DFT $$X[k]$$ and DTFT $$X(e^{j \omega})$$ being periodic by $$N$$ and $$2\pi$$.

Hence, instead of taking the range of $$k$$ in $$[0,N-1]$$, we can also take it in the range: $$[-N/2 , N/2-1]$$ (assuming $$N$$ even for the moment).

So, if you take $$k = -N/2,-N/2+1,...,-1,0,1,2,...,N/2-1$$ as the range of $$k$$ for DFT $$X[k]$$, then the mapping into the negative frequencies become obvious.

Note that, conventional FFT software always take the range of $$k$$ in $$[0,N-1]$$. Therefore, one has to relate them too which is simple: for $$k$$ in $$[-N/2,-1]$$ it corresponds to $$k'$$ in $$[N/2,N-1]$$ in the conventional DFT range, and for $$k$$ in $$[0,N/2-1]$$ it corresponds to the $$k'$$ in $$[0,N/2-1]$$ range in the conventional range.

The inverse DTFT to which the $$\omega$$ refers to when sampling DTFT to create an N-point DFT maybe taken over any $$2\pi$$ interval, since the sampling rate for the DTFT is always normalized to $$\omega_s=2\pi$$:

$$f[n] = \frac{1}{2\pi} \int_{2\pi} F(e^{j\omega})e^{j\omega n} d\omega$$

While we could take the inverse DTFT over $$\omega=0$$ to $$2\pi$$, it is more conventional to use $$\omega$$ over the range $$\omega=-\pi$$ to $$\pi$$ so that its easier to see complex conjugate relationships between poles and zeros. Thus, the inverse DTFT becomes:

$$f[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} F(e^{j\omega})e^{j\omega n} d\omega$$

Also, note due to periodic nature of DTFT:

$$X(e^{j\omega}) = X(e^{j\omega + 2\pi})$$

The range $$\omega=\pi$$ to $$2\pi$$ is an alias of:

the range $$\omega=-\pi$$ to $$0$$

Next, if we consider expressing a discrete frequency $$\omega$$ as a continuous frequency $$\Omega$$ then you need to convert $$\omega$$ into the range $$-\pi$$ to $$\pi$$ because CTFT is not periodic like the DTFT, thus we take the lowest range of values of $$\omega$$ before converting to $$\Omega$$.