Finding a DTFT of a signal

From the last highlighted paragraph:

1. It's perhaps easier to relate a unit delay in the z and s domains; thus \$e^{-sT}\$ delays a continuous signal by \$\small T\$ secs, and \$z^{-1}\$ is the unit delay function in the z-domain, which delays a discrete signal by one sampling increment. Hence \$e^{-sT}\leftrightarrow z^{-1}\$, or \$e^{sT}\leftrightarrow z^{1}=z\$ is often a more convenient form.

2. To go from the s-domain to the frequency domain we use \$s \rightarrow j\omega\$. Using the equivalence in 1., above we may go from the z-domain to frequency domain by: \$\small z\rightarrow e^{st} \rightarrow e^{j\omega T} = cos(\omega T) +j\:sin(\omega T)\$. Now, \$\small \omega T\$ is an angle, which is proportional to \$\small \omega\$, and it's convenient to denote this angle, \$\small \Omega\$, and think of \$\small e^{j\Omega}=cos\:\Omega\:+\:j\:sin\:\Omega\$ as a vector that rotates counter-clockwise from zero radians as frequency increases from \$\small\omega=0\$ rad/sec.

3. Now consider an s-plane root, \$\small s=-\alpha\$. This would transform to a z-plane root: \$\small z \rightarrow e^{-\alpha T}\$, which is real, positive and less than unity in magnitude.

4. Moving on, a complex root in the s-plane: \$\small s=\:-\alpha\:+ j\:\omega\$ transforms to \$\small z \rightarrow e^{-\alpha T\:+ j\:\omega T}=\: e^{-\alpha T}\:\small (cos\:\Omega\:+\:j\:sin\Omega)\$, which is a counter-clockwise rotating vector with radius: \$\small e^{-\alpha T}\$. The conjugate root would rotate clockwise with the same radius. Note that the radius is \$\small \lt 1\$, i.e. the vectors are inside the unit circle.

That's a strange paper.

It seems like the authors didn't really understood that the FFT and the DFT are computing the exact same thing. The FFT is just a smart way to compute the DFT in an efficient manner. Lots of hand-waving in this paper with little mathematics.

In contrast to the fFT, which calculates the entire frequency spectral range at once, the discrete Fourier transform can only evaluate a set of arbitrarily or (if available) optimally selected frequency coordinates with a high resolution.

It is true, that using the discrete Fourier transform you can short-cut and only compute it for frequencies of interest while you are practically forced to compute the complete result using FFT. Also you can't pick an arbitrarily frequency with either the FFT or the DFT.

My guess what they're talking about:

If you take a step back and think about the DFT: It's just a correlation of an input signal with a bunch of sine-waves. The frequency of the sine-waves are chosen to integer multiples of whole sine-periods. The key idea behind this is, that the sine-waves at this frequencies are orthogonal to each other. This makes it possible to decompose and reconstruct all possible input signals without loss and ambiguity.

If you want to improve resolution for a single frequency you can evaluate the correlation with sine-waves of your choice. You'll loose the orthogonal property of the DFT but if you're only interested in the signals spectral magnitude at this specific frequency that's okay. Just don't try to decompose and reconstruct a signal with it because the orthogonality is gone.

If you do so, you're not doing an DFT anymore! You're somewhat in the middle of a discrete sine transform and a discrete cosine transform.

In the paper the authors are interested in peaks within the signal spectrum. What they do is to first do an FFT of the signal to find a first estimate of the peak location. Then they 'zoom' into the spectrum by doing correlation with frequencies that exactly fall between the spectral bins within their window of interest.

Afterwards, then they shrink the window of interest and and evaluate the new inbetween frequencies. Do this 10 times and you'll get the 2^10 frequency resolution improvement.

Btw, in my opinion it is debatable if their resulting peak frequency is an improvement over just doing FFT and find the peak there. Using their zoom technique they certainly generate a nice and smooth looking high resolution spectrum. Curve-fitted to find the peak will certainly yield a peak with very low error. They don't go into any detail to what extend the spectral leakage of their method skews their smooth looking spectrum at the first place.