My professor asked me why you choose $a=0.97$ in a pre-emphasis filter $H(z)=1-\frac{a}{z}$ for your pre-processing step.
If you look at the spectrum of speech sounds, you will notice that less energy is present in the highest frequencies, with an overall decreasing slope. The goal of the pre-emphasis filter is to counter-balance this, and flatten the spectrum. This helps for one or several of the following reasons, depending on your application:
As for the the actual value of $a$, you can try plotting the frequency response of the filter for these different values and/or see where the cutoff frequency is. You could also try to visualize its effect on a small corpus of speech signals - how balanced the spectrum becomes after pre-emphasis. It is largely application-dependent - for example, in some applications, poor microphone quality adds an extra 6dB roll-off of the higher frequencies which need more compensation. In speech recognition, this is the kind of parameter tuned a posteriori to try to get better recognition results - though my experience is that with modern models and recognition techniques it has little influence on performance.
If you're familiar with the concept of "poles" and "zeros" in the z-domain, feel free to skip the next paragraph.
The way to quickly visualize the frequency response of a filter in the z-domain is to "trace" the unit circle, by setting $z = e^{j\omega}$. If a point on the unit circle "trace" is close to a zero, the corresponding frequency will be attenuated. Alternatively, if a point on the unit circle is close to a pole, it will be amplified. By corresponding frequency, I mean $\omega$ when tracing the unit circle by setting $z = e^{jw}$.
In the given form, it should be clear that there will be a "zero" when $a=z$. Setting $a$ to $0.9$ or $0.97$ puts the "zero" at $0.9$ or $0.97$ respectively. This will attenuate the low frequencies that are close $\omega=0$. This filter will attenuate frequencies near $\omega=0$.
The answer to the first two is that setting $a$ to $0.97$ will attenuate said low frequencies more than $a=0.9$.
The answer to the "why" is really dependent upon the application of the filter and what its "processing" step entails if this is a "pre-processing" filter.
Start by finding the frequency response of the pre-emphasis filter. By the way, this is an FIR filter, so there are no filter poles, only zeros.
For the frequency response, substitute $\exp\left(j\frac{2\pi f}{f_s}\right)$ for $z$ (where $f_s$ is your sampling frequency). This yields
$$H(f) = 1 - a\exp\left(-j\frac{2\pi f}{f_s}\right)\quad \text{or}\quad1 - a\left[\cos\left(\frac{2\pi f}{f_s}\right) - j\sin\left(\frac{2\pi f}{f_s}\right)\right]$$
For the purpose of pre-emphasis, you are probably most interested in the magnitude response. This is a complex function, so the magnitude is given by (after massaging a little):
$$\lvert H(f)\rvert = 1 + a^2 - 2a\cos\left(\frac{2\pi f}{f_s}\right)$$
Plot this funciton and you will see that it is a highpass response with a maximum at $1/2$ your sampling frequncy and that the filter coefficient $a$ affects the gain, roll-off and DC component of the "message" signal you are processing. The choice for $a$ depends on the nature of the medium or channel that will be used to store or convey the message signal.
What exactly is the signal that you will be applying pre-emphasis to?
$\alpha = \exp (-2\pi F\Delta t)$, where $\Delta t$ is the sampling period of the sound.
The frequency $F$ above which the spectral slope will increase by 6 dB/octave.
Actually, the pre-emphasis filter is just a 1-order FIR filter.
