Digital Audio Signal Processing Lecture 4 (Notes)

Notes of Digital Audio Signal Processing, Lecture 4.

NOTA: All \(f\) in this blog refers to the normalized frequency.

\(\delta_k\): the feature of time, representing time 'k';
\(x = \sum\limits_{k\in\mathbb{Z}} x[k]\delta_k\) or \(x[n] = \sum\limits_{k\in\mathbb{Z}}x[k]\delta[n-k]\)

\(p_f[n] = e^{2\pi jfn}\): the feature of frequency, and be used to represent \(x[n]\) in terms of frequency \(f\).
\(x[n]\) is a linear combination of \(p_f[n]\) \(\rightarrow\) \(x[n] = \int X(f)p_f[n]\text{d}f\), where \(X(f)\) is the Fourier representation.

\(x[n]p_{f_0}[n] = ae^{j(2\pi f_0n+\theta)}\)
- \(\sum\limits_{n\in\mathbb{Z}} a\cos(2\pi f_0n+\theta) = a\cos\theta\delta(f_0)\),
- \(\sum\limits_{n\in\mathbb{Z}} a\sin(2\pi f_0n+\theta) = a\sin\theta\delta(f_0)\),
- so \[ \begin{equation} \sum\limits_{n\in\mathbb{Z}} ae^{j\theta}p_{f_0}n = ae^{j\theta}\delta(f_0), \end{equation}\label{distribution} \] where \(\delta(f_0)\) is a distribution function

\(x[n]p_{-f}[n] = ae^{j\theta}e^{2\pi j(f_0-f)n}\)
- Looking for a modulated frquency \(f_{modulated} = f_0-f\).
- So we have, \[ \begin{equation}\sum\limits_{n\in\mathbb{Z}} x[n]p_{-f}[n] = ae^{j\theta}\delta(f_0-f),\end{equation}\] meaning there is a contribution of \(x[n]\) only when \(f_0 = f\).
Go for, \(\sum\limits_{n\in\mathbb{Z}} x[n]p_{-f}[n]\) = \(\sum\limits_{n\in\mathbb{Z}} x[n]\overline{p_f[n]} = \mathbf{x}\cdot\mathbf{p_f} = \mathbf{p_f}^H\cdot\mathbf{x}\), assuming \(\mathbf{x}\) is a real vector (signal).
- Meaning projecting signal \(\mathbf{x}\) onto the coordinate \(\mathbf{p_f}\),
- where it first does the modulation (\(e^{-2\pi jfn}\)) and then the summation (\(\sum\))
- Properties of the Euclidean Inner Product: \(\mathbf{u}\cdot\mathbf{v} = \overline{\mathbf{v}\cdot\mathbf{u}}\)
Fourier transform and inverse Fourier transform
- \(X(f)=\sum\limits_{n\in\mathbb{Z}}x[n]e^{-2\pi jfn}\): the "-" sign comes from the conjugate of \(\mathbf{p_f}\) during the dot products
- \(x[n]=\int X(f)e^{+2\pi jfn}\text{d}f\): the "+" sign because these is the linear combination of \(\mathbf{p_f}\)

Fourier Transform: \(X(f) = \int x(t)e^{-2\pi jFt}\text{d}t\).
Discrete-time Fourier Transform: \(X(f) = \sum\limits_{n\in \mathbb{Z}}x[n]e^{-2\pi jfn}\).
The frequency is real and continuous.
\(X(f) \in \mathbb{C}\) is periodic and complex.
Though it is easy to prove, but why does sampling make signal periodic in frequency domain?
- \(X(f) = X(f+1)\)
Convergence: \(|X(f)|\). (norm)

Time-domain \(x[n]\)	Frequency-domain \(X(f)\)
Impulse: \(\delta[n]\)	\(1\)
Damped exponential: \(a^n u[n]\)	\(\dfrac{1}{1-ae^{-2\pi jf}}\)
Rectangular function: \(r_N[n]=u[n]-u[n-N]\)	\(e^{-\pi jf(N-1)}\dfrac{\sin(\pi fN)}{\sin(\pi f)}\)

NOTA: For the rectangular function, the \(\dfrac{\sin(\pi fN)}{\sin(\pi f)}\) in the \(X(f)\) is not the sinc function. Instead, \(\dfrac{\sin(\pi f)}{\pi f} = \text{sinc}(\pi f)\), the sinc function, which is non-periodic, is the Fourier transform of the rectangular function instead of its DTFT. The sampling makes the \(X(f)\) periodic in frequency domain as shown in the table.