Convolution

The index of the samples and the word "time" will be used interchangeably.
$\underline{\textbf{x}_{n}}$ is the input signal of $L$ samples taken as a runtime vector of $Q$ samples. This runtime vector is a column vector of $Q$ samples¹ composed of:

$$\underline{\textbf{x}_{n}}^{t} = ( x[n], x[n-1], x[n-2], ..., x[n-Q+1] )$$

Which are $Q$ samples of the signal x(t), sampled at a frequency $1/T_s$:

FIR filtering of the signal $x[n]$ with filter of $Q$ coefficients² $h_{0}^*,h_{1}^*,...,h_{Q-1}^*$ where $^*$ denotes complex conjugate can be then written in vectorial notation in this way:

$$y[n] = \sum_{k=0}^{L+Q} h[k]^*x[n-k] = h[n]^* \star x[n] = \underline{\textbf{h}}^t \underline{\textbf{x}}_n$$

Where $\star$ is the discrete convolution.

$$y[n] = \begin{pmatrix}h_{0},h_{1},...,h_{Q-1} \end{pmatrix}^... ...x[n]\\ x[n-1]\\ x[n-2]\\ ...\\ x[n-L+1] \end{pmatrix}$$

Or putting all the equations at once to form a vector of $L$ outputs:

$$( y[n], y[n-1], y[n-2], ..., y[n-L+1] ) =$$

$$\begin{pmatrix}h_{0},h_{1},...,h_{Q-1} \end{pmatrix}^* \begi... ...ots\\ x[n-Q+1] & x[n-Q] & ... & x[n-Q-L+2]\\ \end{pmatrix}$$

$$\underline{\textbf{y}}^t = \underline{\textbf{h}}^h ( \under... ...}_0,\underline{\textbf{x}}_1,...,\underline{\textbf{x}}_{L-1})$$

$$\underline{\textbf{y}}^t_n = \underline{\textbf{h}}^h \underline{\underline{{\textbf{X}}}}_{n}$$

The scalar product of two deterministic energy-type signals of $L$ samples:

$$<\underline{\textbf{h}},\underline{\textbf{x}}> = \sum_{k=0}... ... h[k]^* x[k] = \underline{\textbf{h}}^h\underline{\textbf{x}}$$

And for two power-type signals:

$$<\underline{\textbf{h}},\underline{\textbf{x}}> = \lim_{N\to\... ...nfty}\frac{1}{N} \underline{\textbf{h}}^h\underline{\textbf{x}}$$

Note that in this case $\underline{\textbf{x}}$ doesn't have dependence with the sample index $n$.