Multimedia Signal Processing
Convolution and Impulse Response

Signal

digitized signal

Time

• By time sampling and quantization (rounding), an analog signal can be converted into a digital signal
• The advantage of digital signals is that they can be easily processed with a digital signal processor or a computer

$x(t)$

$x[n]$

Zeit $t$

• Mathematically expressed: The analog time signal $\mathrm{x}(t)$ is converted during sampling into a discrete-time signal $\mathrm{x}[n]$ with $n \in \mathbb{Z}$
• Since $n$ is an element from the set of integers, the function $\mathrm{x}[n]$ is always discrete in time
• A more detailed description of the digitization process follows in later chapters

• Of very great importance is the so-called discrete-time unit impulse sequence (or "unit impulse" for short)
  $\mathrm{\delta}[n] = \begin{cases} 1 & \,\,:\,\, n = 0 \\ 0 & \,\,:\,\, n \ne 0 \\ \end{cases}$

$\mathrm{\delta}[n]$

$n$

• The notation with $\mathrm{\delta}[n]$ is chosen because the equivalent signal in analog signal processing is the Dirac delta function $\mathrm{\delta}(t)$

• The discrete-time unit step $\mathrm{u}[n]$ is defined by:
  $\mathrm{u}[n] = \begin{cases} 1 & \,\,:\,\, n \ge 0 \\ 0 & \,\,:\,\, n \lt 0 \\ \end{cases}$

$\mathrm{u}[n]$

$n$

• Is there a mathematical relationship between the unit step and the unit impulse?
• Yes, the integration (or in the discrete-time case, the summation):
  $\mathrm{u}[n] = \sum\limits_{i=-\infty}^n \mathrm{\delta}[i]$

• A discrete-time rectangular pulse with pulse width $P$ is generated by:
  $\mathrm{x}[n] = \begin{cases} 1 & \,\,:\,\, |n| < P/2 \\ 0.5 & \,\,:\,\, |n|=P/2 \\ 0 & \,\,:\,\, |n|> P/2 \\ \end{cases}$
• The figure shows a rectangular pulse with a pulse width $P=9$:

$\mathrm{x}[n]$

$n$

• The case $|n| = P/2$ can only occur for even $P$, e.g. $P=10$. In this case the value $0.5$ ensures that the pulse width is still $P$.

• A discrete-time Gaussian pulse with the standard deviation $\sigma$ is generated by:
  $\mathrm{x}[n] = e^{- 0.5 \, (n / \sigma)^2} $
• The figure shows a Gauss pulse with standard deviation $\sigma=4$:

$\mathrm{x}[n]$

$n$

• A time-discrete triangle pulse with pulse width $P$ is generated by:
  $\mathrm{x}[n] = \begin{cases} 1.0 - 2.0 \, (n / P) & \,\,:\,\, |n| \le P/2 \\ 0 & \,\,:\,\, |n| > P/2 \\ \end{cases}$
• The figure shows a triangle pulse with a pulse width $P=9$:

$\mathrm{x}[n]$

$n$

• A discrete-time sine signal can be generated as follows:
  $\mathrm{x}[n] = A \sin\left(2\pi\frac{n+M}{W}\right) $
• The figure shows a sine wave with wavelength $W=16$, displacement $M=0$, and amplitude $A=1$:

$\mathrm{x}[n]$

$n$

• A discrete-time triangle wave can be generated by:
  $\mathrm{x}[n] = A \left(2.0 \frac{(n+M) \, \bmod \, W}{W} - 1.0\right) $
  where $\bmod$ denotes the modulo operation.
• The figure shows a triangle wave with wavelength $W=16$, displacement $M=0$, and amplitude $A=1$:

$\mathrm{x}[n]$

$n$

• A time-discrete square wave can be generated by:
  $\mathrm{x}[n] = \begin{cases} A & :\quad (n+M) \, \bmod \, W \ge W /2 \\ -A & :\quad (n+M) \, \bmod \, W < W /2\\ \end{cases}$
• The figure shows a square wave for wavelength $W=16$, displacement $M=0$, and amplitude $A=1$:

$\mathrm{x}[n]$

$n$

• The discrete-time convolution of two signals $\mathrm{a}[n]$ and $\mathrm{b}[n]$ is defined by:
  $\mathrm{f}[n] = \sum\limits_{k = -\infty}^{\infty} \mathrm{a}[k] \, \,\mathrm{b}[n-k]$
• To calculate $\mathrm{f}[n]$, the sum must be evaluated for different $n$. To this end, the signal $\mathrm{b}$ is mirrored at the $y$-axis and then shifted by $n$ to the right.
• This means, the sum is calculated for different offsets $n$ of the mirrored signal $\mathrm{b}$.
• For the short notation of the convolution, the operator "$\ast$" is used:
  $\mathrm{f}[n] = \mathrm{a}[n] \ast \mathrm{b}[n] = \sum\limits_{k = -\infty}^{\infty} \mathrm{a}[k] \, \,\mathrm{b}[n-k]$

• Commutativity
  $\mathrm{a}[n] \ast \mathrm{b}[n] = \mathrm{b}[n] \ast \mathrm{a}[n]$
• Associativity
  $\mathrm{a}[n] \ast (\mathrm{b}[n] \ast \mathrm{c}[n]) = (\mathrm{a}[n] \ast \mathrm{b}[n] ) \ast \mathrm{c}[n]$
• Distributivity
  $\mathrm{a}[n] \ast (\mathrm{b}[n] + \mathrm{c}[n]) = \mathrm{a}[n] \ast \mathrm{b}[n] + \mathrm{a}[n] \ast \mathrm{c}[n]$

$\mathrm{f}[n] = \sum\limits_{k = -\infty}^{\infty} \mathrm{a}[k] \, \,\mathrm{b}[n-k]$

$\mathrm{a}[k]$

$\mathrm{b}[k]$

$\mathrm{c}[n]$

$\mathrm{f}[n] = \sum\limits_{k = -\infty}^{\infty} \mathrm{a}[k] \, \,\mathrm{b}[n-k]$

Konstruktion:

$\mathrm{a}[k]$

$\mathrm{b}[0-k]$

$\mathrm{a}[k]$

$\mathrm{b}[1-k]$

$\mathrm{f}[0]$

$\mathrm{f}[n]$

$\mathrm{f}[1]$

$\mathrm{f}[n]$

$\mathrm{f}[n] = \sum\limits_{k = -\infty}^{\infty} \mathrm{a}[k] \, \,\mathrm{b}[n-k]$

Konstruktion:

$\mathrm{b}[5-k]$

$\mathrm{b}[7-k]$

$\mathrm{f}[5]$

$\mathrm{f}[7]$

• Linear time invariant systems (LTI systems) have the following defining properties:
  • Linearity: Let $\mathrm{y}[n]_k$ be the reaction of the system to the input sequence $\mathrm{x}[n]_k$. Then it must applies to the linear combination of the input signals
    $\mathrm{x}[n] = s_1 \, \mathrm{x}[n]_1 + s_2 \, \mathrm{x}[n]_2 + s_3 \, \mathrm{x}[n]_3 + \dots$
    that the reaction of the system is
    $\mathrm{y}[n] = s_1 \, \mathrm{y}[n]_1 + s_2 \, \mathrm{y}[n]_2 + s_3 \, \mathrm{y}[n]_3 + \dots$
    or written differently with $\mathrm{y}[n]_k = \mathrm{S}\left(\mathrm{x}[n]_k\right)$:
    $\mathrm{S}\left(\sum\limits_k \, s_k \mathrm{x}[n]_k\right) = \sum\limits_k s_k \, \mathrm{S}\left(\mathrm{x}[n]_k\right)$
  • Time invariance: A shift of the input sequence by $M$ always results in a corresponding shift in the output sequence:
    $\mathrm{S}\left(\mathrm{x}[n+M]\right) = \mathrm{y}[n+M] \quad \forall \,M \in \mathbb{Z}$

• The impulse response $\mathrm{h}[n]$ of an LTI system is defined as the response to the unit impulse $\mathrm{\delta}[n]$:
  $\mathrm{h}[n] = \mathrm{S}\left(\mathrm{\delta}[n])\right) $
  with
  $\mathrm{\delta}[n] = \begin{cases} 1 & \,\,:\,\, n = 0 \\ 0 & \,\,:\,\, n \ne 0 \\ \end{cases}$
• For example:
  

  $n$

  $n$

  $\mathrm{\delta}[n]$

  $\mathrm{h}[n]$

  System

  $\mathrm{S}$

• Because of the superposition principle, for LTI systems the impulse response $\mathrm{h}[n]$ is sufficient to determine the response of the system to any input
• By using the unit impulse, any input sequence $\mathrm{x}[n]$ can also be written as
  $\mathrm{x}[n] = \sum\limits_k \mathrm{x}[k] \, \mathrm{\delta}[n-k]$
• With $\mathrm{h}[n] = \mathrm{S}\left(\mathrm{\delta}[n])\right) $ and the superposition principle the following holds for the output signal $\mathrm{y}[n]$ of the system:
  $\begin{array}{ll} \mathrm{y}[n] & = \mathrm{S}\left(\sum\limits_k \mathrm{x}[k] \, \mathrm{\delta}[n-k] \right) \\ & = \sum\limits_k \mathrm{x}[k] \, \mathrm{S}\left(\mathrm{\delta}[n-k] \right) \\ & = \sum\limits_k \mathrm{x}[k] \, \mathrm{h}[n-k] \\ & = \mathrm{x}[n] \ast \mathrm{h}[n] \\ \end{array} $
• This means, the response of an LTI system to any input can be determined by convolving the input with the system's impulse response