Graphical auto-correlation, cross-correlation and convolution

Signals (from the figure):

• $x_1(t) = 2$ for $1 \le t \le 2$, zero elsewhere (rectangular pulse).
• $x_2(t) = 2(2 - t)$ for $1 \le t \le 2$, zero elsewhere (jumps to 2 at $t=1$, falls linearly to 0 at $t=2$).

Definitions used (real signals): $R_{11}(\tau) = \int x_1(t)x_1(t+\tau)\,dt$, $R_{12}(\tau) = \int x_1(t)x_2(t+\tau)\,dt$, $y(t) = \int x_1(\lambda)x_2(t-\lambda)\,d\lambda$.

Auto-correlation

Graphically, keep $x(t)$ fixed, slide a copy by $\tau$ (left for $\tau > 0$), multiply and find the area of the overlap. Overlap exists only for $|\tau| < 1$.

Rectangle $x_1$: for $0 \le \tau \le 1$ the overlap is $1 \le t \le 2-\tau$, of width $1-\tau$, height $2\times2 = 4$:

$$R_{11}(\tau) = \int_{1}^{2-\tau} 4\,dt = 4(1-|\tau|), \quad |\tau| \le 1$$

It is a triangle of peak $R_{11}(0) = 4$ = energy of $x_1$, base $-1$ to $1$.

Triangle $x_2$: for $0 \le \tau \le 1$,

$$\begin{aligned} R_{22}(\tau) &= \int_{1}^{2-\tau} 2(2-t)\cdot 2(2-t-\tau)\,dt \\ &= \frac{4}{3} - 2|\tau| + \frac{2}{3}|\tau|^3, \quad |\tau| \le 1 \end{aligned}$$

Peak $R_{22}(0) = 4/3$ = energy of $x_2$; it is even, smooth and reaches 0 at $\tau = \pm1$ (e.g. $R_{22}(0.5) = 5/12$).

Cross-correlation $R_{12}(\tau)$

Slide $x_2$ relative to $x_1$; overlap again only for $|\tau| < 1$.

• $0 \le \tau \le 1$ ($x_2$ moved left, its low tail overlaps): $R_{12}(\tau) = \int_{1}^{2-\tau} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau)^2$
• $-1 \le \tau \le 0$ (its tall front edge overlaps): $R_{12}(\tau) = \int_{1-\tau}^{2} 2\cdot 2(2-t-\tau)\,dt = 2(1-\tau^2)$

Values: $R_{12}(-1) = 0$, $R_{12}(-0.5) = 1.5$, $R_{12}(0) = 2$, $R_{12}(0.5) = 0.5$, $R_{12}(1) = 0$. The curve is not symmetric (cross-correlation is not even); its peak is at $\tau = 0$ with value 2.

Convolution $y(t) = x_1(t) * x_2(t)$

Fold $x_2(\lambda)$ to $x_2(t-\lambda)$ (now non-zero for $t-2 \le \lambda \le t-1$, rising from 0 to 2) and slide it over $x_1(\lambda)$ on $[1,2]$. Output exists from $1+1 = 2$ to $2+2 = 4$.

• $t < 2$: no overlap, $y = 0$.
• $2 \le t \le 3$ (partial overlap $1 \le \lambda \le t-1$):

$$y(t) = \int_{1}^{t-1} 2\cdot 2(2-t+\lambda)\,d\lambda = -2t^2 + 12t - 16 = 2(t-2)(4-t)$$

• $3 \le t \le 4$ (overlap $t-2 \le \lambda \le 2$):

$$y(t) = \int_{t-2}^{2} 2\cdot 2(2-t+\lambda)\,d\lambda = 2(t-4)^2$$

• $t > 4$: $y = 0$.

Check: $y(2) = 0$, $y(3) = 2$ from both pieces (continuous), $y(4) = 0$; area $= 2 \times 1 = 2$ = (area of $x_1$) $\times$ (area of $x_2$) = $2 \times 1$.

 y(t)
  2 |          *
    |       *     *
  1 |     *          *
    |   *               *
  0 +--*------------------*----> t
       2         3        4

Answer: $R_{11} = 4(1-|\tau|)$; $R_{22} = \frac43 - 2|\tau| + \frac23|\tau|^3$; $R_{12} = 2(1-\tau^2)$ for $-1\le\tau\le0$ and $2(1-\tau)^2$ for $0\le\tau\le1$; $y(t) = 2(t-2)(4-t)$ for $2\le t\le3$ and $2(t-4)^2$ for $3\le t\le4$ (all zero elsewhere).