## Q. 8.1

Laplace transform of a noncausal exponential signal

Find the Laplace transform of $x(t) = e^{−t} u(t) + e^{2t} u(−t)$.

## Verified Solution

The Laplace transform of this sum is the sum of the Laplace transforms of the individual terms $e^{−t}$ u(t) and $e^{2t}$ u(−t). The ROC of the sum is the region in the s plane that is common to the two ROCs. From Table 8.1

$e^{-t} \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s+1}, \quad \sigma>-1$

and

$e^{2t} u(-t)\xleftrightarrow{\mathcal{L}}-\frac{1}{s-2}, σ<2.$

In this case, the region in the s plane that is common to both ROCs is −1 < σ < 2 and

$e^{−t} u(t) + e^{2t} u(-t)\xleftrightarrow{\mathcal{L}}\frac{1}{s+1}-\frac{1}{s-2}, −1 < σ < 2$

(Figure 8.12). This Laplace transform has poles at s = −1 and s = +2 and two zeros at infinity.

 Table 8.1 Some common Laplace-transform pairs $δ (t) \overset{\mathcal{L}}{\longleftrightarrow } 1 , \text{All} σ$ $u (t) \overset{\mathcal{L}}{\longleftrightarrow } 1/s , σ > 0$ $– u (-t) \overset{\mathcal{L}}{\longleftrightarrow } 1/s^2 , σ < 0$ $\text{ramp}(t) = t u (t) \overset{\mathcal{L}}{\longleftrightarrow } 1/s^2 , σ > 0$ $\text{ramp}(-t) = -t u (-t) \overset{\mathcal{L}}{\longleftrightarrow } 1/s^2 , σ < 0$ $e^{-αt} u (t) \overset{\mathcal{L}}{\longleftrightarrow } 1/(s+ α), σ > -α$ $-e^{-αt} u (-t) \overset{\mathcal{L}}{\longleftrightarrow } 1/(s+ α), σ < -α$ $t^n u (t) \overset{\mathcal{L}}{\longleftrightarrow }n!/s^{n+1}, σ > 0$ $-t^n u (-t) \overset{\mathcal{L}}{\longleftrightarrow }n!/s^{n+1}, σ < 0$ $te^{-αt} u (t) \overset{\mathcal{L}}{\longleftrightarrow } 1/(s+ α)^2, σ > -α$ $-te^{-αt} u (-t) \overset{\mathcal{L}}{\longleftrightarrow } 1/(s+ α)^2, σ < -α$ $t^n e^{-αt} u (t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{n!}{(s+ α)^{n+1}}, σ > -α$ $-t^n e^{-αt} u (-t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{n!}{(s+ α)^{n+1}}, σ < -α$ $\sin (ω_0t) u (t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{ω_0}{s^2+ω_0^2}, σ > 0$ $-\sin (ω_0t) u (-t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{ω_0}{s^2+ω_0^2}, σ < 0$ $\cos (ω_0t) u (t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{s}{s^2+ω_0^2}, σ > 0$ $-\cos (ω_0t) u (-t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{s}{s^2+ω_0^2}, σ < 0$ $e^{-αt} \sin (ω_0t) u (t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{ω_0}{(s+α)^2+ω_0^2}, σ > -α$ $-e^{-αt} \sin (ω_0t) u (-t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{ω_0}{(s+α)^2+ω_0^2}, σ < -α$ $e^{-αt} \cos (ω_0t) u (t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{s+α}{(s+α)^2+ω_0^2}, σ > -α$ $-e^{-αt} \cos (ω_0t) u (-t) \overset{\mathcal{L}}{\longleftrightarrow } \frac{s+α}{(s+α)^2+ω_0^2}, σ < -α$ $e^{-α|t|} \overset{\mathcal{L}}{\longleftrightarrow } \frac{1}{s+α} – \frac{1}{s-α}= – \frac{2 α}{s^2-α^2}, -α < σ < α$