Question 8.1: Laplace transform of a noncausal exponential signal Find the......
Laplace transform of a noncausal exponential signal
Find the Laplace transform of x(t) = e^{−t} u(t) + e^{2t} u(−t).
Step-by-Step
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>-1and
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 | |
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