Response of an LTI system

Find the response y(t) of an LTI system

(a) With impulse response $h(t) = 5e^{-4t}$ u(t) if excited by x(t) = u(t)

(b) With impulse response $h(t) = 5e^{-4t}$ u(t) if excited by x(t) = u(−t)

(c) With impulse response $h(t) = 5e^{4t}$ u(−t) if excited by x(t) = u(t)

(d) With impulse response $h(t) = 5e^{4t}$ u(−t) if excited by x(t) = u(−t)

Question

Response of an LTI system

Find the response y(t) of an LTI system

(a) With impulse response [latex]h(t) = 5e^{-4t}[/latex] u(t) if excited by x(t) = u(t)

(b) With impulse response [latex]h(t) = 5e^{-4t}[/latex] u(t) if excited by x(t) = u(−t)

(c) With impulse response [latex]h(t) = 5e^{4t}[/latex] u(−t) if excited by x(t) = u(t)

(d) With impulse response [latex]h(t) = 5e^{4t}[/latex] u(−t) if excited by x(t) = u(−t)

Accepted Answer

[latex]\begin{aligned}(a) \mathrm{h}(t)=5 e^{-4 t} \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{H}(s)=\frac{5}{s+4}, \sigma>-4 \ \mathrm{x}(t)=\mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{X}(s)=1 / s, \sigma>0 \end{aligned}[/latex]

Therefore

[latex]Y(s) = H(s) X(s)=\frac{5}{s(s+4)}, σ > 0[/latex]

Y(s) can be expressed in the partial-fraction form

[latex]\begin{gathered} \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s+4}, \sigma>0 \ \mathrm{y}(t)=(5 / 4)\left(1-e^{-4 t} ight) \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s+4}, \sigma>0 \end{gathered}[/latex]

(Figure 8.15)

h(t) = [latex]5e^{−4t}[/latex] u(t), x(t) = u(t) h(t) = [latex]5e^{−4t}[/latex] u(t), x(t) = u(−t)

h(t) = [latex]5e^{4t}[/latex] u(-t), x(t) = u(t) h(t) = [latex]5e^{4t}[/latex] u(-t), x(t) = u(−t)

(b) [latex]\mathrm{x}(t)=\mathrm{u}(-t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{X}(s)=-1 / s, \sigma<0[/latex]

[latex]\begin{aligned} & \mathrm{Y}(s)=\mathrm{H}(s) \mathrm{X}(s)=-\frac{5}{s(s+4)}, \quad-4<\sigma<0 \ & \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s+4}, \quad-4<\sigma<0 \ & \mathrm{y}(t)=(5 / 4)\left[e^{-4 t} \mathrm{u}(t)+\mathrm{u}(-t) ight] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s+4}, \quad-4<\sigma<0 \ & \end{aligned}[/latex]

(Figure 8.15)

[latex]\begin{aligned} (c) & \mathrm{h}(t)=5 e^{4 t} \mathrm{u}(-t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{H}(s)=-\frac{5}{s-4}, \quad \sigma<4 \ & \mathrm{Y}(s)=\mathrm{H}(s) \mathrm{X}(s)=-\frac{5}{s(s-4)}, \quad 0<\sigma<4 \ & \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s-4}, \quad 0<\sigma<4 \ & \mathrm{y}(t)=(5 / 4)\left[\mathrm{u}(t)+e^{4 t} \mathrm{u}(-t) ight] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s+4}, \quad 0<\sigma<4 \end{aligned}[/latex]

(Figure 8.15)

[latex] \begin{aligned} (d) & \mathrm{Y}(s)=\mathrm{H}(s) \mathrm{X}(s)=\frac{5}{s(s-4)}, \quad \sigma<0 \ & \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s-4}, \quad \sigma<0 \ & \mathrm{y}(t)=(5 / 4)\left[\mathrm{u}(-t)-e^{4 t} \mathrm{u}(-t) ight] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s-4}, \quad \sigma<4 \end{aligned}[/latex]

Question 8.8: Response of an LTI system Find the response y(t) of an LTI s......

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