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

Response of an LTI system

Find the response y(t) of an LTI system

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

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

(c) With impulse response h(t)=5e4th(t) = 5e^{4t} u(−t) if excited by x(t) = u(t)

(d) With impulse response h(t)=5e4th(t) = 5e^{4t} u(−t) if excited by x(t) = u(−t)

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(a) h(t)=5e−4tu(t)⟷LH(s)=5s+4,σ>−4x(t)=u(t)⟷LX(s)=1/s,σ>0\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}

Therefore

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

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

Y(s)=5/4s−5/4s+4,σ>0y(t)=(5/4)(1−e−4t)u(t)⟷LY(s)=5/4s−5/4s+4,σ>0\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}\right) \mathrm{u}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s+4}, \sigma>0 \end{gathered}

(Figure 8.15)

h(t) = 5e−4t5e^{−4t} u(t), x(t) = u(t)     h(t) = 5e−4t5e^{−4t} u(t), x(t) = u(−t)

 

h(t) = 5e4t5e^{4t} u(-t), x(t) = u(t)     h(t) = 5e4t5e^{4t} u(-t), x(t) = u(−t)

(b)  x(t)=u(−t)⟷LX(s)=−1/s,σ<0\mathrm{x}(t)=\mathrm{u}(-t) \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{X}(s)=-1 / s, \sigma<0

Y(s)=H(s)X(s)=−5s(s+4),−4<σ<0Y(s)=−5/4s+5/4s+4,−4<σ<0y(t)=(5/4)[e−4tu(t)+u(−t)]⟷LY(s)=−5/4s+5/4s+4,−4<σ<0\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)\right] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s+4}, \quad-4<\sigma<0 \\ & \end{aligned}

(Figure 8.15)

(c) h(t)=5e4tu(−t)⟷LH(s)=−5s−4,σ<4 Y(s)=H(s)X(s)=−5s(s−4),0<σ<4Y(s)=5/4s−5/4s−4,0<σ<4y(t)=(5/4)[u(t)+e4tu(−t)]⟷LY(s)=5/4s−5/4s+4,0<σ<4\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)\right] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=\frac{5 / 4}{s}-\frac{5 / 4}{s+4}, \quad 0<\sigma<4 \end{aligned}

(Figure 8.15)

(d) Y(s)=H(s)X(s)=5s(s−4),σ<0Y(s)=−5/4s+5/4s−4,σ<0y(t)=(5/4)[u(−t)−e4tu(−t)]⟷LY(s)=−5/4s+5/4s−4,σ<4 \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)\right] \stackrel{\mathcal{L}}{\longleftrightarrow} \mathrm{Y}(s)=-\frac{5 / 4}{s}+\frac{5 / 4}{s-4}, \quad \sigma<4 \end{aligned}
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