Problems

From the definition, \[\begin{aligned} F(s) &= \int_0^\infty 4 t\, e^{-3t} e^{-st}\, dt \\ &= 4 \int_0^\infty t\, e^{-(s+3)t}\, dt. \end{aligned}\] Let \(\alpha = s + 3\). We evaluate \(I = \int_0^\infty t\, e^{-\alpha t}\, dt\) using integration by parts. Let \(u = t\) and \(dv = e^{-\alpha t}\, dt\), so \(du = dt\) and \(v = -\frac{1}{\alpha} e^{-\alpha t}\). Then \[\begin{aligned} I &= \left[-\frac{t}{\alpha} e^{-\alpha t}\right]_0^\infty + \frac{1}{\alpha}\int_0^\infty e^{-\alpha t}\, dt. \end{aligned}\] The boundary term vanishes at both limits (at \(t = 0\) because \(t = 0\); at \(t \to \infty\) because the exponential decays faster than \(t\) grows, provided \(\text{Re}(\alpha) > 0\)). The remaining integral evaluates to \(1/\alpha\), so \[\begin{aligned} I = \frac{1}{\alpha^2}. \end{aligned}\] Substituting back \(\alpha = s + 3\), \[\begin{aligned} F(s) = 4 I = \frac{4}{(s+3)^2}. \end{aligned}\]

Region of convergence. We required \(\text{Re}(\alpha) > 0\) for the boundary term to vanish, i.e., \(\text{Re}(s + 3) > 0\). Therefore the ROC is \[\begin{aligned} \text{Re}(s) > -3 \quad (\text{i.e., } \sigma > -3). \end{aligned}\]

(a) By linearity and the table, \[\begin{aligned} F(s) &= \frac{5}{s} - \frac{3}{s+4}. \end{aligned}\]

(b) By linearity and the table, \[\begin{aligned} G(s) &= \frac{3}{s^2 + 9} + 7. \end{aligned}\]

(c) From the table, \(\mathcal{L}\{e^{-at}\sin(\omega t)\, u_s(t)\} = \omega / ((s+a)^2 + \omega^2)\). The time-multiplication property gives \(\mathcal{L}\{t\, f(t)\} = -dF/ds\). With \(a = 1\) and \(\omega = 2\), \[\begin{aligned} \mathcal{L}\{e^{-t}\sin(2t)\, u_s(t)\} &= \frac{2}{(s+1)^2 + 4} \end{aligned}\] and so \[\begin{aligned} H(s) &= -\frac{d}{ds}\left[\frac{2}{(s+1)^2 + 4}\right] \\ &= \frac{4(s+1)}{\left[(s+1)^2 + 4\right]^2}. \end{aligned}\]

(a) By linearity and the table, \[\begin{aligned} F(s) &= \frac{3s}{s^2 + 4} + \frac{4}{s + 1}. \end{aligned}\]

(b) Expand \(g(t) = 2t\, e^{-3t}\, u_s(t) + e^{-3t}\, u_s(t)\). By linearity and the table (\(\mathcal{L}\{t\, e^{-at}\, u_s(t)\} = 1/(s+a)^2\)), \[\begin{aligned} G(s) &= \frac{2}{(s+3)^2} + \frac{1}{s+3} = \frac{s + 5}{(s+3)^2}. \end{aligned}\]

(c) From the table, \(\mathcal{L}\{e^{-at}\cos(\omega t)\, u_s(t)\} = (s+a)/((s+a)^2 + \omega^2)\). With \(a = 2\) and \(\omega = 5\), \[\begin{aligned} H(s) &= \frac{s + 2}{(s+2)^2 + 25}. \end{aligned}\]

The first-derivative property states \[\begin{aligned} \mathcal{L}\left\{\frac{df}{dt}\right\} = s F(s) - f(0). \end{aligned}\] Let \(g(t) = df/dt\), so \(G(s) = sF(s) - f(0)\). Then \[\begin{aligned} \mathcal{L}\left\{\frac{d^2 f}{dt^2}\right\} &= \mathcal{L}\left\{\frac{dg}{dt}\right\} \\ &= s G(s) - g(0) \\ &= s\left[s F(s) - f(0)\right] - \dot{f}(0) \\ &= s^2 F(s) - s f(0) - \dot{f}(0). \end{aligned}\]

Let \(g(t) = \int_0^t f(\tau)\, d\tau\). Then by the fundamental theorem of calculus, \(\dot{g}(t) = f(t)\), and \(g(0) = \int_0^0 f(\tau)\, d\tau = 0\).

Applying the first-derivative property to \(g\), \[\begin{aligned} \mathcal{L}\{\dot{g}(t)\} = s G(s) - g(0) = s G(s). \end{aligned}\] But \(\dot{g}(t) = f(t)\), so \(\mathcal{L}\{\dot{g}(t)\} = F(s)\). Therefore \[\begin{aligned} s G(s) = F(s) \quad \Longrightarrow \quad G(s) = \frac{F(s)}{s}. \end{aligned}\]

Expand \(F(s)\) in partial fractions: \[\begin{aligned} F(s) = \frac{5s + 13}{(s+1)(s+3)} = \frac{A}{s+1} + \frac{B}{s+3}. \end{aligned}\] Solving: \[\begin{aligned} A &= \left.(5s+13)\cdot\frac{1}{s+3}\right|_{s=-1} = \frac{8}{2} = 4, \\ B &= \left.(5s+13)\cdot\frac{1}{s+1}\right|_{s=-3} = \frac{-2}{-2} = 1. \end{aligned}\] Therefore \[\begin{aligned} F(s) = \frac{4}{s+1} + \frac{1}{s+3} \end{aligned}\] and from the Laplace transform table, \[\begin{aligned} f(t) = 4 e^{-t} + e^{-3t}, \quad t \ge 0. \end{aligned}\]

With three distinct real poles, the partial fraction expansion is \[\begin{aligned} \frac{3s + 7}{(s+1)(s+2)(s+5)} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+5}. \end{aligned}\] Using the cover-up method: \[\begin{aligned} A &= \left.\frac{3s+7}{(s+2)(s+5)}\right|_{s=-1} = \frac{4}{(1)(4)} = 1, \\ B &= \left.\frac{3s+7}{(s+1)(s+5)}\right|_{s=-2} = \frac{1}{(-1)(3)} = -\frac{1}{3}, \\ C &= \left.\frac{3s+7}{(s+1)(s+2)}\right|_{s=-5} = \frac{-8}{(-4)(-3)} = -\frac{2}{3}. \end{aligned}\] Therefore \[\begin{aligned} G(s) = \frac{1}{s+1} - \frac{1/3}{s+2} - \frac{2/3}{s+5} \end{aligned}\] and from the Laplace transform table, \[\begin{aligned} g(t) = e^{-t} - \frac{1}{3} e^{-2t} - \frac{2}{3} e^{-5t}, \quad t \ge 0. \end{aligned}\]

First, perform partial fraction expansion. The denominator \(s^2 + 4s + 13\) has complex roots \(s = -2 \pm 3j\), so we keep it as a quadratic term: \[\begin{aligned} Y(s) = \frac{A}{s} + \frac{Bs + C}{s^2 + 4s + 13}. \end{aligned}\] Multiplying through by \(s(s^2+4s+13)\), \[\begin{aligned} 10 = A(s^2+4s+13) + (Bs+C)s. \end{aligned}\] Setting \(s = 0\): \(10 = 13A\), so \(A = 10/13\).

Comparing \(s^2\) coefficients: \(0 = A + B\), so \(B = -10/13\).

Comparing \(s^1\) coefficients: \(0 = 4A + C\), so \(C = -40/13\).

Therefore \[\begin{aligned} Y(s) &= \frac{10/13}{s} + \frac{(-10/13)s + (-40/13)}{s^2 + 4s + 13} \\ &= \frac{10}{13}\cdot\frac{1}{s} - \frac{10}{13}\cdot\frac{s + 4}{(s+2)^2 + 9} \\ &= \frac{10}{13}\cdot\frac{1}{s} - \frac{10}{13}\cdot\frac{(s+2) + 2}{(s+2)^2 + 9} \\ &= \frac{10}{13}\cdot\frac{1}{s} - \frac{10}{13}\cdot\frac{s+2}{(s+2)^2 + 9} - \frac{20}{13}\cdot\frac{1}{(s+2)^2 + 9}. \end{aligned}\] Rewriting the last term to match the table entry for \(e^{-at}\sin(\omega t)\): \[\begin{aligned} Y(s) &= \frac{10}{13}\cdot\frac{1}{s} - \frac{10}{13}\cdot\frac{s+2}{(s+2)^2 + 9} - \frac{20}{39}\cdot\frac{3}{(s+2)^2 + 9}. \end{aligned}\] Taking the inverse Laplace transform from the table, \[\begin{aligned} y(t) = \frac{10}{13} - \frac{10}{13} e^{-2t}\cos(3t) - \frac{20}{39} e^{-2t}\sin(3t), \quad t \ge 0. \end{aligned}\]

The denominator has complex roots (\(s = -1 \pm 3j\)), so we complete the square: \[\begin{aligned} s^2 + 2s + 10 = (s+1)^2 + 9. \end{aligned}\] Now manipulate the numerator to match the table forms \(\frac{s-a}{(s-a)^2 + \omega^2}\) and \(\frac{\omega}{(s-a)^2 + \omega^2}\): \[\begin{aligned} \frac{3s + 1}{(s+1)^2 + 9} &= \frac{3(s+1) - 2}{(s+1)^2 + 9} \\ &= 3\cdot\frac{s+1}{(s+1)^2 + 9} - \frac{2}{3}\cdot\frac{3}{(s+1)^2 + 9}. \end{aligned}\] Taking the inverse Laplace transform from the table, \[\begin{aligned} x(t) = 3 e^{-t}\cos(3t) - \frac{2}{3} e^{-t}\sin(3t), \quad t \ge 0. \end{aligned}\]

Taking the Laplace transform of both sides, using the differentiation property, \[\begin{aligned} s Y(s) - y(0) + 5 Y(s) &= \frac{10}{s}. \end{aligned}\] Substituting \(y(0) = 2\) and solving for \(Y(s)\), \[\begin{aligned} Y(s) &= \frac{10}{s(s+5)} + \frac{2}{s+5}. \end{aligned}\] The first term is the forced response and the second is the free response.

Expanding the forced response in partial fractions: \[\begin{aligned} \frac{10}{s(s+5)} = \frac{2}{s} - \frac{2}{s+5}. \end{aligned}\] So \[\begin{aligned} Y(s) = \frac{2}{s} - \frac{2}{s+5} + \frac{2}{s+5} = \frac{2}{s}. \end{aligned}\] Inverse transforming, \[\begin{aligned} y(t) = 2\, u_s(t). \end{aligned}\] This makes sense: \(y(0) = 2\) is already the steady-state value \(10/5 = 2\), so the system stays put.

Taking the Laplace transform of both sides, using the differentiation property, \[\begin{aligned} s Y(s) - y(0) + 3 Y(s) &= \frac{9}{s}. \end{aligned}\] Substituting \(y(0) = 0\) and solving for \(Y(s)\), \[\begin{aligned} Y(s) &= \frac{9}{s(s+3)}. \end{aligned}\] This is entirely forced response (the free response is zero because \(y(0) = 0\)).

Expanding in partial fractions: \[\begin{aligned} \frac{9}{s(s+3)} = \frac{A}{s} + \frac{B}{s+3}. \end{aligned}\] Cover-up: \(A = 9/3 = 3\), \(B = 9/(-3) = -3\).

Therefore \[\begin{aligned} y(t) = 3 - 3e^{-3t} = 3(1 - e^{-3t}), \quad t \ge 0. \end{aligned}\] This is the classic first-order step response: it starts at \(y(0) = 0\) and exponentially approaches the steady-state value \(9/3 = 3\) with time constant \(\tau = 1/3\).

Taking the Laplace transform of both sides, \[\begin{aligned} s^2 Y(s) - s y(0) - \dot{y}(0) + 4[s Y(s) - y(0)] + 3 Y(s) = \frac{6}{s}. \end{aligned}\] With zero initial conditions, \[\begin{aligned} (s^2 + 4s + 3) Y(s) = \frac{6}{s}. \end{aligned}\] Solving for \(Y(s)\), \[\begin{aligned} Y(s) = \frac{6}{s(s^2 + 4s + 3)} = \frac{6}{s(s+1)(s+3)}. \end{aligned}\] This is entirely forced response (the free response is zero because all initial conditions are zero).

Expanding in partial fractions: \[\begin{aligned} \frac{6}{s(s+1)(s+3)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+3}. \end{aligned}\] Solving: \(A = 6/3 = 2\), \(B = 6/((-1)(2)) = -3\), \(C = 6/((-3)(-2)) = 1\).

Therefore \[\begin{aligned} Y(s) = \frac{2}{s} - \frac{3}{s+1} + \frac{1}{s+3} \end{aligned}\] and \[\begin{aligned} y(t) = 2 - 3e^{-t} + e^{-3t}, \quad t \ge 0. \end{aligned}\]

Taking the Laplace transform with zero initial conditions, \[\begin{aligned} (s^2 + 2s + 10) Y(s) = \frac{20}{s}. \end{aligned}\] Solving for \(Y(s)\), \[\begin{aligned} Y(s) = \frac{20}{s(s^2 + 2s + 10)}. \end{aligned}\] This is entirely forced response. Partial fraction expansion: \[\begin{aligned} Y(s) = \frac{A}{s} + \frac{Bs + C}{s^2 + 2s + 10}. \end{aligned}\] Multiplying through by \(s(s^2 + 2s + 10)\), \[\begin{aligned} 20 = A(s^2 + 2s + 10) + (Bs + C)s. \end{aligned}\] Setting \(s = 0\): \(20 = 10A\), so \(A = 2\).

Comparing \(s^2\) coefficients: \(0 = A + B\), so \(B = -2\).

Comparing \(s^1\) coefficients: \(0 = 2A + C\), so \(C = -4\).

Therefore \[\begin{aligned} Y(s) &= \frac{2}{s} - \frac{2s + 4}{s^2 + 2s + 10}. \end{aligned}\] Completing the square, \(s^2 + 2s + 10 = (s+1)^2 + 9\), so \[\begin{aligned} \frac{2s + 4}{(s+1)^2 + 9} &= \frac{2(s+1) + 2}{(s+1)^2 + 9} = 2\cdot\frac{s+1}{(s+1)^2 + 9} + \frac{2}{3}\cdot\frac{3}{(s+1)^2 + 9}. \end{aligned}\] Taking the inverse Laplace transform, \[\begin{aligned} y(t) = 2 - 2e^{-t}\cos(3t) - \frac{2}{3}e^{-t}\sin(3t), \quad t \ge 0. \end{aligned}\]

Taking the Laplace transform, \[\begin{aligned} s^2 Y(s) - s(1) - (-1) + 2[sY(s) - 1] + 5Y(s) = \frac{10}{s}. \end{aligned}\] Simplifying, \[\begin{aligned} (s^2 + 2s + 5)Y(s) = \frac{10}{s} + s + 1. \end{aligned}\] Therefore \[\begin{aligned} Y(s) = \underbrace{\frac{10}{s(s^2+2s+5)}}_{Y_\text{fo}(s)} + \underbrace{\frac{s + 1}{s^2+2s+5}}_{Y_\text{fr}(s)}. \end{aligned}\]

Forced response. Partial fraction expansion: \[\begin{aligned} \frac{10}{s(s^2+2s+5)} = \frac{2}{s} - \frac{2s + 4}{s^2+2s+5}. \end{aligned}\] Completing the square, \(s^2 + 2s + 5 = (s+1)^2 + 4\), so \[\begin{aligned} \frac{2s+4}{(s+1)^2+4} = \frac{2(s+1) + 2}{(s+1)^2+4} = 2\cdot\frac{s+1}{(s+1)^2+4} + \frac{2}{(s+1)^2+4}. \end{aligned}\] Thus \[\begin{aligned} y_\text{fo}(t) = 2 - 2e^{-t}\cos(2t) - e^{-t}\sin(2t), \quad t \ge 0. \end{aligned}\]

Free response. \[\begin{aligned} \frac{s+1}{(s+1)^2 + 4} = \frac{s+1}{(s+1)^2 + 4} \end{aligned}\] which is already in table form, giving \[\begin{aligned} y_\text{fr}(t) = e^{-t}\cos(2t), \quad t \ge 0. \end{aligned}\]

Total response. \[\begin{aligned} y(t) = 2 - e^{-t}\cos(2t) - e^{-t}\sin(2t), \quad t \ge 0. \end{aligned}\]

Taking the Laplace transform, \[\begin{aligned} s^2 Y(s) - s(1) - 0 + 5[sY(s) - 1] + 6Y(s) = 0. \end{aligned}\] Simplifying, \[\begin{aligned} (s^2 + 5s + 6)Y(s) = s + 5. \end{aligned}\] Factoring the denominator, \((s+2)(s+3)\), gives \[\begin{aligned} Y(s) = \frac{s + 5}{(s+2)(s+3)}. \end{aligned}\] Partial fraction expansion via cover-up: \[\begin{aligned} A &= \left.\frac{s+5}{s+3}\right|_{s=-2} = \frac{3}{1} = 3, \\ B &= \left.\frac{s+5}{s+2}\right|_{s=-3} = \frac{2}{-1} = -2. \end{aligned}\] Therefore \[\begin{aligned} Y(s) = \frac{3}{s+2} - \frac{2}{s+3} \end{aligned}\] and \[\begin{aligned} y(t) = 3e^{-2t} - 2e^{-3t}, \quad t \ge 0. \end{aligned}\] As a sanity check: \(y(0) = 3 - 2 = 1\) \(\checkmark\) and \(\dot{y}(0) = -6 + 6 = 0\) \(\checkmark\).

The denominator factors as \(s^2 + 6s + 9 = (s+3)^2\), a repeated root. The partial fraction expansion is \[\begin{aligned} \frac{s+7}{(s+3)^2} = \frac{A}{s+3} + \frac{B}{(s+3)^2}. \end{aligned}\] Multiplying through by \((s+3)^2\), \[\begin{aligned} s + 7 = A(s+3) + B. \end{aligned}\] Setting \(s = -3\): \(4 = B\). Comparing \(s^1\) coefficients: \(1 = A\).

Therefore \[\begin{aligned} H(s) = \frac{1}{s+3} + \frac{4}{(s+3)^2} \end{aligned}\] and from the Laplace transform table, \[\begin{aligned} h(t) = e^{-3t} + 4t\, e^{-3t}, \quad t \ge 0. \end{aligned}\]

With a third-order repeated root at \(s = -1\), the partial fraction expansion takes the form \[\begin{aligned} \frac{2s+5}{(s+1)^3} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}. \end{aligned}\] Multiplying through by \((s+1)^3\), \[\begin{aligned} 2s + 5 = A(s+1)^2 + B(s+1) + C. \end{aligned}\] Setting \(s = -1\): \(3 = C\). Comparing \(s^2\) coefficients: \(0 = A\). Comparing \(s^1\) coefficients: \(2 = 2A + B\), so \(B = 2\).

Check on \(s^0\): \(5 = A + B + C = 0 + 2 + 3 = 5\). \(\checkmark\)

Therefore \[\begin{aligned} G(s) = \frac{2}{(s+1)^2} + \frac{3}{(s+1)^3}. \end{aligned}\] Using the table entry \(\mathcal{L}^{-1}\{n!/(s+a)^{n+1}\} = t^n e^{-at}\), or equivalently \(\mathcal{L}^{-1}\{1/(s+a)^{n+1}\} = t^n e^{-at}/n!\), \[\begin{aligned} g(t) = 2t\, e^{-t} + \frac{3}{2} t^2 e^{-t}, \quad t \ge 0. \end{aligned}\]

Final value theorem. The poles of \(sY(s)\) are at \(s = -2\) and \(s = -6\), both in the left half-plane, so the theorem applies: \[\begin{aligned} \lim_{t\to\infty} y(t) = \lim_{s\to 0} s Y(s) = \lim_{s\to 0} \frac{12}{(s+2)(s+6)} = \frac{12}{12} = 1. \end{aligned}\]

Verification. Partial fraction expansion: \[\begin{aligned} \frac{12}{s(s+2)(s+6)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+6}. \end{aligned}\] Cover-up method: \(A = 12/12 = 1\), \(B = 12/((-2)(4)) = -3/2\), \(C = 12/((-6)(-4)) = 1/2\).

So \[\begin{aligned} y(t) = 1 - \frac{3}{2} e^{-2t} + \frac{1}{2} e^{-6t}, \quad t \ge 0. \end{aligned}\] As \(t \to \infty\), the exponentials vanish and \(y \to 1\), confirming the final value theorem result.

Initial value theorem. Since the degree of the numerator of \(sY(s)\) does not exceed the degree of the denominator, the theorem applies: \[\begin{aligned} y(0^+) = \lim_{s\to\infty} s Y(s) = \lim_{s\to\infty} \frac{2s^2 + 3s}{s^2 + 5s + 4} = 2. \end{aligned}\]

Verification. Partial fraction expansion: \[\begin{aligned} \frac{2s + 3}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4}. \end{aligned}\] Cover-up method: \(A = \frac{-2+3}{-1+4} = \frac{1}{3}\), \(B = \frac{-8+3}{-4+1} = \frac{5}{3}\).

So \[\begin{aligned} y(t) = \frac{1}{3} e^{-t} + \frac{5}{3} e^{-4t}, \quad t \ge 0. \end{aligned}\] At \(t = 0\): \(y(0) = \frac{1}{3} + \frac{5}{3} = 2\), confirming the initial value theorem result.

Taking the Laplace transform with zero initial conditions, \[\begin{aligned} (s^2 + 6s + 8)Y(s) = (3s + 1)U(s). \end{aligned}\] With \(U(s) = 4/s\), \[\begin{aligned} Y(s) = \frac{4(3s+1)}{s(s^2+6s+8)} = \frac{4(3s+1)}{s(s+2)(s+4)}. \end{aligned}\] Partial fraction expansion: \[\begin{aligned} \frac{4(3s+1)}{s(s+2)(s+4)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+4}. \end{aligned}\] Cover-up: \[\begin{aligned} A &= \frac{4(1)}{(2)(4)} = \frac{1}{2}, \\ B &= \frac{4(-6+1)}{(-2)(2)} = 5, \\ C &= \frac{4(-12+1)}{(-4)(-2)} = -\frac{44}{8} = -\frac{11}{2}. \end{aligned}\] Therefore \[\begin{aligned} y(t) = \frac{1}{2} + 5e^{-2t} - \frac{11}{2}e^{-4t}, \quad t \ge 0. \end{aligned}\]

The region of convergence of a Laplace transform \(F(s)\) is the set of values of the complex variable \(s = \sigma + j\omega\) for which the Laplace integral \(\int_0^\infty f(t) e^{-st}\,dt\) converges (i.e., has a finite value). Typically, the ROC takes the form \(\text{Re}(s) > \sigma_0\) for some \(\sigma_0\), meaning it is a half-plane in the \(s\)-plane.

The ROC matters because the expression for \(F(s)\) is only valid within the ROC. Two different time-domain functions can have the same algebraic \(F(s)\) expression but different ROCs, so the ROC is needed to uniquely identify \(f(t)\).

The connection to the Fourier transform is direct: the Fourier transform of \(f(t)\) equals \(F(s)\) evaluated on the imaginary axis, i.e., \(F(j\omega)\), provided the imaginary axis lies within the ROC (i.e., \(\sigma_0 < 0\) for one-sided signals, or equivalently \(\sigma_0 \le 0\)). If the ROC does not include \(\sigma = 0\), the Fourier transform of the original signal does not exist in the ordinary sense.