Unit D6: The Laplace Transform

For $f(t) = 1$ we integrate directly, valid for $s > 0$:

For $f(t) = e^{at}$ the exponentials combine, and convergence requires $s > a$:

对 $f(t)=1$，直接积分（$s>0$）：

对 $f(t)=e^{at}$，指数合并，收敛条件 $s>a$：

We compute $\mathcal{L}\{t\}$ directly from the definition rather than quoting the table, valid for $s > 0$. Integrate by parts with $u = t$ and $dv = e^{-st}\,dt$, so $du = dt$ and $v = -\tfrac{1}{s}e^{-st}$:

For $s > 0$ the boundary term vanishes at both ends: at $t = 0$ the factor $t$ is zero, and as $t \to \infty$ the exponential $e^{-st}$ dominates the linear growth, so $t\,e^{-st} \to 0$. The remaining integral is $\mathcal{L}\{1\} = 1/s$, giving

This matches $n!/s^{n+1}$ with $n = 1$, and the same integration-by-parts step repeated $n$ times produces the full power rule $\mathcal{L}\{t^{n}\} = n!/s^{n+1}$.

不查表，直接由定义计算。令 $u=t$，$dv=e^{-st}dt$ 分部积分（$s>0$）：

与 $n!/s^{n+1}$（$n=1$）吻合；重复分部积分即得 $\mathcal{L}\{t^{n}\}=n!/s^{n+1}$。

The existence theorem gives a sufficient condition, not a necessary one, so it is worth seeing where it fails. Consider $f(t) = e^{t^{2}}$. For any fixed constant $a$, the ratio

since $t^{2} - at \to \infty$. No choice of $M$ and $a$ can satisfy $|f(t)| \le M e^{at}$ for all large $t$, so $f$ is not of exponential order. Indeed the integral $\int_{0}^{\infty} e^{-st} e^{t^{2}}\, dt$ diverges for every $s$ because the integrand $e^{t^{2}-st} \to \infty$. By contrast $f(t) = t^{100}$ is of exponential order, because any polynomial is eventually dominated by $e^{t}$, so its transform exists for $s > 0$.

存在性定理给出充分条件。考察 $f(t)=e^{t^{2}}$：对任意 $a$，$e^{t^{2}-at}\to\infty$，故 $f$ 非指数阶，变换不存在。反之，$t^{100}$ 是指数阶，其变换在 $s>0$ 存在。

Suppose $|f(t)| \le M e^{at}$. For $s > a$ we bound the transform by pulling the absolute value inside the integral:

As $s \to \infty$ the right side tends to $0$, so every transform of an exponential-order function decays to $0$. A consequence: a rational function with numerator degree at least the denominator degree (for instance a constant or a polynomial in $s$) cannot be the Laplace transform of such a function.

由 $\mathcal{L}\{\sin 5t\}=\dfrac{5}{s^{2}+25}$，第一平移定理（$a=2$）将 $s$ 换为 $s-2$：

Start from $\mathcal{L}\{\sin 5t\} = \dfrac{5}{s^{2}+25}$. Apply the first shift theorem with $a = 2$, replacing $s$ by $s-2$:

由 $\mathcal{L}\{\sin 5t\}=\dfrac{5}{s^{2}+25}$，第一平移定理（$a=2$）将 $s$ 换为 $s-2$：

Use multiplication by $t$ with $F(s) = \dfrac{s}{s^{2}+9}$ and $n = 1$:

Differentiating the quotient gives

$t$ 乘法法则：$\mathcal{L}\{t\cos 3t\}=-\dfrac{d}{ds}\dfrac{s}{s^{2}+9}$。对商式求导：

Linearity lets us split the bracket, then the first shift theorem with $a = -1$ acts on each piece by sending $s \mapsto s + 1$. Start from the unshifted transforms $\mathcal{L}\{t^{2}\} = 2/s^{3}$ and $\mathcal{L}\{\cos 2t\} = s/(s^{2}+4)$. Replacing $s$ by $s+1$ in each:

Adding the two results gives the full transform:

Notice the order of operations: apply linearity first to break the sum apart, then apply the shift to each transform separately. The shift acts on the whole transform of each summand, including the numerator, which is why the cosine numerator becomes $s+1$ rather than staying $s$.

线性性拆括号，再用第一平移定理（$a=-1$，$s\to s+1$）：

Rather than two separate integration-by-parts marathons, compute both at once through the complex exponential. For $s > 0$,

valid because $|e^{-(s-ib)t}| = e^{-st} \to 0$. Rationalize by multiplying top and bottom by the conjugate $s + ib$:

Since $e^{ibt} = \cos bt + i\sin bt$ and the transform is linear, the real part is $\mathcal{L}\{\cos bt\}$ and the imaginary part is $\mathcal{L}\{\sin bt\}$. Reading off the two parts gives both table entries simultaneously:

拆分为余弦和正弦两项（$b=2$）：

Split into a cosine piece and a sine piece:

Each term is a table entry with $b = 2$:

拆分为余弦和正弦两项（$b=2$）：

Write $\dfrac{s+7}{(s-1)(s+2)} = \dfrac{A}{s-1} + \dfrac{B}{s+2}$. Clearing denominators gives $s+7 = A(s+2) + B(s-1)$. Set $s = 1$: $8 = 3A$, so $A = \tfrac{8}{3}$. Set $s = -2$: $5 = -3B$, so $B = -\tfrac{5}{3}$. Therefore

令 $A(s+2)+B(s-1)=s+7$。$s=1$：$A=8/3$；$s=-2$：$B=-5/3$。故

A repeated factor $(s+1)^{2}$ needs two terms in the decomposition, one for each power up to the multiplicity:

Clear denominators: $s = A(s+1)(s-2) + B(s-2) + C(s+1)^{2}$. Set $s = -1$: $-1 = B(-3)$, so $B = \tfrac{1}{3}$. Set $s = 2$: $2 = C(9)$, so $C = \tfrac{2}{9}$. To find $A$, compare $s^{2}$ coefficients: $0 = A + C$, so $A = -\tfrac{2}{9}$. Now invert term by term, using $\mathcal{L}^{-1}\{1/(s-a)\} = e^{at}$ and $\mathcal{L}^{-1}\{1/(s-a)^{2}\} = t e^{at}$:

重因子 $(s+1)^{2}$ 需两项：$A/(s+1)+B/(s+1)^{2}+C/(s-2)$。求解得 $B=-1/3$，$C=2/9$，$A=-2/9$，故

When a denominator is an irreducible quadratic such as $s^{2} + 2s + 5$, do not force real partial fractions. Complete the square to expose a shifted sine or cosine. Take

Match the shift $a = -1$ and frequency $b = 2$. Rewrite the numerator in terms of $(s+1)$ so the cosine and sine patterns appear cleanly:

The first piece inverts to $e^{-t}\cos 2t$ and the second, with the constant $2 = b$, to $e^{-t}\sin 2t$. Hence

The general moral: an irreducible quadratic always pairs with the shifted sine and cosine family, and completing the square is the bridge from the algebra to the table.

代入 $y(0)=1$，$y'(0)=0$ 取变换，整理得代数方程，部分分式后取逆变换得 $y(t)$。

Transform every term, substituting the initial data:

Solve for $Y$ and invert:

代入初始条件取变换：$(s^{2}+4)Y=2s$，故

Transform every term. Use $\mathcal{L}\{y''\} = s^{2}Y - s y(0) - y'(0)$, $\mathcal{L}\{y'\} = sY - y(0)$, and $\mathcal{L}\{e^{3t}\} = 1/(s-3)$:

Collect the $Y$ terms and the constants. The characteristic polynomial factors as $s^{2}-3s+2 = (s-1)(s-2)$:

Hence $Y(s) = \dfrac{s^{2}-6s+10}{(s-1)(s-2)(s-3)}$. Decompose: $\dfrac{A}{s-1}+\dfrac{B}{s-2}+\dfrac{C}{s-3}$. Cover-up at $s=1$: $\dfrac{1-6+10}{(-1)(-2)} = \dfrac{5}{2}=A$. At $s=2$: $\dfrac{4-12+10}{(1)(-1)} = \dfrac{2}{-1}=-2=B$. At $s=3$: $\dfrac{9-18+10}{(2)(1)} = \dfrac{1}{2}=C$. Inverting,

The terms $e^{t}$ and $e^{2t}$ are the homogeneous part fixed by the initial data, and $\tfrac{1}{2}e^{3t}$ is the particular response to the forcing, all produced in one pass.

代入 $y(0)=1$，$y'(0)=0$ 取变换，整理得代数方程，部分分式后取逆变换得 $y(t)$。

Transform: $sY - y(0) + 2Y = 6/s$, that is $(s+2)Y = 1 + 6/s = (s+6)/s$. So

Cover-up at $s = 0$: $A = 6/2 = 3$. At $s = -2$: $B = 4/(-2) = -2$. Therefore

As a check, $y(0) = 3 - 2 = 1$ and as $t \to \infty$ the solution settles to the steady state $y = 3$, the value where $y' = 0$ forces $2y = 6$.

代入 $y(0)=1$：$(s+2)Y=(s+6)/s$。部分分式：$A=3$，$B=-2$，故 $y(t)=3-2e^{-2t}$。

Apply the definition and integrate by parts with $u = e^{-st}$, $dv = y'(t)\,dt$:

If $y$ is of exponential order then $e^{-st}y(t) \to 0$ as $t \to \infty$ for large $s$, so the boundary term is $0 - y(0) = -y(0)$. The remaining integral is $sY(s)$, giving $\mathcal{L}\{y'\} = sY(s) - y(0)$. Iterating the rule on $y''=(y')'$ and substituting the first-derivative formula produces $\mathcal{L}\{y''\} = s^{2}Y - s\,y(0) - y'(0)$, and induction gives $\mathcal{L}\{y^{(n)}\} = s^{n}Y - s^{n-1}y(0) - \cdots - y^{(n-1)}(0)$.

$f(t)=t^{2}$，$\mathcal{L}\{t^{2}\}=2/s^{3}$。第二平移定理（$c=2$）：

Here $f(t-2) = (t-2)^{2}$, so $f(t) = t^{2}$ with $\mathcal{L}\{t^{2}\} = 2/s^{3}$. Apply the second shift theorem with $c = 2$:

$f(t)=t^{2}$，$\mathcal{L}\{t^{2}\}=2/s^{3}$。第二平移定理（$c=2$）：

The factor $e^{-3s}$ means a delay of $c = 3$. The remaining $F(s) = 1/(s^{2}+1)$ inverts to $f(t) = \sin t$. By the inverse second shift theorem we replace $t$ by $t - 3$ and gate with the step:

$F(s)=1/(s^{2}+1)$ 逆变换为 $\sin t$；$e^{-3s}$ 表示延迟 $c=3$：

The second shift theorem needs the function written as $f(t-c)$, but here the factor is $t^{2}$, not $(t-1)^{2}$. Rewrite $t = (t-1)+1$ and expand so every appearance of $t$ becomes a power of $t-1$:

Now the gated function is $u_{1}(t)\big[(t-1)^{2} + 2(t-1) + 1\big]$, a sum of three terms each in clean $t-1$ form. With $c = 1$ and the table entries $\mathcal{L}\{t^{2}\}=2/s^{3}$, $\mathcal{L}\{t\}=1/s^{2}$, $\mathcal{L}\{1\}=1/s$, the second shift theorem multiplies each by $e^{-s}$:

令 $t=(t-1)+1$ 展开：$t^{2}=(t-1)^{2}+2(t-1)+1$，分别对每项用第二平移定理（$c=1$）：

Rewriting by hand, as in Example 5.3, is the algebra behind a compact rule. Start from the definition, where the step $u_c$ cuts the lower limit to $c$:

Substitute $\tau = t - c$, so $t = \tau + c$ and $dt = d\tau$, and the limits become $\tau \in [0,\infty)$:

This packages the rewrite automatically: to transform $u_c(t) g(t)$, shift the argument forward to $g(t+c)$, transform, and multiply by $e^{-cs}$. Applying it to $g(t)=t^2$, $c=1$ reproduces Example 5.3 since $g(t+1)=(t+1)^2 = t^2+2t+1$.

$(s^{2}+1)Y=e^{-\pi s}$，故 $Y=e^{-\pi s}/(s^{2}+1)$。t 平移定理：

Transform both sides, using $\mathcal{L}\{\delta(t-\pi)\} = e^{-\pi s}$:

Since $1/(s^{2}+1)$ inverts to $\sin t$, the second shift theorem with $c = \pi$ gives

The system is at rest until $t = \pi$, then the impulse sets it oscillating.

$(s^{2}+1)Y=e^{-\pi s}$，故 $Y=e^{-\pi s}/(s^{2}+1)$。t 平移定理：

Define the boxcar $d_{\varepsilon}(t-c)$ equal to $1/\varepsilon$ on $[c, c+\varepsilon]$ and $0$ elsewhere, so its total area is $1$. Its transform is

As $\varepsilon \to 0^{+}$, the factor $(1 - e^{-\varepsilon s})/(\varepsilon s) \to 1$ (its first order Taylor expansion is $1 - \tfrac{\varepsilon s}{2} + \cdots$). Hence $\mathcal{L}\{d_{\varepsilon}(t-c)\} \to e^{-cs}$, which we adopt as the definition $\mathcal{L}\{\delta(t-c)\} = e^{-cs}$.

$(s+1)^{2}Y=3e^{-s}$，$1/(s+1)^{2}$ 逆变换为 $te^{-t}$，t 平移得

With $y(0)=y'(0)=0$ and $\mathcal{L}\{\delta(t-1)\} = e^{-s}$, transform both sides:

The factor $1/(s+1)^{2}$ inverts to $t e^{-t}$ (repeated factor at $a=-1$). The $e^{-s}$ is a delay of $c=1$, so by the inverse second shift theorem we replace $t$ by $t-1$ and gate with the step:

Before $t=1$ the system sits at rest. The unit impulse at $t=1$ injects momentum, and the critically damped response $(t-1)e^{-(t-1)}$ rises then decays back to zero. The impulse produces a jump in $y'$, not in $y$ itself, which is why $y$ stays continuous through $t=1$.

$(s+1)^{2}Y=3e^{-s}$，$1/(s+1)^{2}$ 逆变换为 $te^{-t}$，t 平移得

Write the product as $F(s)G(s)$ with $F = 1/s^{2}$ and $G = 1/(s^{2}+1)$, so $f(t) = t$ and $g(t) = \sin t$. Then

Integrating by parts gives

$f(t)=t$，$g(t)=\sin t$，卷积：

Transforming with $y(0) = y'(0) = 0$ gives $(s^{2}+1)Y = G(s)$, so

The factor $1/(s^{2}+1)$ is the transform of $\sin t$, the impulse response. By the convolution theorem,

a single formula valid for any forcing $g$.

$Y=G(s)/(s^{2}+1)$，冲激响应为 $\sin t$，卷积定理给出

Take $f(t) = e^{t}$ and $g(t) = 1$. From the definition,

Check against the convolution theorem. Here $F(s) = 1/(s-1)$ and $G(s) = 1/s$, so the product is

using partial fractions. Inverting gives $e^{t} - 1$, matching the direct integral. The two routes agreeing is the convolution theorem in action: a product of transforms is the transform of a convolution.

直接积分：$(e^{t}*1)(t)=\int_{0}^{t}e^{\tau}d\tau=e^{t}-1$。变换验证：$\mathcal{L}\{1/(s(s-1))\}=\mathcal{L}\{e^{t}-1\}$，两者一致。

Write out the transform of the convolution as a double integral over the triangular region $0 \le \tau \le t < \infty$:

Both functions are of exponential order, so the double integral converges absolutely for large $s$ and Fubini's theorem permits swapping the order of integration. Fix $\tau$ and let $t$ range over $[\tau, \infty)$. Substitute $\sigma = t - \tau$, so $t = \sigma + \tau$, $dt = d\sigma$, and $\sigma \in [0, \infty)$:

The integrals separate because $e^{-s(\sigma+\tau)} = e^{-s\tau} e^{-s\sigma}$, and the two factors are exactly $F(s)$ and $G(s)$. Hence $\mathcal{L}\{f*g\} = F(s) G(s)$. The same change of variables, applied to the inner integral alone, also proves commutativity $f*g = g*f$.

将卷积的变换展开为三角形区域 $0le aule t