Question 1: https://blog.yexca.net/en/archives/198
Question 2: https://blog.yexca.net/en/archives/201
Question 3: https://blog.yexca.net/en/archives/200
Question 4: This article
Question 5: https://blog.yexca.net/en/archives/203

Introduction

This entire question focuses on the Laplace transform, with the final part being an application. Without knowledge of Laplace transforms, it’s difficult to solve quickly. While you might manage the first two major sections and the first three parts of the third section if you work fast (which is actually quite a lot), the last two questions are impossible if you don’t understand the underlying principles. (In the final part, I wrote about how to solve it without knowing the specific scenario, but it’s hard to say if one could react like that under exam pressure.)

The copyright of the questions belongs to the Institute of Science Tokyo. They are cited here for reference only, with no commercial intent.

Background

The Laplace transform $F(s)=\mathcal{L}[f(t)]$ of a twice-differentiable real function $f(t)$ with a real variable $t(\ge0)$ is defined as follows:

Here, $s$ is a complex variable with a positive real part. Answer the following questions. You may use the following Laplace transform identities:

where $\alpha$ is a real constant and $f'(t)=\frac{\mathrm{d}}{\mathrm{d}t}f(t)$.

This section defines the Laplace transform and provides three key properties.

1

Show that the following equations (a–c) hold. Assume $\alpha$ is a real constant and $\beta$ is a non-zero real constant.

a) $\mathcal{L}[\cos(\beta t)]=\dfrac{s}{s^2+\beta^2}$

b) $\mathcal{L}[e^{-\alpha t}\cos(\beta t)] = \dfrac{s+\alpha}{(s+a)^2+\beta^2}$

c) $\mathcal{L}[e^{-\alpha t}\sin(\beta t)]=\dfrac{\beta}{(s+a)^2+\beta^2}$

Solution

a

Substitute into the definition:

Using the corollary of Euler’s formula:

Which is:

The proof is straightforward, just substitute back:

$$ > \begin{align} > \cos\theta > &= \frac{e^{i\theta}+e^{-i\theta}}{2} \\ > &= \frac{(\cos\theta + i\sin\theta)+(\cos(-\theta) + i\sin(-\theta))}{2} \\ > &= \frac{(\cos\theta + i\sin\theta)+(\cos\theta - i\sin\theta)}{2} \\ > &= \frac{2\cos\theta }{2} \\ > &= \cos\theta > \end{align} > $$

Substitute into the original equation:

Integrating yields:

Integration process:

$$ > \begin{align} > \int_0^\infty e^{-at}\mathrm{d}t > &= \frac{1}{-a}e^{-at}\mid_0^\infty \\ > &= 0-\frac{1}{-a} \\ > &= \frac{1}{a} > \end{align} > $$

Simplify by finding a common denominator:

Q.E.D.

b

From $(4.2)$:

And from part (a):

Therefore:

Q.E.D.

c

Since:

Taking the Laplace transform of both sides:

From $(4.3)$:

So:

Meaning:

From part (a):

Substituting this in:

Solving for $F(s)$:

Thus:

While reviewing, I realized a simpler method I thought of initially works too. The method above looks cooler, so I kept it. Here is the shorter version:

$$ > \begin{align} > \mathcal{L}[\frac{\mathrm{d}}{\mathrm{d}t}\cos(\beta t)] > &= \mathcal{L}[-\beta\sin(\beta t)]\\ > &= -\beta \mathcal{L}[\sin(\beta t)]\\ > &= sF(s) - \cos(0) \\ > &= \dfrac{s^2}{s^2+\beta^2} - 1 \\ > &= \dfrac{-\beta^2}{s^2+\beta^2} > \end{align} > $$

Therefore:

$$ > \mathcal{L}[\sin(\beta t)] = \dfrac{\beta}{s^2+\beta^2} > $$

This means there are essentially three ways to solve this (including the direct integration used in part a). It felt great to solve it.

From $(4.2)$:

Therefore:

2

For the following differential equation, answer parts a–d:

Where $\eta$ is a real constant and $f(t)$ is a twice-differentiable real function of $t(\ge 0)$. The initial conditions are $f(0)=1, f'(0)=1$. Let $f'(t)=\frac{\mathrm{d}}{\mathrm{d}t}f(t)$ and $F(s)=\mathcal{L}[f(t)]$.

a) Take the Laplace transform of both sides of equation $(4.5)$ and express $F(s)$ as a function of $s$.

b) When $\eta=0$, solve equation $(4.5)$ using the Laplace transform and express $f(t)$ as a function of $t$.

c) When $\eta=2$, solve equation $(4.5)$ using the Laplace transform and express $f(t)$ as a function of $t$.

d) When $\eta=2$, find the value of $\lim_{t \to \infty}f(t)$ and sketch the graph of $f(t)$.

Solution

a

Using $(4.3)$ and $(4.4)$, take the Laplace transform of both sides:

Rearranging:

So:

b

Substitute $\eta=0$:

Let:

Matching coefficients:

Solving gives $A=1, B=0, C=1$. Substituting back:

Applying the inverse Laplace transform:

c

Substitute $\eta=2$:

Let:

Matching coefficients:

Solving gives $A=1, B=0, C=1$. Thus:

Applying the inverse Laplace transform:

d

From part (c), when $\eta=2$:

Therefore:

Since:

The graph of $f(t)$ starts at 1, exhibits damped oscillation, and eventually stabilizes at 1.

It oscillates while converging to 1 as $t \to \infty$ (a typical damped oscillation).

The graph looks like this:

Code for the plot (using a headless Linux environment, requiring image saving):

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import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt

mpl.use('Agg') # For headless environments

# Define function f(t) = 1 + e^(-t) * sin(t)
t = np.linspace(0, 10, 1000)  # Time range [0, 10]
f_t = 1 + np.exp(-t) * np.sin(t)

# Plotting
plt.figure(figsize=(8, 5))
plt.plot(t, f_t, label=r"$f(t) = 1 + e^{-t}\sin(t)$", color='b')
plt.axhline(y=1, color='r', linestyle='--', label=r"$f(t) \to 1$ (limit value)")

# Formatting
plt.title(r"Function $f(t) = 1 + e^{-t}\sin(t)$", fontsize=14)
plt.xlabel("$t$", fontsize=12)
plt.ylabel("$f(t)$", fontsize=12)
plt.legend(fontsize=10)
plt.grid(alpha=0.3)
plt.tight_layout()

# Display
plt.show()

# Save image
plt.savefig("./1.png")

3

The differential equation for the single-degree-of-freedom spring-mass-damper system shown in Figure 4.1 can be expressed as:

Where position $x(t)$ and external force $p(t)$ are twice-differentiable real functions of time $t(\ge 0)$. The origin for $x(t)$ is the natural length of the spring, and the positive direction for $x(t)$ and $p(t)$ is the direction in which the spring extends. Mass $m$, spring constant $k$, and damping coefficient $\mu$ are positive constants. Below, we consider the rewritten version of equation $(4.6)$:

Defined as $\omega_0=\sqrt{k/m}, \gamma=\mu/(2\sqrt{mk}), q(t)=p(t)/m$. Let $X(s)=\mathcal{L}[x(t)]$ and $Q(s)=\mathcal{L}[q(t)]$. The initial conditions are $x(0)=0, x'(0)=0$. Answer parts a–e:

• Figure 4.1: Single-degree-of-freedom spring-mass-damper system

a) Take the Laplace transform of both sides of equation $(4.7)$, calculate the transfer function $H(s)=X(s)/Q(s)$, and express $H(s)$ as a function of $s$ using $\gamma$ and $\omega_0$.

b) In the transfer function $H(s)$, replace $s$ with $i\omega$ to calculate $Y(\omega)=20\log_{10}\mid H(i\omega)\mid$. Express $Y(\omega)$ as a real function of $\omega$, where $\omega$ is a positive real number and $i$ is the imaginary unit.

c) Let $\hat{\omega}$ be the value of $\omega$ that maximizes $Y(\omega)$. Express $\hat{\omega}$ using $\gamma$ and $\omega_0$.

d) What is the phenomenon where $Y(\omega)$ reaches its maximum at $\omega=\hat{\omega}$ called?

e) What is the range of $\gamma$ for which the phenomenon in (d) occurs?

Solution

a

Take the Laplace transform of both sides:

Using $(4.3)$, $(4.4)$ and the given initial conditions:

Divide by $Q(s)$:

So:

b

Substitute $i\omega$ for $s$:

The magnitude is:

Thus:

c

$Y(\omega)$ is maximized when $(\omega_0^2-\omega^2)^2+(2\gamma\omega_0\omega)^2$ is minimized. Taking the derivative with respect to $\omega$:

Setting the derivative to zero:

Or:

Which gives:

Since $\omega$ is a positive real number:

d

Resonance.

Resonance occurs when the frequency of an external force is near the natural frequency of the system, resulting in maximum response amplitude.

e

For $\hat{\omega}$ to be a positive real number:

Which means:

Since $\mu, m, k > 0$, we have $\gamma > 0$, so:

Reference

Euler’s Formula

Laplace Transform Symbols in LaTeX

https://blog.csdn.net/ydyang1126/article/details/77247654

Common Laplace and Inverse Laplace Transform Identities