Properties of Laplace Transformations

The Laplace transform is a powerful mathematical tool used extensively in engineering and physics, particularly in solving differential equations. It transforms a function of time, $f(t)$, into a function of a complex variable, $s$. Understanding its properties is crucial for its effective application. We will explore these properties in detail, providing examples to illustrate their use.

1. Linearity Property

The Laplace transform is a linear operator. This means that if we have two functions, $f_1(t)$ and $f_2(t)$, and their respective Laplace transforms are $F_1(s)$ and $F_2(s)$, then the Laplace transform of a linear combination of these functions is the same linear combination of their transforms.

Mathematically, if $L\{f_1(t)\} = F_1(s)$ and $L\{f_2(t)\} = F_2(s)$, then for any constants $a$ and $b$:

$L\{a f_1(t) + b f_2(t)\} = a L\{f_1(t)\} + b L\{f_2(t)\} = a F_1(s) + b F_2(s)$

This property is fundamental because it allows us to break down complex functions into simpler ones whose Laplace transforms we already know or can easily compute.

Example:

Let's find the Laplace transform of $f(t) = 3e^{-2t} + 5\sin(4t)$.

We know the standard Laplace transforms:

• $L\{e^{at}\} = \frac{1}{s-a}$
• $L\{\sin(bt)\} = \frac{b}{s^2+b^2}$

Using the linearity property:

$L\{3e^{-2t} + 5\sin(4t)\} = 3 L\{e^{-2t}\} + 5 L\{\sin(4t)\}$

Substituting $a = -2$ and $b = 4$:

$L\{e^{-2t}\} = \frac{1}{s - (-2)} = \frac{1}{s+2}$

$L\{\sin(4t)\} = \frac{4}{s^2+4^2} = \frac{4}{s^2+16}$

Therefore,

$L\{3e^{-2t} + 5\sin(4t)\} = 3 \left(\frac{1}{s+2}\right) + 5 \left(\frac{4}{s^2+16}\right) = \frac{3}{s+2} + \frac{20}{s^2+16}$

2. First Shifting Property (s-domain shift)

This property states that if the Laplace transform of a function $f(t)$ is $F(s)$, then the Laplace transform of $e^{at}f(t)$ is $F(s-a)$. This is also known as the frequency shifting property.

Mathematically:

$L\{e^{at} f(t)\} = F(s-a)$

This property is extremely useful for finding the Laplace transform of functions that are exponentially multiplied. It essentially shifts the function's transform in the s-domain.

Example:

Let's find the Laplace transform of $f(t) = e^{-3t} \cos(2t)$.

We know the Laplace transform of $\cos(bt)$ is $\frac{s}{s^2+b^2}$. So, for $\cos(2t)$, $L\{\cos(2t)\} = \frac{s}{s^2+2^2} = \frac{s}{s^2+4}$. Here, $F(s) = \frac{s}{s^2+4}$.

Now, we need to find $L\{e^{-3t} \cos(2t)\}$. Using the first shifting property with $a = -3$:

$L\{e^{-3t} \cos(2t)\} = F(s - (-3)) = F(s+3)$

Substitute $(s+3)$ for $s$ in $F(s) = \frac{s}{s^2+4}$:

$L\{e^{-3t} \cos(2t)\} = \frac{s+3}{(s+3)^2+4} = \frac{s+3}{s^2+6s+9+4} = \frac{s+3}{s^2+6s+13}$

3. Second Shifting Property (t-domain shift)

This property deals with the effect of shifting a function in the time domain. If the Laplace transform of $f(t)$ is $F(s)$, then the Laplace transform of $f(t-a)u(t-a)$ is $e^{-as}F(s)$, where $u(t-a)$ is the unit step function, defined as:

$u(t-a) = \begin{cases} 0 & \text{if } t < a \\ 1 & \text{if } t \ge a \end{cases}$

Mathematically:

$L\{f(t-a)u(t-a)\} = e^{-as}F(s)$, for $a \ge 0$.

This property is crucial for finding the Laplace transform of functions that are defined piecewise or that start at a later time. The $e^{-as}$ term represents the delay introduced by shifting the function.

Example:

Let's find the Laplace transform of a function that is zero for $t < 2$ and equals $t^2$ for $t \ge 2$. This function can be written as $f(t) = (t-2)^2 u(t-2)$.

First, let $g(t) = t^2$. Its Laplace transform is $G(s) = L\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}$.

Our function is $f(t) = g(t-2)u(t-2)$, where $g(t) = t^2$. Here, $a=2$.

Using the second shifting property:

$L\{f(t)\} = L\{(t-2)^2 u(t-2)\} = e^{-2s} G(s)$

Substituting $G(s) = \frac{2}{s^3}$:

$L\{f(t)\} = e^{-2s} \left(\frac{2}{s^3}\right) = \frac{2e^{-2s}}{s^3}$

4. Multiplication by $t$ Property

This property states that if $L\{f(t)\} = F(s)$, then the Laplace transform of $t f(t)$ is given by:

$L\{t f(t)\} = -\frac{d}{ds} F(s) = -F'(s)$

If we multiply by $t^n$, the property extends to:

$L\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$

This property is useful for finding Laplace transforms of functions that involve $t$ as a multiplier, such as $t \sin(t)$ or $t^2 e^{-t}$. It relates the Laplace transform to the derivative of the original transform.

Example:

Let's find the Laplace transform of $f(t) = t \cos(t)$.

We know $L\{\cos(t)\} = \frac{s}{s^2+1}$. So, $F(s) = \frac{s}{s^2+1}$.

Using the multiplication by $t$ property ($n=1$):

$L\{t \cos(t)\} = -\frac{d}{ds} \left(\frac{s}{s^2+1}\right)$

We need to compute the derivative of $F(s)$ using the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$. Let $u=s$ and $v=s^2+1$. Then $u'=1$ and $v'=2s$.

$\frac{d}{ds} \left(\frac{s}{s^2+1}\right) = \frac{(1)(s^2+1) - (s)(2s)}{(s^2+1)^2} = \frac{s^2+1 - 2s^2}{(s^2+1)^2} = \frac{1-s^2}{(s^2+1)^2}$

Therefore,

$L\{t \cos(t)\} = - \left(\frac{1-s^2}{(s^2+1)^2}\right) = \frac{s^2-1}{(s^2+1)^2}$

5. Division by $t$ Property

If $L\{f(t)\} = F(s)$ and the limit $\lim_{t \to 0} \frac{f(t)}{t}$ exists, then the Laplace transform of $\frac{f(t)}{t}$ is given by:

$L\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty} F(\sigma) d\sigma$

This property is useful for functions that have $t$ in the denominator, like $\frac{\sin(t)}{t}$. It involves integrating the function's transform in the s-domain.

Example:

Let's find the Laplace transform of $f(t) = \frac{e^{2t}-e^{3t}}{t}$.

First, consider $g(t) = e^{2t}-e^{3t}$. Its Laplace transform is:

$G(s) = L\{e^{2t}\} - L\{e^{3t}\} = \frac{1}{s-2} - \frac{1}{s-3}$

Now, we apply the division by $t$ property to $f(t) = \frac{g(t)}{t}$:

$L\left\{\frac{e^{2t}-e^{3t}}{t}\right\} = \int_{s}^{\infty} \left(\frac{1}{\sigma-2} - \frac{1}{\sigma-3}\right) d\sigma$

Let's evaluate the integral:

$\int \left(\frac{1}{\sigma-2} - \frac{1}{\sigma-3}\right) d\sigma = \ln|\sigma-2| - \ln|\sigma-3| = \ln\left|\frac{\sigma-2}{\sigma-3}\right|$

Now, apply the limits of integration:

$\left[\ln\left|\frac{\sigma-2}{\sigma-3}\right|\right]_{s}^{\infty} = \lim_{\sigma \to \infty} \ln\left|\frac{\sigma-2}{\sigma-3}\right| - \ln\left|\frac{s-2}{s-3}\right|$

As $\sigma \to \infty$, $\frac{\sigma-2}{\sigma-3} \to 1$. So, $\lim_{\sigma \to \infty} \ln\left|\frac{\sigma-2}{\sigma-3}\right| = \ln(1) = 0$.

Therefore, the result is:

$0 - \ln\left|\frac{s-2}{s-3}\right| = -\ln\left(\frac{s-2}{s-3}\right) = \ln\left(\frac{s-3}{s-2}\right)$ (assuming $s>3$ for the argument to be positive and well-defined)

6. Differentiation in the s-domain

This property is identical to the Multiplication by $t$ property and is often stated separately for clarity. It states that if $L\{f(t)\} = F(s)$, then $L\{t^n f(t)\} = (-1)^n \frac{d^n F(s)}{ds^n}$.

The core idea is that differentiation with respect to $s$ in the Laplace domain corresponds to multiplication by $-t$ in the time domain.

Example:

Consider $f(t) = t e^{at}$.

We know $L\{e^{at}\} = \frac{1}{s-a}$. Let $F(s) = \frac{1}{s-a}$.

Using the property $L\{t f(t)\} = - \frac{d}{ds} F(s)$:

$L\{t e^{at}\} = - \frac{d}{ds} \left(\frac{1}{s-a}\right)$

The derivative of $\frac{1}{s-a}$ with respect to $s$ is $-\frac{1}{(s-a)^2}$.

$L\{t e^{at}\} = - \left(-\frac{1}{(s-a)^2}\right) = \frac{1}{(s-a)^2}$

7. Integration in the s-domain

This property is identical to the Division by $t$ property and is often stated in relation to differentiation. It states that if $L\{f(t)\} = F(s)$, then $L\{\frac{f(t)}{t}\} = \int_{s}^{\infty} F(\sigma) d\sigma$.

The core idea is that integration with respect to $s$ in the Laplace domain corresponds to division by $t$ in the time domain.

8. Laplace Transform of Derivatives

This is a cornerstone property for solving differential equations. If $L\{f(t)\} = F(s)$, then the Laplace transform of the first and second derivatives of $f(t)$ are:

First Derivative:

$L\{f'(t)\} = s F(s) - f(0)$

Here, $f(0)$ is the initial value of the function $f(t)$ at $t=0$.

Second Derivative:

$L\{f''(t)\} = s^2 F(s) - s f(0) - f'(0)$

Here, $f'(0)$ is the initial value of the derivative $f'(t)$ at $t=0$.

For higher-order derivatives, the general formula is:

$L\{f^{(n)}(t)\} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - f^{(n-1)}(0)$

These properties allow us to convert a differential equation in the time domain into an algebraic equation in the s-domain, which is generally easier to solve.

Example: Solving a Differential Equation

Solve the differential equation $y'' + 4y = 0$ with initial conditions $y(0)=1$ and $y'(0)=0$.

Let $L\{y(t)\} = Y(s)$.

Using the Laplace transform properties for derivatives:

$L\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)$

Substitute the initial conditions $y(0)=1$ and $y'(0)=0$:

$L\{y''(t)\} = s^2 Y(s) - s(1) - 0 = s^2 Y(s) - s$

The Laplace transform of $4y(t)$ is $4Y(s)$ by linearity.

Taking the Laplace transform of the entire equation:

$L\{y''(t) + 4y(t)\} = L\{0\}$

$(s^2 Y(s) - s) + 4Y(s) = 0$

Now, solve for $Y(s)$:

$Y(s)(s^2 + 4) = s$

$Y(s) = \frac{s}{s^2+4}$

To find the solution $y(t)$, we need to find the inverse Laplace transform of $Y(s)$. We recognize this form:

$L\{\cos(bt)\} = \frac{s}{s^2+b^2}$

With $b=2$, we have $L\{\cos(2t)\} = \frac{s}{s^2+4}$.

Therefore, the solution is $y(t) = \cos(2t)$.

9. Laplace Transform of Integrals

Similar to the derivative property, there's a property for integrals. If $L\{f(t)\} = F(s)$, then the Laplace transform of the integral of $f(t)$ is:

$L\left\{\int_{0}^{t} f(\tau) d\tau\right\} = \frac{F(s)}{s}$

This property is useful when dealing with integral equations or when solving differential equations that involve integrals.

Example:

Find the Laplace transform of $\int_{0}^{t} e^{-2\tau} d\tau$.

Let $f(t) = e^{-2t}$. We know $L\{f(t)\} = F(s) = \frac{1}{s-(-2)} = \frac{1}{s+2}$.

Using the integral property:

$L\left\{\int_{0}^{t} e^{-2\tau} d\tau\right\} = \frac{F(s)}{s} = \frac{1/ (s+2)}{s} = \frac{1}{s(s+2)}$

10. Convolution Property

The convolution of two functions $f(t)$ and $g(t)$, denoted by $(f * g)(t)$, is defined as:

$(f * g)(t) = \int_{0}^{t} f(\tau) g(t-\tau) d\tau$

The convolution property of the Laplace transform states that the Laplace transform of the convolution of two functions is the product of their individual Laplace transforms:

$L\{(f * g)(t)\} = L\{f(t)\} \cdot L\{g(t)\} = F(s)G(s)$

Conversely, if $H(s) = F(s)G(s)$, then $h(t) = L^{-1}\{H(s)\} = (f * g)(t)$. This property is particularly useful for finding the inverse Laplace transform of a product of two functions.

Example:

Find the inverse Laplace transform of $H(s) = \frac{1}{(s-1)(s-2)}$.

We can identify $F(s) = \frac{1}{s-1}$ and $G(s) = \frac{1}{s-2}$.

The inverse Laplace transforms are $f(t) = L^{-1}\{F(s)\} = e^t$ and $g(t) = L^{-1}\{G(s)\} = e^{2t}$.

Using the convolution property, $h(t) = L^{-1}\{F(s)G(s)\} = (f * g)(t)$.

$h(t) = \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} e^{\tau} e^{2(t-\tau)} d\tau$

$h(t) = \int_{0}^{t} e^{\tau} e^{2t} e^{-2\tau} d\tau = e^{2t} \int_{0}^{t} e^{-\tau} d\tau$

Evaluating the integral:

$e^{2t} \left[-e^{-\tau}\right]_{0}^{t} = e^{2t} (-e^{-t} - (-e^0)) = e^{2t} (-e^{-t} + 1)$

$h(t) = e^{2t} - e^{2t}e^{-t} = e^{2t} - e^{t}$

Alternatively, using partial fraction decomposition for $H(s)$:

$H(s) = \frac{1}{(s-1)(s-2)} = \frac{A}{s-1} + \frac{B}{s-2}$

$1 = A(s-2) + B(s-1)$

Setting $s=1$: $1 = A(1-2) \implies 1 = -A \implies A = -1$.

Setting $s=2$: $1 = B(2-1) \implies 1 = B \implies B = 1$.

So, $H(s) = -\frac{1}{s-1} + \frac{1}{s-2}$.

The inverse Laplace transform is $h(t) = -e^t + e^{2t}$, which matches the convolution result.

11. Periodic Functions Property

If $f(t)$ is a periodic function with period $T$, its Laplace transform is given by:

$L\{f(t)\} = \frac{1}{1-e^{-sT}} \int_{0}^{T} e^{-st} f(t) dt$

The integral term $\int_{0}^{T} e^{-st} f(t) dt$ is the Laplace transform of one period of the function. This property simplifies the process of finding the Laplace transform of repeating waveforms like square waves or sawtooth waves.

Example:

Find the Laplace transform of a square wave with amplitude $A$ and period $T$, which is $A$ for $0 \le t < T/2$ and $-A$ for $T/2 \le t < T$.

The integral over one period is:

$\int_{0}^{T} e^{-st} f(t) dt = \int_{0}^{T/2} A e^{-st} dt + \int_{T/2}^{T} (-A) e^{-st} dt$

$= A \left[\frac{e^{-st}}{-s}\right]_{0}^{T/2} - A \left[\frac{e^{-st}}{-s}\right]_{T/2}^{T}$

$= -\frac{A}{s} (e^{-sT/2} - e^0) + \frac{A}{s} (e^{-sT} - e^{-sT/2})$

$= -\frac{A}{s} e^{-sT/2} + \frac{A}{s} + \frac{A}{s} e^{-sT} - \frac{A}{s} e^{-sT/2}$

$= \frac{A}{s} (1 - 2e^{-sT/2} + e^{-sT})$

We can factor the term in the parenthesis: $1 - 2e^{-sT/2} + e^{-sT} = (1 - e^{-sT/2})^2$.

So, the integral is $\frac{A}{s} (1 - e^{-sT/2})^2$.

Now, applying the periodic function property:

$L\{f(t)\} = \frac{1}{1-e^{-sT}} \left(\frac{A}{s} (1 - e^{-sT/2})^2\right)$

$= \frac{A}{s} \frac{(1 - e^{-sT/2})^2}{1 - e^{-sT}}$

Since $1 - e^{-sT} = (1 - e^{-sT/2})(1 + e^{-sT/2})$, we can simplify:

$L\{f(t)\} = \frac{A}{s} \frac{(1 - e^{-sT/2})^2}{(1 - e^{-sT/2})(1 + e^{-sT/2})} = \frac{A}{s} \frac{1 - e^{-sT/2}}{1 + e^{-sT/2}}$