Signals and Systems

9.5 Properties of the Laplace Transform

Imron Rosyadi

9.5 Properties of the Laplace Transform

Leveraging Transform Domain Relationships

Just like with the Fourier Transform, the Laplace Transform has a rich set of properties that simplify analysis and provide deeper insights. These properties are invaluable for:

• Deriving new transform pairs.
• Analyzing LTI systems.
• Solving differential equations.
• Checking calculations.

We will explore key properties and their implications on both the signal and its ROC.

Introduce the section by highlighting the utility of properties in Signals and Systems, drawing a parallel to Fourier Transform properties. Emphasize that understanding these properties is crucial for practical applications.

Linearity of the Laplace Transform

If \(x_1(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X_1(s)\) with ROC \(R_1\), and \(x_2(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X_2(s)\) with ROC \(R_2\),

Then, for constants \(a, b\):

\[ a x_{1}(t)+b x_{2}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} a X_{1}(s)+b X_{2}(s) \tag{9.82} \]

ROC: At least contains \(R_1 \cap R_2\). The ROC can sometimes be larger than the intersection, especially due to pole-zero cancellation.

Tip

If \(R_1 \cap R_2\) is empty, then the linear combination \(ax_1(t) + bx_2(t)\) generally does not have a Laplace Transform.

Explain that linearity holds for both the transform and the inverse transform. The critical point here is the ROC; it’s at least the intersection, meaning it can expand. This expansion is usually due to cancellations.

Example 9.13: Linearity and ROC Extension

Consider \(x(t) = x_1(t) - x_2(t)\) where:

\(X_1(s)=\frac{1}{s+1}\), with \(\operatorname{Re}\{s\}>-1\) (\(R_1\)) \(X_2(s)=\frac{1}{(s+1)(s+2)}\), with \(\operatorname{Re}\{s\}>-1\) (\(R_2\))

The intersection \(R_1 \cap R_2\) is \(\operatorname{Re}\{s\}>-1\).

Now, \(X(s) = X_1(s) - X_2(s)\):

\[ X(s)=\frac{1}{s+1}-\frac{1}{(s+1)(s+2)} = \frac{(s+2)-1}{(s+1)(s+2)} = \frac{s+1}{(s+1)(s+2)} = \frac{1}{s+2} \tag{9.86} \]

The pole at \(s=-1\) is canceled by a zero at \(s=-1\).

Note

Due to the pole-zero cancellation, the ROC for \(X(s)\) extends to \(\operatorname{Re}\{s\} > -2\), which is larger than the intersection of \(R_1\) and \(R_2\).

Clearly show how the pole at -1 and zero at -1 cancel in the combined \(X(s)\). The Mermaid diagram visually represents the poles and ROCs for \(X_1(s)\), \(X_2(s)\), and the resulting \(X(s)\), highlighting the expanded ROC due to cancellation.

Time Shifting

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ x\left(t-t_{0}\right) \stackrel{\mathcal{L}}{\longleftrightarrow} e^{-s t_{0}} X(s) \tag{9.87} \]

ROC: Remains \(R\).

• Shifting a signal in the time domain corresponds to multiplying its Laplace Transform by a complex exponential \(e^{-st_0}\).
• The ROC is unchanged because the convergence behavior of \(x(t-t_0)\) is the same as \(x(t)\).

Explain that time shifting doesn’t change the exponential growth/decay rate, hence the ROC remains the same. This is a common property across Fourier and Laplace transforms.

Shifting in the s-Domain

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ e^{s_{0} t} x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X\left(s-s_{0}\right) \tag{9.88} \]

ROC: \(R + \operatorname{Re}\{s_0\}\). The ROC of \(X(s-s_0)\) is the ROC of \(X(s)\) shifted by \(\operatorname{Re}\{s_0\}\).

Special Case: Modulation If \(s_0 = j\omega_0\) (pure imaginary), then \(e^{j\omega_0 t}x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s-j\omega_0)\). - This corresponds to a shift in the \(s\)-plane parallel to the \(j\omega\)-axis. - The ROC is unchanged in terms of its real axis boundaries.

Illustrate how multiplying by an exponential in the time domain shifts the entire s-plane representation. Emphasize the ROC shift, which is crucial. Explain the modulation special case and its significance in communication systems.

Time Scaling

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ x(a t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{|a|} X\left(\frac{s}{a}\right) \tag{9.90} \]

ROC: \(aR\). The ROC is also scaled by \(a\).

• If \(a > 0\), the ROC is scaled. For \(0 < a < 1\), it’s compressed; for \(a > 1\), it’s expanded.
• If \(a < 0\) (time reversal), the ROC undergoes a reversal about the \(j\omega\)-axis as well as scaling.

Time Reversal (\(a=-1\)):

\[ x(-t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(-s), \quad \text { with } \mathrm{ROC}=-R \tag{9.91} \]

Explain that time compression/expansion affects the rate of change, thus scaling the s-domain representation and its ROC. Time reversal is a specific case that also flips the ROC.

Conjugation

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ x^{*}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X^{*}\left(s^{*}\right) \tag{9.93} \]

ROC: \(R\).

Important Consequence for Real Signals: If \(x(t)\) is a real signal, then \(X(s) = X^*(s^*)\). This implies that if \(X(s)\) has a pole or zero at \(s_0\), it must also have a pole or zero at the complex conjugate point \(s_0^*\).

Explain the property and its direct implication for real-world signals. Most physical signals are real, so their Laplace transforms always exhibit conjugate symmetry for poles and zeros.

Convolution Property

If \(x_1(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X_1(s)\) with ROC \(R_1\), and \(x_2(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X_2(s)\) with ROC \(R_2\), then:

\[ x_{1}(t) * x_{2}(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X_{1}(s) X_{2}(s) \tag{9.95} \]

ROC: Containing \(R_1 \cap R_2\). Similar to linearity, the ROC can be larger if pole-zero cancellation occurs in the product \(X_1(s)X_2(s)\).

Important

This is a cornerstone for Linear Time-Invariant (LTI) system analysis. Convolution in the time domain becomes multiplication in the \(s\)-domain, greatly simplifying system response calculations.

Emphasize the profound importance of this property for LTI systems. It transforms a complex convolution operation into a simple multiplication, which is why frequency/s-domain analysis is so powerful.

Differentiation in the Time Domain

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ \frac{d x(t)}{d t} \stackrel{\mathcal{L}}{\longleftrightarrow} s X(s) \tag{9.98} \]

ROC: Containing \(R\). The ROC can be larger if \(X(s)\) has a first-order pole at \(s=0\) that is canceled by the multiplication by \(s\).

Example: - \(u(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s}\), \(\operatorname{Re}\{s\}>0\) - \(\frac{d}{dt}u(t) = \delta(t) \stackrel{\mathcal{L}}{\longleftrightarrow} s \cdot \frac{1}{s} = 1\), ROC is the entire \(s\)-plane.

Explain how differentiation in time translates to multiplication by \(s\) in the Laplace domain. Highlight the ROC behavior, especially the potential for expansion when a pole at the origin is canceled, as seen with the unit step and impulse.

Differentiation in the s-Domain

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ -t x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{d X(s)}{d s} \tag{9.100} \]

ROC: \(R\).

This property is useful for finding transforms of signals multiplied by \(t^n\).

Explain this less intuitive property: multiplication by time in the time domain corresponds to differentiation in the \(s\)-domain. This is particularly useful for signals like \(t \cdot e^{-at}u(t)\).

Example 9.14: Using s-Domain Differentiation

Let’s find the Laplace transform of \(x(t)=t e^{-at} u(t)\).

We know that \(e^{-at} u(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s+a}\), with \(\operatorname{Re}\{s\}>-a\).

Applying the differentiation in the s-domain property:

\[ t e^{-a t} u(t) \stackrel{\mathcal{L}}{\longleftrightarrow} -\frac{d}{d s}\left[\frac{1}{s+a}\right] = - \left( -\frac{1}{(s+a)^2} \right) = \frac{1}{(s+a)^{2}} \tag{9.102} \]

ROC: \(\operatorname{Re}\{s\}>-a\).

Repeated application yields:

\[ \frac{t^{n-1}}{(n-1) !} e^{-a t} u(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{(s+a)^{n}} \tag{9.104} \]

Walk through the derivation. Emphasize how a complex time-domain signal (multiplied by t) becomes a simpler derivative in the s-domain.

Interactive Demo: \(t e^{-at} u(t)\)

Observe the time-domain signal \(x(t) = t e^{-at} u(t)\) and its relation to \(e^{-at} u(t)\). Vary the parameter ‘a’ and see how the signals change.

viewof param_a = Inputs.range([0.1, 5.0], {value: 1, step: 0.1, label: "Parameter 'a'"});

Encourage students to manipulate the ‘a’ parameter and observe the behavior of both signals. They should see how \(t e^{-at} u(t)\) initially grows before decaying, compared to the direct exponential decay. This visually reinforces the mathematical relationship.

Example 9.15: Inverse Transform with Multiple Poles

Consider \(X(s)=\frac{2 s^{2}+5 s+5}{(s+1)^{2}(s+2)}\), with \(\operatorname{Re}\{s\}>-1\).

Using partial-fraction expansion (PFE) for multiple-order poles (as discussed in the appendix or advanced PFE techniques):

\[ X(s)=\frac{2}{(s+1)^{2}}-\frac{1}{(s+1)}+\frac{3}{s+2} \tag{9.105} \]

Since the ROC is \(\operatorname{Re}\{s\}>-1\) (to the right of all poles), all terms correspond to right-sided signals.

• From Eq. (9.104): \(\frac{2}{(s+1)^2} \stackrel{\mathcal{L}^{-1}}{\longleftrightarrow} 2 t e^{-t} u(t)\)
• From Example 9.1: \(-\frac{1}{s+1} \stackrel{\mathcal{L}^{-1}}{\longleftrightarrow} -e^{-t} u(t)\)
• From Example 9.1: \(\frac{3}{s+2} \stackrel{\mathcal{L}^{-1}}{\longleftrightarrow} 3 e^{-2t} u(t)\)

Combining these, the inverse transform is:

\[ x(t)=\left[2 t e^{-t}-e^{-t}+3 e^{-2 t}\right] u(t) \]

Highlight how the previously derived transform pair for \(t e^{-at}u(t)\) is directly applied here. This demonstrates the practical utility of understanding transform properties for solving more complex inverse Laplace transform problems.

Integration in the Time Domain

If \(x(t) \stackrel{\mathcal{L}}{\longleftrightarrow} X(s)\) with ROC \(R\), then:

\[ \int_{-\infty}^{t} x(\tau) d \tau \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s} X(s) \tag{9.106} \]

ROC: Containing \(R \cap \{\operatorname{Re}\{s\}>0\}\).

• This property is the inverse of time-domain differentiation.
• It can be derived using the convolution property, as \(\int_{-\infty}^{t} x(\tau) d \tau = x(t) * u(t)\).
• Since \(u(t) \stackrel{\mathcal{L}}{\longleftrightarrow} \frac{1}{s}\) with \(\operatorname{Re}\{s\}>0\), the ROC of the integral is the intersection of \(R\) and \(\operatorname{Re}\{s\}>0\).

Connect this property to both differentiation and convolution. Explain why the ROC for integration includes the \(\operatorname{Re}\{s\}>0\) constraint from the unit step function.

Initial and Final-Value Theorems

These theorems allow us to find the initial and final values of a time-domain signal directly from its Laplace Transform, under certain conditions.

Initial-Value Theorem: (Conditions: \(x(t)=0\) for \(t<0\); no impulses or higher-order singularities at \(t=0\))

\[ x\left(0^{+}\right)=\lim _{s \rightarrow \infty} s X(s) \tag{9.110} \]

Final-Value Theorem: (Conditions: \(x(t)=0\) for \(t<0\); \(x(t)\) has a finite limit as \(t \rightarrow \infty\))

\[ \lim _{t \rightarrow \infty} x(t)=\lim _{s \rightarrow 0} s X(s) \tag{9.111} \]

Tip

These theorems are excellent tools for checking the correctness of Laplace transform calculations or the steady-state behavior of systems without explicitly finding \(x(t)\).

Clearly state the conditions for each theorem, as they are crucial for correct application. Emphasize their practical use in verifying results and understanding system behavior.

Summary of Laplace Transform Properties

Property	Signal \(x(t)\)	Laplace Transform \(X(s)\)	ROC
Linearity	\(ax_1(t) + bx_2(t)\)	\(aX_1(s) + bX_2(s)\)	At least \(R_1 \cap R_2\)
Time Shifting	\(x(t-t_0)\)	\(e^{-st_0}X(s)\)	\(R\)
Shifting in \(s\)-Domain	\(e^{s_0 t}x(t)\)	\(X(s-s_0)\)	\(R + \operatorname{Re}\{s_0\}\)
Time Scaling	\(x(at)\)	\(\frac{1}{|a|}X(\frac{s}{a})\)	\(aR\)
Conjugation	\(x^*(t)\)	\(X^*(s^*)\)	\(R\)
Convolution	\(x_1(t) * x_2(t)\)	\(X_1(s)X_2(s)\)	At least \(R_1 \cap R_2\)
Differentiation (Time)	\(\frac{d}{dt}x(t)\)	\(sX(s)\)	At least \(R\)
Differentiation (\(s\)-Domain)	\(-tx(t)\)	\(\frac{d}{ds}X(s)\)	\(R\)
Integration (Time)	\(\int_{-\infty}^{t} x(\tau) d\tau\)	\(\frac{1}{s}X(s)\)	At least \(R \cap \{\operatorname{Re}\{s\}>0\}\)
Initial-Value Theorem	\(x(0^+)\) (for \(x(t)=0, t<0\))	\(\lim_{s \to \infty} sX(s)\)	N/A
Final-Value Theorem	\(\lim_{t \to \infty} x(t)\) (for \(x(t)=0, t<0\), finite limit)	\(\lim_{s \to 0} sX(s)\)	N/A

Review the table quickly, highlighting the main takeaway for each property. Emphasize that these properties form the foundation for applying the Laplace Transform to complex engineering problems.

Some Laplace Transform Pairs

TABLE 9.2 LAPLACE TRANSFORMS OF ELEMENTARY FUNCTIONS

Transform pair	Signal	Transform	ROC
1	\(\delta(t)\)	1	All \(s\)
2	\(u(t)\)	\(\frac{1}{s}\)	\(\operatorname{Re}\{s\}>0\)
3	\(-u(-t)\)	\(\frac{1}{s}\)	\(\operatorname{Re}\{s\}<0\)
4	\(\frac{t^{n-1}}{(n-1) !} u(t)\)	\(\frac{1}{s^{n}}\)	\(\operatorname{Re}\{s\}>0\)
5	\(-\frac{t^{n-1}}{(n-1) !} u(-t)\)	\(\frac{1}{s^{n}}\)	\(\operatorname{Re}\{s\}<0\)
6	\(e^{-\alpha t} u(t)\)	\(\frac{1}{s+\alpha}\)	\(\operatorname{Re}\{s\}>-\alpha\)
7	\(-e^{-\alpha t} u(-t)\)	\(\frac{1}{s+\alpha}\)	\(\operatorname{Re}\{s\}<-\alpha\)
8	\(\frac{t^{n-1}}{(n-1) !} e^{-\alpha t} u(t)\)	\(\frac{1}{(s+\alpha)^{n}}\)	\(\operatorname{Re}\{s\}>-\alpha\)
9	\(-\frac{t^{n-1}}{(n-1) !} e^{-\alpha t} u(-t)\)	\(\frac{1}{(s+\alpha)^{n}}\)	\(\operatorname{Re}\{s\}<-\alpha\)
10	\(\delta(t-T)\)	\(e^{-s T}\)	All \(s\)
11	\(\left[\cos \omega_{0} t\right] u(t)\)	\(\frac{s}{s^{2}+\omega_{0}^{2}}\)	\(\operatorname{Re}\{s\}>0\)
12	\(\left[\sin \omega_{0} t\right] u(t)\)	\(\frac{\omega_{0}}{s^{2}+\omega_{0}^{2}}\)	\(\operatorname{Re}\{s\}>0\)
13	\(\left[e^{-\alpha t} \cos \omega_{0} t\right] u(t)\)	\(\frac{s+\alpha}{(s+\alpha)^{2}+\omega_{0}^{2}}\)	\(\operatorname{Re}\{s\}>-\alpha\)
14	\(\left[e^{-\alpha t} \sin \omega_{0} t\right] u(t)\)	\(\frac{\omega_{0}}{(s+\alpha)^{2}+\omega_{0}^{2}}\)	\(\operatorname{Re}\{s\}>-\alpha\)
15	\(u_{n}(t)=\frac{d^{n} \delta(t)}{d t^{n}}\)	\(s^{n}\)	All \(s\)
16	\(u_{-n}(t)=\underbrace{u(t) * \cdots * u(t)}_{n \text { times }}\)	\(\frac{1}{s^{n}}\)	\(\operatorname{Re}\{s\}>0\)