Signals & System Lecture 10

Signals & Systems - Lecture 10

Laplace Transformation

Definition: The Laplace Transformation is a mathematical technique used to transform a time-domain signal into a complex frequency-domain signal. The transformation variable is denoted by ( s ), where ( s = \sigma + j\omega ). This transformation is particularly useful for analyzing linear time-invariant (LTI) systems and is best suited for continuous signals.

In the Laplace transformation:

• Derivatives in the time domain are multiplied by ( s ).
• Integrations are divided by ( s ).

The Laplace Transform is given by: $L[x(t)]=X(s)={\int }_{-\infty }^{\infty }x(t)\cdot e^{-st}dt$

For signals defined for ( t \geq 0 ), the Laplace Transform simplifies to: $L[x(t)]=X(s)={\int }_0^{\infty }x(t)\cdot e^{-st}dt$

This version, considering all values of ( t ), is known as the bilateral Laplace Transform or the two-sided Laplace Transform.

Region of Convergence (ROC): The Region of Convergence (ROC) is defined as the set of all values of ( s ) for which the Laplace integral converges: ${\int }_{-\infty }^{\infty }x(t)\cdot e^{-st}dt$ The ROC is crucial for determining the stability and causality of the system.

Example: Laplace Transform of the Unit Step Signal

Consider the unit step signal: $x(t)=u(t)$

By definition, the Laplace Transform is: $X(s)={\int }_{-\infty }^{\infty }u(t)\cdot e^{-st}dt$

Since ( u(t) = 0 ) for ( t < 0 ) and ( u(t) = 1 ) for ( t \geq 0 ), the integral limits reduce to: $X(s)={\int }_0^{\infty }{e}^{-st}dt$

Integrating, we get: $X(s)={\left[\frac{e^{-st}}{-s}\right]}_0^{\infty }$

Evaluating at the bounds: $X(s)=\left[\frac{e^{-\infty }}{-s}-\frac{e^0}{-s}\right]=\left[0-\frac{1}{-s}\right]$

Therefore, the Laplace Transform of the unit step signal is: $X(s)=\frac{1}{s},\text{for }\Re (s)>0$

Ramp Signal

Certainly! Let’s complete the Laplace Transform calculation for the signal ( x(t) = r(t) ), where ( r(t) ) is the ramp function.

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Given ( x(t) = r(t) ), which is ( r(t) = t \cdot u(t) ), where ( u(t) ) is the unit step function.

By definition of the Laplace Transform: $X(s)={\int }_{-\infty }^{\infty }x(t)\cdot e^{-st}dt$

Substituting ( x(t) = t \cdot u(t) ): $X(s)={\int }_{-\infty }^{\infty }t\cdot u(t)\cdot e^{-st}dt$

Since ( u(t) = 0 ) for ( t < 0 ) and ( u(t) = 1 ) for ( t \geq 0 ), this reduces to: $X(s)={\int }_0^{\infty }t\cdot e^{-st}dt$

To solve this integral, we use integration by parts. Let:

• ( u = t ) and ( dv = e^{-st} , dt ).

Then:

• ( du = dt ) and ( v = \int e^{-st} , dt = \frac{-e^{-st}}{s} ).

Now, apply integration by parts: $X(s)={\left[\frac{-t\cdot e^{-st}}{s}\right]}_0^{\infty }+\frac{1}{s}{\int }_0^{\infty }{e}^{-st}dt$

Evaluate the first term: $X(s)={\left[\frac{-t\cdot e^{-st}}{s}\right]}_0^{\infty }=\left(0-\frac{0}{s}\right)=0$

For the second term, integrate: $\frac{1}{s}{\int }_0^{\infty }{e}^{-st}dt=\frac{1}{s}\cdot {\left[\frac{-e^{-st}}{s}\right]}_0^{\infty }=\frac{1}{s}\cdot \left(0-\frac{-1}{s}\right)=\frac{1}{s^2}$

Thus, the Laplace Transform of the ramp function ( x(t) = r(t) = t \cdot u(t) ) is: $X(s)=\frac{1}{s^2}$

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x(t)=e^at u(t)

Complex Frequency

When dealing with higher-order differential equations, transforming these equations into the ( s )-domain using the Laplace Transform simplifies the analysis by converting differential equations into algebraic equations. This significantly eases the process of solving and analyzing dynamic systems.

Example from Classical Mechanics

Consider the differential equation representing a damped harmonic oscillator: $F=Ma+Bv+kx=0$

Here:

• ( kx ) is the spring’s restoring force.
• ( Bv ) is the damping force.

Taking the differential equation: $M\frac{d^{2}x}{d{t}^2}+B\frac{dx}{dt}+kx=0$

Applying the Laplace Transform, we convert the equation into the ( s )-domain: $M{s}^{2}X(s)+BsX(s)+kX(s)=0$

This is a second-order algebraic equation in ( X(s) ), which can be easily solved for system behavior analysis.

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References:

• Date: 2024.08.29
• Time: 09:08

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Notes on Laplace Transform

Definition:

• Laplace Transform is used to transform a signal from the time domain to a complex frequency domain (s-domain).
• The Laplace Transform helps convert differential equations in the time domain into algebraic equations in the frequency domain, making analysis and problem-solving easier.

Application:

• In the time domain, higher-order differential equations represent the system dynamics. These can be complex and difficult to solve.
• When transforming into the s-domain, the differential equation simplifies to an algebraic form that is easier to manage.

Example:

• A second-order system equation such as: $$m\frac{d^{2}x}{d{t}^2}+B\frac{dx}{dt}+Kx=0$$ is transformed into: $$M{s}^{2}X(s)+BsX(s)+KX(s)=0$$ Here, the terms $M{s}^2$, $Bs$, and $K$ are algebraic equivalents in the s-domain, making the equation simpler to solve.

Formula:

• The Laplace Transform of a function $x(t)$ is given by: $$X(s)={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$$ This transformation holds for all values of $t$.

Notes:

• It is also called Bilateral Laplace Transform (L.T.), as it integrates over all values of time (from $-\infty$ to $+\infty$).

This page provides a concise summary of Laplace Transforms, highlighting their utility in simplifying differential equations and aiding in system analysis in the s-domain.

Notes on Laplace Transform

Unilateral or One-sided Laplace Transform:

For $t\ge 0$:

$$\mathcal{L}[x(t)]=X(s)={\int }_0^{\infty }x(t)e^{-st}dt$$

This is the unilateral or one-sided Laplace Transform.

Define Region of Convergence (ROC):

The Region of Convergence (ROC) is the set of values of $s$ for which the Laplace Transform integral:

$${\int }_{-\infty }^{\infty }x(t)e^{-st}dt$$

converges. The value of $s$ for which this integral converges is called the ROC.

Example Problem:

Find the Laplace transform of the following signal and draw its ROC:

1. Unit-step signal:
   • $x(t)=u(t)$
   By definition of the Laplace Transform: $$X(s)={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$$

This page explains the concept of unilateral Laplace Transform and introduces the Region of Convergence (ROC), along with a practical example of calculating the Laplace transform of a unit-step signal.

Notes on Laplace Transform - Continued

Calculation of Laplace Transform for Unit-step Signal:

The Laplace transform of the unit-step signal $x(t)=u(t)$ is given by:

$$X(s)={\int }_{-\infty }^{\infty }u(t)e^{-st}dt$$

This can be simplified as:

$$X(s)={\int }_0^{\infty }{e}^{-st}dt=\frac{e^{-st}}{-s}{|}_0^{\infty }$$

Evaluating the limits:

$$X(s)=\left[\frac{e^{-\infty }}{-s}-\frac{e^0}{-s}\right]=\frac{0-1}{-s}=\frac{1}{s}$$

Thus, the Laplace transform of a unit-step function is:

$$X(s)=\frac{1}{s}$$

Region of Convergence (ROC):

For the unit-step function, the ROC is for $s>0$.

• The shaded region in the $s$-plane indicates the ROC, which lies to the right of the pole at $s=0$.

Important Observation:

• If the signal is on the right side, the ROC will also be on the right side of the $s$-plane.

General Expression for $s$:

$$s=\sigma +j\Omega$$

This page completes the example by providing the full solution for the Laplace transform of a unit-step signal, along with a visual representation of the ROC in the $s$-plane.

x = 0
x > 0
\text{Real axis: } \sigma \quad (set \sigma label on x-axis)
\text{Imaginary axis: } j\Omega \quad (set j\Omega label on y-axis)

Complete Notes and Desmos Graph for Laplace Transforms

2. Ramp Signal:

The function is defined as:

$$x(t)=r(t)=t\cdot u(t)$$

Taking the Laplace Transform (LT):

$$\mathcal{L}[x(t)]={\int }_0^{\infty }t\cdot e^{-st}dt$$

Solving this integral:

$$X(s)=\frac{1}{s^2}$$

The ROC for this is $s>0$, which corresponds to shading the right half of the complex $s$-plane.

Desmos Graph for Ramp Signal:

x > 0

This shades the right side of the $s$-plane where $s>0$, indicating the ROC for the ramp signal.

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3. Exponential Signal:

The function is defined as:

$$x(t)=e^{2t}\cdot u(t)$$

For this example, we take $a=2$.

Taking the Laplace Transform:

$$\mathcal{L}[x(t)]={\int }_0^{\infty }{e}^{(2-s)t}dt=\frac{1}{s-2}$$

The ROC for this signal is $s>2$, meaning the region of convergence lies to the right of the line $x=2$ on the complex plane.

Desmos Graph for Exponential Signal (with $a=2$):

x > 2

This will shade the right side of the $s$-plane where $s>2$, indicating the ROC for the exponential signal with $a=2$.

4. Anti-causal Unit Step Signal

Definition and Signal:

For an anti-causal unit step signal, the function is defined as:

$$x(t)=u(-t)$$

This signal is non-zero for negative values of $t$, and it is represented by a step at $t=0$ with values of 1 for $t<0$.

Laplace Transform:

By the definition of the Laplace Transform:

$$X(s)={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$$

Substituting $x(t)=u(-t)$:

$$X(s)={\int }_{-\infty }^0{e}^{-st}dt={\left[\frac{e^{-st}}{-s}\right]}_{-\infty }^0$$

Evaluating the limits:

$$X(s)=\frac{1}{-s}\left[e^0-e^{-\infty }\right]=\frac{1}{-s}\cdot [1-0]=-\frac{1}{s}$$

Thus, the Laplace transform for the anti-causal unit step signal is:

$$X(s)=-\frac{1}{s},\text{for}\sigma <0$$

Region of Convergence (ROC):

The ROC for this anti-causal signal is for $s<0$, as the signal is only non-zero in the negative time domain.

Important Note:

• Magic Point: If a signal is anti-causal, the ROC will be left-sided, meaning the ROC includes the negative half of the $s$-plane.

This example demonstrates how the Laplace Transform of an anti-causal signal differs from the causal case, and how the ROC covers the left side of the complex $s$-plane.

x < 0

Summary and Desmos Graphs for Laplace Transformations

2. Exponential Signal with Negative Shift

The function is:

$$x(t)=e^{-3t}u(-t)$$

Taking the Laplace transform:

$$\mathcal{L}[x(t)]={\int }_{-\infty }^0{e}^{(-3-s)t}dt$$

After solving:

$$X(s)=\frac{1}{s+3}$$

The ROC is for $\sigma <-3$.

Desmos Graph:

x < -3

This shades the left side of the $s$-plane where $\sigma <-3$, indicating the region of convergence (ROC).

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3. Exponential Signal with Positive Shift

The function is:

$$x(t)=e^{bt}u(t)$$

Taking the Laplace transform:

$$\mathcal{L}[x(t)]={\int }_0^{\infty }{e}^{(b-s)t}dt$$

After solving:

$$X(s)=\frac{1}{s-b}$$

The ROC is for $\sigma >b$.

Desmos Graph:

For $b=5$ as an example:

x > 5

This shades the right side of the $s$-plane where $\sigma >5$, indicating the ROC.

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4. Exponential Signal with a Positive and Negative Shift

The function is:

$$x(t)=e^{2t}u(-t)$$

Taking the Laplace transform:

$$\mathcal{L}[x(t)]={\int }_{-\infty }^0{e}^{(2-s)t}dt$$

After solving:

$$X(s)=\frac{1}{s-2}$$

The ROC is for $\sigma <2$.

Desmos Graph:

x < 2

This shades the left side of the $s$-plane where $\sigma <2$, indicating the ROC for this signal.

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These graphs and summaries provide a visual representation of the Laplace transform and the region of convergence (ROC) for various signals, both causal and anti-causal.