Unit 3: Laplace Transform - Subjective Questions

Definition: The Laplace transform of a function is defined by

where is a complex variable, usually written as .

Conditions for existence:

• should be piecewise continuous on every finite interval in .
• should be of exponential order; that is, there must exist constants , , and such that for .

Significance:

• It converts differential equations in the time domain into algebraic equations in the -domain.
• Initial conditions can be incorporated directly.
• It is useful for analyzing robot actuator dynamics, control systems, and transient responses.

Standard Laplace transforms:

1. For the unit constant function,
   

2. For , where is a non-negative integer,
   

3. For the exponential function,
   

4. For the sine function,
   

5. For the cosine function,
   

These formulas form the basic transform table used in differential-equation and control-system calculations.

First shifting theorem: If

then

For example,

Second shifting theorem: If is the unit-step function and , then

For example, since

we have

The first theorem shifts the transform variable, while the second theorem represents a delay in time.

Let and .

Linearity property:

Differentiation with respect to :

More generally,

Integration property:
If

then

These properties allow complicated transforms to be obtained from standard transform formulas without evaluating the defining integral every time.

Using the property

we obtain the following results.

For ,

Therefore,

For ,

Therefore,

Hence,

and

Let .

For the first derivative,

Using integration by parts,

Applying the same result to the second derivative gives

Thus,

In general,

This property is particularly important because it includes initial conditions directly in the transformed equation.

Let

Then and . Taking the Laplace transform of both sides,

Since ,

Therefore,

For ,

Hence,

Indeed, the integral equals , whose transform is also .

Write the rational function in partial-fraction form:

Multiplying by gives

Comparing coefficients,


Subtracting the first equation from the second gives , and hence .

Therefore,

Using

we obtain

First complete the square in the denominator:

Rewrite the numerator in terms of :

Thus,

Using the standard inverse transforms,

and

we get

and

Therefore,

Convolution theorem: If

then

Write

The corresponding inverse transforms are

Therefore,

Let . The integral becomes

Hence,

Taking the Laplace transform of the differential equation,

Using the initial conditions,

Therefore,

Collecting terms,

Thus,

Let

Solving gives and . Hence,

Taking the inverse transform,

This solution satisfies both and .

Let . Taking transforms,

Using the initial conditions,

Thus,

so

Use the standard result

For ,

Therefore,

which gives

Consider a general equation

with initial conditions specified at .

Procedure:

• Take the Laplace transform of every term.
• Replace the transforms of derivatives using
  
• Substitute all initial conditions.
• Collect the terms containing and solve algebraically for .
• Decompose into standard forms using partial fractions, completing the square, or convolution.
• Apply the inverse Laplace transform to obtain .

The main advantage is that the initial conditions are incorporated automatically, making the method suitable for transient and forced-response analysis in robotic systems.

Bilateral Laplace transform:

It considers the function over the entire time axis.

Unilateral Laplace transform:

It considers the function from the initial time onward.

Differences:

• The bilateral transform is useful for signals defined for both positive and negative time.
• The unilateral transform is more convenient for causal systems and initial-value problems.
• The unilateral transform includes initial conditions naturally in derivative formulas.

For example,

In robotics, actuator positions, velocities, and currents are usually specified at an initial time, and the system is analyzed for . Therefore, the unilateral Laplace transform is generally preferred.

Decompose the function as

Multiplying through by the denominator gives

Comparing coefficients,



Thus,

Since

rewrite the numerator:

Therefore,

Taking the inverse transform,

Let .

Initial value theorem:

This theorem gives the value of the function immediately after the initial time without finding the inverse transform.

Final value theorem:

The final value theorem is valid only if all poles of lie in the open left half of the -plane, except possibly a simple pole at the origin. In other words, the system response must approach a finite steady value.

For example, if

then

The theorem cannot be used for sustained oscillations or unstable responses.

Let and .

Taking Laplace transforms of the first equation,

so

For the second equation,

so

From the second equation,

Substitute this into the first equation:


Therefore,

Also,

so

Hence, the solution is

Let and .

Transforming the first equation,

which gives

Transforming the second equation,

which gives

The algebraic system is

The determinant is

Solving,

Rewrite the numerators:

Therefore,

Consider a system of first-order equations written as

where is the vector of unknown functions and is a constant coefficient matrix.

Taking the Laplace transform,

Rearranging,

where is the identity matrix.

Hence,

Steps:

• Transform each differential equation.
• Insert the initial conditions.
• Form the algebraic matrix equation.
• Find or solve the simultaneous algebraic equations.
• Take inverse transforms component by component.

This approach is systematic and is well suited to coupled robot motions, such as interconnected joint or actuator models.

For a periodic function with period ,

Here, and over one period. Therefore,

Evaluate the integral by integration by parts:

Hence,

The periodic-function formula is useful when a robot input, such as a repeated command or cyclic actuator signal, repeats after a fixed time interval.