# 2-Modeling in the Frequency Domain

```Modeling in Frequency Domain
Common Laplace Transform
Dirac Delta Function
Unit Step Function
Laplace Transform
Example: Find the Laplace Transform of
Solution:
Inverse Laplace Transform
Laplace Transforms Theorem
Example: Find the Inverse Laplace Transform
of
Solution:
Since the Laplace Transform of tu(t) is 1/s2 and the given has a
shift on the domain, we use the Frequency Shift Theorem,
Coursework # 1
Partial Fraction Expansion
Case 1: Roots of the Denominator are
real and distinct
Example: Find the Inverse Laplace Transform of the given function
using Partial Fraction Expansion
The number of the denominator will dictate how many sum of
terms will be. Since there are two denominator factor, there will
also be two terms.
We will temporary label the numerators of the two terms as K1 and
K2
Next is to find value of K1 and K2, we will multiply first the whole
expression by (s+1).
To find K1, let s = -1 to eliminate the last term.
At s=-1
Do the same thing for K2, multiply the expansion by (s+2)
and substitute s=-2 and this will yield to K2=-2
`
So the partial fraction expansion of F(s) is
`
Lastly is to find the Inverse Laplace Transform of the
expansion.
Since
,
Exercise: Given the following differential
equation use the Laplace transform to solve for
y(t) if all initial conditions are zero.
Case 2: Roots of the Denominator are
real and repeated
Example: Find the Inverse Laplace Transform of the given function
using Partial Fraction Expansion
The number of the terms will be the same for case # 1. In this
example, there are three denominators, there will also be three
sum of terms.
The exponent of the repeated denominator is 2, so the exponent of
one term with the repeated denominator will start at 1, and the
other term will be 2.
The process of getting K1, K2 and K3 will be the same for case 1.
Following the process, this will yield to the following values.
K1=2, K2=-2 &amp; K3=-2
The process of getting K1, K2 and K3 will be the same for case 1.
Following the process, this will yield to the following values.
K1=2, K2=-2 &amp; K3=-2
Case 3: Roots of the Denominator are
complex and imaginary
Example: Find the Inverse Laplace Transform of the given function
using Partial Fraction Expansion
In this example, there will be two sum of terms since there are two
factor denominators.
K1 can be obtained with same process with previous cases. So in
this case, K1 will be 3/5
For K2 and K3, we will multiply both sides by the whole
denominator.
Arranging terms..
To get K2 and K3, we will balance the coefficients by equating like
terms..
In this case,
This will result to K2=-3/5 and K3=-6/5
The last term can be express as shown by completing its square.
Exercise
Transfer Function
Control System
Input R(s)
Transfer Function
G(s)
Output C(s)
A transfer function represents the relationship between the output signal of
a control system and the input signal.
A block diagram is a visualization of the control system which uses blocks to
represent transfer function and arrows for various inputs and outputs.
Control System
Input R(s)
Transfer Function
G(s)
Output C(s)
In the context of Laplace Transform, transfer function is defined as the ratio
of the Laplace transform of the output and the Laplace transform of the input
Example: Find the transfer function represented
by the following differential equation, assume
zero initial condition
Solution:
Taking the Laplace Transform on both sides by using the
differentiation theorem
Then the transfer function G(s) is..
Getting the Transfer Function in
- Electrical Network
Example:
Using KVL loop,
Next, we convert this expressions in terms of vc(t)
by using
and
Then we come up to this expression after substituting,
Taking the Laplace Transform on both sides,
Rearranging the terms and simplifying,
Getting the Transfer Function in
- Translational Mechanical System
Force F(s)
Transfer Function
G(s)
Displacement X(s)
Block Diagram of a Translational Mechanical System
Example:
Draw first the free body diagram with all the present forces acting on
the system
Assume equilibrium condition, so all the forces present must be equal
to zero.
Next we take the Laplace Transform on both sides, assume zero initial
conditions
Factoring out X(s),