Signal & Linear systemChapter 4 Frequency - Domain
Analysis : Laplace Transform
Basil Hamed
4.1 The Laplace TransformMotivation for the Laplace TransformCT Fourier transform enables us to do a lot of things, e.g.
o Analyze frequency response of LTI systemso Samplingo Modulation
Why do we need yet another transform?One view of Laplace Transform is as an extension of the
Fourier transform to allow analysis of broader class of signals and systems
In particular, Fourier transform cannot handle large (and important) classes of signals and unstable systems, i.e. when
Basil Hamed 2
4.1 The Laplace Transform The Laplace transform transforms the
problem(D.EQ) from time domain to frequency domain.
Then the solution of the original D.EQ is arrived at, by obtaining the inverse transforms.
One of the problem that we faced using Fourier transform is many of the signals do not have Fourier transform.[ ex. exp(t)u(t), tu(t), and other time signals that are not absolutely integral]
The difficulty could be resolved by extending the Fourier transform so that x(t) is expressed as sum of complex exponentials, exp(-st) where
Basil Hamed 3
4.1 The Laplace Transformexp(), exp( satisfies the absolute integrable For
Ex. Given Find the frequency domaini. Fourier Transform the Fourier transform
does not exist.
ii. Laplace Transform
Basil Hamed 4
4.1 The Laplace TransformLaplace transform is the tool to map signals and system behavior from the time-domain into the frequency domain.
Basil Hamed 5
For a signal x(t), its Laplace transform is defined by:
BilateralThis general definite is known as two-sided (or bilateral) Laplace Transform.
4.1 The Laplace TransformThe one sided (unilateral) Laplace transform:
Ex Given Find X(s) Solution
Basil Hamed 6
The above signal has Laplace transform only if thusX(s) exists only if
4.1 The Laplace TransformThe range of values for the complex variable S for which the Laplace transform converges is called the Region of Convergence (ROC)
Basil Hamed 7
π₯2 (π‘ )=βπβππ‘π’ (βπ‘ )π₯1 (π‘ )=πβππ‘π’ (π‘ )
π π {π }>βππ π {π }<βπ
4.1 The Laplace TransformEx. Given Find X(s) Solution
Note that X(s) for the two previous examples are the same the only distinguish is ROC
Basil Hamed 8
Therefore, in order for the Laplace transform to be unique foreach signal x(t). The ROC must be specified as part of the
transform
4.1 The Laplace Transform
Basil Hamed 9
4.1 The Laplace TransformEx. Given Find X(s)Solution
Basil Hamed 10
Re{s}>-2 Re{s}<1
-2<Re{s}<1
4.1 The Laplace TransformEx. Given Find X(s) Solution
Ex. Find the Laplace transform of Ξ΄(t) and u(t).
Basil Hamed 11
4.1 The Laplace TransformEx. Find the Laplace transform of and cos Ο0t u(t).
Basil Hamed 12
4.2 Properties of Laplace transform
Basil Hamed 13
Ex. Find Laplace Transform of
s
4.2 Properties of Laplace transform
Basil Hamed 14
πΌπ π₯ (π‘ )β π (π ) ,π hπππ₯ (π‘βπ‘ 0 )π’(π‘β π‘0)βπβπ‘ 0π π (π )
Ex. Find X(s)
4.2 Properties of Laplace transform
Basil Hamed 15
Shifting in the S Domain
πΌπ π₯ (π‘ )β π (π ) ,π hππππ 0π‘ π₯ (π‘)β π (π βπ 0)
Ex. Find
πΏ [πβππ‘ π πππ0π‘ ]= π0
(π +π)2+π02
From Laplace Table we have
4.2 Properties of Laplace transform
Basil Hamed 16
Time ScalingπΌπ π₯ (π‘ )β π (π ) ,π hπππ₯ (πΌπ‘)β 1πΌ π (
π πΌ )
Ex Find L{u()}, L{u()}=(1/) 1/s/ =1/s The result is expected, since u(t)=u(t) for>0
Differentiation & Integration in the Time Domain
β«0
π‘
π₯ (π )d Ο β 1π π (π )
4.2 Properties of Laplace transform
Ex. Find y(t)Solution
Basil Hamed 17
4.2 Properties of Laplace transform
Basil Hamed 18
Differentiation in The S-Domain
Ex Given r(t)=t u(t), Find R(s)Solution R(s)=-
Ex. Find Solution
4.2 Properties of Laplace transform
Basil Hamed 19
Convolution Then Ex given Find h(t)
Solution Y(s)=X(s)H(s) H(s)=Y(s)/X(s)
4.2 Properties of Laplace transform
Basil Hamed 20
Initial-Value Theorem π₯ (0 )= limπ β β
π π (π )
This property is useful, since it allows us to compute the initial value of the signal x(t) directly from the Laplace transform X(s) without having to find the inverse x(t)Ex Given Find x(0)
π₯ (0 )= limπ β β
π π (π )= limπ ββ
β3 π 3+2π π 3+π 2+3 π +2
= limπ β β
β3+2/π 2
1+1/ π +3/π 2+2/ π 3 =β3
4.2 Properties of Laplace transform
Basil Hamed 21
Final-Value Theorem
limπ ββ
π₯(π‘)= limπ β0
π π (π )
Final-value Theorem exists only if the system is stable
Final-value Theorem is useful in many applications such as control theory, where we may need to find the final value(steady-state value) of the output of the system without solving for time domain
4.2 Properties of Laplace transform
Ex. Given Solution
Basil Hamed 22
limπ ββ
π (π‘ )=limπ β 0
π πΉ (π )= limπ β 0
π 5π (π 2+π +2)
=52
Ex. Given Solution System is unstable so there is no final value
StabilityStability conditions for an LTIC system Asymptotically stable if and only if all the poles of H(s) are in
left-hand plane (LHP). The poles may be repeated or non-repeated.
Unstable if and only if either one or both of these conditions hold
(i) at least one pole of H(s) is in right-hand plane (RHP)(ii) repeated poles of H(s) are on the imaginary axis
A system is said to be βmarginally stableβ if it has at least one distinct pole on the jΟ axis but no repeated poles on jΟ Marginally breaks
Basil Hamed 23
Stability In most applications we desire a stable system We can easily check for stability by looking to see where the
systemβs poles areExample i. ii. Solution i. All poles are on LHP system is stable
ii. One pole on RHP system is unstable
Basil Hamed 24
Inverse Laplace TransformThe function X(s) has to be a proper rational function to find the inverse of Laplace transform.The basic procedure is to express X(s) as a summation of terms whose inverse Laplace transform are available in a table.There are four general forms of solving the partial fraction; the roots of D(s) are either:1. Real and Distinct2. Complex and Distinct3. Real and Repeated4. Complex and Repeated
Basil Hamed 25
Inverse Laplace Transform
Ex. Find x(t)Solution:
Basil Hamed 26
Real simple Poles
π 1(π )
Inverse Laplace Transform
Basil Hamed 27
Inverse Laplace Transform
Ex. Solution
Basil Hamed 28
Repeated Real Poles
Inverse Laplace Transform
Ex. find x(t)Solution
Basil Hamed 29
Simple Complex Poles
Inverse Laplace Transform
Basil Hamed 30
Inverse Laplace TransformX(s) contains distinct complex roots:
Basil Hamed 31
Inverse Laplace Transform
Ex. Solution:
Basil Hamed 32
Repeated Complex Poles
Inverse Laplace Transform
Basil Hamed 33
4.3 Solution of Differential & Integro-Differential Equations
The Laplace transform of differential equation is an algebraic equation that can be readily solved for Y(s). Next we take the inverse Laplace transform of Y(s) to find the desired solution y(t)
Basil Hamed 34
4.3 Solution of Differential & Integro-Differential Equations
Example 4.10 P. 371 Solve the following second-order linear differential equation:
y (0) = 2, (0) =1and input x (t ) =.Solution
Basil Hamed 35
Time Domain Laplace (Frequency) Domain
4.3 Solution of Differential & Integro-Differential Equations
Basil Hamed 36
4.3 Solution of Differential & Integro-Differential Equations
Zero-input & Zero-state ResponsesThe Laplace transform method gives the total response, which include zero-input and zero state components. It is possible to separate the two components if we so desire.Letβs think about where the terms come from:
Basil Hamed 37
Input term
Initial condition term
4.3 Solution of Differential & Integro-Differential Equations
Basil Hamed 38
4.4 Analysis of Electrical Networks
How to compute T.F for circuit one way to find the T.F of the circuit is to compute its differential equation and then take its Laplace transformHowever, it is generally simpler to compute T.F directly.Transfer Function:T.F is defined as the s-domain ratio of the output to the input
Basil Hamed 39
Input
Output
4.4 Analysis of Electrical Networks
Weβve seen that the system outputβs LT is:
Basil Hamed 40
So, if the system is in zero-state then we only get the second term:
βSystem effect in zero-state case is completely set by the transfer function
4.4 Analysis of Electrical NetworksPoles and Zeros of a system Given a system with Transfer Function:
Basil Hamed 41
We can factor B(s) and A(s): (Recall: A(s) = characteristic polynomial)
Pole-Zero PlotThis gives us a graphical view of the systemβs behavior
4.4 Analysis of Electrical Networks
Basil Hamed 42
Example
4.4 Analysis of Electrical Networks
Basil Hamed 43
Time Domain S- Domain
πΌ= ππ πΏ +
πΌ0
π π (π )=πΏ [ππΌ (π )βπ (0 )]
4.4 Analysis of Electrical Networks
Example: given the Circuit shown , find y(t)Solution:Apply Laplace TransformThe total voltage in the loop is
Basil Hamed 44
π¦ (π‘ )=β5πβπ‘ cos [2 π‘+26 . 6Β° ]π’ (π‘)
4.4 Analysis of Electrical Networks
Exercise 4.4-1P 482Find the zero state response , if the input voltage is . Find TF, write differential eq relating to x(t)Solution Loop Eq;
Basil Hamed 45
Gramer rule yields
4.4 Analysis of Electrical Networks
Exercise 4.4-4 P 482 Find the loop currents for the input x(t) as shown in Figure below
Basil Hamed 46
Solution: The Loop Eq. are
4.4 Analysis of Electrical Networks
Gramerβs rule yields
Basil Hamed 47
4.5 Block Diagrams
β’ Large systems may consist of an enormous number of components or elements. Analyzing such systems all at once could be next to impossible. In such cases, it is convenient to represent a system by suitably interconnected subsystems.
β’ Each subsystem can be characterized in terms of its input-output relationships.
Basil Hamed 48
4.5 Block Diagrams
49
H(s)X(s) Y(s)
H1(s) + H2(s)X(s) Y(s)H1(s)
X(s) Y(s)H2(s)
=
H1(s)X(s) Y(s)H2(s) H1(s)H2(s)X(s) Y(s)=W(s)
G(s) 1 + G(s)H(s)X(s) Y(s)G(s)X(s) Y(s)
H(s)
- =E(s)
4.5 Block DiagramsExample: A basic feedback system consisting of block find TF
50
More on this later in Control Course feedback