Signal & Linear system
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Transcript of Signal & Linear system
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Signal & Linear systemChapter 4 Frequency - Domain
Analysis : Laplace Transform
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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
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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
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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
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4.1 The Laplace TransformLaplace transform is the tool to map signals and system behavior from the time-domain into the frequency domain.
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For a signal x(t), its Laplace transform is defined by:
BilateralThis general definite is known as two-sided (or bilateral) Laplace Transform.
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4.1 The Laplace TransformThe one sided (unilateral) Laplace transform:
Ex Given Find X(s) Solution
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The above signal has Laplace transform only if thusX(s) exists only if
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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)
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𝑥2 (𝑡 )=−𝑒−𝑎𝑡𝑢 (−𝑡 )𝑥1 (𝑡 )=𝑒−𝑎𝑡𝑢 (𝑡 )
𝑅𝑒 {𝑠 }>−𝑎𝑅𝑒 {𝑠 }<−𝑎
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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
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Therefore, in order for the Laplace transform to be unique foreach signal x(t). The ROC must be specified as part of the
transform
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4.1 The Laplace Transform
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4.1 The Laplace TransformEx. Given Find X(s)Solution
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Re{s}>-2 Re{s}<1
-2<Re{s}<1
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4.1 The Laplace TransformEx. Given Find X(s) Solution
Ex. Find the Laplace transform of δ(t) and u(t).
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4.1 The Laplace TransformEx. Find the Laplace transform of and cos ω0t u(t).
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4.2 Properties of Laplace transform
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Ex. Find Laplace Transform of
s
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4.2 Properties of Laplace transform
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𝐼𝑓 𝑥 (𝑡 )↔ 𝑋 (𝑠 ) ,𝑇 h𝑒𝑛𝑥 (𝑡−𝑡 0 )𝑢(𝑡− 𝑡0)↔𝑒−𝑡 0𝑠 𝑋 (𝑠 )
Ex. Find X(s)
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4.2 Properties of Laplace transform
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Shifting in the S Domain
𝐼𝑓 𝑥 (𝑡 )↔ 𝑋 (𝑠 ) ,𝑇 h𝑒𝑛𝑒𝑠0𝑡 𝑥 (𝑡)↔ 𝑋 (𝑠−𝑠0)
Ex. Find
𝐿 [𝑒−𝑎𝑡 𝑠𝑖𝑛𝜔0𝑡 ]= 𝜔0
(𝑠+𝑎)2+𝜔02
From Laplace Table we have
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4.2 Properties of Laplace transform
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Time Scaling𝐼𝑓 𝑥 (𝑡 )↔ 𝑋 (𝑠 ) ,𝑇 h𝑒𝑛𝑥 (𝛼𝑡)↔ 1𝛼 𝑋 (
𝑠𝛼 )
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
∫0
𝑡
𝑥 (𝜏 )d τ ↔ 1𝑠 𝑋 (𝑠)
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4.2 Properties of Laplace transform
Ex. Find y(t)Solution
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4.2 Properties of Laplace transform
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Differentiation in The S-Domain
Ex Given r(t)=t u(t), Find R(s)Solution R(s)=-
Ex. Find Solution
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4.2 Properties of Laplace transform
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Convolution Then Ex given Find h(t)
Solution Y(s)=X(s)H(s) H(s)=Y(s)/X(s)
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4.2 Properties of Laplace transform
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Initial-Value Theorem 𝑥 (0 )= lim𝑠→ ∞
𝑠𝑋 (𝑠)
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)
𝑥 (0 )= lim𝑠→ ∞
𝑠𝑋 (𝑠 )= lim𝑠 →∞
−3 𝑠3+2𝑠𝑠3+𝑠2+3 𝑠+2
= lim𝑠→ ∞
−3+2/𝑠2
1+1/ 𝑠+3/𝑠2+2/ 𝑠3 =−3
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4.2 Properties of Laplace transform
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Final-Value Theorem
lim𝑠 →∞
𝑥(𝑡)= lim𝑠→0
𝑠𝑋 (𝑠)
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
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4.2 Properties of Laplace transform
Ex. Given Solution
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lim𝑠 →∞
𝑓 (𝑡 )=lim𝑠→ 0
𝑠𝐹 (𝑠 )= lim𝑠→ 0
𝑠 5𝑠(𝑠2+𝑠+2)
=52
Ex. Given Solution System is unstable so there is no final value
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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 “marginally stable” if it has at least one distinct pole on the jω axis but no repeated poles on jω Marginally breaks
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Stability In most applications we desire a stable system We can easily check for stability by looking to see where the
system’s poles areExample i. ii. Solution i. All poles are on LHP system is stable
ii. One pole on RHP system is unstable
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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
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Inverse Laplace Transform
Ex. Find x(t)Solution:
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Real simple Poles
𝑋 1(𝑠 )
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Inverse Laplace Transform
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Inverse Laplace Transform
Ex. Solution
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Repeated Real Poles
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Inverse Laplace Transform
Ex. find x(t)Solution
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Simple Complex Poles
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Inverse Laplace Transform
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Inverse Laplace TransformX(s) contains distinct complex roots:
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Inverse Laplace Transform
Ex. Solution:
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Repeated Complex Poles
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Inverse Laplace Transform
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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)
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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
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Time Domain Laplace (Frequency) Domain
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4.3 Solution of Differential & Integro-Differential Equations
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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.Let’s think about where the terms come from:
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Input term
Initial condition term
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4.3 Solution of Differential & Integro-Differential Equations
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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
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Input
Output
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4.4 Analysis of Electrical Networks
We’ve seen that the system output’s LT is:
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So, if the system is in zero-state then we only get the second term:
⇒System effect in zero-state case is completely set by the transfer function
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4.4 Analysis of Electrical NetworksPoles and Zeros of a system Given a system with Transfer Function:
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We can factor B(s) and A(s): (Recall: A(s) = characteristic polynomial)
Pole-Zero PlotThis gives us a graphical view of the system’s behavior
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4.4 Analysis of Electrical Networks
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Example
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4.4 Analysis of Electrical Networks
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Time Domain S- Domain
𝐼= 𝑉𝑠𝐿 +
𝐼0
𝑠𝑉 (𝑠 )=𝐿 [𝑆𝐼 (𝑠 )−𝑖 (0 )]
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4.4 Analysis of Electrical Networks
Example: given the Circuit shown , find y(t)Solution:Apply Laplace TransformThe total voltage in the loop is
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𝑦 (𝑡 )=√5𝑒−𝑡 cos [2 𝑡+26 . 6° ]𝑢 (𝑡)
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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;
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Gramer rule yields
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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
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Solution: The Loop Eq. are
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4.4 Analysis of Electrical Networks
Gramer’s rule yields
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4.5 Block Diagrams
• 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.
• Each subsystem can be characterized in terms of its input-output relationships.
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4.5 Block Diagrams
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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)
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4.5 Block DiagramsExample: A basic feedback system consisting of block find TF
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