CS代写 Revision of Laplace Transforms Algebra of Block Diagrams

Revision of Laplace Transforms Algebra of Block Diagrams

 Motivation
 Complex numbers

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 Definition of Laplace Transform  Properties of Laplace Transform  Partial fraction expansion
 Conclusions

Why we love Laplace transforms
 Time derivatives, integrals and convolutions become algebraic operations in s-domain Much, much simpler to analyse and design systems – and write them down concisely.
 Different functions in time-domain yield different functions in s-domain  No loss of information going from one domain to the other
 Leads to crucial concepts of transfer functions, poles, zeros etc

Complex numbers
 Complex numbers consist of real and imaginary parts:

Complex numbers in complex plane

Operations with complex numbers
 Addition of complex numbers:
 Multiplication of complex numbers:  Exponential of a complex number:

Laplace transform

Motivation (possible design steps)
Time domain: Complex (Laplace) domain.
Step 1: Modelling
Step 2: Design specifications
Step 4: analysis/design
Overshoot, Undershoot, Rise time, Setting time, Steady-state error
Locations of poles
and zeroes of transfer function in the complex plane
Locations of poles in the complex plane
Step 6: Verification, interpretation

Laplace transform
 Laplace transform for signal is
 Inverse Laplace transform of Y(s) is
 Region of convergence is for which where ,

Taken from “Control Systems Design”, Goodwin, Graebe & from “Control Systems Design”, Goodwin, Graebe & :
 It is useful to understand at least a few proofs of items in the tables of Laplace transforms.
 You need to understand all proofs that are given in slides at the end of this lecture.
 It is essential to know how to use Laplace tables/rules in solving problems.

 Determine the unit step response of the following system, assuming the initial condition y(0)=10:
 Step 1: take Laplace transform of both sides:
(we used the linearity of LT and formulas for the derivative and for the unit step signal)

 Step 2: solve for Y(s) – this involves algebraic calculations!
 Step 3: rewrite Y(s) noting that y(0)=10 and using “partial fraction expansion”:
 Step 4: take the inverse Laplace transform

Example continued
 Suppose that assuming the same input and initial condition, we now want to calculate
 One way to do this is to go through all previous steps, obtain y(t) and then calculate:

Example continued
 But this is a place where we can use “Final Value Theorem” and calculate the limit directly in complex domain:
 Note that you always need to first verify that the limit exists when using Final Value Theorem (e.g. stability).

 Consider the system
 Assuming zero initial conditions, find:
– Response to unit step signal
– Response to unit impulse signal
– Response to a sine signal with amplitude 2
– Limits as time goes to infinity for all the above

Example revisited
 Recall that in the example in we needed to find the inverse Laplace transform of:
 This term can not be found in LT tables. We need to transform it to its “partial fraction form”:
 Note that in new form, we can find all terms in LT tables. We show how to do this in general.

Partial fraction expansion (non-repeated poles)
 We can represent strictly proper transfer functions with different poles as follows:
where “residues” are computed as follows  Using LT tables, we have

 If we consider
its poles are s=0 and s=-1. Using the
formulas we have just given, we obtain
which confirms what we already shown earlier by direct computation.

Complex conjugate poles
 Consider a typical term in FPE:
 When the pole is complex, then the residue is
too and they appear in conjugate pairs: (if system impulse response is real-valued).

Complex conjugate pole
 Direct calculations yield:
 In the last line we have a real function!

Partial fraction expansion (repeated poles)
 Consider now a transfer function with repeated poles that can be written as follows:

 PFE can be written as follows:
See Ogata “Modern Control Engineering”

Partial fraction expansion via Matlab
 We can write in Matlab:
num=[2 5 3 6]
den=[1 6 11 6] [r,p,k]=residue(num,den)
r =-6.0000 -4.0000 3.0000
p =-3.0000 -2.0000 -1.0000

Example 1 (exponential):

Example 2 (powers):

Property 1 (integration):

Property 2 (time shift):

Property 3 (exponential weighting):

Property 4 (differentiation):

Property 5 (linearity)

Properties 6 & 7 (limits)
NOTE: You need to always check that limits exist before applying formulas! We will see that existence of these limits is related to the notion of “stability”.

Transfer function
 Consider an input-output model:  Assuming zero initial conditions:  This yields:

Property 8 (convolution)

 Transfer functions obtained assuming zero initial conditions.
 Transfer function = Laplace transform of system impulse response.
 Often easier to find the transfer function than the impulse response
 Convolution is difficult to do. But if we go to s- domain it becomes multiplication: easy.

Algebra of block diagrams

Series connection

Parallel connection

Unity feedback connection
Feedforward transfer function Comparator

General feedback connection
Feedforward connection
Comparator + feedback

Two inputs:
Turn off one input and compute the transfer function in the usual manner for the other input.
This is how we will compute “sensitivity functions”.

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