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TRANSCRIPT
Laplace transform
EO2 – Lecture 3
Pavel Máša
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• We know, the Fourier transform due to strict conditions of existence does not exists for number of very common waveforms – even sin function have to be dumped
How to ensure, the transform will exist for many waveforms (almost all physically feasible waveforms)?
We dump it ourselves – we will multiply the waveform byfunction
INTRODUCTION
e¡¾te¡¾t
XE31EO2 - Pavel Máša - Lecture 3
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• We know the Fourier transform
• To extend the set of integrable functions, we dump f (t) function by – But dumping has effect just when t ≥ 0,
negative time „amplifies“ the function,we have to introduce condition t ≥ 0
– In circuit analysis we study tasks „what happens after...“, so this condition is not limiting– The history of the circuit is described by initial conditions (voltage across capacitor,
current passing inductor at t = 0).
• Direct Laplace transform
• When σ = 0 the Laplace transform come into Fourier transform
FROM FOURIER TRANSFORM TO LAPLACE TRANSFORM
F(¾; j!) =
Z +1
0
f (t)e¡¾te¡j!t dt =
Z +1
0
f (t)e¡(¾+j!)t dt =
Z +1
0
f (t)e¡pt dtF(¾; j!) =
Z +1
0
f (t)e¡¾te¡j!t dt =
Z +1
0
f (t)e¡(¾+j!)t dt =
Z +1
0
f (t)e¡pt dt
e¡¾t
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Note – fonts
Two different fonts are used for typesetting of Fourier and Laplace transform
• Consequently, we can meet with two different symbols for typing of
– Fourier transform
– Laplace transform
Note
• Instead of p operator s is sometimes used
• p / s is sometimes called complex frequency
Lff (t)g
F ff (t)g
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• For dumping function we defined necessary condition t ≥ 0• But this condition is not sufficient to ensure the (Fourier) integral of the function converges
– When σ < 0 (or certain number), it does not dump, but „amplifies“, depending up to properties of f(t) function
The Laplace transform of the function f(t) exists for all complex numbers such that
The part of p plane satisfying this condition is called region of convergence (don’t confuse with region of stability, with lays in left part of p plane, left from the region of convergence (and left from imaginary axis)!!! – poles diverge)
REGION OF CONVERGENCE
e¡¾t
f(t)e¡¾t
¾ > ¾min
(BIBO) stability• BIBO – Bounded‐Input Bounded‐Output – If a
system is BIBO stable, then the output will be bounded for every input to the system that is bounded.
• Passive circuit is always stable, if it contains non zero resistivity, zero resistivity – limit of stability
• Active circuit (containing some amplifier) need to have feedback
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Fourier transform (frequency characteristic)
pole (tends to ∞)
P (p) =1
1 + pRCP (p) =
1
1 + pRC
EXAMPLE– 1 POLE IN P‐PLANE
R = 100 ÐR = 100 Ð
C = 1 mFC = 1 mF
pp = ¡10pp = ¡10pole
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200160-200 120-150 80
sigma
-100 40-50 00-200
200
600
1 000
1 400
1 800
50
2 200
-40
2 600
3 000
3 400
3 800
4 200
4 600
5 000
100-80
omega
-120-160-200
zeroes
poles
poles
3‐D view of p‐plane, side view
the upper view of the same p‐plane
P (p) =pC(pL + R)
p2LC + pRC + 1P (p) =
pC(pL + R)
p2LC + pRC + 1
Zeroes: p01 = 0; p02 = ¡100p01 = 0; p02 = ¡100
Poles: pp1;2 = ¡50 § 86:6jpp1;2 = ¡50 § 86:6j
R = 10 ÐR = 10 Ð
L = 0:1 HL = 0:1 H
C = 1 mFC = 1 mF
EXAMPLE – ZEROES AND POLES IN P‐PLANE
Fourier transform – imaginary axis XE31EO2 - Pavel Máša - Lecture 3
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property time domain Laplace transform(frequency domain)
Linearity
Time shifting
Differentiation
integration
convolution
SELECTED PROPERTIES OF LAPLACE TRANSFORM
df (t)
dtpF (p)¡f (0+)Z t
0
f (¿ ) d¿1
pF (p)
f(t) ¤ g(t) =
Z t
0
f (¿ )g(t¡ ¿ ) d¿
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Time domain Frequency domainRegion of
convergence
Dirac delta function 1
Unit step function(DC voltage connected at
time t = 0)
Exponential decay
TABLE OF SELECTED LAPLACE TRANSFORMS
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Time domain Frequency domainRegion of
convergence
Exponentially‐decayingsine wave
Exponentially‐decayingcosine wave
Phase‐shifted sine wave
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• We start from inverse Fourier transform
• Using inverse Fourier transform we will evaluate dumped function
• Moving dumping on right side of equation
• By substitution of variables and integral limits we got Laplace transform
So far as it is possible, we don’t use the inverse Laplace transform integral directly, but we try to use transform properties and known transforms from the Table
INVERSE LAPLACE TRANSFORM
f(t) =1
2¼
Z +1
¡1F(j!)ej!td!f(t) =
1
2¼
Z +1
¡1F(j!)ej!td!
f (t)e¡¾t =1
2¼
Z +1
¡1F(p)ej!t d!f (t)e¡¾t =
1
2¼
Z +1
¡1F(p)ej!t d!
f (t) =1
2¼
Z +1
¡1F(p)e¾tej!t d! =
Z +1
¡1F(p)ept d!f (t) =
1
2¼
Z +1
¡1F(p)e¾tej!t d! =
Z +1
¡1F(p)ept d!
¡1 fF (p)g = f (t) =1
2¼j
Z ¾+1
¾¡1F(p)ept dp¡1 fF (p)g = f (t) =
1
2¼j
Z ¾+1
¾¡1F(p)ept dp
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• The resulting network function (e.g. transfer function P(p)) / variable (voltage transform U(p), current transform I(p), …) will be rational function, ratio of two polynomial functions
• First, by possible polynomial division we have to ensure, the degree of polynomial P(p) in nominator is smaller than degree of polynomial Q(p) in denominator; at once we can factor out the highest order polynomial coefficient in denominator
• We will compute both polynomial roots in nominator (zeros) and denominator (poles) of the function
• Now, the partial fraction decomposition procedure depends on pole type
1. Simple real roots
THE PROCEDURE OF INVERSE LAPLACE TRANSFORM EVALUATION
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2. Repeated real roots with root multiplicity α, β, γ
3. The pair of complex conjugate roots
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• Now we have left to find coefficients A, B, …1. General method (compare coefficients at same powers)
1. Function F’(p) (after partial fraction decomposition) multiply by initial denominator
2. Compare coefficients with the same power of p in nominator of original function
1.
2.
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2. Single pole method
it is not universal tool, when the polynomial has roots with multiplicity, it can be used only with root of highest power, other coefficients must be calculated using other methods
In the function F’(p) we will replace variable p by value of the root pi.• Bracket, containing root pi must be removed (it is zero).
Mathematical description:
Example:
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If (σ = 0) and without initial conditions Fourier, if moreover sine waveform source sinusoidal steady state
→
→
→
→
→
→
OPERATIONAL CHARACTERISTICS OF TWO PORTS
Transform of derivative – multiplication by p (and initial condition)
Transform of an integral – division by p
Kirchhoff's laws in operational form
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INDUCTOR AND CAPACITOR EQUIVALENT CIRCUIT DIAGRAMS
initial conditionvoltage source
operationalimpedance
initial conditioncurrent source
operationaladmittance
initial conditioncurrent source
operationaladmittance
operationalimpedance
initial conditionvoltage source
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EQUIVALENT CIRCUIT AND TERMINALS OF REAL CIRCUIT ELEMENT
The capacitor was charged at 10 V. Find the current passing the capacitor.
parallel equivalent circuit
series equivalent circuit
• Ohm’s law
• current divider
??? But the current passing capacitor should be the same???
This is actual capacitor!!!
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In time t = 0 an inductor was passed by the current iL(0) = 2 A. Find the Laplace transform of current passing the inductor when t > 0 and voltage transform at t > 0.
parallel equivalent circuit
series equivalent circuit
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When the initial conditions are zero, the operational characteristics will be analogical to those in Fourier transform or sinusoidal steady state
• Impedance and admittance, including the input one of two ports
• Transfer function (voltage, current, …)
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The integrating network in the figure is excited by rectangular pulse in second figure. Compute the waveform of output voltage. The capacitor has zero voltage at the time of connection of source (zero initial condition).
• To find the solution we will use table of Laplace transforms• rectangular pulse is superposition of two unit step functions multiplied by Um
1.
2.
3.
Statement in the square bracket will be temporary omitted (it is information about time delay, transformed later)
The transform of the square bracket are two unit step functions, the second is time shifted by t0
t0
0
Um
t
EXAMPLE – THE SAME AS IN THE LAST LECTURE
P (p) =1
1 + pRCP (p) =
1
1 + pRC
U2(p) = U1(p) ¢ P (p) =Um
p
1
1 + pRC
£1¡ e¡pt0
¤U2(p) = U1(p) ¢ P (p) =
Um
p
1
1 + pRC
£1¡ e¡pt0
¤
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