laplace and fourier transformations -...

Laplace andFourier Trans-formations

Paweł Ptaszek

Process Dynamics

Laplace and Fourier Transformations

Paweł Ptaszek

Cracov University of Technology

May 22, 2013

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Paweł Ptaszek

Process Dynamics

The Laplace transformation of a function of time f (t) consistsof “operating on” the function by multiplaying it by e−st andintegrating with respect to time t from 0 to infinity. Theoperation of Laplace transforming will be indicated by thenototation:

L[f (t)] ≡∫ ∞0f (t)e−stdt

where:L = Laplace transform operators = Laplace transform variableIn integrating between the definite limits of 0 and infinity we“integrate out” the time variable t and are left with a newqantity that is a function of s. We will use the notation:

L[f (t)] ≡ F (s)

The variable s is a complex number.2 / 33


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Process Dynamics

Laplace transformation converts function from the time domain(where t is the independent variable) into the Laplace domain(where s is the independent variable).

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Process Dynamics

Linearity property

L[f1(t) + f2(t)] = L[f1(t)] + L[f2(t)]

L[f1(t) + f2(t)] =

∫ ∞0

[f1(t) + f2(t)]e−stdt

=

∫ ∞0f1(t)e−stdt +

∫ ∞0f2(t)e−stdt

= L[f1(t)] + L[f2(t)] = F1(s) + F2(s)

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Process Dynamics

Laplace transformation of important functions

Step functionf (t) = Ku(t)

where K is a constant and u(t) is the unit step functiondefine as:

u(t) = 1 for t > 0

u(t) = 0 for t ¬ 0

Laplace transforming this function gives

L[Ku(t)] ≡∫ ∞0

[Ku(t)]e−stdt = K∫ ∞0e−stdt

L[Ku(t)] =

[−Kse−st

]t=∞t=0

= −Ks

[0− 1] =Ks

= [0− 0]−[Ks2e−st

]t=∞t=0

= K(

1s2

) 5 / 33


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Process Dynamics


Rampf (t) = Kt

where K is a constantLaplace transforming this function gives

L[Kt] ≡∫ ∞0

[Kt]e−stdt

We use integrating by parts

L[Kt] = K∫ ∞0te−stdt =

[−Ktse−st

]t=∞t=0

+

∫ ∞0

Kse−stdt

L[Kt] = [0− 0]−[Ks2e−st

]t=∞t=0

= K(

1s2

)6 / 33


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Process Dynamics


Sinef (t) = sin(ωt)

where ω is a frequencyLaplace transforming this function gives

L[sin(ωt)] ≡∫ ∞0

[sin(ωt)]e−stdt

We use Euler’s formula

sin(ωt) =e jωt − e−jωt

2j

L[sin(ωt)] =

∫ ∞0

e jωt − e−jωt

2je−stdt

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Process Dynamics


Sine (continuation)

L[sin(ωt)] =

∫ ∞0

e jωt − e−jωt

2je−stdt

=12j

[−e−(s−jω)t

s − jω+e−(s+jω)t

s + jω

]t=∞t=0

=12j

[1s − jω

− 1s + jω

]

L[sin(ωt)] =ω

s2 + ω2

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Process Dynamics


Exponentialf (t) = e−at

where a is a contantLaplace transforming this function gives

L[e−at ] ≡∫ ∞0e−ate−stdt

L[e−at ] =

[ −1s + a

e−(s+a)t]t=∞t=0

=1s + a

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Process Dynamics


Exponential multiplied by time

f (t) = te−at

where a is a contantLaplace transforming this function gives

L[te−at ] ≡∫ ∞0te−ate−stdt =

∫ ∞0te−(a+s)tdt

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Process Dynamics


Exponential multiplied by timeWe use integrating by parts

L[te−at ] =

∫ ∞0te−(a+s)tdt

=

[−te−(s+a)t

s + a

]t=∞t=0

+

∫ ∞0

e−(s+a)t

s + adt

= [0− 0]−[

1(s + a)2

e−(s+a)t]t=∞t=0

L[te−st ] =1

(s + a)2

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Exponential multiplied by time (general case)

L[tne−at ] =n!

(s + a)n+1

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Process Dynamics


Impulse (Dirac delta function δ(t))

δ(t) =du(t)dt

u(t) = limτ→0

[1− e−t/τ ]

L[δ(t)] = L[ddt{ limτ→0

(1− e−t/τ )}]

= limτ→0L[

1τe−t/τ

]L[δ(t)] = lim

τ→0

[1

τs + 1

]= 1

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Process Dynamics

Laplace transformation of mathematical operations

Differentation with respect to time

L[dxdt

]=

∫ ∞0

(dxdt

)e−stdt

We use integrating by parts∫ ∞0

(dxdt

)e−stdt =

[xe−st

]t=∞t=0 +

∫ ∞0sxe−stdt

= 0− xt=0 + s∫ ∞0x(t)e−stdt

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Process Dynamics


Differentation ... (continuation)

L[dxdt

]= sX (s)− xt=0

The operation of differentiation in th time domain is replacedby multiplication by s in the Laplace domain, minus an initialcondition.

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Process Dynamics


Differentation ... (continuation)

L

[d2xdt2

]= s2X (s)− sxt=0 −

(dxdt

)t=0

In general case dNxdtN is replaced by sN , when all initial

conditions are zero.

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Process Dynamics


Differentation ... (continuation)The Nth-order differential equation becomes an Nth-orderalgebraic equation ( when all initial conditions are zero)

aNdNxdtN

+ aN−1dN−1xdtN−1

+ · · ·+ a1dxdt

+ a0x = m(t)

aNsNX (s)+aN−1s

N−1X (s)+· · ·+a1sX (s)+a0X (s) = M(s)

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Process Dynamics


Integration

L[∫x(t)dt

]=

∫ ∞0

(∫x(t)dt

)e−stdt

We use integrating by parts

L[∫x(t)dt

]=

[−1se−st

∫xdt]t=∞t=0

+1s

∫ ∞0x(t)e−stdt

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Process Dynamics


Integration ... (continuation)

L[∫x(t)dt

]=

1sX (s) +

1s

(∫x(t)dt

)t=0

The operation of integration is equivalent to division by sin the Laplace domain, using zero initial conditions. Thus,integration is the inverse of differentiation.

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Convolution

f1(t) ∗ f2(t) =

∫ t0f1(t − τ)f2(τ)dτ

f1(t) ∗ f2(t) = f2(t) ∗ f1(t)

[f1(t) ∗ f2(t)] ∗ f3(t) = f1(t) ∗ [f2(t) ∗ f3(t)]

[f1(t) + f2(t)] ∗ f3(t) = f1(t) ∗ f3(t) + f2(t) ∗ f3(t)

L[f1(t)]L[f2(t)] = L[f1(t) ∗ f2(t)]

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Process Dynamics

Transfer function

x(t) = Km(t)∫ ∞0x(t)e−stdt = K

∫ ∞0m(t)e−stdt

X (s) = KM(s)

Now we have transfer function (operator transmittance)

X (s)M(s)

= K

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Process Dynamics

Inversion of Laplace tranforms

L−1[F (t)] = f (t)

f (t) =1

2πj

∫ α+jω

α−jωestF (s)ds

F (s) = F1(s) + F2(s) + · · ·+ FN(s)

f (t) = L−1[F1(s)] + L−1[F2(s)] + · · ·+ L−1[FN(s)]

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F (s) =Z (s)P(s)

Z (s) = Mth order polynomial in sP(s) = Nth order polynomial in sFactoring the denominator into its roots (or zeros) gives

F (s) =Z

(s − p1)(s − p2)(s − p3) · · · (s − pN)

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F (s) =As − p1

+Bs − p2

+Cs − p3

+ · · ·+ Ws − pN

A = lims→p1

[(s − p1)F (s)]

B = lims→p2

[(s − p2)F (s)]

C = lims→p3

[(s − p3)F (s)]

...

W = lims→pN

[(s − pN)F (s)]

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Process Dynamics

Fourier tarnsformation

The Fourier transformation is special case of Lapacetransformation when s = jω

X (jω) =

∫ +∞

−∞x(t)e−jωtdt

As result we obtain complex Fourier spectrum of our signalx(t). This means that Fourier transform decompose signal onhis different frequency components (very simple approach).Some times Fourier spectrum is called amplitude spectrum.

Fourier spectrum is complex number, and we have real partRe[X (jω)] and imaginary part Im[X (jω)].

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Process Dynamics


The Fourier transformation is special case of Lapacetransformation when s = jω

X (jω) =

∫ +∞

−∞x(t)e−jωtdt

As result we obtain complex Fourier spectrum of our signalx(t). This means that Fourier transform decompose signal onhis different frequency components (very simple approach).Some times Fourier spectrum is called amplitude spectrum.Fourier spectrum is complex number, and we have real partRe[X (jω)] and imaginary part Im[X (jω)].

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also we can show amplitude

|X (jω)| =√Re[X (jω)]2 + Im[X (jω)]2

and phase

argX (jω) = arc tg(Im[X (jω)]

Re[X (jω)]

)= θ(ω)

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Process Dynamics


Inverse transform - synthesis of signal:

x(t) =

∫ +∞

−∞X (jω)e jωtdt

X (jω) - “weight” of frequency

Examples we will discuss on exercise :)

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Process Dynamics


Inverse transform - synthesis of signal:

x(t) =

∫ +∞

−∞X (jω)e jωtdt

X (jω) - “weight” of frequencyExamples we will discuss on exercise :)

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Process Dynamics

Frequency domain dynamics

The frequency response of most processes is defined as thesteadystate behavior of the system when forced by a sinusoidalinput.Forced function - sine wave:

m(t) = m̄ · sin(ωt)

The perion of one complete cycle is T but better quantity isfrequency

ω =1T

dim(ω) = Hz

ω =2πT

dim(ω) =rads

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On outputx(t) = x̄ · sin(ωt + θ)

x(t) - output resulting from the forced function (sine wave offrequency ωx̄ - amplitude of x(t)θ - phase angle in radians (phase shift angle)

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Process Dynamics


Making the substitutions s = jω gives a complex numberG (jω) = Re[G (jω)] + j · Im[G (jω)] that has the following:

A magnitude |G (jω)| that is the same as the magnituderatio M that would be obtained by forcing the sysytemwith a sine wave input of frequency ω

|G (jω)| =√

(Re[G (jω])2 + (Im[G (jω])2 = M =x̄m̄

A phase angle or argument argG (jω), that is equal to thephase angle θ that would be obtained when forcing thesystem with a sine wave of freguency ω

argG (jω) = arc tg(Im[G (jω)]

Re[G (jω)]

)= θ(ω)

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In Euler’s form (polar form), the complex number G (jω) isrepresented as

G (jω) = |G (jω)|e j ·argG(jω)

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Frequency domain dynamics - representation

Nyquist plots (polar plot) - is generated by ploting thecomplex number G (jω) in a two dimensional diagramwhose ordinate is the imaginary part of G (jω) and whoseabscissa is the real part of G (jω). The real and imaginaryparts of G (jω) at a specific value of frequency ω1 defibe apoint in this coordinate system.

Bode plots - reqiure that two curves be plotted instead ofone. The two curves show how magnitude ratio and phaseangle vary with frequency. The magnitude ratio is plottedagainst the log of frequency on log − log plot. In this caselog modulus is defined

L ≡ 20 log10|G (jω)|

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Process Dynamics


Nyquist plots (polar plot) - is generated by ploting thecomplex number G (jω) in a two dimensional diagramwhose ordinate is the imaginary part of G (jω) and whoseabscissa is the real part of G (jω). The real and imaginaryparts of G (jω) at a specific value of frequency ω1 defibe apoint in this coordinate system.

Bode plots - reqiure that two curves be plotted instead ofone. The two curves show how magnitude ratio and phaseangle vary with frequency. The magnitude ratio is plottedagainst the log of frequency on log − log plot. In this caselog modulus is defined

L ≡ 20 log10|G (jω)|32 / 33


Paweł Ptaszek

Process Dynamics


Nichols plots - it is a single curve in a coordinate systemwith phase as the abscissa and log modulus as theordinate. Frequency is a parameter along the curve.

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