Analog
Design
Kevin
Aylward B.Sc
Laplace
Transfer Functions
Back to Contents
Abstract
This paper forms an introduction to Laplace Transfer functions as it
applies to electronic design. The mathematics is kept to the barest minimum. In
practice, for electronic design, the Laplace Transform is used in such a manner
that the mathematical basis is usually forgotten. So, for the most part, one
can gloss over the mathematics and just use the results, which are trivial to
apply in practice.
In reality, there is no need to ever solve the equations. Today this is
done in Spice simulation programs. However, to design effectively, one requires
a conceptual model of what a transfer function is, and its key features.
Overview
The following are the principle concepts for the introduction of
Laplace Transforms.
1 The basic equations of electronics are differential equations with
respect to time.
2 Differential equations are harder to solve then algebraic equations.
In short, Laplace Transforms convert differential equations into
algebraic equations. These algebraic equations are then solved. The solutions
to these equations are then inverted to obtain the time response of an
electronic circuit.
In addition, because of the Laplace Transform's relation to the Fourier
Transform, the steady state frequency response can also be obtained from the
Laplace transformed equations.
Laplace Transform
The Laplace transform is defined by:
g(s)=
∫
0
∞
f(t)
e
−st
dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiaacIcacaWGZbGaaiykaiabg2da9maapehabaGaamOzaiaacIcacaWG0bGaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadohacaWG0baaaaqaaiaaicdaaeaacqGHEisPa0Gaey4kIipakiaadsgacaWG0baaaa@47CE@
If f(t)=0 for t<0 and s=jω then g(jω)=ℑ(f(t)) where ℑ is the fourier transform of f(t)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@81EC@
As far as electronics goes, the most important
results from the definition, are the following.
Laplace transform of a derivative of function
Consider:
h(s)=
∫
0
∞
f'(t)
e
−st
dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9maapehabaGaamOzaiaacEcacaGGOaGaamiDaiaacMcacaWGLbWaaWbaaSqabeaacqGHsislcaWGZbGaamiDaaaaaeaacaaIWaaabaGaeyOhIukaniabgUIiYdGccaWGKbGaamiDaaaa@487A@
Integrating by parts.
h(s)=
[
f(t)
e
−st
]
0
∞
+s
∫
0
∞
f(t)
e
−st
dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9maadmaabaGaamOzaiaacIcacaWG0bGaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadohacaWG0baaaaGccaGLBbGaayzxaaWaa0baaSqaaiaaicdaaeaacqGHEisPaaGccqGHRaWkcaWGZbWaa8qCaeaacaWGMbGaaiikaiaadshacaGGPaGaamyzamaaCaaaleqabaGaeyOeI0Iaam4CaiaadshaaaaabaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aOGaamizaiaadshaaaa@5539@
h(s)=0+sg(s)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9iaaicdacqGHRaWkcaWGZbGaam4zaiaacIcacaWGZbGaaiykaaaa@3FF9@
h(s)=sg(s)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9iaadohacaWGNbGaaiikaiaadohacaGGPaaaaa@3E5D@
That is the Laplace transform of the
derivative of a function is the Laplace Transform of the function multiplied by s. In general the nth
derivative of a function has a Laplace Transform that is multiplied by sn.
Consider:
h(s)=
∫
0
∞
f
i
(t)
e
−st
dt, where
f
i
(t) is the integral of f(t)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9maapehabaGaamOzamaaBaaaleaacaWGPbaabeaakiaacIcacaWG0bGaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadohacaWG0baaaaqaaiaaicdaaeaacqGHEisPa0Gaey4kIipakiaadsgacaWG0bGaaiilaiaabccacaqG3bGaaeiAaiaabwgacaqGYbGaaeyzaiaabccacaWGMbWaaSbaaSqaaiaadMgaaeqaaOGaamikaiaadshacaWGPaGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGPbGaaeOBaiaabshacaqGLbGaae4zaiaabkhacaqGHbGaaeiBaiaabccacaqGVbGaaeOzaiaabccacaWGMbGaaiikaiaadshacaGGPaaaaa@6857@
Integrating by parts.
h(s)=−
1
s
[
f
i
(t)
e
−st
]
0
∞
+
1
s
∫
0
∞
f
(t)
e
−st
dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9iabgkHiTmaalaaabaGaaGymaaqaaiaadohaaaWaamWaaeaacaWGMbWaaSbaaSqaaiaadMgaaeqaaOGaaiikaiaadshacaGGPaGaamyzamaaCaaaleqabaGaeyOeI0Iaam4CaiaadshaaaaakiaawUfacaGLDbaadaqhaaWcbaGaaGimaaqaaiabg6HiLcaakiabgUcaRmaalaaabaGaaGymaaqaaiaadohaaaWaa8qCaeaacaWGMbWaaSbaaSqaaaqabaGccaGGOaGaamiDaiaacMcacaWGLbWaaWbaaSqabeaacqGHsislcaWGZbGaamiDaaaaaeaacaaIWaaabaGaeyOhIukaniabgUIiYdGccaWGKbGaamiDaaaa@5A0E@
h(s)=−
f
i
(0)
s
e
−st
+
1
s
∫
0
∞
f
(t)
e
−st
dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9iabgkHiTmaalaaabaGaamOzamaaBaaaleaacaWGPbaabeaakiaacIcacaaIWaGaaiykaaqaaiaadohaaaGaamyzamaaCaaaleqabaGaeyOeI0Iaam4CaiaadshaaaGccqGHRaWkdaWcaaqaaiaaigdaaeaacaWGZbaaamaapehabaGaamOzamaaBaaaleaaaeqaaOGaaiikaiaadshacaGGPaGaamyzamaaCaaaleqabaGaeyOeI0Iaam4CaiaadshaaaaabaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aOGaamizaiaadshaaaa@54C0@
For simplicity, assume that fi(0) = 0, then:
h(s)=
g(s)
s
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiaacIcacaWGZbGaaiykaiabg2da9maalaaabaGaam4zaiaacIcacaWGZbGaaiykaaqaaiaadohaaaaaaa@3E6D@
That is the transform of the integral of a function is the transform of
the function divided by s.
Note: In most analysis the initial condition of fi(0) =0 is
of no consequence. It can be added in if required.
Application
Consider a series circuit consisting of a voltage source, resistor
capacitor and inductor.
V
i
=IR+L
di
dt
+
1
c
∫
idt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBaaaleaacaWGPbaabeaakiabg2da9iaadMeacaWGsbGaey4kaSIaamitamaalaaabaGaamizaiaadMgaaeaacaWGKbGaamiDaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaWGJbaaamaapeaabaGaamyAaiaadsgacaWG0baaleqabeqdcqGHRiI8aaaa@476A@
Now multiply the equation by e-st and integrate
∫
0
∞
V
i
e
−st
dt
=
∫
0
∞
I
Re
−st
dt
+
∫
0
∞
L
di
dt
e
−st
dt
+
∫
0
∞
1
c
e
−st
(
∫
idt
)dt
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@72B4@
But because of the relations derived above:
V
¯
i
(s)=
I
¯
(s)R+sL
I
¯
(s)+
1
sc
I
¯
(s)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOvayaaraWaaSbaaSqaaiaadMgaaeqaaOGaaiikaiaadohacaGGPaGaeyypa0JabmysayaaraGaaiikaiaadohacaGGPaGaamOuaiabgUcaRiaadohacaWGmbGabmysayaaraGaaiikaiaadohacaGGPaGaey4kaSYaaSaaaeaacaaIXaaabaGaam4CaiaadogaaaGabmysayaaraGaaiikaiaadohacaGGPaaaaa@4C06@
Were the bars represent the Laplace Transforms of the applied voltage
and current.
V
¯
i
(s)=
I
¯
(s)(R+sL+
1
sc
)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOvayaaraWaaSbaaSqaaiaadMgaaeqaaOGaaiikaiaadohacaGGPaGaeyypa0JabmysayaaraGaaiikaiaadohacaGGPaGaaiikaiaadkfacqGHRaWkcaWGZbGaamitaiabgUcaRmaalaaabaGaaGymaaqaaiaadohacaWGJbaaaiaacMcaaaa@46F1@
V
¯
i
(s)=
I
¯
(s)(R+sL+
1
sc
)
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOvayaaraWaaSbaaSqaaiaadMgaaeqaaOGaaiikaiaadohacaGGPaGaeyypa0JabmysayaaraGaaiikaiaadohacaGGPaGaaiikaiaadkfacqGHRaWkcaWGZbGaamitaiabgUcaRmaalaaabaGaaGymaaqaaiaadohacaWGJbaaaiaacMcaaaa@46F1@
V
¯
i
(s)
I
¯
(s)
=R+sL+
1
sc
MathType@MTEF@5@5@+=feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqaqFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaceWGwbGbaebadaWgaaWcbaGaamyAaaqabaGccaGGOaGaam4CaiaacMcaaeaaceWGjbGbaebacaGGOaGaam4CaiaacMcaaaGaeyypa0JaamOuaiabgUcaRiaadohacaWGmbGaey4kaSYaaSaaaeaacaaIXaaabaGaam4Caiaadogaaaaaaa@45A8@
This is a transfer function. It gives the ration of input voltage to
output current in terms of their Laplace transforms. In this particular case,
the transfer function is the input impedance of the RLC circuit, in terms of
the variable s.
In general, one simple treats inductors and capacitors in the same
manner as resistors, by noting that the Laplace impedance of a capacitor is
1/sc and the impedance of an inductor is sl.
© Kevin Aylward 2013
All rights reserved
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reproduced
providing that this source is acknowledged.
Website last modified 30th August
2013
www.kevinaylward.co.uk