signal very important
TRANSCRIPT
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Fourier Series Fourier Transform
Laplace Transform
Applications of Laplace Transform Z-Transform
BSC Modul 4: Advanced Circuit Analysis
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Fourier Series Fourier Transform Laplace Transform
Applications of Laplace Transform Z-Transform
BSC Modul 4: Advanced Circuit Analysis
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
The Fourier series of a periodic functionf(t) is arepresentation that resolves f(t) into a dc component andan ac component comprising an infinite series of harmonicsinusoids.
Given a periodic function f(t) = f(t+nT)where n is aninteger and T is the period of the function.
where 0=2/T is called the fundamental frequency inradians per second.
ac
n
n
dc
tnbtnaatf
=
++=1
0000 )sincos()(
Trigonometric Fourier Series (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
)2()(and
21,0
10,1)( +=
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Three types of symmetry
1.Even Symmetry : a function f ( t ) if its plot issymmetrical about the vertical axis.
In this case,
)()( tftf =
0
)cos()(
4
)(2
2/
0 0
2/
00
==
=
n
T
n
T
b
dttntfTa
dttfT
a
Typical examples of even periodic function
Symmetry Considerations (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
2.Odd Symmetry : a function f ( t ) if its plot isanti-symmetrical about the vertical axis.
In this case,
)()( tftf =
=
=
2/
00
0
)sin()(4
0
T
n dttntf
T
b
a
Typical examples of odd periodic function
Symmetry Considerations (2)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
3.Half-wave Symmetry : a function f ( t ) if
)()
2
( tfT
tf =
=
=
=
evenanfor,0
oddnfor,)sin()(4
evenanfor,0
oddnfor,)cos()(4
0
2/
00
2/
00
0
T
n
T
n
dttntfTb
dttntfTa
a
Typical examples of half-wave odd periodic functions
Symmetry Considerations (3)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example 1
Find the Fourier series expansion of f(t) givenbelow.
=
=
1 2sin
2cos1
12)(
n
tnn
ntf
Ans:
Symmetry Considerations (4)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example 2
Determine the Fourier series for the half-wavecosine function as shown below.
=
==1
2212,cos
14
2
1)(
k
knntn
tf
Ans:
Symmetry Considerations (5)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Steps for Applying Fourier Series
1.Express the excitation as a Fourier series.
2.Transform the circuit from the time domain to the
frequency domain.3.Find the response of the dc and ac components in
the Fourier series.
4.Add the individual dc and ac responses using the
superposition principle.
Circuit Applications (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example
Find the response v0( t ) of the circuit below whenthe voltage source vs( t ) is given by
( ) 12,sin12
2
1
)(1 =+=
=kntnntv n
s
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Amplitude spectrum of the output
voltage
Solution
Phasor of the circuit
For dc component, (n=0 or n=0), Vs = => Vo = 0
For nth harmonic,
In time domain,
s0 V25
2V
nj
nj
+=
)5
2tan(c425
4)(1
1
220
=
+
=k
ntnosn
tv
s22
1
0 V425
5/2tan4V,90
2V
n
n
nS
+
==
Circuit Applications (3)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Given:
The average power is
The rms value is
=
=+=+=
1
Im0mdc
1
0ndc )cos(II)(and)cos(VV)(mn
Vn tmtitntv
)(1
222
0
= ++= nnnrms baaF
= += 1nndcdc )cos(IV2
1
IVPn
nn
Average Power and RMS Values (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example
Determine the average power supplied to thecircuit shown below ifi( t )= 2+ 10cos(t + 10)+ 6cos(3t+ 35) A
Answer: 41.5W
Average Power and RMS Values (2)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
The exponential Fourier series of a periodic function f ( t ) describes the spectrum off ( t ) in terms of the amplitude andphase angle of ac components at positive and negativeharmonic.
The plots of magnitude and phase ofcnversus n0 are called
the complex amplitude spectrum and complex phasespectrum off ( t ) respectively.
==T tjn
n TdtetfT
c0
0 /2where,)(1 0
==
n
tjnn
oectf )(
Exponential Fourier Series (1)
B i i S d Ci i Th
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Michael E.Auer 01.11.2011 BSC04
The complex frequency spectrum of the function
f(t)=et, 0
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Michael E.Auer 01.11.2011 BSC04
Filter are an important component of electronics and communications
system.
This filtering process cannot be accomplished without the Fourier seriesexpansion of the input signal.
For example,
(a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes
only the dc component when c
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Michael E.Auer 01.11.2011 BSC04
(a) Input and output spectra of a bandpass filter, (b) the bandpass filter
passes only the dc component when
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Fourier Series
Fourier Transform Laplace Transform
Applications of Laplace Transform Z-Transform
BSC Modul 4: Advanced Circuit Analysis
Basics in Systems and Circuits Theory
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
It is an integral transformation of f(t) from the time domainto the frequency domain F()
F() is a complex function; its magnitude is called theamplitude spectrum, while its phase is called the phase
spectrum.Given a function f(t), its Fourier transform denoted by F(),is defined by
= )()( dtetfF
tj
Definition of Fourier Transform (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example 1:
Determine the Fourier transform of a singlerectangular pulse of wide and height A, as shownbelow.
Definition of Fourier Transform (2)
2sin
2
2
2/
2/
)(
2/2/
2/
2/
cA
j
eeA
ej
A
dtAeF
jj
tj
tj
=
=
=
=
Solution:
Amplitude spectrum of the
rectangular pulse
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
Example 2:
Obtain the Fourier transform of theswitched-on exponential function asshown.
Definition of Fourier Transform (3)
Solution:
ja
dte
dteedtetfF
etuetf
tja
tjjattj
at
at
+=
=
==
==
+
1
)()(
Hence,
0t,0
0t,)()(
)(
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
[ ] )()()()( 22112211 FaFatfatfaF +=+
Linearity:
IfF1() and F2() are, respectively, the FourierTransforms off1(t) and f2(t)
Example:
[ ] ( ) ( )[ ] [ ])()(21
)sin( 00000
+==
jeFeFjtF
tjtj
Properties of Fourier Transform (1)
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Basics in Systems and Circuits Theory
Michael E.Auer 01.11.2011 BSC04
[ ] constantais,)(1
)( aa
Fa
atfF
=
Time Scaling:
IfF() is the Fourier Transforms off(t), then
If |a|>1, frequency compression, or time expansion
If |a|
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y y
Michael E.Auer 01.11.2011 BSC04
[ ] )()(0
0
FettfFtj
=
Time Shifting:
IfF() is the Fourier Transforms off( t ) , then
Example:
[ ]
jetueF
j
t
+=
1
)2(
2
)2(
Properties of Fourier Transform (3)
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y y
Michael E.Auer 01.11.2011 BSC04
)()( 00
= FetfF tj
Frequency Shifting (Amplitude Modulation):IfF() is the Fourier Transforms off( t ) , then
Example:
[ ] )(21
)(2
1)cos()( 000 ++= FFttfF
Properties of Fourier Transform (4)
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Michael E.Auer 01.11.2011 BSC04
)()( sFjtudt
df
F =
Time Differentiation:IfF() is the Fourier Transforms off( t ) , then theFourier Transform of its derivative is
Example:
( ) jatuedtd
Fat
+=
1)(
Properties of Fourier Transform (5)
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Michael E.Auer 01.11.2011 BSC04
)()0(
)(
)(
Fj
F
dttfF
t
=
Time Integration:IfF() is the Fourier Transforms off( t ) , then theFourier Transform of its integral is
Example:
[ ] )(1
)( +=
jtuF
Properties of Fourier Transform (6)
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Michael E.Auer 01.11.2011 BSC04
Fourier transforms can be applied to circuits with non-sinusoidalexcitation in exactly the same way as phasor techniques beingapplied to circuits with sinusoidal excitations.
By transforming the functions for the circuit elements into thefrequency domain and take the Fourier transforms of theexcitations, conventional circuit analysis techniques could beapplied to determine unknown response in frequency domain.
Finally, apply the inverse Fourier transform to obtain theresponse in the time domain.
Y() = H()X()
Circuit Application (1)
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Michael E.Auer 01.11.2011 BSC04
Example:
Find v0( t ) in the circuit shown below for
vi( t ) =2e -3tu(t )
Circuit Application (2)
Solution:
)()(4.0)(givesansformFourier trinversetheTaking
)5.0)(3(
1)(V
Hence,
21
1
)(V
)(V)(iscircuittheoffunctiontransferThe
3
2)(VissignalinputtheofansformFourier trThe
35.0
0
0
i
0
i
tueetv
jj
jH
j
tt =
++
=
+==
+
=
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Michael E.Auer 01.11.2011 BSC04
Fourier Series Fourier Transform
Laplace Transform
Applications of Laplace Transform Z-Transform
BSC Modul 4: Advanced Circuit Analysis
Basics in Systems and Circuits Theory
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Michael E.Auer 01.11.2011 BSC04
It is an integral transformation of f(t) from the timedomain to the complex frequency domain F(s)
Given a function f(t), its Laplace transform denoted byF(s), is defined by
Where the parameter s is a complex number
[ ] ==
0)()()( dtetftfLsF st
Definition of Laplace Transform
js += , real numbers
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Michael E.Auer 01.11.2011 BSC04
When one says "the Laplace transform" without qualification,the unilateral or one-sided transform is normally intended.The Laplace transform can be alternatively defined as thebilateral Laplace transformor two-sided Laplace transform byextending the limits of integration to be the entire real axis. If
that is done the common unilateral transform simply becomesa special case of the bilateral transform.
The bilateral Laplace transform is defined as follows:
Bilateral Laplace Transform
[ ]
== )()()( dtetftfLsF st
Basics in Systems and Circuits Theory
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Determine the Laplace transform of each of the following functions shown:
Examples of Laplace Transforms (1)
a) The Laplace Transform of unit step, u(t) is given by
[ ]
=== 01
1)()( sdtesFtuLst
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Michael E.Auer 01.11.2011 BSC04
b) The Laplace Transform of exponential function, e-atu(t), a>0is given by
[ ]
+===
0
1)()(
sdteesFtuL
stt
Examples of Laplace Transforms (2)
c) The Laplace Transform of impulse function, (t) is given by
[ ]
===0
1)()()( dtetsFtuL st
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Michael E.Auer 01.11.2011 BSC04
Examples of Laplace Transforms (3)
ssF
1)( =
+=
ssF
1)( 1)( =sF
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Michael E.Auer 01.11.2011 BSC04
Table of Selected Laplace Transforms (1)
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Michael E.Auer 01.11.2011 BSC04
[ ] )()()()( 22112211 sFasFatfatfaL +=+
Linearity:IfF1(s) and F2(s) are, respectively, the LaplaceTransforms off1( t ) and f2( t )
Example:
[ ] ( )22
)(
2
1)()cos(
+
=
+=
s
stueeLtutL
tjtj
Properties of Laplace Transform (1)
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Michael E.Auer 01.11.2011 BSC04
[ ] )(1)(a
sFa
atfL =
Scaling:
IfF(s) is the Laplace Transforms off( t ) , then
Example:
[ ] 224
2)()2sin(
+
=s
tutL
Properties of Laplace Transform (2)
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Michael E.Auer 01.11.2011 BSC04
[ ] )()()( sFeatuatfL as=
Time Shift:
IfF(s) is the Laplace Transforms off( t ) , then
Example:
[ ] 22)())(cos(
+=
sseatuatL as
Properties of Laplace Transform (3)
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Michael E.Auer 01.11.2011 BSC04
)()()( asFtutfeL at +=
Frequency Shift:
IfF(s) is the Laplace Transforms off(t), then
Example: [ ]22)(
)()cos(
++
+=
as
astuteL
at
Properties of Laplace Transform (4)
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Michael E.Auer 01.11.2011 BSC04
)0()()( =
fssFtu
dt
dfL
Time Differentiation:
IfF(s) is the Laplace Transforms off(t), then theLaplace Transform of its derivative is
Properties of Laplace Transform (5)
Time Integration:
IfF(s) is the Laplace Transforms off( t ) , then theLaplace Transform of its integral is
)(1
)(0
sFs
dttfLt
=
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Michael E.Auer 01.11.2011 BSC04
)(lim)0( ssFfs
=
Initial and Final Values:
The initial-value and final-value properties allow usto find f(0) and f() off(t) directly from its Laplacetransform F(s).
Initial-value theorem
)(lim)(0
ssFfs= Final-value theorem
Properties of Laplace Transform (6)
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Michael E.Auer 01.11.2011 BSC04
In principle we could recover f(t) from F(s) via
But, this formula isnt really useful.
sesFj
tf
j
j
std)(
2
1)(
+
=
The Inverse Laplace Transform (1)
= xF
1
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Michael E.Auer 01.11.2011 BSC04
Suppose F(s) has the general form of
The finding the inverse Laplace transform ofF(s) involves two steps:
1.Decompose F(s) into simple terms using partialfraction expansion.
2.Find the inverse of each term by matching entriesin Laplace Transform Table.
)(
)()(
sD
sNsF =
The Inverse Laplace Transform (2)
numerator polynomial
denominator polynomial
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Michael E.Auer 01.11.2011 BSC04
ExampleFind the inverse Laplace transform of
Solution:4
6
1
53)(
2
+
+
+
=
sss
sF
0t),()2sin(353(
4
6
1
53)(
2
111
+=
++
+
=
tute
sL
sL
sLtf
t
The Inverse Laplace Transform (3)
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Michael E.Auer 01.11.2011 BSC04
The Laplace transform is useful in solving linearintegro-differential equations.
Each term in the integro-differential equation istransformed into s-domain.
Initial conditions are automatically taken intoaccount.
The resulting algebraic equation in the s-domain canthen be solved easily.
The solution is then converted back to time domain.
Application to Integro-differential Equations (1)
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Michael E.Auer 01.11.2011 BSC04
Example:
Use the Laplace transform to solve the differentialequation
Given: v(0) = 1; v(0) = -2
)(2)(8)(6)(2
2
tutvdt
tdvdt
tvd =++
Application to Integro-differential Equations (2)
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Michael E.Auer 01.11.2011 BSC04
Solution:Taking the Laplace transform of each term in the givendifferential equation and obtain
[ ] [ ]
)()21(4
1)(
Transform,LaplaceinverseBy the
42)(
2424)()86(
havewe,2)0(';1)0(ngSubstituti
2
)(8)0()(6)0(')0()(
42
41
21
412
2
2
tueetv
ssssV
s
ss
sssVss
vv
ssVvssVvsvsVs
tt ++=
++
++=
++=++=++
===++
Application to Integro-differential Equations (3)
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Michael E.Auer 01.11.2011 BSC04
Fourier Series Fourier Transform
Laplace Transform
Applications of Laplace Transform Z-Transform
BSC Modul 4: Advanced Circuit Analysis
Basics in Systems and Circuits Theory
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Michael E.Auer 01.11.2011 BSC04
Steps in Applying the Laplace Transform:1.Transform the circuit from the time domain to
the s-domain
2.Solve the circuit using nodal analysis, mesh
analysis, source transformation, superposition, orany circuit analysis technique with which we arefamiliar
3.Take the inverse transform of the solution and
thus obtain the solution in the time domain.
Circuit Element Models (1)
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Michael E.Auer 01.11.2011 BSC04
Assume zero initial condition forthe inductor and capacitor,
Resistor : V(s)=RI(s)
Inductor: V(s)=sLI(s)
Capacitor: V(s) = I(s)/sCThe impedance in the s-domain
is defined as Z(s) = V(s)/I(s)
The admittance in the s-domainis defined as Y(s) = I(s)/V(s)
Time-domain and s-domain representations of passive
elements under zero initial conditions.
Circuit Element Models (2)
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Michael E.Auer 01.11.2011 BSC04
Circuit Element Models (3)
Non-zero initial condition for the
inductor and capacitor,
Resistor : V(s)=RI(s)
Inductor: V(s)=sLI(s) + LI(0)
Capacitor: V(s) = I(s)/sC + v(0)/s
Basics in Systems and Circuits Theory
I d E l
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Introductory Example
Charging of a capacitor v
V(0) = 0
Basics in Systems and Circuits Theory
Ci i El M d l E l (1)
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Michael E.Auer 01.11.2011 BSC04
Example 1:
Find v0(t) in the circuit shown below, assuming zero initialconditions.
Circuit Element Models Examples (1)
Basics in Systems and Circuits Theory
Ci it El t M d l E l (2)
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Michael E.Auer 01.11.2011 BSC04
Solution:
Transform the circuit from the time domain to the s-domain:
s
L
tu
3sC1F
31
ssH1
s
1)(
=
=
Apply mesh analysis, on solving for V0(s):
220 )2()4(
2
2
3)(V ++= ss 0V,)2sin(2
3)( 40 =
ttetv t
Inverse transform
Circuit Element Models Examples (2)
Basics in Systems and Circuits Theory
Ci it El t M d l E l (3)
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Michael E.Auer 01.11.2011 BSC04
Example 2:
Determine v0(t) in the circuit shown below, assuming zeroinitial conditions.
V)()21(8:Ans 22 tutee tt
Circuit Element Models Examples (3)
Basics in Systems and Circuits Theory
Ci it El t M d l E l (4)
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Michael E.Auer 01.11.2011 BSC04
Example 3:
Find v0(t) in the circuit shown below. Assume v0(0)=5V.
V)()1510()(v:Ans 20 tueettt +=
Circuit Element Models Examples (4)
Basics in Systems and Circuits Theory
Circuit Element Models Examples (5)
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Example 4:
The switch shown below has been in position bfor a longtime. It is moved to position aat t=0. Determine v(t) for t > 0.
where0,t,I)IV()(v:Ans 0/
00 RCReRtt =>+=
Circuit Element Models Examples (5)
Basics in Systems and Circuits Theory
Circuit Analysis
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Circuit analysis is relatively easy to do in the s-domain.By transforming a complicated set of mathematicalrelationships in the time domain into the s-domain where weconvert operators (derivatives and integrals) into simple
multipliers of s and 1/s.This allow us to use algebra to set up and solve the circuitequations.
In this case, all the circuit theorems and relationships
developed for dc circuits are perfectly valid in the s-domain.
Circuit Analysis
Basics in Systems and Circuits Theory
Circuit Analysis Example (1)
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Example:
Consider the circuit below. Find thevalue of the voltage across thecapacitor assuming that the value
of vs(t)=10u(t) V and assume thatat t=0, -1A flows through theinductor and +5V is across thecapacitor.
Circuit Analysis Example (1)
Basics in Systems and Circuits Theory
Ci it A l i E l (2)
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Michael E.Auer 01.11.2011 BSC04
Solution:Transform the circuit from time-domain (a) into s-domain (b) using LaplaceTransform. On rearranging the terms, we have
By taking the inverse transform, we get
2
30
1
35V1
+
+
=ss
V)()3035()(v 21 tueettt =
Circuit Analysis Example (2)
Basics in Systems and Circuits Theory
Circuit Analysis Example (3)
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Michael E.Auer 01.11.2011 BSC04
Example:
The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t) V.(a) Find V0(s) using the Thevenin theorem. (b) Apply the initial- and final-value theorem to find v0(0) an v0(). (c) Obtain v0(t).
Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c) (3.333+0.6667e-0.3t)u(t) V.
Circuit Analysis Example (3)
Basics in Systems and Circuits Theory
Transfer Functions
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The transfer function H(s) is the ratio of the output responseY(s) to the input response X(s), assuming all the initialconditions are zero.
h ( t ) i s t h e i m p u l se r e sp o n se f u n c t i on .
Four types of gain:
1. H(s) = voltage gain = V0(s)/Vi(s)
2. H(s) = Current gain = I0(s)/Ii(s)
3. H(s) = Impedance = V(s)/I(s)
4. H(s) = Admittance = I(s)/V(s)
)(
)()(
sX
sYsH =
Transfer Functions
Basics in Systems and Circuits Theory
Transfer Functions Example (1)
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Michael E.Auer 01.11.2011 BSC04
Example:The output of a linear system is y(t)=10e-tcos4twhen the input is x(t)=e-tu(t).
Find the transfer function of the system and its impulse response.
Solution:
Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get
Apply inverse transform for H(s), we get
16)1(
44010
16)1(
)1(10
)(
)()(
22
2
++=
++
+==
ss
s
sX
sYsH
)()4sin(40)(10)( tutetth t=
Transfer Functions Example (1)
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Basics in Systems and Circuits Theory
BSC Modul 4: Advanced Circuit Analysis
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Michael E.Auer 01.11.2011 BSC04
Fourier Series Fourier Transform
Laplace Transform Applications of Laplace Transform
Z-Transform
BSC Modul 4: Advanced Circuit Analysis
Basics in Systems and Circuits Theory
Introduction
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Michael E.Auer 01.11.2011 BSC04
In continuous systems Laplace transforms play a unique role. They
allow system and circuit designers to analyze systems and predictperformance, and to think in different terms - like frequencyresponses - to help understand linear continuous systems.
Z-transforms play the role in sampled systems that Laplacetransforms play in continuous systems.
In continuous systems, inputs and outputs are related by differentialequations and Laplace transform techniques are used to solvethose differential equations.
In sampled systems, inputs and outputs are related by difference
equations and Z-transform techniques are used to solve thosedifferential equations.
Basics in Systems and Circuits Theory
Fourier, Laplace and Z-Transforms
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Michael E.Auer 01.11.2011 BSC04
, p
For right-sided signals (zero-valued for negative time index) theLaplace transform is a generalization of the Fourier transform ofa continuous-time signal, and the z-transform is a generalizationof the Fourier transform of a discrete-time signal.
Fourier Transform
Laplace Transform Z-Transform
generalization
continuous-time signals discrete-time signals
Basics in Systems and Circuits Theory
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Michael E.Auer 01.11.2011 BSC04
The Z-transform converts a discretetime-domain signal, which isa sequence of real or complex numbers, into a complexfrequency-domain representation.It can be considered as a discrete-time equivalent of the Laplacetransform.
There are numerous sampled systems that look like the one shown below.
discrete-time signals
Basics in Systems and Circuits Theory
Definition of the Z-Transform
http://en.wikipedia.org/wiki/Discrete_mathematicshttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Sequencehttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Laplace_transformhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Frequency-domainhttp://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Sequencehttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Time-domainhttp://en.wikipedia.org/wiki/Discrete_mathematics -
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Michael E.Auer 01.11.2011 BSC04
Let us assume that we have a sequence, yk.
The subscript "k" indicates a sampled time interval and that ykis thevalue of y(t) at the kth sample instant.
ykcould be generated from a sample of a time function.For example: yk= y(kT), where y(t) is a continuous time function, and T is thesampling interval.
We will focus on the index variable k, rather than the exact time kT, in all that wedo in the following.
[ ] kk
kk zyyZ
==
0
Basics in Systems and Circuits Theory
Z-Transform Example
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Michael E.Auer 01.11.2011 BSC04
k
kayy = 0
p
Given the following sampled signal:
[ ]az
z
z
az
azaaZ
k
kk
kkk
=
=
==
=
=
1
11
00
We get the Z-Transform for y0 = 1
Basics in Systems and Circuits Theory
Z-Transform of Unit Impulse and Unit Step
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Michael E.Auer 01.11.2011 BSC04
Given the following sampled signal Dk:
Dkis zero for k>0, so all those terms are zero.Dkis one for k = 0, so that
[ ] 1=k
DZ
[ ] 1...1 321
=+++=
z
zzzzuZ
k
Given the following sampled signal uk:
ukis one for all k.
Basics in Systems and Circuits Theory
More Complex Example of Z-Transform
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Michael E.Auer 01.11.2011 BSC04
Given the following sampled signal fk:
( ) )sin(bkTekTff akTk
==
[ ]k
k
akT
k
k
kk zbkTezffZ
=
=
==
)sin(00
Finally:
[ ]
+
=*
2
1
cz
z
cz
z
j
fZ k jbTaTec+=where
Basics in Systems and Circuits Theory
S- and Z-Plane Presentation
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Michael E.Auer 01.11.2011 BSC04
Basics in Systems and Circuits Theory
Inverse Z-Transform
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Michael E.Auer 01.11.2011 BSC04
The inverse z-transform can be obtained using one of two methods:
a) the inspection method,b) the partial fraction method.
In the inspection method each simple term of a polynomial in z,H(z), is substituted by its time-domain equivalent.
For the more complicated functions of z, the partial fraction methodis used to describe the polynomial in terms of simpler terms, andthen each simple term is substituted by its time-domain equivalentterm.