3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Lecture 1
1.1. Circuit analysis and synthesis
REPRESENTATION AND
RESPONSE IN CONTINUOUSTIME LTI CIRCUITS
1.1. Circuit analysis and synthesis....................... 1-2
1.2. Circuits (and systems) classification ............ 1-3
1.3. I/O representation in LTI systems.................. 1-4
1.4. Zero-input and zero-state response .............. 1-5
1.5. Natural and forced response........................ 1-10
1.6. Poles and zeros of the network function .... 1-13
An electric circuit is a physical entity that supports electric and
magnetic fields and provides a path for the electric current to flow. A number
of access points, i. e. terminals, can be defined, where excitation signals are
applied and/or response signals are sensed. An electric circuit can be
modelled as an interconnection of ideal elements that satisfies Kirchoff’s
laws. At the same time, each circuit element establishes a relation between
the voltage drop between its terminals and the current intensities flowing
through them, i. e. a constitutive law.
Circuit analysis consists in developing systematic procedures,
following a unified methodology, to mathematically obtain the response
signal of the circuit to any possible excitation signal (see Fig. 1.1(a)). Other
important aspects of circuit analysis are, namely, the study of second order
effects and the estimation of the validity ranges of the results, considering
the inherent limitations of any circuit model.
Circuit synthesis and design consist in the selection of a specific
topology and a particular set of values for the circuit elements, in order to
match a desired relation between the excitation and the response signals
(Fig. 1.1(b)). The synthesis of an electric circuit for the realization of a
particular functionality is based on approximation and realizability.
Approximation consists in substituting the desired functionality —or
behaviour— by a resembling functionality that is realizable with the
available circuit primitives. Also, for a given set of circuit primitives, there
is a limited set of functionalities that are realizable, i. e. that can be
implemented.
1.7. Forced response in sin. steady state .......... 1-16
1.8. Evolution of filters......................................... 1-29
b) SYNTHESIS & DESIGN
a) ANALYSIS
Excitation
(known)
x(t)
Circuit
(known)
Response =?
Excitation
(known)
y(t)?
x(t)
Circuit =?
Elements
value?
Response
(required)
y(t)
Figure 1.1: Illustrating the concepts of analysis, synthesis and design
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
Escuela Superior de Ingenieros, Universidad de Sevilla
1-2
Circuit Analysis and Synthesis
Lecture 1
1.2. Circuits (and systems) classification
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
3º Telecomunicación
Passive/Active
Linear/Nonlinear
Linear circuits are made of linear elements. A linear element has a
linear relation between the voltage drop between its terminals and the
current flowing through them. This relation can be described by integrodifferential linear equations. Any element that does not match this definition
is a nonlinear element. A circuit contanining any nonlinear element is no
longer a linear circuit.
Passive circuits contain only passive elements, namely: resistors,
capacitors and/or inductors. Active circuits contain active elements of any
kind: controlled sources, operational amplifiers, transistors, etc. Active
elements are those able to supply power to the signal path.
Time variant/invariant
Continuous/Discrete time
Time invariant circuits are made only of time invariant elements. The
characteristics of a time invariant are not a function of time, while time
variant elements change with time. Every real device changes with time, but
do not forget that we are dealing with idealized elements.
In a continuous-time circuit, input and output signals are defined at
every time instant (Fig. 1.2(a)).
Lumped/distributed
y(t) = f[x(t)]
(1.1)
In a discrete-time circuit, on the contrary, input and output signals are
defined in a discrete set of time instants —evenly spaced, usually:
y(n) = f[x(n)]
n = …, – 2, – 1, 0, 1, 2, 3, …
(1.2)
Analog/Digital
In an analog circuit, the amplitude of the signals varies within a
continuous set of values, both in continuous and discrete time (see Fig.
1.2(a) and (b)). In a digital circuit, signal amplitudes can only take values of
a restricted set within a particular range (Fig. 1.2(c)).
Continuous time
t
T
0 1
Discrete time
x ( n ) ≡ x ( nT )
n
Digital values
x ( nT )
Analog values
Analog values
(a)
(b)
3
2
1
0
T
1
Discrete time
t
(c)
Figure 1.2: Signal formats in circuits and systems.
1-3
During this course, we will concentrate on linear time-invariant (LTI)
analog continuous-time (CT) circuits under the lumped-circuit assumption.
1.3. I/O representation in LTI systems
x(t)
0
Under the lumped-circuit assumption, the system characteristics are
concentrated into idealized discrete components with no (or negligible)
spatial extent. The voltage across a circuit branch is unambiguosly defined.
Also the current through a circuit branch is well defined, the current inflow
equals the outflow. This will allow to apply Kirchoff’s laws. For this
assumption to be made, the size of the circuit must be considerably smaller
than the wavelength of the signals it processes. Otherwise, the circuit will
have distributed parameters, and a different analysis technique applies.
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
From now on, we will consider a LTI circuit made of an interconnection
of LTI elements and independent sources. Independent voltage or current
sources are not linear devices, nor they remain unchanged with time, but, in
this occasion, their voltages/currents will play the role of the excitation
signals for our circuits. Without loss of generality, we will consider always
that the circuit has a single excitation signal and a single response signal —
single-input single-output circuits. The case of multiple inputs and/or
Escuela Superior de Ingenieros, Universidad de Sevilla
1-4
Circuit Analysis and Synthesis
Lecture 1
outputs can be solved by using the superposition principle, that applies for
linear systems.
Our target will be finding a relation between the input and the output of
the circuit. At the circuit level, each component can be modelled by
idealized circuit elements: resistors, capacitors, opamps, etc. Making use of
their constitutive equations and Kirchoff’s laws, the I/O relation is found. At
a higher abstraction level, circuit blocks can be identified as analog
operators: scaling blocks, integrators, adders, etc. Their interactions are
pictured by a signal flow graph (SFG). Combining the analog operators with
the SFG equations, an I/O relation is found. At any abstraction level, and
always under the lumped-circuit assuption —what means that no
differentiation is present apart from time differentiation— the relation
between the I/O signals of a continuous-time LTI circuit with a single input
and a single output, is represented by an ordinary differential equation
(ODE) with constant coefficients:
N
i
dy
M
j
dx
∑ b i -------i = ∑ a j -------j
i=0
dt
j=0
dt
(1.3)
For a particular circuit, given the I/O points, i. e. the terminals in which
the excitation will be applied and the response will be sensed, an equation of
this form can be found by applying some analythical methods. Depending
on the physical format of the signals, up to six different I/O relations can be
defined. They are illustrated in Fig. 1.3.
1.4. Zero-input and zero-state response
The solution to the differential equation that relates the output signal
with the form of the input will depend both on the excitation and on the
initial state of the circuit. This initial state is conveyed by the initial
conditions of the reactive elements, in other words: the charge contained at
the capacitors and the magnetic flux held by the inductors . The response of
the circuit can be therefore decomposed into a zero-state reponse and a zeroinput reponse.
1-5
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Driving-point functions
(defined in one-port networks)
Admittance
x(t) = v(t)
y(t) = i( t)
Impedance
x(t) = i(t)
y(t) = v(t)
i( t)
+ - Passive elements
v ( t ) - Controlled sources
- Nullators & norators
-
Transfer functions
(defined in two-port networks)
x(t)
- Passive elements
- Controlled sources
- Nullators & norators
Figure 1.3: I/O relations
Short-circuit:
y(t)
y(t) = i( t)
- Transadmittance
x(t) = v(t)
- Current transfer
x(t) = i(t)
y(t) = v(t)
- Transimpedance
x(t) = i( t)
- Voltage transfer
x( t) = v(t )
Open-circuit:
There is a starting point in time, t = 0 , from which the excitation signal
is applied. Every event in the history of the circuit until that point is resumed
in the initial conditions of the reactive elements. From that point on, the
complete response of the circuit will have a part that is related with the
excitation and another part that is related with the initial state. The part
related with the excitation signal is called zero-state response. The part
related with the initial conditions is the zero-input response. If a circuit is
completely at rest, i. e. the initial conditions are zero, the only response that
is present once the excitation signal is applied is the zero-state response. On
the other hand, if no exitation signal is applied, the observed response is just
the zero-input response.
Zero-state and zero-input responses can be obtained analytically from
the ODE representing the I/O relation both in the time domain and in the
Laplace domain.
Time domain
The objective is to obtain, for a given circuit, the zero-state response
for an arbitrary excitation. But, could it be done based on the zero-state
response to a simple elementary excitation? This poses two different
questions: (1) Can an arbitrary excitation signal be described as as sum of
Escuela Superior de Ingenieros, Universidad de Sevilla
1-6
Circuit Analysis and Synthesis
Lecture 1
elementary signals? (2) Can the circuit response be expressed as the sum of
the responses to each excitation component?
The answer to the first question is yes, based on the impulse signal and
the impulse response. Because of the sampling property, any arbitrary input
can be expressed as a convolution of the signal samples with the impulse
(Dirac’s delta function)
∞
x ( t ) = ∫ - x ( τ )δ ( t – τ ) dτ
0
∞
y ( t ) = ∫ - x ( τ )h ( t – τ ) dτ
0
(1.5)
Trying to calculate the impulse response in order to compute the
convolution integral is similar to solve the problem in the Laplace domain.
In order to obtain the zero-state response in the time domain, we will limit
ourselves to solve the ODE in (1.3).
For the calculation of the zero-input response there are two options:
1) To use the initial conditions as the integration constants while
solving the ODE.
2) To consider them as equivalent sources (Fig. 1.4), solve them
separately, and apply the superposition principle.
Laplace (or frequency) domain
iL ( t )
iL ( t )
+
+
–
i (0 ) ≠ 0
L
L
at rest
−
−
iC ( t ) +
–
v (0 ) ≠ 0
c
+
iC ( t )
C
vC ( t )
C
–
i L ( 0 )u o ( t )
vL ( t )
L
vL ( t )
(1.4)
what means that any arbitrary input can be described as a sum of simple signals. The response of the system to an impulse, δ ( t ) , is the impulse response, h ( t ) . Being a linear system, h ( t ) can be obtained from the poles of
the system (this concept will be introduced later). This means that the response to each elementary component of the input is known.
The second question is answered by the superposition principle. Being
a linear circuit and describing the input as a sum of impulses, and knowing
the response of the system to an impulse of any amplitude, the response of
the circuit will be given by the convolution integral:
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
vC ( t )
−
−
at rest
–
v C ( 0 )u o ( t )
Figure 1.4: Equivalent circuit of a reactive element with initial conditions
Let us compute the zero-state response in the Laplace domain.
Consider, without loss of generality, a LTI circuit with a single input and a
single output. It is described by the following differential equation:
N
M
i
j
dy
dx
∑ b i -------i = ∑ a j -------j
i = 0 dt
j = 0 dt
(1.6)
Circuit + Initial conditions
differential
equation
η [ x ( t ), y ( t ) ] = 0
Laplace transform
solve in the
time domain
algebraic
equation
η [ X ( s ), Y ( s ) ] = 0
solve in the
Laplace domain
An alternative methodology consists in applying Laplace transform to
the ODE in (1.3). This converts the differential equation into an algebraic
equation. As X ( s ) = L [ x ( t ) ] is known, Y ( s ) can be computed. Applying
the inverse transform y ( t ) is obtained (Fig. 1.5).
Figure 1.5: Circuit response calculation in the time and Laplace domains
1-7
Escuela Superior de Ingenieros, Universidad de Sevilla
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
y ( t ) for a
Inverse Laplace transform
given x ( t )
Y ( s ) for a
given X ( s )
1-8
Circuit Analysis and Synthesis
Lecture 1
1.5. Natural and forced response
Using the following property of the Laplace transform:
d
L ----- y ( t ) = sY ( s ) – y ( 0 )
dt
Y(s ) = L[ y( t) ] ⇒
(1.7)
and the circuit being at rest:
–
(1)
(0 ) = … = y
–
(1)
(0 ) = … = x
y(0 ) = y
x(0 ) = x
–
(n – 1)
(0 ) = 0
–
(n – 1)
(0 ) = 0
–
–
(1.8)
applying Laplace transform to (1.6) yields:
M
∑ aj s
Y(s) =
j
j=0
------------------ X(s)
N
i
∑ bi s
i=0
= H ( s )X ( s )
(1.9)
Therefore, in the Laplace domain, the zero-state response is the product of
the Laplace transform of the input times the transfer function, H ( s ) , also
known as network function of system function.
A property of the Laplace transform states:
∞
L ∫ - f ( τ )g ( t – τ ) dτ = F ( s )G ( s )
0
(1.10)
then, for a given input source and a given output terminal, the transfer function is the Laplace transform of the impulse response:
H(s) = L[h( t) ]
(1.11)
In order to compute the zero-input response, there are two alternatives:
1) Introducing initial conditions in the Laplace transform of the differential equation, using (1.7).
2) Considering them as equivalent sources and applying superposition, once again.
1-9
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
For given input source and output terminal, excitation signal and initial
state, the system response is a unique waveform. However, it can be
decomposed into two components: the natural response, related with the
poles of the system, and the forced response, related with the poles of the
excitation.
The network function, H ( s ) , relates the Laplace transform of the
excitation signal, X ( s ) , with the Laplace transform of the zero-state
response, Y ( s ) , through:
(1.12)
Y ( s ) = H ( s )X ( s )
If the circuit has not been at rest, additional terms would have appeared in
this equation. For a LTI circuit under the lumped-circuit assumption, H ( s )
is a real rational function (RRF). In the majority of practical occasions, X ( s )
is a RRF too. If both of them are RRF’s, their product, Y ( s ) , is a real rational
function as well:
N(s)
N(s)
Y ( s ) = ----------- = --------------------------------------------------------------(1.13)
D(s)
( s – p 1 ) ( s – p 2 )… ( s – p n )
where p 1, p 2, …, p n are the finite poles of the response. Let us assume that
N ( s ) is of a polynomial of the same or smaller order than D ( s ) . Let us suppose also that all the poles are simple. Y ( s ) can be expanded in a sum of simpler partial fractions:
n
K2
Kn
Ki
K1
- + ------------- + … + ------------- = ∑ -----------Y ( s ) = ------------s – p1 s – p2
s – pn
s – pi
(1.14)
i=1
where K i is the residue of pole p i , that can be obtained by:
Ki = Y ( s ) ( s – pi )
i = 1, 2, …, n
s = pi
(1.15)
If a multiple pole has been present, p i , with multiplicity m , several
summands of the folowing form would have to be added:
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 10
Circuit Analysis and Synthesis
Lecture 1
m
K i, 1
K i, 2
K i, m
K i, j
-----------+ ------------------- + … + -------------------- = ∑ -----------------m
j
s – pi ( s – p )2
(
s
–
p
)
(
s
–
p
)
j=1
i
i
i
(1.16)
Now the coefficients K i, j would have been calculated by:
m
1 d
K i, m – j = --- ------- [ Y ( s ) ( s – p i ) ]
j = 1, 2, …, m – 1
j
j! ds
(1.17)
If p i and K i are complex, their corresponding conjugates will be
present as well. Otheorwise Y ( s ) would not be a rational function with real
coefficients, i. e. a real rational function. Let us resume now considering that
all poles are simple. The wave form of the response in the time domain can
be obtaining by the inverse transform of (1.14):
y ( t ) = K1 e
+ K2 e
p2 t
+ … + Kn e
pn t
n
= ∑ Ki e
pi t
(t > 0)
(1.18)
i=1
jθ i
If we express each complex K i as K i e , and consider that there are k
complex poles and ( n – k ) real poles in Y ( s ) , (1.18) can be rewritten:
k
y ( t ) = ∑ Ki e
i=1
k
--2
= ∑ 2 Ki e
i=1
αi t j ( βi t + θi )
e
n
∑
+
Ki e
αi t
=
i = k+1
αi t
(1.19)
n
cos ( β i t + θ i ) +
∑
Ki e
αi t
i =k+1
Therefore, if a LTI system is excited by a time-variant source whom Laplace
transform is a real rational function, and the response poles are simple, then,
in the time domain, the response appears as a sum of exponentials and sinusoids modulated in amplitude by exponentials.
Besides, the poles of the response in the Laplace domain, Y ( s ) , can be
split into two groups: tranfer function poles and excitation poles. It can be
written that:
1 - 11
3º Telecomunicación
N( s)
Y ( s ) = X ( s )H ( s ) = --------------------------- =
D n ( s )D f ( s )
N( s)
= ------------------------------------------------------------------------------------------------( s – p n1 ) ( s – p n2 )… ( s – p f1 ) ( s – p f2 )…
s = pi
p1 t
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
(1.20)
where subindex n indicates natural poles —belonging to the network function— and subindex f denotes forced poles —those of the excitation signal
transform. If all poles are simple and N ( s ) is of a smaller order than D ( s ) ,
the partial fraction expansion of the response can be written in this way:
K n2
K f2
 K n1
  K f1

Y ( s ) =  --------------- + --------------- + … +  -------------- + -------------- + … =
 s – p n1 s – p n2
  s – p f1 s – p f2

(1.21)
= Yn ( s ) + Yf ( s )
Here, Y n ( s ) represents the natural component of the response, or simply the
natural response. The poles in Y n ( s ) belong to the system, not to the excitation. These poles are natural, characteristic of the system. On the other
side, Y f ( s ) is the forced component of the response, or just the forced response. Its poles belong to the excitation, not to the circuit. The excitation
forces the system to respond in this way. In the time domain, the complete
response is the sum of the natural and forced responses’ waveforms:
y ( t ) = ( K n1 e
+ ( K f1 e
p f1 t
p n1 t
+ K f2 e
+ K n2 e
p f2 t
p n2 t
+ …) +
(1.22)
+ … ) = yn ( t ) + yf ( t )
These components can be related with the zero-state and the zero-input
reponses. In the case of the zero-state, the system at rest at the initial point,
the complete response will have a natural component and a forced
component. In the case of the zero-input response, the response will be
completely natural, because there can not be any pole of the excitation signal
in absence of it.
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 12
Circuit Analysis and Synthesis
Lecture 1
1.6. Poles and zeros of the network function
Stability
For a LTI circuit, under the lumped-circuit assumption, the network
function is a rational function of the complex frequency variable s . It can be
written as a quotient of polynomials in s with real coefficients:
m
m–1
am s + am – 1 s
+ … + a1 s + a0
N(s)
H ( s ) = ----------- = ------------------------------------------------------------------------------------n
n–1
D(s)
+ … + b1 s + b0
bn s + bn – 1 s
(1.23)
The roots of the numerator and the denominator are called the critical
frequencies of the circuit. The dynamic properties of the circuit can be
derived from the study of these critical frequencies. The poles of the network
function, i. e. the roots of the polynomial in the denominator, are called
natural frequencies. If the set of critical frequencies of a circuit is known,
the network function can be written in a factorized form:
m
∏ ( s – zi )
N(s)
i=1
H ( s ) = ----------= K --------------------------n
D(s)
∏ ( s – pi )
(1.24)
i=1
In addition to the critical frequencies, if the order of the polynomials in
the numerator and the denominator differ, m ≠ n , then the network function
is said to have poles or zeroes at the infinity,
am m – n
(1.25)
H ( s ) s → ∞ → ------ s
bn
If m > n , H ( s ) s → ∞ → ∞ , then the function presents m – n poles in ∞ . On
the contrary, if m < n , then H ( s ) s → ∞ → 0 , it has n – m zeroes in ∞ .
For an active circuit, critical frequencies can be anywhere in the
complex plane, and they can have any multiplicity. The only restriction
being that complex critical frequencies must appear in pairs with their
respective conjugates. Otherwise, the network function coefficients would
not be real.
1 - 13
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
An important property of a circuit is stability. A LTI circuit is said to be
stable if the natural response vanishes with time. In other words, the
repsonse of the circuit remains bounded as long as the excitation signal is
bounded too. This type of stability is referred as BIBO stability (boundedinput, bounded-output).
A necessary condition for BIBO stability is that impulse response must
remain bounded. For this to happen, natural frequencies must hold the
following conditions:
a) All the poles of the system must have a real part that is less than
zero, i. e. poles must lie in the open left-half-plane of the s -plane.
Suppose that there is a real pole in the right-half-plane of the s plane, let us say at s = p , where p > 0 . H ( s ) can be expressed as:
k - + -----------N' ( s )
H ( s ) = ---------s – p D' ( s )
(1.26)
realizing the inverse Laplace transform, the impulse response in the
time domain is:
pt
h ( t ) = ke + h' ( t )
(1.27)
which is not bounded, as p > 0 .
Let us consider as well the case of a pair of conjugate complex poles
*
with a positive real part: p = α + jβ and p = α – jβ , with
α, β > 0 . The network function can be written in this way:
jθ
– jθ
ke
ke
N' ( s )
H ( s ) = ----------------------+ ---------------------- + -----------s – α + jβ s – α – jβ D' ( s )
(1.28)
hence, the impulse response in the time domain is:
αt
h ( t ) = 2ke cos ( βt + θ ) + h' ( t )
(1.29)
where the first term is not bounded as α > 0 .
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 14
Circuit Analysis and Synthesis
Lecture 1
b) Poles in the jω -axis must be simple.
s
Consider a multiple pole in s = 0 . H ( s ) can be written in this way:
k
H ( s ) = ----- + H' ( s )
q
s
s
(1.30)
s
the corresponding impulse response is:
q–1
k
h ( t ) = ------------------ t
+ h' ( t )
( q – 1 )!
s
multiplicity greater than 1, the network function has the form :
jθ
– jθ
ke
ke
H ( s ) = ---------------------- + ---------------------- + H' ( s )
q
q
( s + jω )
( s – jω )
(1.32)
2k q – 1
h ( t ) = ------------------ t
cos ( ωt + θ ) + h' ( t )
( q – 1 )!
(1.33)
which is not bounded neither.
Routh-Hurwitz criterion
The Routh-Hurwitz stability criterion is a method for determining
whether or not a system is stable based upon the coefficients in the system's
characteristic equation. It is particularly useful for higher-order systems
because it does not require the polynomial expressions in the transfer
function to be factored. Starting at the characteristic equation:
+ … + an – 1 s + an = 0
(1.34)
the following table is constructed:
1 - 15
n–2
n–3
a2
a4
a6
…
a1
a3
a5
a7
…
B1
B3
B5
…
…
C1
C3
C5
…
…
…
…
…
…
(1.35)
…
B1 = ( a1 a2 – a0 a3 ) ⁄ a1
B3 = ( a1 a4 – a0 a5 ) ⁄ a1
B5 = ( a1 a6 – a0 a7 ) ⁄ a1
…
(1.36)
C1 = ( B1 a3 – a1 B3 ) ⁄ B1
…
and its impulse response in the time domain:
n–1
n–1
a0
where:
If we consider now a pair of conjugate pure imaginary poles with
n
n
…
(1.31)
which is not bounded.
a0 s + a1 s
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
The necessary condition for the system to be stable is that all
coefficients of the denominator polynomial are non null and of the same
sign. A sufficient condition is that there are no changes of signin the first
column of the table. The number of changes indicates the number of roots
of the characteristic equation that lie in the right-half-plane.
1.7. Forced response in sinusoidal steady state
Our interest will be in designing circuits with the poles in the left-halfplane of the s -plane, i. e. with a bounded response, in order to build analog
filters. We are interested as well in studying their response in sinusoidal
steady state (SSS). This is because any periodic signal can be expressed as
a sum (finite or not) of sines and cosines.
Consider the situation depicted in Fig. 1.6. A LTI circuit, with its poles
in the left-half-plane of the s -plane, is excited by a sinusoidal signal with its
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 16
Circuit Analysis and Synthesis
Lecture 1
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
3º Telecomunicación
αt
y f ( t ) = A H ( α + jω ) e sin [ ωt + θ + ∠H ( α + jω ) ]
poles in the
left-half-plane
αt
Ae sin ( ωt + θ )
yn ( t ) + yf ( t )
Figure 1.6: Generic circuit excited with a sinusoidal signal.
amplitude modulated by an exponential. Because of the position of the
poles, the natural response vanishes in time. The response, after an infinitely
long period, will be forced. If the Laplace transform of the input is:
Aω cos θ
A ( s – α ) sin θ
X ( s ) = -------------------------------(1.37)
+ -------------------------------2
(s – α) + ω
2
2
(s – α) + ω
2
(1.38)
the forced response is:
y ( t ) = A H ( jω ) sin [ ωt + θ + ∠H ( jω ) ]
(1.44)
*
kA
k A
Y f ( s ) = ------------------------ + -----------------------s – α – jω s – α + jω
The Bode plots are a powerful graphic tool to visualize how the system
will behave in SSS. Let us consider the factorized form of H ( s ) , and
suppose that the system is in SSS, this is s = jω :
M
We are interested in the forced response, the only observed once the SSS is
reached. This response will contain the poles of the input exclusively:
(1.39)
∏ ( jω – z i )
i=1
H ( jω ) = K ------------------------------N
∏ ( jω – p l )
(1.45)
l=1
where
j [ θ + ∠H ( α + jω ) ]
H ( α + jω ) e
ω cos θ + jω sin θ
k = H ( α + jω ) ---------------------------------------- = ---------------------------- ---------------------------------------- (1.40)
2
2jω
j
The transform of the forced response is therefore:
Let us analyze the contribution of a generic factor ( jω – s i ) , where s i
can be either one of the zeros or one of the poles of the network function. If
it is considered to be a complex number, it will have a real and an imaginary
part, this is s i = α i + jβ i . The factor ( jω – s i ) is also complex:
– αi + j ( ω – βi ) = Mi e
– j  θ + ∠H – π
---

2
H ( α + jω )
e
Y f ( s ) = ---------------------------- A e------------------------------- + ---------------------------------2
s – α – jω
s – α + jω
(1.41)
jθ i
(1.46)
and hence can be written in polar form. The magnitude being:
Mi =
2
αi + ( ω – βi )
2
(1.47)
while the phase is1:
and in the time domain:
1 - 17
We can conclude that a LTI system, with poles in the left-half-plane,
processes an sinusoidal input with an exponentially modulated amplitude,
by modifying:
a) the amplitude of the signal by a factor H ( α + jω )
b) the phase, by adding ∠H ( α + jω ) = arg [ H ( α + jω ) ]
Applying these results to pure sinusoidal input, that is α = 0 :
(1.43)
x ( t ) = A sin ( ωt + θ )
Bode plots
the complete response will be given by:
A ( s – α ) sin θ + Aω cos θ
Y ( s ) = H ( s )X ( s ) = H ( s ) ----------------------------------------------------------2
2
(s – α) + ω
j  θ + ∠H – π
---

2
(1.42)
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 18
Circuit Analysis and Synthesis
Lecture 1
1 – sgn ( – α i )
ω–β
θ i = atan ---------------i + sgn ( ω – β i ) ------------------------------- π
2
–αi
(1.48)
If we apply this notation to the factorized form of H ( jω ) :
H ( jω ) = K e
jθ z1
jθ z2
jθ zm
M z2 e …M zm e
------------------------------------------------------------------------ =
jθ p1
jθ p2
jθ pn
M p1 e M p2 e …M pn e
jθ K M z1 e
M z1 M z2 …M zm j ( θK + θ z1 + θz2 + … + θ zm – θ p1 – θ p2 – … – θpn ) (1.49)
= K -------------------------------------- e
=
M p1 M p2 …M pn
= M ( ω )e
jθ ( ω )
where M ( ω ) and θ ( ω ) are the magnitude and phase of H ( jω ) , respectively. The idea behind is to express both, magnitude and phase, as a sum of contributions of each critical frequency of the circuit. Concerning the phase,
this is already achieved, as the total phase is obtained as the sum of the contributions to the phase of each factor:
M
N
θ H = ∠K + ∑ θ z – ∑ θ p
i
i
i=1
(1.50)
i=1
To obtain a similar expression for the contributions to the magnitude,
we will need to represent it in logarithmic form, thus defining:
M
N
H ( jω ) dB = 20log 10 K + ∑ 20log 10 M z – ∑ 20log 10 M p
i
i
i=1
(1.51)
i=1
The unit employed to express the magnitude of the network function is the
decibel (dB). This unit is commonly employed to measure power transfer.
1. With this definition, the phase of each complex number ( jω – s i ) is within the
limits of the interval [ – π, π ] . The arch-tangent is a multivalued function, and it
π π
is common to take the values within – ---, --- . The second summand in (1.48)
2 2
introduces a correction factor for all the angles obtained to be in the range
[ – π, π ] .
1 - 19
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
3º Telecomunicación
We are dealing with signal amplitudes, while the power is related with the
square of the amplitude. In order to be consistent with this, the decibels of
the magnitude of the network function will be computed by applying
20log 10 ( . ) instead of doing 10log 10 ( . ) .
The exact calculation of the Bode plots, the magnitude and phase plots,
is quite laborious. However, an approximate represntation can be easily
obtained. Observing the factorized form of N ( s ) and D ( s ) , only three
different types of factor can be identified:
a) A constant term K.
b) Factors of the form s + ω o , corresponding to real roots. A particular case will be those factor of the form s , corresponding to a
root in the origin.
2
c) Factors of the form s + as + b , corresponding to a pair of complex conjugate roots.
The magnitude in dB, and also the phase, of a product of factors is equal to
the sum of the contributions of the factors. Representing the Bode plots of
H ( jω ) means combining the representation of the basic terms. Let us review
their elementary contributions to the Bode plots:
a) The constant term K :
Let us consider a function of the form H ( s ) = K , this is H ( jω ) = K
in SSS. The following contribution is found for the magnitude:
(1.52)
H ( jω ) dB = 20log 10 H ( jω ) = 20log 10 K
while for the phase:
1 – sgn K
θ H ( ω ) = –  ---------------------- π


2
(1.53)
The magnitude of 20log 10 K will be positive for K > 1 and negative
for K < 1 . The phase will be 0° , or 0rad , for K > 0 and – 180 ° , or – πrad ,
for K < 0 (see Fig. 1.7).
b) Factor of the form ( s + ω o ) :
Let us consider now H ( s ) = s + ω o , and then H ( jω ) = jω + ω o .
The contribution to the magnitude will be:
2
H ( jω ) dB = 20log 10 ω o + ω
Escuela Superior de Ingenieros, Universidad de Sevilla
2
(1.54)
1 - 20
Circuit Analysis and Synthesis
Lecture 1
20log 10 K
0rad
Phase, rad.
Magnitude, dB
ω
π
θ H ( ω o ) = atan ( 1 ) = --4
K<0
0rad
ω
(1.60)
7.0e+01
6.0e+01
20log 10 K
5.0e+01
Magnitude (dB)
– π rad
Figure 1.7: Magnitude and phase Bode plots of a constant factor.
4.0e+01
3.0e+01
2.0e+01
(1.55)
that are represented in Fig. 1.8. It is interesting to know how these functions
behave at lower and higher frequencies —compare with the pole or zero frequency, ω o . This will help us to easily compute the approximate Bode plot
of less manageable network functions. At frequencies well below ω o :
(1.56)
H ( jω ) dB ≈ 20log 10 ω o
for
ω « ωo
This term will be positive as long as ω o > 1 , and negative if ω o < 1 . At
higher frequencies:
(1.57)
H ( jω ) dB ≈ 20log 10 ω
for
ω » ωo
this is, for frequencies well above, ω o (considered 8.26rad/s in Fig. 1.8), the
magnitude increases 20dB/decade , or 6dB/octave , which is equivalent.
Concerning the phase, at low frequencies:
(1.58)
for
ω « ωo
θH ≈ 0
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
1.0e+01
0.0e+00
1.0e-04 1.0e-03 1.0e-02 1.0e-01 1.0e+00 1.0e+01 1.0e+02 1.0e+03 1.0e+04
Frequency (rad/s)
3.0e+00
2.5e+00
2.0e+00
Phase (rad.)
while that of the phase is:
ω
θ H ( ω ) = atan  ------
ω 
o
1 - 21
(1.59)
Both, in magnitude and phase plots, the qualitative changes occur
around the root frequency, ω o . For ω = ω o we have:
H ( jω o ) dB = 20log 10 ω o + 3dB
ω
ω
K <1
3º Telecomunicación
while for the higher frequencies:
π
for
ω « ωo
θ H ≈ --2
K>0
Phase, rad.
Magnitude, dB
K >1
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
1.5e+00
1.0e+00
5.0e-01
0.0e+00
-5.0e-01
-1.0e+00
1.0e-04 1.0e-03 1.0e-02 1.0e-01 1.0e+00 1.0e+01 1.0e+02 1.0e+03 1.0e+04
Frequency (rad/s)
Figure 1.8: Magnitude and phase plots for a factor of type ( s + ω o ) .
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 22
Lecture 1
It will be of interest to consider the goodness of the approximate
diagrams, those based in the asymptotic behavior. Let us evaluate the
contributions in magnitude and phase for frequencies ω o ⁄ 10 and 10ω o . For
ω o ⁄ 10 :
H ( j0.1ω o ) dB = 20log 10 ω o + 0.043dB
(1.61)
θ H ( 0.1ω o ) = 0.03πrad
θ H ( 10ω o ) = 0.47πrad
and
(1.64)
ω
π
θ H ( ω ) = arctg  ---- = -- 0
2
(1.65)
what is a special case of the curves in Fig. 1.8 (try to mentally displace
log 10 ω o towards – ∞ ). In the case of a pole at the origin, magnitude and
phase will have the inverse shape:
0.0
(1.62)
Notice that the deviation from the asymptotes is relatively larger for the
phase than for the magnitude.
Before studying the following factor, notice that the contribution hereby
described correspond to a zero of the network function in the left-half-plane.
In particular, z = – ω o , where ω o > 0 , ω o = 8.26rad/s in the picture. If we
consider the case of a zero in the right-half-plane of the s -plane, i. e. a factor
of type ( s – ω o ) , with ω o > 0 , the magnitude plot will be the same, while
the phase plot is obtained by multiplying the current plot times – 1 . This is
0rad at lower frequencies and – π ⁄ 2 at higher frequencies. If a real pole in
the left-half-plane is studied, this is, a contribution of type:
1
(1.63)
H ( s ) = -------------s + ωo
then both the magnitude and the phase plots will be the inverse of those depicted in Fig. 1.8 (see Fig. 1.9 this time). If the pole was in the right-halfplane, the magnitude plot would be the same as in the left-half-plane pole,
and the phase plot would be the inverse, this is, the same as in the left-halfplane zero (Fig. 1.8).
b. 2) Factor of the form s :
This is a particular case of the previous factor, obtained by doing
ω o = 0 . It represents a zero, or pole if in the denominator, at the origin. For
the magnitude an phase we have:
3º Telecomunicación
H ( jω ) dB = 20log 10 ω
-10.0
-20.0
Magnitude (dB)
on the other hand:
H ( j10ω o ) dB = 20log 10 ω o + 20.043dB
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
-30.0
-40.0
-50.0
-60.0
-70.0
1.0e-04 1.0e-03 1.0e-02 1.0e-01 1.0e+00 1.0e+01 1.0e+02 1.0e+03 1.0e+04
Frequency (rad/s)
1.0e+00
5.0e-01
0.0e+00
Phase (rad.)
Circuit Analysis and Synthesis
-5.0e-01
-1.0e+00
-1.5e+00
-2.0e+00
-2.5e+00
-3.0e+00
1.0e-04 1.0e-03 1.0e-02 1.0e-01 1.0e+00 1.0e+01 1.0e+02 1.0e+03 1.0e+04
Frequency (rad/s)
Figure 1.9: Magnitude and phase plots for a factor of type ( s + ω o )
1 - 23
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
Escuela Superior de Ingenieros, Universidad de Sevilla
–1
.
1 - 24
Circuit Analysis and Synthesis
and
Lecture 1
H ( jω ) dB = – 20 log 10 ω
(1.66)
ω
π
θ H ( ω ) = – atan  ---- = – -- 0
2
(1.67)
what is a special case of Fig. 1.9.
2
c) Factor of the form s + as + b :
This type of factor corresponds to a pair of conjugate complex roots. Let
us suppose that they are in the left-half-plane of the s -plane. We can express
*
them as: z = – α + jβ and z = – α – jβ , where α, β > 0 . Thus:
2
2
H ( s ) = s + 2αs + α + β
2
(1.68)
which can be re-written in this way:
2
2 ωo
H ( s ) = s + ------ s + ω o
Q
(1.69)
2
2
Here, we have defined the zero, or pole, frequency, ω o = α + β , and
the quality factor, Q = ω o ⁄ ( 2α ) . This gives us an idea of the relative magnitude of the imaginary part of the roots, β , compared with the real part, α .
As in this diagram:
Im(s)
β
ωo
θ
(0,0)
1 Q = -------------2 cos θ
1
--- < Q < ∞
2
Re(s)
α
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
3º Telecomunicación
and the magnitude and phase contributions to the Bode plots will be:
and
ωoω 2
2
2 2
H ( jω ) dB = 20log 10 ( ω o – ω ) +  -----------
 Q 
(1.71)
oω 
 ω
--------- Q 
θ H ( ω ) = atan  -------------------
 ω 2 – ω 2
 o

(1.72)
Fig. 1.10 depicts these curves. Now, the asymptotic behaviour of the
magnitude is given by:
for
ω « ωo
 40log 10 ω o
H ( jω ) dB ≈ 
(1.73)
for
ω » ωo
 40log 10 ω
Hence, at low frequencies, (1.71) can be approximated by 40log 10 ω o . This
term will be positive or negative depending on whether ω o is less or greater
than 1, respectively. At higher frequencies, where s 2 is the dominant term
in H ( s ) , the magnitude increases 40dB/decade , as two poles are present.
The asymptotes for the phase are:
for
ω « ωo
 0rad
(1.74)
θH ( ω ) ≈ 
for
ω » ωo
 πrad
Around ω o , a completely different behaviour, called resonance, is
observed. In order to determine the location of the minimum, let us solve:
ωo ω 2
d ( ω 2 – ω 2 ) 2 +  ----------
(1.75)
= 0
o
 Q 
dω
We end up at:
In sinusoidal steady state, s = jω , then:
ωo ω
2
2
H ( jω ) = ( ω o – ω ) + j ----------Q
1 - 25
1
ω min = ω o 1 – ---------2
2Q
(1.70)
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
ω min = 0
1
if Q > ------2
(1.76)
otherwise
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 26
Circuit Analysis and Synthesis
Lecture 1
2
If ( 1 ⁄ 2Q ) « 1 , then it can be considered that ω min ≅ ω o . The magnitude
of the network function at ω o is:
(1.77)
H ( jω o ) dB = 40log 10 ω o – 20 log 10 Q
therefore Q determines the depth of the minimum. For the phase, we have:
2
 ωo
⁄ Q
π
θ H ( ω o ) = atan  --------------- = --2
 0 
(1.78)
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Notice that:
d
Q
------- [ θ H ( ω ) ]
= 2 -----ω = ωo
dω
ωo
(1.79)
what means that Q determines the slope of the phase at ω = ω o . The bigger
Q , the deeper the minimum and the stepper the slope of the phase around
ω o (Fig. 1.11). For Q > 5 the minimum is practically at ω o . For Q < 1 ⁄ 2 ,
70.0
100.0
60.0
80.0
Magnitude (dB)
Magnitude (dB)
50.0
60.0
40.0
Q=2
30.0
1.0e-01
1.0e+00
1.0e+01
Frequency (rad/s)
Q=10
10.0
1.0e+00
1.0e+02
Q=1
Q=5
20.0
20.0
0.0
1.0e-02
Q=0.707
40.0
1.0e+01
Frequency (rad/s)
3.5e+00
4.0e+00
3.0e+00
Q=1
2.5e+00
2.0e+00
Phase (rad.)
Phase (rad.)
3.0e+00
1.0e+00
Q=0.707
2.0e+00
1.5e+00
1.0e+00
5.0e-01
0.0e+00
-1.0e+00
1.0e-02
0.0e+00
1.0e-01
2
1.0e+00
1.0e+01
Frequency (rad/s)
ωo
1.0e+02
2
Figure 1.10: Bode plots for s + ------ s + ω o .
Q
1 - 27
Q=2
Q=5
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
-5.0e-01
1.0e+00
Q=10
1.0e+01
Frequency (rad/s)
2
ωo
2
Figure 1.11: Bode plots for s + ------ s + ω o with different Q ’s.
Q
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 28
Circuit Analysis and Synthesis
Lecture 1
the function does not exhibit resonance peak and the minimum is the DC
value. If we evaluate H ( jω ) at frequencies:
1
ω 1 = ω o  1 ± -------

2Q
0.0
-20.0
1 2
1
1 2
1
H ( jω 1 ) dB = 40log 10 ω o + 20log 10  ± ---- + ---------- +  ---- ± ---------- (1.81)
Q
 Q
2
2
2Q
4Q
Magnitude (dB)
(1.80)
we have:
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
-40.0
-60.0
-80.0
2
If Q » 1 , 1 ⁄ Q can be negligible compared to 1 ⁄ Q , and hence,
-100.0
1.0e-02
(1.82)
where it can be observed that the magnitude in (1.82) is 3dB above the minimum level. Therefore, ( ω o ⁄ Q ) , the coefficient of s in (1.69), can be seen
as the 3dB bandwidth of a 2nd order band reject filter.
2 –1
2
A factor of this type in the denominator, this is [ s + ( ω o ⁄ Q )s + ω o ] ,
corresponds to a pair of conjugate complex poles in the left-half-plane of the
s -plane. The magnitude and phase contributions will be the inverse than in
the case of the pair of zeros (see Fig. 1.12). If it was the case of a pair of
right-half-plane conjugate complex zeros, they would have the magnitude
contribution of Fig. 1.10, while the phase contribution of Fig. 1.12. If they
were a pair of poles in the right-half-plane, the magnitude contribution
would be that of Fig. 1.12, but now the pahse contribution would be the one
depicted in Fig. 1.10.
1.0e-01
1.0e+00
1.0e+01
Frequency (rad/s)
1.0e+02
1.0e-01
1.0e+00
1.0e+01
Frequency (rad/s)
1.0e+02
1.0e+00
0.0e+00
-1.0e+00
Phase (rad.)
H ( jω 1 ) dB = 40log 10 ω o – 20 log 10 Q + 20log 10 2
-2.0e+00
-3.0e+00
-4.0e+00
1.0e-02
ω
2
2
o
Figure 1.12: bode plot for a factor of type  s + ------ s + ω o

1.8. Evolution of filters
Q

–1
.
Virtually all communication, measurement and instrumentation
equipment contains any kind of signal filtering device. Nowadays, millions
of audio-frequency filters are produced every year. Between 1920 and 1960,
the vast majority of the audio filters were built from discrete RLC
components. During the 50’s the search for a significant cut in production
cost for this type of circuit started. The idea was to replace inductors by
active circuits, as these circuits were cheaper and had a lower physical
profile. This substitution was based on the fact that a circuit consisting in
resistors, capacitors and active elements —transistors, opamps, valves—
can exhibit resonance, as RLC circuit do. Resonance, which implies abrupt
changes in magnitude and phase, a consequence of a large quality factor, is
sought for an improved filter selectivity, at a reasonable filter order. These
RC-active filters remain in the academic context until the arrival of the
operational amplifier (OPAMP), in the mid-60’s. The OPAMP’s were cheap
1 - 29
Escuela Superior de Ingenieros, Universidad de Sevilla
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
1 - 30
Circuit Analysis and Synthesis
Lecture 1
active components of a good quality. At the very first moment, the physical
profile was not greatly reduced, but for the same cost, amplification was
provided concurrently with filtering. At the early 70’s, production cost
dropped to half with the arrival of the first integration technologies (Hybrid
IC’s). Advances in integration technologies reduced both the cost and size
of filters. Nowadays, continuous-time active filters are completely
integrated in VLSI technology. There are also analog sampled filters based
on switched-capacitors, and digital filters based on digital siganl processors
(DSP’s).
Passive filters. Fig. 1.13(a) depicts an example of a passive filter. It is
composed of resistors, inductors and capacitors. Concretely, this is a
lowpass filter. This type of filter was extensively employed, but inductors
were too bulky and tgherefore expensive. In addition, reproducibility is
poor, this requires manual manufacturing and selection by measurements.
On the other hand, their components have a low sensitivity and can carry
large currents. They are still employed in power supply sources.
Active filters. Advances in microelectronic technology has quite significantly reduce the size of resistors and capacitors. It is not the case, however,
of the inductors. These are expensive, with important parasitics and nonlinearities, and can not be properly isolated. Before the transistor, the available
active elements were the triode lamps. They were expensive, unreliable,
sizable and power consuming. Since the appearing of the first reliable and
cheap opamps, inductors have been progressively substituted by RC-active
circuits. The reduction on the production cost and the circuit sizes has
continued as integration technologies evolve. Several design techniques
have been developed for active filters. One of them is the emulation of
passive filter prototypes. Other alternatives are the simulation of elements or
the cascade of biquads (Fig. 1.13(b)). Active filters avoid the use of
inductors. In addition, they feature more functionalities and provide
amplification. If they are integrated monolithically, production can be fully
automated, reducing cost and increasing accuracy.
Digital filters. Advances in microprocessors permit the realization of
the majority of the filtering functions in the digital domain but in real-time.
The advantages of digital circuits are, principally, improved noise inmunity
and implementation robustness. The circuit in Fig. 1.14 realizes the same
function as the circuit in Fig. 1.13(a).
x[n]
=
Delay
x ( nT )
z–1
a
a
3
a
2
z –1
x(t) = s(t) + n(t)
t
1
3
--2
x(t)
4
--3
1
--2
b
y(t) ≅ s( t)
1
3
1
z–1
b
2
z–1
b
1
Scaling
b
0
y( t) ≅ s( t)
t
(a)
R⁄2
C
y[n]
=
y ( nT )
Figure 1.14: Digital filter example.
C
C
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Figure 1.13: Passive (a) and active (b) filter realization.
The input to this filter are the samples, x(nT), of signal x(t). The output
are the samples y(nT) of the desired response. The filter is composed of shift
registers, multipliers and adders. It is a minicomputer that is digitally
simulating the operations ocurring in the analog system. In this context,
digital simulation is the design of a digital system that realizes the same
operation as a given analog system. The digital filter of Fig. 1.14 is a digital
1 - 31
Escuela Superior de Ingenieros, Universidad de Sevilla
R
x(t)
R
C
2R
C
Rα y ( t )
(b)
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
1 - 32
Circuit Analysis and Synthesis
Lecture 1
simulation of the analog filter of Fig. 1.13(a).
In this course we will concentrate on continuous-time passive and
active filters. Active filters present several advantages if compared with
their predecessors:
1) Size and weight has been reduced.
2) Reliability has increased with the production automation.
3) Mass production costs have dropped.
4) Components of a high quality, improved performance.
5) Active filter and digital circuitry can be built on the same chip.
6) Design and tuning is simple.
7) More functionalities and filtering functions.
8) Provide signal amplification.
They have also some disadvantages compared with passive filters:
1) Active components has a limited bandwidth. Passive filters can
be used up to much higher frequency ranges.
2) Passive filters are less sensitive to the environmental conditions.
3) Active filters require a power supply.
4) Signals amplitude is limited by distortion. Besides, resistors and
active devices generate noise. The dynamic range in active circuits is usually smaller than in passive circuits.
Solved problems
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Solution:
In the time-domain:
These are the constitutive equations for these elements:
v
i R = ----cR
ic = C
dv c
L
dt
di L
dt
(1.83)
= vc
besides, nodal analysis yields:
iR + iL + ic = 0
(1.84)
Substituting the constitutive relations in (1.83) the following differential
equation in i L results:
2
LC
d iL
dt
2
L di L
+ --+ iL = 0
R dt
(1.85)
The solution to this equation have the form:
iL ( t ) = k1 e
α1 t
+ k2 e
α2 t
(1.86)
2
with
1
4R C
1
α 1, 2 = – ----------- ± ----------- 1 – -------------2RC 2RC
L
if
α1 ≠ α2
(1.87)
or, alternatively:
1.1.- Consider the circuit in the diagram. The initial conditions are
v c ( 0 ) = V o and i L ( 0 ) = I o . Obtain the zero-input response for the
inductor’s current.
n1
iR
R
n2
iC
C
iL
L
i L ( t ) = ( k 1 + k 2 t )e
with
1α = – ---------2RC
αt
if
(1.88)
α1 = α2 = α
(1.89)
the values of k 1 and k 2 are obtained from the initial conditions, i L ( 0 ) = I o
and ( di L ⁄ dt ) ( 0 ) = V o ⁄ L :
–α2 Io + Vo ⁄ L
α1 Io – Vo ⁄ L
k 1 = --------------------------------and
k 2 = -----------------------------(1.90)
α1 – α2
α1 – α2
for (1.86), or:
1 - 33
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 34
Circuit Analysis and Synthesis
k1 = Io
Lecture 1
k2 = –α Io + Vo ⁄ L
and
(1.91)
for (1.88).
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
( s + 1 ⁄ RC )I o + V o ⁄ L
k1
k2
I L ( s ) = ---------------------------------------------------- = ----------- + -----------------2
s – α ( s – α )2
s + s ⁄ RC + 1 ⁄ LC
if α 1 = α 2 = α . Now:
1α = – ---------2RC
In the frequency domain:
The constitutive equations are now:
Vc ( s )
I R ( s ) = -----------R
k1 = Io
and thus



I c ( s ) = sCV c ( s ) – CV o 

sLI L ( s ) – LI o = V c ( s ) 
(1.99)
k2 = –α Io + Vo ⁄ L
and
iL ( t ) = k1 e
α1 t
+ k2 e
α2 t
from (1.95)
(1.101)
or:
i L ( t ) = ( k 1 + k 2 t )e
αt
from (1.98)
(1.102)
1.2.- Compute the step response for the circuit in the figure.
iL
and the following algebraic equation results:
L
L
s LCI L ( s ) + s --- I L ( s ) – LCI o + I L ( s ) – --- I o – CV o = 0
R
R
2
solving in I L ( s ) , yields:
k1
k2
( s + 1 ⁄ RC )I o + V o ⁄ L
- = -------------I L ( s ) = ---------------------------------------------------+ ------------2
s – α1 s – α2
s + s ⁄ RC + 1 ⁄ LC
–α2 Io + Vo ⁄ L
and thus k 1 = --------------------------------α1 – α2
and
(1.96)
α1 Io – Vo ⁄ L
k 2 = -----------------------------α1 – α2
(1.97)
C
L
–
vC ( 0 ) = 0
Solution:
Now applying the KCL yields:
i R + i L + i c = Eu o ( t )
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
(1.103)
resulting in the following ODE:
2
LC
or, alternatively:
1 - 35
R
–
iL ( 0 ) = 0
In the time domain:
2
1
1
4R C
α 1, 2 = – ----------- ± ----------- 1 – -------------2RC 2RC
L
Eu o ( t )
(1.94)
if α 1 ≠ α 2 (1.95)
(1.100)
Calculating the inverse Laplace transform we obtain i L ( t ) :
(1.92)
where I R ( s ) = L [ i R ( t ) ] , I c ( s ) = L [ i c ( t ) ] , I L ( s ) = L [ i L ( t ) ] and
V c ( s ) = L [ v c ( t ) ] . These equations include the initial conditions of the reactive elements. Applying Kirchoff’s current law (KCL):
(1.93)
IR ( s ) + Ic ( s ) + IL ( s ) = 0
where
(1.98)
L di L
+ --+ i L = Eu o ( t )
2
R dt
dt
d iL
Escuela Superior de Ingenieros, Universidad de Sevilla
(1.104)
1 - 36
Circuit Analysis and Synthesis
Lecture 1
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
This is an inhomogeneus equation. Its solution is the combination of the solution of the homogeneus equation (that obtained in the previous exercise)
and a particular solution. Thus:
i L ( t ) = Eu o ( t ) + k 1 e
α1 t
+ k2e
α2 t
if
α1 ≠ α2
(1.105)
Es - + -----1 - ( s ) = -------- s 2 + ------I

RC LC L
sLC
2
(1.106)
k
k1
k2
E ⁄ LC
I L ( s ) = ------------------------------------------------------ = ----o- + -------------+ -------------2
s s – α1 s – α2
s ( s + s ⁄ RC + 1 ⁄ LC )
αt
(1.113)
if α 1 ≠ α 2 , where:
2
1
4R C
1
α 1, 2 = – ----------- ± ----------- 1 – -------------2RC 2RC
L
or, alternatively:
i L ( t ) = Eu o ( t ) + ( k 1 + k 2 t ) e
(1.112)
Solving for I L ( s ) :
where:
1
1
4R C
α 1, 2 = – ----------- ± ----------- 1 – -------------2RC 2RC
L
3º Telecomunicación
if α 1 = α 2 = α
(1.107)
and therefore:
α2 E
k 1 = -----------------α1 – α2
with
1α = – ---------2RC
(1.114)
(1.108)
–α1 E
k 2 = -----------------α1 – α2
and
(1.115)
or, alternatively:
Once more, constants k 1 and k 2 are derived from the initial conditions,
being now i L ( 0 ) = 0 and ( di L ⁄ dt ) ( 0 ) = 0 . This results in:
α2 E
–α1 E
(1.109)
and
k 2 = -----------------k 1 = -----------------α1 – α2
α1 – α2
where
for (1.105), and:
k1 = –E
and then k 1 = – E
k 2 = αE
and
(1.110)
k
k1
k2
I L ( s ) = ----o- + ----------- + -----------------s s – α ( s – α )2
1α = – ---------2RC
if α 1 = α 2 = α
(1.116)
(1.117)
and
k 2 = αE
(1.118)
By obtaining the inverse transform of I L ( s ) we arrive again to (1.105)
and (1.107), respectively.
for (1.107).
In the frequency domain:
Now, considering that L [ Eu o ( t ) ] = E ⁄ s , the KCL yields:
E
I R ( s ) + I c ( s ) + I L ( s ) = --s
Substituting the constitutive relations of each branch:
1.3.- Consider the– t ⁄ Tcircuit of the figure. The input signal is
1
v1 ( t ) = Vo e
. Find the values for R , C and K that make the
response v 2 ( t ) to be an exponential with a time constant smaller than
T1 .
1 - 37
Escuela Superior de Ingenieros, Universidad de Sevilla
(1.111)
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
1 - 38
Circuit Analysis and Synthesis
Lecture 1
C
v1 ( t )
R
Kv 1 ( t )
Solution:
The Laplace transform of the excitation signal is:
sV o
V i ( t ) = ----------------2
2
s + ωo
+
v2 ( t )
−
Solution:
Analyzing the circuit, the following equation is found:
V 2 – kV 1
-------------------- + sC ( V 2 – V 1 ) = 0
R
the network function results:
V2( s )
s + k ⁄ RCH ( s ) = ------------= ---------------------V1( s )
s + 1 ⁄ RC
while the transform of the input signal is:
Vo
V 1 ( s ) = -------------------s + 1 ⁄ T1
C
+
(1.119)
vi ( t )
+
-
L
R
vo ( t )
_
(1.120)
(1.121)
(1.122)
In order to have a single exponential we need that k ⁄ RC = 1 ⁄ T 1 . This
will cause pole-zero cancellation. For the time constant of the remaining
exponential to be less than T 1 , it will be required that RC < T 1 , therefore:
RC
k<1
-------- = T 1
(1.123)
k
1.4.- Propose a circuit with a null forced response to the excitation signal
v i ( t ) = V o cos ω o t . What can we use it for?
1 - 39
(1.124)
For the forced response to be null, transmission zeros at that frequency will
be required, this is at s = ± jω o . This circuit will implement them:
then, the Laplace transform of the response is:
Vo
s + k ⁄ RC
V 2 ( s ) = H ( s )V 1 ( s ) = ----------------------- ⋅ --------------------s + 1 ⁄ RC s + 1 ⁄ T 1
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
It has the following transfer function:
2
Vo ( s )
s + 1 ⁄ LC
H ( s ) = ------------= ---------------------------------------------2
Vi ( s )
s + s ⁄ RC + 1 ⁄ LC
(1.125)
Selecting L and C to hold ω o = 1 ⁄ LC , the excitation poles cancel with
the network function zeros. The response will not have any of the excitation
poles. Therefore it results in a null forced response. Not strange as this circuit is a 2nd order symmetric band rejection filter. Let us re-write H ( s ) :
2
2
s + ωo
H ( s ) = -------------------------------------2
2
s + 2αω o + ω o
(1.126)
where ω o = 1 ⁄ LC and α = 1 ⁄ 2RC . Plotting H ( jω )
H ( jω ) dB
0
ωo
ω
2
2
ωo – ω
H ( jω ) = --------------------------------------------------2
2
2 2
( ω o – ω ) + 4α ω
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 40
Circuit Analysis and Synthesis
Lecture 1
Being Y ( s ) = H ( s )X ( s ) , several alternatives are possible:
a) Trivial case: H ( s ) = 1 . The response will be completely forced.
It requires X ( s ) to be equal to Y ( s )
2
b) H ( s ) = 1 ⁄ ( s + 1 ) y X ( s ) = 1 ⁄ ( s + 1 ) , it means
A transmission zero is seen at ω o .
This is an alternative circuit:
+
R
vi ( t )
+
-
C
vo ( t )
L
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
2
1
s + ------LC
H ( s ) = -------------------------------2 R
1
s + --- s + ------L
LC
R
+
x ( t ) = V o senω o t
_
y(t)
+
-
RC = 1seg
C
_
2
1.5.- Find a circuit and an excitation signal that results in the following
zero-state response:
y ( t ) = V1e
–( t ⁄ τ )
– V 2 cos ( ω o t + θ ) ,
Solution:
Let us obtain the Laplace transform of the response, in the first place:
V1
–( t ⁄ τ )
] = ---------------------L [ V1e
s + (1 ⁄ τ)
(1.127)
s – ωo
----------------L [ V 2 cos ( ω o t + θ ) ] = L [ V 2 cos ω o t – V 2 senω o t ] = V 2
2
2
s + ωo
s – ωo
V1
Y ( s ) = ---------------------- – V 2 -----------------2
2
s + (1 ⁄ τ)
s +ω
(1.128)
o
substituting the values:
(1.129)
1
1
1 s–1
1
Y ( s ) = --- ⋅ ----------- – --- ⋅ -------------- = – ----------------------------------2
2 s + 1 2 s2 + 1
(s + 1)(s + 1)
1 - 41
x ( t ) = Vo e
–( t ⁄ τ )
LC = 1seg
where V 1 = 0.5V , τ = 1s , V 2 = ( 1 ⁄ 2 )V , ω o = 1rad/s and
θ = π ⁄ 4.
thus
c) H ( s ) = 1 ⁄ ( s + 1 ) y X ( s ) = 1 ⁄ ( s + 1 ) , this time
(1.130)
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
+
+
-
2
L
y(t)
C
_
Although in practice, any real inductor will have losses and this
H ( s ) will not be realizable with passive elements.
d) X ( s ) = 1 . The input is an impulse, the response will be completely natural. This is obvious because it will be the impulse
response. H ( s ) is required to match Y ( s ) . This network function
is not realizable though as a passive filter.
1.6.- The current i ( t ) in Fig. 1 has the waveform shown in Fig. 2.
(a)Obtain and draw the location of the poles and zeros of the input
signal.
(b)Which value of T p cancels the natural component of the
response, v ( t ) ?
(c)Let T p be the value of T p already obtained. Find and draw v ( t )
0
for T p < T p , T p = T p and T p > T p .
0
0
Escuela Superior de Ingenieros, Universidad de Sevilla
0
1 - 42
Circuit Analysis and Synthesis
Lecture 1
1Ω
i( t)
a
1F
v(t)
a–1
−
Tp
Figure 1
Figure 2
t
Solution:
In the time domain, i ( t ) is expressed as:
i ( t ) = au o ( t ) – u o ( t – T p )
– sT p
)⁄s
(1.131)
(1.132)
(a) Let us first find the location of the poles. It is easy to identify that the
only pole is at the origin: p 1 = 0 . For the zeros, the following equation must
be solved:
a–e
–zk Tp
(1.133)
= 0
where z k can be complex, z k = α k + jβ k . Then (1.133) can be re-written:
e
–αk Tp
( cos β k T p – j sin β k T p ) = a
(1.134)
Because a is real, sin β k T p must be zero. It means that β k T p must be 0 or
kπ , being k an integer number. There is an infinite number of zeros. Their
imaginary parts score:
kπ
(1.135)
β k = ± -----where
k = 0, 1, 2, …, ∞
Tp
On the other hand, the real part of the infinite zeros comes from here:
e
–αk Tp
cos β k T p = a
(1.136)
As a is positive, only the even multiples of π will be selected from the infinite set of β k T p ’s already calculated. Then from (1.135) we will only keep
1 - 43
Summarizing, the zeros of the excitation are at
1 1 2πk
with
k = 0, 1, 2, …, ∞
z k = ------ ln --- ± j --------Tp a
Tp
and there is a single zero at
in the Laplace domain, it corresponds to:
I(s) = (a – e
3º Telecomunicación
these: k = 0, 2, 4, … . For them, cos β k T p = 1 , and so:
1
α k = ------ ln 1--now with
k = 0, 2, 4, …, ∞
Tp a
i( t )
+
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
z2
z1 ′
(1.138)
p1 = 0
Im ( s )
z3
z1
z0
(1.137)
p1
Re ( s )
z2 ′
Notice that:
1 1
α = ------ ln --- < 0
Tp a
because:
a>1
(b) In order to find the value of T p that cancels the natural component
of ther response, v ( t ) , from the time instant t = T p . This differential
equation describes the evolution of the circuit:
dv ( t )
( t )- = i ( t )
(1.139)
C ------------ + v-------dt
R
in the Laplace domain:
 sC + --1- V ( s ) = I ( s )

R
and thus
( s )- = ---------------------1⁄C H(s) = V
---------I ( s ) s + 1 ⁄ RC
(1.140)
The Laplace transform fo the response, considering that the circuit is at
rest, as nothing has been said:
– sT p
1
(1.141)
)
V ( s ) = H ( s )I ( s ) = ----------------------------------- ( a – e
sC ( s + 1 ⁄ RC )
The only way to cancel the real pole that H ( s ) has at s = – 1 ⁄ RC is by using the real zero z 0 = ( 1 ⁄ T p ) ln ( 1 ⁄ a ) of the excitation. If T p is the value
0
of T p that makes this cancellation possible, then:
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 44
Circuit Analysis and Synthesis
Lecture 1
1
1-------- ln 1--- = – ------Tp a
RC
this is
T p = RC ln a
(1.142)
0
0
(c) Let us obtain first v ( t ) for a generic T p . This is its transform:
– sT p
1⁄C
a–e
a ⁄ C - – ------------------------------e
V ( s ) = ---------------------------------- = -----------------------------sC ( s + 1 ⁄ RC )
s ( s + 1 ⁄ RC ) s ( s + 1 ⁄ RC )
– sT p
(1.143)
– sTp
The term e
postpones the waveform corresponding to the second term
by T p . We are interested in expanding the fraction 1 ⁄ s ( s + 1 ⁄ RC ) :
1
RC
RC
------------------------------- = -------- – ----------------------(1.144)
s ( s + 1 ⁄ RC )
s
s + 1 ⁄ RC
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Here, the horizontal asymptote of the exponential, that is v ( t ) = aR for the
first piece, changes to v ( t ) = ( a – 1 )R after t = T p .
Consider now that T p = T p . In this conditions, exp ( T p ⁄ RC ) = a
0
0
and v ( T p ) = ( a – 1 )R . The form of the response is now:
0
– t ⁄ RC
)
 aR ( 1 – e
v(t) = 
( a – 1 )R

for
for
0 < t < Tp
t > Tp
(1.147)
and therefore:
v(t)
aR
( a – 1 )R
In this way, the inverse transform of V ( s ) is:
v ( t ) = aR ( 1 – e
– t ⁄ RC
)u ( t ) – R [ 1 – e
– ( t – T p ) ⁄ RC
]u ( t – T p )
or, better:
– t ⁄ RC

)
for
0 < t < Tp
aR ( 1 – e

v(t) = 
 ( a – 1 )R – ( a – e Tp ⁄ RC )Re – t ⁄ RC
for t > T p

Tp
(1.145)
(1.146)
Let us particularize v ( t ) for the different cases. Let us suppose in the
first place that T p < T p . We have obtained that exp ( T p ⁄ RC ) = a . As
0
0
T p < T p , the term exp ( T p ⁄ RC ) , in the second piece of v ( t ) , is smaller
0
than a . Besides, v ( T p ) < ( a – 1 )R , therefore the response has thye
following shape:
v(t)
aR
0
t
Finally, if T p > T p , then v ( T p ) > ( a – 1 )R . The second summand in
0
the second piece in (1.146) is positive, and decays with time. que se va
haciendo cada vez má pequeño a medida que pasa el tiempo. The shape of
the repsonse will be:
v(t)
aR
( a – 1 )R
Tp
0
Tp
t
1.7.- Calculate the frequency response (Bode plot) of a circuit which has the
following network function:
( a – 1 )R
2
Tp Tp
0
1 - 45
t
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
s +1
H ( s ) = --------------------------------2
s + s0.01 + 1
Escuela Superior de Ingenieros, Universidad de Sevilla
(1.148)
1 - 46
Circuit Analysis and Synthesis
Lecture 1
Solution:
Observing H ( s ) , there is a pair of zeros at s = ± j , with an infinite
quality factor, and a pair of conjugate complex poles with pole frequency
ω o = 1rad/s and quality factor Q = 100 . Symbolically:
H ( jω ) dB
1
0
log 10 ω
2
2
s + ωz
H ( s ) = --------------------------------2
2 ω0
s + ------ s + ω 0
Q
(1.149)
if we do s = jω :
2
2
For the phase ∠H ( jω ) :
1 – sgn ( ω o – ω )
 ωω o ⁄ Q
∠H ( jω ) = ------------------------------------- π – atan  -------------------
2
 ω2 –ω2 
0
(1.150)
∠H ( jω )
∠H ( jω )
0
2

ω 
H ( jω ) dB = 40log 10 ω z + 20log 10  1 – ------ –
2

ωz 
adding them up:
ωz
ω
H ( jω ) dB
H ( jω ) dB
1.8.- Obtain the network function H ( s ) to this asymptotic Bode plot.
Consider only real roots.
40dB/dec
log 10 ω
≈ ω0
|H(jω)|
log 10 ω
0
-20dB/dec
– 40 dB/dec
Substituting ω 0 , Q and ω z by their corresponding values, and adding
the contributions, ends up in:
1 - 47
1
π
– --2
These four summands contribute to H ( jω ) dB in a different manner:
– 40 log 10 ω o
ω
π
+ --2
0
0
log 10 ω
ω
π
– --2
–π
∠H ( jω )
(1.151)
2 2

ω 2
– 40log 10 ω 0 – 20log 10  1 – ω
------ +  -----------

ω 0 Q
 ω 2
ωz
≈ ω0
0
π
For the magnitude in decibels:
40log 10 ω z
(1.152)
that represent the following contributions:
ωz – ω
H ( jω ) = -------------------------------------ω0
2
2
ω 0 – ω + jω -----Q
H ( jω ) dB
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
60dB
-40dB/dec
-20dB
ω1
ω2
arg[H(jω)]
ω1/10 ω1
10ω1 ω2/10 ω2
10ω2
−π/2
−3π/4
−π
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 48
Circuit Analysis and Synthesis
Lecture 1
Solution:
In the first place, the initial slope of of the magnitude plot is – 20dB/dec
(or – 6dB/oct ) indicates the presence of a pole at the origin (un factor 1 ⁄ s ).
This pole explains the initial phase of – π ⁄ 2 .
After that, at ω = ω 1 another – 20dB/dec are added to the slope of the
magnitude, and – π ⁄ 4 is contributed to the phase. Therefore, we can
conclude that there is another pole at s = – ω 1 . It can not be at s = ω 1
because the phase would have been contributed +π ⁄ 4 instead of – π ⁄ 4 .
Following this, there is a pair of ceros at ω 2 . In order to prevent any
change in the phase, each one of these zeros must be in a different side of the
jω -axis. This is, at s = ± ω 2 . Do not forget that they are real.
Up to now:
( s + ω2 ) ( s – ω2 )
(1.153)
H ( s ) = K ---------------------------------------s ( s + ω1 )
From the magnitude plot we have:
lim H ( jω ) dB = 20log 10 K = – 20
(1.154)
and then log 10 K = – 1
(1.155)
ω→∞
thus
K = 0.1
( s + ω2 ) ( s – ω2 )
then, finally H ( s ) = --------------------------------------10s ( s + ω 1 )
(1.156)
Also, there is a gap of 80dB between H ( jω 1 ) dB and H ( jω 2 ) dB . If
the slope of the magnitude plot between ω 1 and ω 2 is – 40dB/dec , then
ω 2 = 100ω 1 . And finally:
( s + 100ω 1 ) ( s – 100ω 1 )
H ( s ) = – ---------------------------------------------------------(1.157)
10s ( s + ω 1 )
The obtained zero-state response for an excitation signal x ( t ) is
–t
–t
y ( t ) = [ e – e cos t ]u o ( t ) . Find x ( t ) .
H ( jω )
1
0
0
1
ω
Solution:
The first thing to do is computing the Laplace transform of the response:
1 - – -------------------------1
1
Y ( s ) = ---------- – --------------------------=
s + 1 2(s + 1 – j) 2(s + 1 + j )
(1.158)
1
= -----------------------------------------------------------------( s + 1 )( s + 1 – j )( s + 1 + j )
Besides, H ( s ) has 3 poles. They are at s = – 1 and s = – 1 ± j , thus:
N(s)
(1.159)
H ( s ) = -----------------------------------------------------------------( s + 1 )( s + 1 – j )( s + 1 + j)
As Y ( s ) = H ( s )X ( s ) , then X ( s ) = 1 ⁄ N ( s ) . so the problem reduces to
calculate N ( s ) . The order of N ( s ) is less than three, because H ( jω ) → 0
when ω → ∞ . It has to be, therefore, 0, 1 or 2. Besides H ( j1 ) = 0 . For this
2
to happen, a factor of the form s + 1 is required in the function’s numerator.
Finally, as H ( j0 ) = 1 , we have:
2
K(s + 1)
-----------------------------------------------2
( s + 1 ) ( s + 2s + 2 )
= K
---- = 1
2
from where
K = 2
(1.160)
s = j0
In this way:
2
2( s + 1)
H ( s ) = -----------------------------------------------2
( s + 1 ) ( s + 2s + 2 )
and
1.9.- The frequency response of a network is plotted in the figure (not to
scale). Function H ( s ) is supposed rational. In this circuit
H ( j0 ) = 1 , H ( j1 ) = 0 and t H ( j∞ ) = 0 . Consider that H ( s )
has the minimum number of simple poles in s = – 1 and s = – 1 ± j .
what means that:
1
x ( t ) = --- sin ( t )u o ( t )
2
1 - 49
Escuela Superior de Ingenieros, Universidad de Sevilla
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
1 X ( s ) = --------------------2
2(s + 1)
(1.161)
(1.162)
1 - 50
Circuit Analysis and Synthesis
Lecture 1
Proposed problems
1.15.-Calculate k in the circuit of the figure, in order to have oscillatory
behaviour. It is a phase-shift oscillator.
1.10.-Propose a system and an excitation having a null forced response.
C V2
Vi
I1
1.11.-Propose a system and an excitation having a null natural response.
R
i1 ( t )
– -tτ
I2
v1 ( t )
1
τ = --- s.
2
−
1.13.-Obtain the forced response of the following circuit:
1F
+
-
1H
R
Ao
A ( s ) = -----------------------------------------------------------------( 1 + sτ 1 ) ( 1 + sτ 2 ) ( 1 + sτ 3 )
an
d y
dt
n
+ an – 1
d
dt
y
+ … + a 0 y = B + A sin ( ωt )
has constant coefficients. Initial conditions are zero.
a) Obtain the network function. Suppose that B + A sin ( ωt ) represents the excitation.
b) Obtain the forced response.
1 - 51
Vi
A(s)
Vo
_
n–1
n–1
(1.163)
where τ 1 = 1s , τ 2 = 1µs , τ 3 = 0.1µs . Applying negative feedback
to the opamp, as in the figure, the response slows down because of the
displacement of the poles. Obtain the dc gain yielding oscillatory
behaviour.
Vo ( t )
1.14.-The following differential equation:
n
R
–k
+
V i ( t ) = Asinωt
I3
R
1.16.-An operational amplifier can be described a model of 3 poles:
+
1C
1Ω
C V
o
C V3
1.12.-Consider the circuit of the figure. The response is a linear combination
of v 1 ( t ) and i 1 ( t ) . Which non-trivial combination results in a null
forced response? Which one results in a null natural response?
Vo e
3º Telecomunicación
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
1.17.-Find the root locus as a function of R . Find the values of R , L and C
that makes this circuit and oscillator. Obtain the oscillation frequency.
iL ( t )
L
vi
L = 1H
+
R
C
Escuela Superior de Ingenieros, Universidad de Sevilla
vC ( t )
R = 1Ω
C = 1F
−
1 - 52
Circuit Analysis and Synthesis
Lecture 1
REPRESENTATION AND RESPONSE IN CONTINUOUS-TIME LTI CIRCUITS
3º Telecomunicación
1.18.-Obtain the network function corresponding to these Bode plots.
|H(jω)|dB
< H(jω)
6 dB
A0
−40 dB/dec
log (ω1)
0
log(ω)
log(ω)
1.19.-Draw the approximate Bode plot for the function:
20s ( s + 100 ) H ( s ) = ----------------------------------( s + 2 ) ( s + 10 )
in the frequency range 0.1rad/s ≤ ω ≤ 1000rad/s .
1 - 53
Curso 2004/05 © Departamento de Electrónica y Electromagnetismo
Escuela Superior de Ingenieros, Universidad de Sevilla
1 - 54