Introduction to Frequency-Domain Analysis of
Continuous-Time, Linear and Time-Invariant Systems
• Time-domain analysis of transient response
• Fourier series of periodic Dirichlet signals
• Bode plots of system frequency-response
• Bilateral Fourier transform for zero-state response
• Unilateral Laplace transform for transient response
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Time-domain analysis of continuous-time LTI systems
• Signals: properties, operations, construction, important signals
• Single-input, single-output systems: properties
• The zero-input response (ZIR): characteristic values and modes
• The zero (initial) state response (ZSR): the impulse response, convolution
• System stability
• The (zero state) system transfer function and eigenresponse
• Resonance and step-response of second-order tuned circuits
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Introduction to signals - discrete and continuous time
• A function x is a mapping of elements of a domain A to a range B , i.e.,
x : A ! B,
in such a way that for all elements t of A, x(t) is a single element of B , i.e.,
8t 2 A, x(t) 2 B.
• The strict range of x is {x(t) | t 2 A} ⇢ B .
• For example x(t) = t2 , t 2 R, has strict range the non-negative real numbers,
R 0 := {z 2 R | z
where “a := b” means true by definition of a.
0} =: [0, 1)
• In this course, we will use the words “function” and “signal” interchangeably.
• Also, we will consider a one-dimensional, ordered domain A for our signals, and associate
that domain with time.
• Note that in image processing, signals (images) will have two-dimensional spatial domains.
• Video signals are three-dimensional (two-dimensional images as a function of time).
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Signal/function types - discrete and continuous time (cont)
• If the domain A has “countably” many values, then x : A ! B is said to be discrete time.
• A set is countable if there is a one-to-one mapping between it and a subset of the set of
integers, Z.
• In this course, we will consider only continuous time-signals x whose domains are subsets
of the real line, R, typically either A = R = ( 1, 1) or A = R 0 = [0, 1), and
hence uncountably infinite in size.
• We will also consider signals that are either real or complex valued, i.e., their range B = R
or B = C.
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A signal operation/transformation - time shift
• For all signals x : R ! R and scalars ⌧ 2 R, define a new signal
8t 2 R, (
⌧
x)(t) = x(t
⌧
x as
⌧ ),
i.e., the signal x delayed by ⌧ seconds (equivalently, advanced by
⌧ seconds).
• An example pulse x is depicted in Figure (a).
•
•
2 x (x delayed by -2 seconds, or advanced by 2 seconds) is depicted in Figure (b); note
that 2 = (
( 2)) = x(1).
2 x)( 1) = x( 1
2
x(3
x (x delayed by 2 seconds) is depicted in Figure (c); note that 2 = (
2) = x(1).
2
x)(3) =
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Periodic and aperiodic signals
• For the following introductory discussion, we will use examples where x : R ! R.
• A signal is said to be periodic if there is a fixed real number T 6= 0 such that
8t 2 R, x(t) = x(t
T ).
• The smallest such T > 0 of a periodic signal x is said to be the period of x.
• We take constant (DC) signals, i.e.,
9c 2 R such that 8t 2 R, x(t) = c,
to be periodic with “fundamental frequency” 1/T = 0 (which motivates us to choose
their period T = 1 by definition).
• A periodic signal with period T will be called T -periodic.
• A signal that is not periodic is said to be aperiodic.
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Periodic signals - examples
• For example, the sinusoid x(t) = sin(5t), t 2 R, has period 2⇡/5.
• The signal (a) in the following figure has period 4.
• The truncated exponential signal (b) is aperiodic.
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Expressing periodicity with the delay operator
• Let’s recall the time-delay operator
functional form.
so as to restate the definition of periodicity in
• A signal x is said to be periodic if there is a real T 6= 0 such that
x =
T
x,
recalling that this statement expresses functional equivalence, i.e.,
8t 2 R, x(t) = (
T
x)(t) := x(t
• So if x is T -periodic (for T > 0) then x is an “eigenfunction” of
T ).
nT
for all n 2 Z.
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Sums of periodic signals
• A periodic signal with period T > 0 “repeats itself” every nT seconds, for any integer
n > 0;
• that is, the pattern of the signal is repeated every consecutive interval of duration nT
seconds.
• Suppose x and y are periodic signals with periods Tx > 0 and Ty > 0, respectively.
• If there are positive integers nx > 0 and ny > 0 such that nx Tx = ny Ty =: T , then the
sum of the signals, x + y , will repeat itself every T seconds, where to be clear,
8t 2 R, (x + y)(t) := x(t) + y(t);
i.e., x + y will also be a periodic signal with period  T .
• The converse statement is also true, leading to:
• If x and y are periodic, then: x + y is periodic if and only if Tx /Ty is rational (= ny /nx
for integers nx, ny > 0).
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Sums of periodic signals - examples
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Sums of periodic signals - examples (cont)
• Consider the following periodic signals: for t 2 R,
x1 (t) = 5 cos(7t + 1),
y1 (t) = 4 sin(3⇡t + ⇡/2),
x2 (t) = 13 sin(12t
p
y2 (t) = 2 sin(5⇡t
1),
⇡/6)
• x1 + x2 is periodic because the period of x1 is 2⇡/7 and that of x2 is 2⇡/12; so the
ratio of their periods, 7/12 (or 12/7), is rational.
• y1 + y2 is periodic because the period of y1 is 2⇡/(3⇡) = 2/3 and that of y2 is
2⇡/(5⇡) = 2/5; so, the ratio of their periods, 5/3 (or 3/5), is rational.
• In these cases, the period of the sum is the least common integer multiple of the components’
periods, e.g., the period of x1 + x2 is 2⇡ , while that of y1 + y2 is 2.
• However, x1 +y2 is aperiodic because the ratio of their periods is (2⇡/7)/(2/5) = 5⇡/7
which is not rational (⇡ = 3.1415... is not a rational number).
• Adding a constant to a periodic signal results in a periodic signal, i.e., if x is T -periodic
and c 2 R, then x + c is T -periodic where (x + c)(t) := x(t) + c, t 2 R.
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Energy signals
• A signal x : R ! R is said to be an energy signal if
Z 1
0 < Ex :=
|x(t)|2 dt < 1.
1
• Note that if x is either the voltage or current signal associated with a resistive load, then
Ex would be proportional to the energy dissipated by the load.
• For example, for parameters A, T, b 2 R with b > 0, if
⇢
Ae b(t T ) if t T
x(t) =
0
if t < T,
then
Ex =
Z
1
1
|x(t)|2 dt = A2
Z
1
e
2b(t T )
dt = A2/(2b);
T
since Ex = A2/(2b) < 1, x is an energy signal.
• The exponential signal x(t) =
7e
5t ,
t 2 R, is not an energy signal because Ex = 1.
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Power signals
• A signal x : R ! R is said to be a power signal if the following limit exists and
Z
1 ⌧ /2
0 < Px := lim
|x(t)|2 dt < 1,
⌧ !1 ⌧
⌧ /2
i.e., the Px is average energy per unit time of x.
• Note that if x was an energy signal, then Px = 0.
• Conversely, if Px > 0, then Ex = 1.
• If a signal x is periodic with period T , then
Px =
where
R
T
1
T
Z
T
|x(t)|2 dt
denote integration over any interval of R of length T , i.e., over any period of x.
• To see why, note that if ⌧ /T is an integer, then the numerator of Px can be written as a
sum of ⌧ /T identical, consecutive integrals over periods of length T .
• Power signals are not necessarily periodic, though all of our examples will be.
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Power versus energy signals - examples
• Fig. (b) is an energy signal with
Z 1
Z
Eg =
g 2 (t)dt =
1
1
4e
2(t+1)
dt =
2e
2(t+1)
1
1
• Fig. (a), a wave with period T = 4, is a power signal with
✓Z 0
◆
Z
Z 2
1 2 2
1
t
Pf =
f (t)dt =
12 dt +
(1 t)2 dt =
4 2
4
4
2
0
1
0
2
= 2.
1
(1
12
2
t)3
=
0
• Adding a non-zero constant c 2 R to (i.e., spatially shifting) any energy signal x gives a
power signal x + c with Px = c2.
• Also, for any T -periodic, power-signal
x and non-zero constant c 2 R, Px+c = Px + c2
R
only if x has zero mean, i.e., T x(t)dt = 0, cf. Parseval’s theorem.
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2
.
3
Power versus energy signals - examples
• Consider a real-valued sinusoid of canonical form (referenced to cosine):
x(t) = A cos(wo t + ), t 2 R,
where the parameter A > 0, A 2 R (i.e., A 2 R>0) is the amplitude, period T =
2⇡/wo 2 R>0 , and phase 2 R (since cosine has period 2⇡ , we can take 2 [ ⇡, ⇡)).
• A sinusoid is a power signal with
Z 2⇡/wo
1
Px =
A2 cos2 (wo t + )dt
2⇡/wo 0
✓
◆
Z
A2 wo 2⇡/wo 1
1
=
cos(2wo t + 2 ) +
dt
2⇡ 0
2
2
✓
◆ 2⇡/wo
✓
◆
A 2 wo
1
1
A 2 wo
⇡
=
sin(2wo t + 2 ) + t
=
0+
2⇡
4wo
2
2⇡
wo
0
A2
=
.
2
p
p
• Thus, the root mean-square (RMS) value of a sinusoid x is Px = A/ 2.
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Periodic extensions of aperiodic signals of finite support
• The support of a signal x is the interval of time over which it is not zero.
• A signal of finite support is sometimes called a pulse.
• In this example, the pulse x has support of duration 4.
• For T
4, the T -periodic extension of x depicted in this figure is
xT (t) :=
1
X
n= 1
x(t
nT ) =
1
X
n= 1
(
nT
x)(t), t 2 R,
i.e., superposition of time-shifted pulses x by all integer (n) multiples of T .
• If x is an energy signal then xT will be a power signal with PxT = Ex /T .
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Neither all signals of finite support are energy signals
nor all periodic signals are power signals
• Consider the signal
x(t) =
⇢
1
1 t
0
0t<1
else
• The support of this signal is of duration 1.
• But note that limt"1 x(t) = 1, i.e., the signal has finite escape time.
• This signal is not an energy signal because Ex = 1.
• Also, any periodic extension of this signal would not be a power signal.
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Bounded signals
• A signal x is bounded if there exists a real M < 1 such that
8t 2 R, |x(t)|  M.
• Obviously, the signal of the previous example is not bounded.
• The sinusoid x(t) = A cos(wo t + ), t 2 R is bounded by M = A > 0.
• The signal x(t) = c + A cos(wo t + ), t 2 R, where the constant c 2 R, is bounded
by M = max{A + c, | A + c|} (note that c may be negative).
p
• Note that the signal x(t) = 1/ t for t
1, and x(t) = 0 else, is a bounded signal
(M = 1) with limt!1 x(t) = 0, but it is
R⌧
– not an energy signal since Ex = lim⌧ !1 1 t 1dt = lim⌧ !1 log(⌧ ) = 1, and
– not a power signal since Px = lim⌧ !1 ⌧
1 log(⌧ /2)
= 0.
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Signal types - discussion
• We will define other types of signals in the following as needed, e.g., causal signals, Dirichlet
signals.
• Such definitions can be extended to C-valued signals by, e.g.,
– simultaneously applying the definition to the both the real and complex components of
the signal, e.g., for a C-valued signal to be periodic, both its real and imaginary parts
need to be periodic with rational ratio of their periods; or
– using modulus instead of absolute value, i.e., for x 2 C,
|x|2 = |Re{x}|2 + |Im{x}|2 .
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Signal operations - time scaling
• Recall the time-delay operator
⌧
for any fixed ⌧ 2 R: (
⌧
x)(t) := x(t
⌧ ) 8t 2 R.
• A signal x : R ! R time-dilated by factor a > 0 (equivalently, time-contracted by factor
1/a) is the signal
(Sa x)(t) := x(t/a), t 2 R.
• In the example for x(t/a), the rectangular pulse portion ends at t/a = 2, i.e., at t =
2a = 4 for a = 2, and t = 2a = 1 for a = 1/2.
• Also, note how the y-intercept does not change since, obviously, 0 = 0/a for all a > 0.
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Time inversion
• A signal x : R ! R time-inverted (or time-reflected) is the signal
(S
1
x)(t) := x( t), t 2 R.
• Basically, the time-inverted signal of x is x’s reflection in the y-axis, as the following example
illustrates.
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Combinations of temporal (domain) operations
• Given a signal x : R ! R, how would one plot {x( 2t + 2), t 2 R}?
• Note that
x( 2t + 2) = (
2
x)( 2t) = (S1/2
2
x)( t) = (S
1 S1/2
2
x)(t), t 2 R.
• So, beginning with the signal x (Fig. (a) below), do the following:
(b) time delay by -2 (advance by 2), then
(c) time dilate by 1/2 (contract by 2), then
(d) time reflect
• To verify one point: by Fig. (d) 1 = x( 2t + 2)|t=1 = x(0) = 1 by Fig. (a).
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Combinations of temporal (domain) operations (cont)
• It’s easy to verify that time dilation and reflection (S) operations generally commute; for
example
S
1 S0.5
= S0.5 S
• But time-dilation/reflection S and time-shift
1
=S
0.5 .
operations generally do not commute.
• In the previous example, Fig. (e) depicts
2 S 0.5 x which is obviously di↵erent from
S 0.5
2(t + 2) 6= 2t + 2.
2 x depicted in Fig. (d); this is simply because
• We can also deal with combined temporal operations by first factoring, e.g.,
x( 2t + 2) = x( 2(t
1)) = (S
0.5
x)(t
1) = (
1 S 0.5
x)(t), t 2 R.
• So, we would do the time-scale/reflection operations before the time-shift, but obviously
with di↵erent parameters than used without first factoring as above.
• In summary for this example, we’ve shown
1 S 0.5
= S
0.5
2
6=
2 S 0.5 .
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Spatial (range) operations
• For a signal x : R ! R:
– A spatial increase (upward shift) by constant b 2 R leads to the signal
(x + b)(t) = x(t) + b, t 2 R.
– A spatial amplification (dilation) by factor a 2 R leads to the signal
(ax)(t) = a · x(t), t 2 R.
– A spatial reflection (in the time axis) leads to the signal
x(t), t 2 R.
• As with temporal operations, spatial shift and spatial dilation operations generally do not
commute.
• If combinations of spatial and temporal operations are in play, then any spatial operation
will commute with any temporal operation.
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Spatial (range) operations - example
• For the example pulse x of Fig. (a),
3x(t) + 5 can be plotted by
(b) first spatially reflecting and amplifying x by 3 to get
3x(t),
(c) then shift up by 5.
• Alternatively, since 3x + 5 =
spatially reflect and amplify by 3.
3(x
5/3), one can first shift x down by 5/3, then
• It would be a mistake to first shift by 5 then scale by -3 as this would result in
3x 15 6= 3x + 5.
3(x+5) =
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Even and odd signals
• With the notion of time-inversion, we can define other signal attributes.
• A signal x : R ! R is even if x(t) = x( t) 8t 2 R.
• A signal x is odd if x(t) =
x( t) 8t 2 R.
• Clearly, {cos(t), t 2 R} is an even signal, {sin(t), t 2 R} is an odd signal,
and {3e 5t, t 2 R} is neither.
• Any signal can be written as a sum of even and odd signals through the identity
x(t) + x( t)
x(t) x( t)
x+S 1x
x
+
8t 2 R, i.e., x =
+
2
2
2
where xe(t) := 12 (x(t) + x( t)) is even and xo(t) := 12 (x(t) x( t)) is odd.
x(t) =
• For example, we can write
+ e5t
e 5t e5t
+3
= 3cosh(5t) + ( 3sinh(5t)) 8t 2 R,
2
2
where sinh (hyperbolic sine) is odd () -3sinh is odd) and cosh is even.
3e
5t
= 3
e
5t
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S
2
1x
,
Even and odd signal decomposition - graphical example
• For the sub-domain 1  t  1:
x(t) = 1 + t, x( t) = 1
t,
xe (t) = (1 + t + 1 t)/2 = 1, and
xo (t) = (1 + t (1 t))/2 = t.
• For the sub-domain 1  t  2: x(t) =
2, x( t) = 0, xe (t) = (2 + 0)/2 =
1, and xo (t) = (2 0)/2 = 1.
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Even and odd signal - properties
• It’s easy to show that:
– If x is odd (respectively, even) then cx is odd (respectively, even) for all scalars c 2 R.
– The sum of odd functions is odd.
– The sum of even functions is even.
– The product of even functions is even.
– The product of odd functions is even.
– The product of an even and an odd function is odd.
• For example, if f and g are arbitrary odd functions, then 8t 2 R,
(f g)( t) = f ( t)g( t) = ( f (t))( g(t)) = f (t)g(t) = (f g)(t),
i.e., f g is even.
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Some important signals - polynomials
• This course will focus on certain families of signals and combinations made from them.
• x is a polynomial of degree or order n 2 {0, 1, 2, ...} if 8t 2 R,
x(t) = an tn + an
n
X
=
ak t k ,
1t
n 1
+ ... + a1 t + ao
k=0
where we will take the scalar coefficients ak 2 R 8k 2 {0, 1, 2, ..., n}.
• By the fundamental theorem of algebra, x has n roots in C, i.e., there are n solutions tk
to x(t) = 0 (allowing for multiple, identical roots) so that
x(t) = an (t t1 )(t t2 )...(t
n
Y
= an
(t tk ).
tn )
k=1
• When all coefficients are real, all roots of x are either real or come in complex-conjugate
pairs.
• Obviously, the constant signal x = a0 is a polynomial of order zero.
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Exponential signals
• Real-valued exponential signals have the form x(t) = Aebt , t 2 R, where A, b 2 R.
• Periodic, complex-valued
exponential signals have the form x(t) = Aej(wt+ ), t 2 R,
p
where j :=
1 and A > 0, w, 2 R and the period is 2⇡/w.
• In electrical engineering, we don’t use i for
p
1 as i typically symbolizes electrical current.
• Since the signal {ejv , v 2 R} has period 2⇡ , we can take the phase
2 [ ⇡, ⇡).
• By the Euler-De Moivre identity,
x(t) = Aej(wt+
)
= A cos(wt + ) + jA sin(wt + ), t 2 R.
• Here, the constant (or “direct current” (DC)) signal obviously corresponds to w = 0
(again, infinite period).
• Also, here w is the real-valued frequency – we will consider complex-valued frequencies
(s 2 C) when we study transient response by unilateral Laplace transform.
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Exponential signals (cont)
• For the frequency s = +jw 2 C, consider the complex exponential function x : R ! C:
x(t) = Aest+j
= Ae t ej(wt+ ) , t 2 R.
• Note the real part of this signal is a sinusoid with exponential envelope:
Re{x}(t) = Ae t cos(wt + ), t 2 R.
• In the figure, e 2t cos(10⇡t) is plotted with its envelope ±e
the period T = 2⇡/(10⇡) = 0.2, and = 2 < 0.
2t ,
i.e., w = 10⇡ so that
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The unit-step function, u
• The unit-step will be denoted by u:
u(t) =
⇢
1 if t 0
0 else (t < 0)
• Note that u(0) = 1 is indicated by the solid period in the figure; and just before time 0,
u(0 ) = 0.
• The unit step obviously has a single jump discontinuity of unit height at the time-origin.
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The unit impulse (Dirac delta function),
• Suppose we wish to di↵erentiate the unit-step function, u, i.e., compute
u̇ = du/dt = D u, where we have defined the time-derivative operator
D :=
d
dt
• Clearly, (D u)(t) = 0 for all t 6= 0, but what happens at the jump discontinuity, t = 0?
• Consider approximating the derivative (D u)(0) by a di↵erence quotient
" (t)
:=
u(t)
u(t
"
")
"
=
u
"
u
"
:
, t 2 R,
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The unit impulse (Dirac
• As " ! 0,
"
function),
(cont)
is taller and thinner but with constant area = 1, i.e.,
Z 1
8" > 0,
" (t) = 1.
1
• That is, as " ! 0, the support of
at the origin diverges " 1.
• So as " ! 0,
"
"
disappears, but its area remains constant as its value
converges to the unit impulse , graphically depicted as
"
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The unit impulse,
(cont)
• Since the unit step is constant except at the origin,
Du =
,
i.e., the unit impulse is the derivative at a unit jump discontinuity.
• We will see later how this peculiar function is crucial to our understanding of linear systems.
• For now, we will examine some of its important properties.
• We’ve already ascertained that
8t 2 R,
where the second fact follows from
containing zero in their interior,
8t 6= 0,
Z t
(t) = 0
(⌧ )d⌧
= u(t)
1
= D u and also implies that, for all intervals I ⇢ R
Z
(⌧ )d⌧
= 1.
I
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Unit impulse,
- ideal sampling property
• Since 8t 6= 0, (t) = 0, if the signal x is continuous at the origin, then
(x )(t) = x(t) (t) = x(0) (t)
where x(0) (t) is an impulse but a unit impulse only if x(0) = 1.
• By definition, the impulse “↵ ” for scalar ↵ 2 R satisfies
Z 1
↵ (t)dt = ↵.
1
• The ideal sampling property of the unit impulse is: For all signals x continuous at the origin,
Z 1
x(t) (t)dt = x(0).
1
• By a simple change of variables of integration: for all signals x continuous at time T 2 R,
Z 1
x(t) (t T )dt = x(T ).
1
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Unit impulse,
- jump discontinuity example
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Signal construction - example
• For ⌧ > , note the function:
(
u
⌧
u)(t) = u(t
)
u(t
⌧) =
⇢
1 if  t < ⌧
0 else,
• That is,
u
⌧ u is a rectangular pulse of unit height and support [ , ⌧ ); as such,
it is a “window” function on this interval.
• This signal x is linear (x(t) = t + 1) just on the time-interval [ 1, 1], and a constant
(x(t) = 2) just on [1, 2].
• So, x(t) = (t + 1)(u(t + 1)
u(t
1)) + 2(u(t
1)
u(t
2)), t 2 R.
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Dirichlet signals
• In this course, we will focus on signals that are Dirichlet so as to study signals and linear
systems in a frequency domain.
• A signal x : R ! R:
– Satisfies the weak Dirichlet condition if it is absolutely integrable, i.e., if
Z 1
|x(t)|dt < 1,
1
or just absolutely integrable over any period T if x is periodic, i.e.,
Z
|x(t)|dt < 1.
T
– Satisfies the strong Dirichlet conditions if over any finite interval of time: (i) it has a
finite number of jump discontinuities, and (ii) it has a finite number of minima and
maxima.
• Note how the strong Dirichlet condition (ii) is not possessed by the bounded signal x(t) =
cos(1/t), t 2 R as it has infinitely many extrema in any interval containing the origin
(t = 0).
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Systems - single input, single output (SISO)
• In the figure, f is an input signal that is being transformed into an output signal, y , by the
depicted system (box).
• To emphasize this functional transformation, and clarify system properties, we will write
the output signal (i.e., system “response” to the input f ) as
y
= Sf,
where, again, we are making a statement about functional equivalence:
8t 2 R, y(t) = (Sf )(t).
• Again, Sf is not S “multiplied by” f , rather a functional transformation of f .
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SISO systems (cont)
• The n signals xk , k 2 {1, 2, ..., n}, are the internal states of the system.
• We will see that states can be taken as “outputs of integrators,” e.g., the states of a linear
circuit can be taken as the currents through inductors and voltages across capacitors.
• In discussing the following properties of systems, we will generally consider Dirichlet, continuoustime, complex-valued input signals, f : R ! C.
• But if there is a finite origin of time (e.g., f : [0, 1) ! C) the following properties pertain
to systems with state variables that are initially zero (8k, xk (0) = 0).
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Linear systems
• A linear system S has the following two properties:
– (Homogeneity/Scaling) For all input signals f and scalars a,
S(af ) = aSf,
where we remind that, by definition, 8t 2 R, (af )(t) = a · f (t) and “·” is just scalar
multiplication.
– (Superposition) For all signals f1 and f2,
S(f1 + f2 ) = Sf1 + Sf2 ,
where we remind that, by definition, 8t 2 R, (f1 + f2)(t) = f1(t) + f2(t) and
“+” on the right-hand-side is just scalar summation.
• Implicitly, when defining such system properties, we are restricting our attention to signals
f for which Sf : R ! C is a well-defined (output) signal, restricting our attention to
Dirichlet signals in particular.
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Linear systems - examples
• Suppose a system is defined by the input-output relationship,
Z t
8t 2 R,
y(t) =
f (⌧ )d⌧ = (Sf )(t).
1
• This system is linear owing to the linearity property of the integration operator (and commutativity of scalar multiplication and addition).
• Scaling: For all input signals f , scalars a, and time t 2 R,
Z t
Z t
(Saf )(t) = 3
af (⌧ )d⌧ = a3
f (⌧ )d⌧ = a(Sf )(t),
i.e., S(af ) = aSf .
1
1
• Superposition: For all input signals f1 , f2 , and time t 2 R,
Z t
Z
(S(f1 + f2 ))(t) = 3
(f1 (⌧ ) + f2 (⌧ ))d⌧ = 3
1
= (Sf1 + Sf2 )(t),
t
f1 (⌧ )d⌧ + 3
1
Z
t
f2 (⌧ )d⌧
1
i.e., S(f1 + f2) = Sf1 + Sf2 - by the absolute integrability property of Dirichlet signals,
the integral of the sum (of integrands) is the sum of the integrals (of each integrand).
43
Linear systems - examples (cont)
• Suppose a system is defined by the input-output relationship,
8t 2 R,
y(t) = 3f (t) + 7;
• or just y = 3f + 7 = Sf where “7” here is understood to be the constant signal = 7
for all time.
• Clearly, in general,
S(f1 + f2 ) = (f1 + f2 ) + 7 6= f1 + 7 + f2 + 7 = Sf1 + Sf2 .
• So, this is not a linear system because it does not have the superposition property.
• One can easily check that the scaling property also fails, i.e., that:
for all signals f 6= 0 and scalars a such that
S(af ) 6= aSf.
• This system is said to be affine.
• For a signal, we will use the adjectives affine and linear interchangeably.
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Disproving system linearity - discussion
• The logical negation of a universal statement is an existential one (i.e., there is a counterexample).
• Also recall De Morgan’s rule: for all statements p and q ,
⇠ (p and q) = (⇠ p) or (⇠ q),
where ⇠ here represents logical negation.
• So, to disprove linearity, we either need only find a scalar and input-signal for which the
scaling property does not hold or a pair of input signals for which the superposition property
does not hold.
• In the previous example of an affine system, both properties more generally do not hold,
i.e., not just for a single counter-example.
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Linear systems - examples (cont)
• Consider the system whose (zero initial-state) output y to input f are related by
3Dy + y
= f2
• To check whether this system is linear, consider arbitrary input signals f1 and f2 and scalars
↵1 and ↵2 .
• For k 2 {1, 2}, let yk be the system output to fk , i.e., 3 D yk + yk = fk2 .
• Thus, by linearity of the derivative D operator,
3 D(↵1 y1 + ↵2 y2 ) + (↵1 y1 + ↵2 y2 ) = ↵1 (3 D y1 + y1 ) + ↵2 (3 D y2 + y2 )
= ↵1 f12 + ↵2 f22
6= (↵1 f1 + ↵2 f2 )2 ,
i.e., ↵1f12 + ↵2f22 6= (↵1f1 + ↵2f2)2 for arbitrary signals fk and scalars ↵k .
• For example, if f1 ⌘ 1 ⌘ f2 (the constant signal 1) and ↵1 = 1 = ↵2 , then
↵1 f12 + ↵2 f22 = 2 6= 4 = (↵1 f1 + ↵2 f2 )2 .
• So, this system is not linear.
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Time-invariant systems
• Recall the time-shift (delay) operator
8t 2 R,
⌧
(
f,
⌧
f )(t) = f (t
⌧ ).
• A system S is time-invariant if for all ⌧ 2 R and Dirichlet f : R ! C,
i.e., if S and
⌧
S
commute.
⌧
f
=
⌧ Sf,
• That is, if y = Sf , then the system S is time-invariant if for all signals f and delays ⌧ ,
the response to ⌧ f (i.e., “f (t ⌧ )”) is ⌧ y (i.e., “y(t ⌧ )”).
• Or even more simply, a delay in the input results in the same delay in the output.
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Time-invariant systems - examples
• For example, consider again the integrator system
Z
8t 2 R, (Sf )(t) =
• For all Dirichlet f and ⌧ 2 R:
8t 2 R, (S
⌧
f )(t) =
=
=
Z
Z
Z
t
f (r)dr.
1
t
(
⌧
f )(r)dr
1
t
f (r
1
t ⌧
⌧ )d r
f (r0 )dr0 ,
r0 := r
⌧
1
= (Sf )(t ⌧ )
= ( ⌧ Sf )(t).
• Thus, this system is time invariant.
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Time-invariant systems - examples (cont)
• Consider again the affine system defined by y = Sf = 3f + 7.
• Clearly, for all signals f : R ! C and delays ⌧ 2 R: 8t 2 R,
S(
⌧
f )(t) = 3(
⌧
f )(t) + 7 = 3f (t
⌧) + 7 =
⌧ (Sf )(t)
= y(t
⌧ ),
where y := Sf .
• So, this system is time-invariant.
• For the system 8t 2 R, (Sf )(t) = 3f (t) + 7t: if ⌧ 6= 0 then
S(
⌧
f )(t) = 3(
⌧
f )(t) + 7t = 3f (t
⌧ ) + 7t 6= 3f (t
⌧ ) + 7(t
⌧) = (
• That is, there are cases (in fact, all but ⌧ 6= 0) where time-shift in the input does not lead
to the same-time-shift in the output.
• So, this system is time varying, i.e., not time invariant.
• One can similarly show that the system 8t 2 R, (Sf )(t) = 3tf (t) + 7 is also timevarying.
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Time-invariant systems - examples (cont)
• Consider the system whose output to input f is y(t) = (Sf )(t) = f (t2 ), 8t 2 R.
• To check for time-invariance, take an arbitrary input f and scalar T and let g =
i.e., g(t) = f (t T ), 8t 2 R.
• The output to g is (Sg)(t) = g(t2 ) = f (t2
• Also, (
T y)(t)
= f ((t
T ), 8t 2 R.
T )2 ), 8t 2 R.
• So, this system is not time-invariant because for arbitrary f, T ,
(S
T f )(t)
Tf,
= f (t2
T ) 6= f ((t
T )2 ) = (
T Sf )(t),
8t 2 R.
• For example, if f (t) = t, 8t 2 R, and T = 1, then
(S T f )(t) = t2 1 6= (t 1)2 = ( T Sf )(t), 8t 6= 1.
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⌧ Sf )(t).
Causal signals and systems
• A signal f is causal if f (t) = 0 for all t < 0.
• Note that for any signal x : R ! R, f = xu is a causal signal.
• A system S is said to be causal if: for all causal input signals f , the output signal (response)
Sf = y is also causal.
• As we will be focusing on time-invariant (and linear) systems, the choice of time origin will
be arbitrary, and we will typically choose 0.
• If the system has non-zero response at negative time, it is said to be anticipating the causal
signal which does not “arrive” till time 0.
• So, non-causal systems are said to be anticipative.
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Causal systems - examples
• Consider again, the integrator system 8t 2 R, (Sf )(t) =
Rt
1
f (r)dr.
• Take arbitrary causal f , f (t) = 0 for all t < 0.
Rt
Rt
• Thus, for all t < 0, (Sf )(t) = 1 f (r)dr = 1 0dr = 0.
• So, this system is causal.
• Now consider the example 8t 2 R, (Sf )(t) = 3f 2 (t + 2).
• If f is causal, then for t =
1 note that
(Sf )( 1) = 3f 2 ( 1 + 2) = 3f 2 (1)
• So, (Sf )( 1) = 0 only if f (1) = 0, which may not be the case for an arbitrary causal
signal.
• So, for arbitrary causal f and t < 0, (Sf )(t) 6= 0 (e.g., when f (1) 6= 0).
• Hence this system is not causal.
• One can easily show that this system is also not linear, but it is time invariant.
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Causal systems - examples (cont)
• Consider the system 8t 2 R, (Sf )(t) = 3tf (t) + 7.
• If f is causal, then for all t < 0,
(Sf )(t) = 3tf (t) + 7 = 3t · 0 + 7 = 7 6= 0.
• So this system is also not causal.
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Stateless/memoryless systems
• In a memoryless (or stateless) system, 8t 2 R and input signals f , the output at time t,
y(t), depends on the input f only through f (t).
• For example, the system 8t 2 R, y(t) = (Sf )(t) = 3tf (t) + 7 is obviously a memoryless system.
Rt
• But in the integrator system 8t 2 R, y(t) = (Sf )(t) = 1 f (r)dr, clearly the output
y(t) generally depends on f (r) for r 6= t (here, for all r  t).
• So this integrator system is stateful.
• We will see that stateful systems have associated internal state signals (again, we will denote
them with x).
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Canonical representation of a SISO LTIC system by ODE
• In this course, we focus on linear and time-invariant (LTI) SISO systems S in continuous
time.
• If y = Sf represents the response of the system y to input f , then we will represent the
system with the following ordinary di↵erential equation (ODE) relating y and f :
dn y
dn 1 y
dy
dm f
dm 1 f
df
+
a
...
+
a
+
a
y
=
b
+
b
... + b1
+ b0 f,
m
n
1
1
0
m
1
d tn
d tn 1
dt
dtm
dtm 1
dt
with integers n, m 0, and the scalar coefficients ak , bk 2 R, 8k.
• Note that we can take the coefficient an = 1 without loss of generality, i.e., one can divide
a system equation by the coefficient of dny/dtn if it is not 1 to get a system equation in
the above form.
• Also note that this system equation is a functional equivalence in that it holds for all time,
i.e., one can stipulate “(t)” for all participating signals giving an equation that would hold
for all t 2 R.
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Compact ODE forms of a SISO system equation
• We will write such system equations more compactly in two ways.
• First, using the time-derivative operator
D :=
d
,
dt
we can write the system equation as
D n y + an
1D
n 1
y... + a1 D y + a0 y
= bm D m f + bm
1D
m 1
f... + b1 D f + b0 f.
• Note that D 0 is just the identity map: 8f, D0 f = f .
• Furthermore, we can define the system polynomials (with real coefficients, an = 1),
Q(s) :=
n
X
a k sk
and P (s) :=
k=0
m
X
k=0
bk sk , 8s 2 C.
• So, the system equation can be written simply as
Q(D)y = P (D)f
where P is a polynomial of degree m and Q, the characteristic polynomial of the system,
has degree n.
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This SISO system is LTIC
• The system with input f and output y (the response to f ) satisfying
D n y + an
1D
n 1
y... + a1 D y + a0 y
= bm D m f + bm
1D
m 1
f... + b1 D f + b0 f
is Linear, Time-Invariant and Causal (LTIC) because:
– the time-derivative operators Dk are LTIC for all integers k
0, and
– the coefficients bk , ak of the polynomials P and Q are all assumed to be constant,
i.e., no time-varying coefficients.
• In the following, we will explicitly “solve” this system (for y in terms of f ) and more clearly
demonstrate that it is LTIC.
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SISO Canonical ODE - resistive circuit example
• Consider this simple, ideal resistive circuit.
• By Kirchho↵’s current law (KCL) at node y (i.e., zero is the sum of the currents referenced
into the node between the resistors) and Ohm’s law (current in terms of node voltages),
0 y
= 0,
R1
R2
where “0” on the left-hand-side is the ground voltage.
f
y
+
• In canonical form, the input-output relationship is the voltage divider:
y
=
R2
f.
R1 + R2
• Obviously, this system is memoryless/stateless.
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SISO Canonical ODE - passive RC circuit example
• In this passive RC circuit, the KCL nodal equation at y is
f
y
R
+ C D(0
y) +
• In canonical form,
Dy +
0
y
R
1
y
RC
+ C D(0
y) = 0.
1
f.
2RC
=
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SISO Canonical ODE - passive RC circuit example (cont)
• We now solve the system D y + (RC) 1 y = (2RC)
explicitly in terms of the input f and initial conditions.
1f ,
i.e., we find the output y
• Using the co-factor method, assume the output is of the form
y(t) = A(t)e
and substitute it into the system equation: 8t
t/(RC)
0,
1
1
1
y+
y =
f (t)
RC
RC
2RC
1 t/(RC)
) (D A)(t) = f (t)
e
2RC Z
t
1 r/(RC)
) A(t) = A(0 ) +
f (r)
e
dr
2RC
0
Z t
1
) y(t) = A(0 )e t/(RC) +
f (r)
e
2RC
0
(D A)(t)e
t/(RC)
(t r)/(RC)
dr
• where 0 is the origin of time, and clearly this system is stateful.
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SISO Canonical ODE - passive RC circuit example (cont)
• If the initial condition y(0 ) is given just before the origin of time, then clearly A(0 ) =
y(0 ), i.e.,
Z t
1
8t 0 , y(t) = y(0 )e t/(RC) +
f (r)
e (t r)/(RC)dr.
2RC
0
• We will see how this form of the solution is important for transient analysis as it separates
out
– the portion of the output {y(t) | t 0} due only to the initial condition, here simply
the constant signal y(0 ), is called the Zero-Input Response (ZIR), and
Rt
1
– the portion due only to the input, here 0 f (r) 2RC
e (t r)/(RC)dr, t 0 - this is
called the Zero-State Response (ZSR) or zero initial-state response.
• Note for the above example, if input f ⌘ 0, then the voltage source is a short circuit and
the energy stored in the capacitor will eventually dissipate, i.e., ZIR! 0 over time.
• For the example of Fig. 1.26 of Lathi p. 78:
check that the system equation is D y = R D f + f /C so that 0 is the characteristic
value - see equ. (1.36a). Here if f ⌘ 0, then the current source is an open circuit so the
capacitor’s energy does not dissipate, i.e., ZIR is constant - see (1.42).
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SISO Canonical ODE - active RC circuit example
• Recall that with an op-amp (having a
power supply not shown), a relatively lowpowered input signal can control a separately high-powered output signal.
• Recall that an ideal op-amp
– draws zero current from its two (±)
input leads, and
– has zero di↵erential input voltage,
– i.e., has infinite input impedance.
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SISO Canonical ODE - active RC circuit example (cont)
• So, for this circuit, the voltage at the negative input terminal of the op-amp is also f .
• Thus, KCL nodal equation at the “ ” input terminal is
) vA
vA f
+ C1 D(y f ) = 0
R1
= f + R1 C1 D f R1 C1 D y.
• Also, KCL at node A is
f
vA
y vA
+
+ C2 D(0
R1
R2
vA ) = 0.
• Substituting vA into KCL at A and then rewriting in canonical form gives:
✓
◆
1
1
1
1
D2 y +
+
Dy +
y
C2 R1
R2
R C R C1
✓
✓ 2 2 1◆◆
1
1
1
1
1
= D2f +
+
+
Df +
f.
R1 C1
C2 R1
R2
R2 C2 R1 C1
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Time-Domain Analysis of LTIC SISO Systems - Introduction
• Consider a LTIC SISO system with input f : [0, 1) ! R and output y : [0, 1) ! R,
i.e., with time origin 0.
• We will now study solutions of general LTIC SISO systems of the form
with initial conditions (D
n 1
Q(D)y = P (D)f,
y)(0 ), (Dn 2 y)(0 ), ..., (D y)(0 ), y(0 ), where
– D is the time derivative operator;
– Q is the system’s
characteristic polynomial of degree n
P
Q(D) = nk=0 ak D k , an = 1;
1 with real coefficients:
– P is a polynomial of degree m with real coefficients: P (D) =
Pm
k=0 bk D
k;
• By “solution” we mean to find the entire signal y : [0, 1) ! R given:
– a causal input signal f with zero initial conditions, (Dk f )(0 ) = 0 for all integers
k 0 (so f typically specified with a step-function u factor);
– the n initial conditions; and
– the model of the system itself, P, Q.
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Approach to solution: ZIR and ZSR
• The total solution y to P (D)f = Q(D)y and the given initial conditions is a sum of two
parts:
– the ZSR, yZS, which solves
P (D)f = Q(D)yZS
with zero initial conditions, i.e., (Dk y)(0 ) = 0 8 integers k
0; and
– the ZIR, yZI, which solves
0 = Q(D)yZI
with the given initial conditions.
• The total response y of the system to f and the given initial conditions is, by linearity,
y
= yZI + yZS .
• We will determine the ZIR by finding the characteristic modes of the system.
• We will determine the ZSR by convolution of the input with the impulse response.
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Total response - example
• Recall that this RC circuit with input f and output y is described by
1
1
y =
f
RC
2RC
i.e., Q(D) = D +1/(RC) with degree n = 1, and P (D) = 1/(2RC) with degree
m = 0.
Dy +
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Total response - example
• Also recall that the total-response is
y(t) = y(0 )e
i.e., 8t
0,
t/(RC)
+
Z
t
f (r)
0
1
e
2RC
yZI (t) = y(0 )e t/(RC) ,
Z t
1
yZS (t) =
f (r)
e
2RC
0
(t r)/(RC)
(t r)/(RC)
dr 8t
0,
dr.
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Comment: Solution by homogeneous and particular components
• You may have learned how to find the total response by summing
– any “particular” solution yp satisfying P (D)f = Q(D)yp irrespective of initial conditions and possibly including characteristic modes, and
– a generic “homogeneous” solution yh satisfying 0 = Q(D)yh consisting only of n
characteristic modes, with n free parameters/coefficients.
• The n free parameters of the homogeneous solution are determined by applying the n initial
conditions to the total response, y = yp + yh.
• The di↵erence of this approach is simply that
– yp may be the ZSR together with a portion of the ZIR, and
– yh is the remaining portion of the ZIR.
• That is, the ZSR is the special “particular” solution with zero initial conditions, and the
ZIR is the special “homogeneous” solution that accounts for the initial conditions by itself.
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ZIR - the characteristic values
• To solve for the ZIR, i.e., solve
Q(D)y = 0 given (Dn 1 y)(0), (Dn 2 y)(0), ..., (D y)(0), y(0),
recall that exponential functions, {est | t 2 R} for frequency s 2 C, are eigenfunctions of
di↵erential operators of the form Q(D) for a polynomial Q.
• That is, for scalar s 2 C, if we substitute y = {est | t 2 R}, we get
8t 2 R, (Q(D)y)(t) = Q(D)est = Q(s)est .
• So, to solve Q(s)est = 0 for all time t
period, we require
0, where est is not always zero over this time
Q(s) = 0, the characteristic equation of the system.
• If s is a root of the characteristic polynomial Q of the system,
– s would be a characteristic value of the system, and
– the signal {est | t
8t 0.
0} is a characteristic mode of the system, i.e., Q(D)est = 0
• Since Q has degree n, there are n roots of Q in C, each a system characteristic value.
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ZIR - the characteristic values (cont)
• Though there may be some repeated roots of the characteristic polynomial Q, there will
always be n di↵erent, linearly independent characteristic modes, µk , i.e.,
8t
0,
n
X
k µk (t)
k=1
= 0 , 8k, scalars
• So, by system linearity, we will be able to write
8t
0, yZI (t) =
n
X
k
= 0.
ck µk (t),
k=1
for scalars ck 2 C that are found by considering the given initial conditions
(Dl y)(0 ) = (Dl
n
X
ck µk )(0 ) =
k=1
i.e., n equations in n unknowns (ck ).
n
X
k=1
ck (Dl µk )(0 ), for 0  l  n
1,
• The linear independence of the modes implies linear independence of these n equations in
ck , and so they have a unique solution.
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ZIR - the case of n di↵erent and real characteristic values
• If there are n di↵erent roots of Q in R, s1 , s2 , ..., sn , then there are n characteristic modes:
for k 2 {1, 2, ...n},
µk (t) = esk t , t
• Therefore,
0, yZI (t) =
8t
• Since 8k 2 {1, 2, ..., n}, l
n
X
0.
ck e sk t .
k=1
0,
D l µk = slk µk
and µk (0) = 1,
the n unknown scalars ck 2 R can be solved using the n equations:
(Dl y)(0 ) =
n
X
k=1
ck slk , for 0  l  n
1.
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ZIR - the case of n di↵erent and real characteristic values - example
• For this series RLC circuit with input f and output y , KVL gives
Z
1 t
8t 0, Ri(t) + L(D i)(t) + vc (0) +
i(⌧ )d⌧ = f (t).
C 0
• We can di↵erentiate this equation and substitute i = y/R to get
D y + LR 1 D2 y + (RC) 1 y = D f
) D2 y + RL 1 D y + (LC) 1 y = RL 1 D f,
i.e., P (s) = RL 1s and the characteristic polynomial of this second-order (n = 2)
circuit is
Q(s) = s2 + RL
1
s + (LC)
1
.
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ZIR - the case of n di↵erent and real char. values - example (cont)
• For example, if R = 700⌦, L = 0.1H, and C = 1µF, then
Q(s) = s2 + 7000s + 107 = (s + 5000)(s + 2000).
• So, the characteristic values are -5000, -2000.
• The characteristic modes are exp( 5000t), exp( 2000t) 8t
• The ZIR is of the form
8t
5000t
0 , yZI (t) = c1 e
• If the given initial conditions are
(D y)(0 ) =
+ c2 e
2000t
0.
.
2000V/s, y(0 ) = 10V,
then
2000 =
5000c1 + 2000c2 = (D yZI )(0 )
10 = c1 + c2 = yZI (0 )
• Thus, c1 =
6 and c2 = 16, i.e.,
8t
0 , yZI (t) =
6e
5000t
+ 16e
2000t
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ZIR - the case of not-real characteristic values
• The characteristic polynomial Q may have non-real roots, but such roots come in complexconjugate pairs because Q’s coefficients ak are all real.
• For example, if the characteristic polynomial is
Q(s) = s3 + 5s2 + 11s + 15 = (s + 3)(s2 + 2s + 5)
then the characteristic values (Q’s roots) are
3,
1 ± 2j
recalling j =
p
1.
• Because we have three di↵erent characteristic values 2 C, we can specify three corresponding characteristic modes,
e
3t
yZI (t) = c1 e
3t
, e(
1+2j)t
, e(
1 2j)t
1+2j)t
+ c3 e (
and construct the ZIR as
+ c2 e (
1 2j)t
8t
0
.
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ZIR - not-real characteristic values with real characteristic modes
• By the Euler-De Moivre identity,
0, yZI (t) = c1 e
8t
3t
+ 2(c2 + c3 )e
t
cos(2t) + 2j(c2
c 3 )e
t
sin(2t)
• Because all initial conditions are real and yZI is real valued, c2 and c3 will be complex
conjugates, i.e.,
c2 = c3 2 C.
• So, we can write
8t
0 , yZI (t) = c1 e 3t + c̃2 e t cos(2t) + c̃3 e t sin(2t), where
c̃2 = (c2 + c3 ) = 2Re{c2 } 2 R
c̃3 = 2j(c2 c3 ) = 2Im{c2 } 2 R
i.e., just two “degrees of freedom” with these two coefficients to accommodate the initial
conditions, so not an “underspecified” scenario.
• Thus, instead of e(
1±2j)t ,
the following real-valued characteristic modes can be used
e
t
t
cos(2t), e
sin(2t)
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ZIR - not-real char. values with real char. modes (cont)
• So for t
0 , if the ZIR is of the form
yZI (t) = c1 e
) (D yZI )(t)
) (D2 yZI )(t)
=
=
3t
+ c̃2 e
t
cos(2t) + c̃3 e
3c1 e 3t + c̃2 [ e t cos(2t) 2e
9c1 e 3t + c̃2 [ 3e t cos(2t) + 4e
t
t
t
sin(2t),
sin(2t)] + c̃3 [ e t sin(2t) + 2e t cos(2t)],
sin(2t)] + c̃3 [ 3e t sin(2t) 4e t cos(2t)],
and the initial conditions are
(D2 y)(0 ) = 3, (D y)(0 ) = 10, y(0 ) = 5,
then the three equations to solve for the three unknown constants c1, c̃2, c̃3 are
5 = c1 + c̃2 = yZI (0 )
10 =
3c1 c̃2 + 2c̃3 = (D yZI )(0 )
3 = 9c1 3c̃2 4c̃3 = (D2 yZI )(0 )
whose solution is
c1 = 6, c̃2 =
1, c̃3 = 13.5.
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ZIR - not-real char. values with real char. modes (cont)
• In summary, for the case of a complex-conjugate pair of characteristic values ↵ ± j , with
↵, 2 R, we can use the real-valued modes (with two di↵erent associated real coefficients),
e↵t cos( t), e↵t sin( t)
8t
0,
instead of complex-valued modes (with associated complex-conjugate coefficients)
e(↵±j
)t
0,
8t
in either case, the pair of coefficients give two degrees of freedom to meet the given initial
conditions.
• Restricting the span to real-valued signals, we can use these two real-valued modes because
span{e(↵+j
)t
, e(↵
j )t
} = span{e↵t cos( t), e↵t sin( t)}, t 2 [0, 1).
• Generally, linear combinations of characteristic modes span all possible real-valued ZIR
signals.
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ZIR - the case of repeated characteristic values
• Consider the case where at least one characteristic value is of order > 1, i.e., repeated
roots of the characteristic polynomial, Q.
• For example, Q(D) = (D +7)3 (D
single root at 2.
• Again, {e
7t , t
• Also, {te
7t , t
0} is a characteristic mode because Q(D)e
(D +7)e
2
(D +7) te
2) has a triple (twice repeated) root at
7t
7t
=
=
=
=
=
2
7t
⌘ 0 follows from
7t
⌘ 0 follows from
= D e 7t + 7e 7t
=
7e 7t + 7e 7t
= 0 8t 0.
0} is a characteristic mode because Q(D)te
7t
7 and a
(D +14 D +49)te
D2 te 7t + 14 D te 7t + 49te 7t
( 14e 7t + 49te 7t ) + 14(e 7t 7te
( 14 + 14)e 7t + (49 98 + 49)te
0 8t 0.
7t
) + 49te
7t
7t
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ZIR - the case of repeated characteristic values (cont)
• Similarly, t2 e
7t
is also a characteristic mode because (D +7)3t2e
7t
⌘ 0.
• Note that without three such linearly independent characteristic modes {e 7t , te 7t , t2 e 7t ;
t 0} for the twice-repeated (triple) characteristic value -7, the initial conditions will create
an “overspecified” set of n equations involving fewer than n “unknown” coefficients (ck )
of the linear combination of modes forming the ZIR.
• For this example,
yZI (t) = c0 e
7t
+ c1 t e
• If the given initial conditions are, say,
(D3 y)(0 ) = 10, (D2 y)(0 ) =
7t
+ c2 t 2 e
7t
+ c3 e2t , t
0
.
5, (D y)(0 ) = 6, y(0 ) = 12,
the four equations to solve for the four unknown coefficients ck are:
yZI (0 ) = c0 + c3
(D yZI )(0 ) =
7c0 + c1 + 2c3
(D2 yZI )(0 ) = 49c0 14c1 + 2c2 + 4c3
(D3 yZI )(0 ) =
343c0 + 147c1 42c2 + 8c3
= 12
= 6
=
5
= 10
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ZIR - general case of repeated characteristic values
• In general, a set of k linearly independent modes corresponding to a characteristic value
s 2 C repeated k 1 times (i.e., of multiplicity k) are,
tk
1 st
e , tk
2 st
e , ..., test, est,
for t
• Also, if s = a ± jw are characteristic values repeated k
w 6= 0, we can use the 2k real-valued modes
0.
1 times, with a, w 2 R and
tr eat cos(wt), tr eat sin(wt), for r 2 {0, 1, 2, ..., k
1}.
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Zero State Response - the impulse response
• Let h be a LTI system’s ZSR to the unit impulse, .
• At this point, it’s difficult to see why h should be a well-defined signal, let alone a crucial
one to the study of LTI systems.
• In the following, we will derive h by inverse Fourier or inverse Laplace transform of the
system transfer function H = P/Q.
• For a procedure to compute h directly from P (D) = Q(D)h see, for example, Sec. 2.3
of Lathi.
• For now, we will assume that h is known for a given LTI system and demonstrate how it
can be used to derive the ZSR to an arbitrary (Dirichlet) input signal f by the process of
convolution.
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ZSR - LTI property and convolution with impulse response
• Consider a LTI system with impulse response h and recall the approximate impulse function,
"
:= "
1
(u
" u),
where 0 < " ⌧ 1.
• For arbitrary integer k, consider a short interval of time [k", (k + 1)") and suppose
over this interval the input signal f , i.e., {f (t), k"  t < (k + 1)"}, is smooth enough
to be well approximated by a constant f (k"), i.e., for k"  t < (k + 1)",
f (t) ⇡ f (k")(u(t
where "f (k") "(t
k")
u(t
(k + 1)")) = f (k") " (t
k") ",
k") is zero outside of this short interval of time.
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ZSR - LTI property and convolution with impulse response (cont)
• Piecing together these non-overlapping approximations leads to a Riemann approximation
of f over R:
f (t) ⇡
1
X
k= 1
f (k") " (t
k")", 8t 2 R
• Note that as " # 0 (i.e., " # d⌧ , a di↵erential), this approximation becomes more precise
tending to a Riemann integral that is the sampling property of the unit impulse:
Z 1
f (t) =
f (⌧ ) (t ⌧ )d⌧, 8t 2 R.
1
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ZSR - LTI property and convolution with impulse response (cont)
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ZSR - LTI property and convolution with impulse response (cont)
• Again, we have constructed a Riemann approximation of the input signal as a linear combination of delayed approximations of the impulse:
f
⇡
1
X
f (k")
"
k" "
k= 1
• For the (zero initial states) LTI system under consideration, let the ZSR of
• So, by time-invariance, the zero-state response to
k" "
is
"
be h".
k" h" .
• Thus, by linearity of the system, the approximate ZSR to f is
yZS ⇡
i.e., 8t, yZS(t) ⇡
1
X
k= 1
1
X
f (k")
f (k") (
k" h"
"
k" h" )(t)
"
k= 1
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ZSR - LTI property and convolution with impulse response (cont)
• As " # 0, this Riemann approximate integral converges to the precise ZSR to f is 8t,
Z 1
yZS (t) =
f (⌧ )( ⌧ h)(t)d⌧
1
Z 1
=
f (⌧ )h(t ⌧ )d⌧
1
=: (f ⇤ h)(t), 8t 2 R,
i.e., the convolution of the input f and (zero state) impulse response
h = lim h" .
"#0
• Note how the convolution is a consequence of the LTI property (of the zero-state system).
• Also note that the convolution operator (⇤) is a bilinear mapping of two signals (here, f
and h) to a third signal (y = f ⇤ h).
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ZSR - impulse response - example
• Recall that the total response y to this (LTIC) circuit with causal input f and initial
condition y(0 ) is 8t 0,
Z t
1
t/(RC)
y(t) = y(0 )e
+
f (r)
e (t r)/(RC)dr
2RC
0
Z 1
1
t/(RC)
= y(0 )e
+
f (r)
e (t r)/(RC)u(t r)dr
2RC
0
• Clearly, the ZSR portion is yZS = f ⇤h with impulse response h(t) = (2RC)
t 2 R, and f (t) = 0 for t < 0 (i.e., f = f u) since f is causal.
1 e t/(RC) u(t),
87
Total response - example - circuit with switches thrown at 0
• The circuit is operational for time t 2 R, i.e., including negative time when the switch on
the right is closed and the one on the left is open.
• At time zero, the two switches are simultaneously thrown.
• Prior to time zero, the circuit was in DC steady state (DC battery of A volts).
• So, y(0 ) = 0 and vc (0 ) = A (capacitor has infinite DC impedance).
• Since vc is a state of the circuit (it’s the output of an integrator (of y )), it’s continuous at
0 so that vc(0 ) = vc(0) = A.
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Total response - example - circuit with switches thrown at 0 (cont)
• But y is not a state variable and at time 0, y(0) = (f (0)
continuous at 0, i.e., y(0 ) 6= y(0), unless f (0) = A.
• By KVL, 8t
• Thus, R
0, f (t) = vc (t) + Ry(t) = vc (0) + C
1Df
= D y + (RC)
1y.
1
vc (0))/R; so y will not be
Rt
0
y(⌧ )dt + Ry(t).
• Exercise: Show by the co-factor method and then IBP that the total response is:
Z t
t/(RC)
8t 0, y(t) = y(0 )e
+
R 1 (D f )(⌧ )e (t ⌧ )/(RC) d⌧
0
✓
◆
Z t
1
1
(t ⌧ )/(RC)
= y(0 )e t/(RC) +
f (⌧ )
(t ⌧ )
e
d⌧
R
R2 C
0
Z 1
= y(0 )e t/(RC) +
f (⌧ )h(t ⌧ )d⌧,
0
where the impulse response h(t) =
1
R
(t)
1
e t/(RC)u(t).
R2 C
• Note: The total response will be discontinuous at time 0 when f (0) 6= A, i.e., y(0) =
y(0 ) + f (0)/R, because h has an impulse component (again, f (0 ) := 0).
89
Convolution of causal signals
• Generally, if
– the system is causal, so h is a causal signal, and
– the origin of time is zero, so that we can also assume the input f (t) = 0 for t < 0
(i.e., f is a causal signal too),
then
(f ⇤ h)(t) =
Z
1
f (⌧ )h(t
1
⌧ )d ⌧ =
Z
t
f (⌧ )h(t
0
⌧ )d⌧, 8t
0.
• Here, since h is causal, the integrand is zero for ⌧ > t.
• And since f is causal, is zero for ⌧ < 0.
• Here, we integrate from 0 so that integrals involving impulses
biguous, i.e., we’re not “splitting hairs”.
at the origin are unam-
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Discussion: the ZSR has zero initial conditions
• We’ve identified the ZSR yZS to a causal input f of a LTIC system with (causal) impulse
response h as the convolution,
Z t
yZS (t) =
f (⌧ )h(t ⌧ )d⌧, t 0.
0
• Here, we assume that the input f has zero “initial conditions,”
(Dk f )(0 ) = 0 for all integers k
0.
• So, clearly, yZS (0 ) = 0 and by di↵erentiating the above display, we get that (Dk yZS )(0 ) =
0 for all integers k > 0 as well.
• But yZS may not be continuous at time 0.
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ZSR and ZIR - op-amp circuit example
• For another example, consider the above circuit input f , output y , and vc (0) = 2V.
• By KCL at “-” and vc , respectively,
y
f
1
• So, vc = 3f
+
vc
f
2
= 0 and
f
vc
2
+ 0.5 D(0
vc ) = 0.
(1)
2y and by substitution,
Q(D)y = D y + y
=
3
D f + f = P (D)f.
2
(2)
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ZSR and ZIR - op-amp circuit example (cont)
• The system characteristic polynomial is Q(D) = D +1, so that the system characteristicvalue (system pole, root of Q) is -1, and characteristic mode is e t.
• Thus, yZI (t) = y(0 )e t , t
0 , where (by KCL at “-” op-amp input at time 0 )
3
1
3
1
f (0 )
vc (0 ) = 0
2=
2
2
2
2
noting vc(0 ) = vc(0) since vc is a continuous state variable.
y(0 ) =
1;
• For the ZSR, if the forced response yF is known, one might be tempted to simply write
yZS (t) = yF (t) + ce t u(t),
and solve for the coefficient c of the characteristic mode by using yZS(0) = 0.
• However, we’ll see later that when the degree of P equals that of Q (as in this example),
and the input f (0) 6= 0, then yZS will be discontinuous at time 0 too.
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Op-amp circuit example: ZSR discontinuous at time 0
• The “transfer function” for this op-amp circuit,
H(s) =
P (s)
1.5s + 1
=
= 1.5
Q(s)
s+1
0.5
,
s+1
is proper but not strictly proper, where b1 = 1.5 is the leading coefficient of P .
• We’ll see later that the impulse response h of this system has an impulse component
(indicating direct coupling between input and output),
h(t) = 1.5 (t)
• Thus, yZS = h ⇤ f = bn f + (h
0.5e t u(t).
bn ) ⇤ f ,
– where h bn corresponds to a “strictly proper” system P̃ /Q = P/Q bn = H bn,
– and so (h
apply,
bn ) ⇤ f is continuous at time 0 and so zero initial (at time 0) conditions
– but yZS is discontinuous at time zero if f (0) 6= 0.
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Op-amp circuit example: ZSR discontinuous at time 0 (cont)
• For example, if the input f (t) = e 2t u(t) then f is discontinuous at time 0 and the
forced response (or “eigenresponse” since f is exponential) of this system is
yF (t) = H( 2)f (t) = 2f (t).
• If we take yZS (t) = 2f (t) + ce t u(t) and find c using yZS (t) = 0, we get c =
• However, c =
2.
0.5 for the true ZSR.
• One can verify this by finding the true ZSR by convolution, yZS = h ⇤ f (or by using the
unilateral Laplace transform which we will later discuss in detail).
0,
• Indeed by linearity, for all t
yZS (t) = (f ⇤ h)(t)
= e 2t u(t) ⇤ (1.5 (t) 0.5e t u(t))
Z t
= 1.5e 2t 0.5
e 2⌧ e (t ⌧ )d⌧ = 1.5e
2t
0.5e t (1
e t)
0
= 2e
2t
0.5e t .
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Graphical convolution - example 1
• We now describe how to compute convolutions with the help of graphical illustrations for
the di↵erent cases of integration.
• For example, let’s compute f ⇤ h where
f (t) = 2e t u(t) and h(t) = u(t)
• Again, (f ⇤ h)(t) :=
R1
1
f (⌧ )h(t
u(t
3), 8t 2 R.
⌧ )d ⌧ .
• For this example, we will see graphically that there are three cases for time t (which is fixed
when integrating over the dummy variable of integration ⌧ ).
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Graphical convolution - example 1 - case 1
• Case 1: When t < 0,
f (⌧ )h(t
⌧ ) = 0, 8⌧ 2 R.
• So, simply because the integrand is constantly zero in this case,
Z 1
(f ⇤ h)(t) =
f (⌧ )h(t ⌧ )d⌧ = 0.
1
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Graphical convolution - example 1 - case 2
• Case 2: When t
0 and t
3 < 0 (i.e., 0  t < 3),
Z 1
(f ⇤ h)(t) =
f (⌧ )h(t ⌧ )d⌧
1
Z t
=
f (⌧ )h(t ⌧ )d⌧
0
Z t
=
2e ⌧ · 1d⌧ = 2
2e t .
0
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Graphical convolution - example 1 - case 3
• Case 3: When t
3
0 (i.e., t
(f ⇤ h)(t) =
=
=
Z
Z
Z
3),
1
f (⌧ )h(t
⌧ )d ⌧
f (⌧ )h(t
⌧ )d ⌧
1
t
t 3
t
2e
t 3
⌧
· 1d⌧ = 2e
(t 3)
2e t .
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Graphical convolution - example 1 - summary
• In summary,
8
< 0
2 2e t
(f ⇤ h)(t) =
: 2e (t 3)
2e
t
t<0
0t<3
3t
• Note that f ⇤ h is continuous even though both f and h have/For jump discontinuities.
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Convolution is commutative
• Convolution is commutative, i.e.,
8f, h, h ⇤ f = f ⇤ h,
• This is proved by simple change of variable of integration (⌧ 0 = t
⌧ ).
• To illustrate this, let’s recompute the previous example instead as
Z 1
h(⌧ )f (t ⌧ )d⌧, 8t 2 R.
(h ⇤ f )(t) =
1
• Again, we have the same three cases in t.
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Graphical convolution - example 1b - case 1
• Case 1: When t < 0,
h(⌧ )f (t
⌧ ) = 0, 8⌧ 2 R.
• Thus, (h ⇤ f )(t) = 0.
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Graphical convolution - example 1b - case 2
• Case 2: When 0  t < 3,
Z
(h ⇤ f )(t) =
Z
=
Z
1
h(⌧ )f (t
⌧ )d ⌧
1
t
h(⌧ )f (t
⌧ )d ⌧
0
t
1 · 2e (t ⌧ ) d⌧
Z t
t
= 2e
e⌧ d⌧ = 2e t(et
=
0
1) = 2
2e t .
0
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Graphical convolution - example 1b - case 3
• Case 3: When t
3,
(h ⇤ f )(t) =
=
=
Z
1
h(⌧ )f (t
⌧ )d ⌧
1
3
Z
h(⌧ )f (t
Z0 3
0
1 · 2e
⌧ )d ⌧
(t ⌧ )
d⌧ = 2e
t
Z
3
e⌧ d⌧ = 2e3
t
2e
t
0
• In summary, we confirm that f ⇤ h = h ⇤ f for this example.
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Graphical convolution - example 2
• Now consider a graphical convolution example where each signal is of finite span.
• Suppose we wish to compute f ⇤ h where
f (t) = (2
t)(u(t)
u(t
2)) and h(t) = u(t + 1)
u(t
2), 8t 2 R.
• Note that h is not causal.
• Graphically, we will see that there are five cases for t.
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Graphical convolution - example 2 - case 1
• Case 1: When t + 1 < 0 (i.e., t <
1), f (⌧ )h(t
• Thus,
(f ⇤ h)(t) =
Z
= 0.
1
⌧ ) = 0, 8⌧ 2 R.
f (⌧ )h(t
⌧ )d ⌧
1
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Graphical convolution - example 2 - case 2
• Case 2: When 0  t + 1 < 2 (i.e., 1  t < 1),
Z 1
(f ⇤ h)(t) =
f (⌧ )h(t ⌧ )d⌧
=
=
1
t+1
Z
f (⌧ )h(t
Z0t+1
0
(2
⌧ )d ⌧
⌧ ) · 1d⌧ = 2
0.5(1
t)2
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Graphical convolution - example 2 - case 3
• Case 3: When t + 1
2 and t
2 < 0 (i.e., 1  t < 2),
Z 2
(f ⇤ h)(t) =
f (⌧ )h(t ⌧ )d⌧
Z0 2
=
(2 ⌧ ) · 1d⌧
0
= 2
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Graphical convolution - example 2 - case 4
• Case 4: When 0  t
2 < 2 (i.e., 2  t < 4),
Z 2
(f ⇤ h)(t) =
f (⌧ )h(t
=
Z
⌧ )d ⌧
t 2
2
(2
t 2
= 0.5(4
⌧ ) · 1d⌧
t)2
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Graphical convolution - example 2 - case 5
• Case 5: When t
2
2 (i.e., t
4), f (⌧ )h(t
⌧ ) = 0 8⌧ 2 R.
• Thus,
(f ⇤ h)(t) = 0.
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Convolution - example 2 - summary
• In summary for this example, we again have a continuous
8
>
> 0
>
< 2 0.5(1 t)2
2
(f ⇤ h)(t) =
>
2
>
>
: 0.5(4 t)
0
f ⇤ h:
t< 1
1t<1
1t<2
2t<4
4t
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Convolution - additive finite support-lengths property
• Also note that in the previous example,
– the support of f ⇤ h (time interval over which f ⇤ h is non-zero) is of length 5,
– while that of f is 2 and that of h is 3.
• Convolution has a general property that, 8f, h, the length of support of f ⇤ h is the sum
of that of f and that of h.
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Convolution - other important properties
• One can directly show that convolution is a bi-linear mapping from pairs of (Dirichlet)
signals (f, h) to signals (yZS), consistent with convolution’s commutative property and the
(zero state) system with impulse response h being LTI;
• that is, 8 signals f, g, h and scalars ↵,
2 C,
(↵f + g) ⇤ h = ↵(f ⇤ h) + (g ⇤ h)
• By changing order of integration (Fubini’s theorem), one can easily show that convolution
is associative, i.e., 8 signals f, g, h,
(f ⇤ g) ⇤ h = f ⇤ (g ⇤ h),
cf. next slide.
• We’ll use these properties when composing more complex systems from simpler ones.
• By the ideal sampling property, the identity signal for convolution is the impulse ,
i.e., 8 signals f ,
f⇤
=
⇤f = f
• By just changing variables of integration, we can show how to exchange time-shift with
convolution, i.e., 8 signals f, h : R ! C and times ⌧ 2 R,
(
⌧f)
⇤h =
⌧ (f
⇤ h) (or just associativity & commutativity with g =
⌧
).
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Convolution - proof of associative property
• 8 signals f, g, h and 8t 2 R,
Z 1
((f ⇤ g) ⇤ h)(t) =
(f ⇤ g)(⌧ )h(t ⌧ )d⌧
1
Z 1
Z 1
=
f ( )g(⌧
)d h(t ⌧ )d⌧
1
1
Z 1
Z 1
=
g(⌧
)h(t ⌧ )d⌧ f ( )d , by Fubini’s theorem
1
1
Z 1
Z 1
=
g(⌧ 0 )h((t
) ⌧ 0 )d ⌧ 0 f ( ) d , ⌧ 0 = ⌧
1
1
Z 1
=
(g ⇤ h)(t
)f ( )d
1
= (f ⇤ (g ⇤ h))(t).
• Q.E.D.
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Total response - second-order, series RLC circuit example
• Recall that for this series RLC circuit with R = 700⌦, L = 0.1H, and C = 1µF:
D2 y + RL 1 D y + (LC) 1 y
(D +2000)(D +5000)y
= RL 1 D f
= 7000 D f
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Total response - second-order, series RLC circuit example
• We’ll show later that the impulse response for this circuit is
✓
◆
14000 2000t
35000 5000t
h(t) =
e
+
e
u(t), t 2 R.
3
3
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Total response - second-order, series RLC circuit example (cont)
• If the input is, say, f (t) = 5e
1000t u(t),
t 2 R, then the ZSR yZS = f ⇤ h is:
8t < 0, yZS (t) = 0
Z t
8t 0, yZS (t) =
h(⌧ )f (t ⌧ )d⌧
0
Z t
14000 2000⌧
35000 5000⌧
=
(
e
+
e
)5e 1000(t
3
3
0
Z t
7000
1000t
(2e 1000⌧ 5e 4000⌧ )d⌧
=
5e
3
0
✓
7000
2
5
=
5e 1000t
(1 e 1000t )
(1
3
1000
4000
7
70 2000t 175 5000t
=
· 5e 1000t +
e
e
4
3
12
⌧)
d⌧
e
4000t
)
◆
• Note how the ZSR consists of a forced response plus characteristic modes.
• Re. the forced (eigen-) response, note that the transfer function
H( 1000) =
7
4
where H(s) =
P (s)
7000s
=
.
Q(s)
(s + 2000)(s + 5000)
117
Total response - second-order, series RLC circuit example (cont)
• Recall if the initial conditions are (D y)(0 ) =
the ZIR is
8t
2000V/s and y(0 ) = 10V, then
0 , yZI (t) = 16e
2000t
6e
5000t
.
• Combining with the the ZSR just derived, the total response of the system (to both initial
conditions and input) is: 8t 0 ,
y(t) = yZI (t) + yZS (t)
n
X
=
ck µk (t) + (h ⇤ f )(t)
k=1
= 16e
2000t
6e
5000t
+
✓
7
· 5e
4
1000t
+
70
e
3
2000t
175
e
12
5000t
◆
u(t)
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Total response - discussion of existence and uniqueness
• The existence of the ZIR follows from the linear independence of characteristic modes.
• Similarly, the existence of the ZSR follows from the assumption that the input and impulse
response are Dirichlet signals, so that the convolution is well defined.
• When the initial conditions are (respectively, input signal is) zero, the unique solution for
the ZIR (respectively, ZSR) is the zero signal.
• Given this, it’s easy to prove by contradiction the uniqueness of the ZIR and ZSR in general.
• For example, assume that yZI and ỹZI are two di↵erent ZIRs for the same system (Q) and
the same initial conditions:
– Note that the signal yZI
– That is, yZI
ỹZI is the ZIR for Q with zero initial conditions.
ỹZI = 0, a contradiction.
– So, the original assumption of the existence of two di↵erent ZIRs is false.
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System stability - ZIR - asymptotically stable
• Consider a SISO system with input f and output y .
• Recall that the ZIR yZI is a linear combination of the system’s characteristic modes, where
the coefficients depend on the initial conditions.
• A system is said to be asymptotically stable if for all initial conditions,
lim yZI (t) = 0.
t!1
• So, a system is asymptotically stable if and only if all of its characteristic values have strictly
negative real part.
• For example, if the characteristic polynomial Q(s) = (s + 5)(s + 7), then
– the system’s characteristic values (roots of Q) are
– the ZIR is of the form yZI(t) = c1e
7t
+ c2 e
5t ,
7,
t
5 (strictly negative and real),
0 ,
– So, yZI(t) ! 0 as t ! 1 for all c1, c2, and
– hence is asymptotically stable.
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System stability - bounded signals
• A signal y is said to be bounded if
9 M < 1 s.t. 8t 2 R, |y(t)|  M ;
otherwise y is said to be unbounded.
• For example, y(t) = 3e
5t u(t),
t 2 R is bounded (can use M = 3).
• Also, 3 cos(5t), t 2 R, is bounded (again, can use M = 3).
• But both 3e2t cos(5t) and 3e2t are unbounded; for the latter example, obviously
lim 3e2t = 1.
t!1
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System stability - ZIR - marginally stable
• A system is said to be marginally stable if it is not asymptotically stable but yZI is always
(for all initial conditions) bounded.
• A system is marginally stable if and only if
– it has no characteristic values with strictly positive real part,
– it has at least one purely imaginary characteristic value, and
– all purely imaginary characteristic values are not repeated.
• That is, a marginally stable system
– has some characteristic modes of the form cos(wt) or sin(wt),
– while the rest are of the form tk e
– for real ↵ > 0, frequency w
↵t cos(wt)
or tk e
↵t sin(wt),
0 and integer degree k
0 (and time t
0).
• For example, if the characteristic
polynomial is Q(s) = s(s2 + 2)(s + 3), with charp
acteristic
values
0
,
±j
2
and
3, then the system is marginally stable with modes 1,
p
p
cos( 2t), sin( 2t) and e 3t , the first three of which are bounded but do not tend to
zero as time t ! 1.
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System stability - ZIR - unstable
• A system that is neither asymptotically nor marginally stable (i.e., a system with unbounded
modes) is said to be unstable.
• For example, if the characteristic polynomial
is Q(s) = (s2 + 2)2(s + 3) then the purely
p
imaginary
p characteristic
p values ±j 2 are repeated, and hence the two additional modes
t sin( 2t), t cos( 2t) are unbounded - so this system is unstable.
• Similarly, the system with Q(s) = s2 (s2 + 2)(s + 3) is unstable owing to the repeated
characteristic value at 0 with associated modes 1 and t, the latter unbounded.
• For another example, the system with Q(s) = (s2 + 2)(s 3) is unstable owing to the
(unstable) characteristic value at 3 with associated unbounded mode e3t.
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System stability - ZIR - series RLC circuit example
• If the output signal is redefined as the capacitor voltage, then P (s) = (LC)
but the characteristic polynomial remains
1
= wo2
Q(s) = s2 + 2⇣wo s + wo2 , where
p
– resonant frequency wo = 1/ LC , and
p
– damping factor ⇣ = (R/2) C/L.
– The characteristic values (roots of Q) of the system are
✓
◆
q
2
wo ⇣ ± ⇣
1
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System stability - ZIR - series RLC circuit example (cont)
Cases regarding the discriminant ⇣ 2
1 (assuming fixed L, C > 0):
• ⇣ = 0 (R = 0): two purely imaginary characteristic values ±jwo , i.e., marginally stable.
p
• 0 < ⇣ < 1 (0 < R < 2 L/C): two complex conjugate characteristic values of strictly
p
negative real part and magnitude wo, c± := wo(⇣ ± j 1 ⇣ 2), i.e., asymptotically
stable, underdamped.
p
• ⇣ = 1 (R = 2 L/C ): a repeated (double) real characteristic value at wo < 0,
i.e., asymptotically stable, critically damped.
p
• 1 < ⇣ < 1 (R > 2 L/C ): two di↵erent real strictly negative char. values, r± :=
p
wo (⇣ ± ⇣ 2 1), i.e., asymptotically stable, overdamped.
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System stability - ZSR - BIBO stable
• A SISO system is said to be Bounded Input, Bounded Output (BIBO) stable if
8 bounded input signals f , the ZSR yZS is bounded.
• A necessary condition for BIBO stability is absolute integrability of the impulse response,
Z 1
|h(⌧ )|d⌧ < 1.
0
• To see why: For arbitrary bounded input f (by Mf with 0  Mf < 1): 8t
0,
|yZS (t)| = |(f ⇤ h)(t)|
Z t
=
f (t ⌧ )h(⌧ )d⌧
Z 0t

|f (t ⌧ )h(⌧ )|d⌧ (by the triangle inequality)
Z0 t

Mf |h(⌧ )|d⌧
0 Z
1
 Mf
|h(⌧ )|d⌧ =: My < 1,
0
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System stability - ZSR - BIBO stable
• The condition of absolute integrability of the impulse response,
Z 1
|h(⌧ )|d⌧ < 1,
0
is also necessary for, and hence equivalent to, BIBO stability.
• When we study LTI systems in the frequency domain and derive the impulse response, we
will readily see how asymptotic stability (of ZIR) is equivalent to BIBO stability (of ZSR).
• Example: If the system has char. values ±jwRh of multiplicity 1, and the input is f (t) =
t
cf cos(wf t)u(t), then yZS (t) has the term 0 ch cos(wh ⌧ )cf cos(wf (t ⌧ ))d⌧
Z
ch cf t
=
cos((wh wf )⌧ + wf t) + cos((wh + wf )⌧ wf t) d⌧ ;
2 0
so if wf 6= wh, then yZS is bounded, else yZS(t) has the term 0.5chcf t cos(wf t) which
is unbounded - i.e., marginally stable system is not BIBO.
• For another example, see Ex. 2.6-4 of Lathi.
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ZSR - the transfer function, H
• Recall that for any polynomial Q and frequency s 2 C (including s = jw, w 2 R),
Q(D)est = Q(s)est , 8t
0.
• So, if we guess that yp (t) = H(s)est , t 0, is a “particular” solution of
Q(D)y = P (D)f in response to input f (t) = est u(t), we get by substitution that
8t
0, s 2 C, H(s)Q(s)est = Q(D)y = P (D)f = P (s)est
) H(s) = P (s)/Q(s).
• The “rational polynomial” H = P/Q is known as the system’s transfer function and will
figure prominently in our study of frequency-domain analysis.
• So, the ZSR (forced response + characteristic modes) would be of the form:
yZS (t) = H(s)est + linear combination of char. modes, t
k
again with (D yZS)(0 ) := 0 8 integers k
0,
0.
• Recall for the previous series RLC circuit example with input f (t) = 5e
1000t ,
P (s)
7000
=
, s2C
Q(s)
(s + 2000)(s + 5000)
yZS (t) = H( 1000)f (t) + linear combination of char. modes, t
t
0:
H(s) =
0.
128
ZSR - impulse response h, transfer function H , and eigenresponse
• yZS (t) = H(s)Aest+j + linear combination of char. modes, with t
0 and A >
0, 2 R, is the ZSR to input f (t) = Aest+j u(t), with (Dk yZS )(0 ) := 0 for all
integers k 0.
• The eigenresponse is a special case of the forced response for exponential inputs.
• If s = jw for w 2 R (i.e., input f is sinusoidal) and the system is asymptotically stable,
then the ZSR tends to the steady-state eigenresponse of the system:
y(t) ⇠ H(jw)Aej(wt+
)
as t ! 1.
• Since y = h ⇤ f , we get that as t ! 1 for a causal and asymptotically stable system,
✓Z t
◆
Z t
j(w(t ⌧ )+ )
jw⌧
yZS (t) =
h(⌧ )Ae
d⌧ =
h(⌧ )e
d⌧ Aej(wt+ ) ⇠ H(jw)Aej(wt+ ),
0
0
Z 1
)
h(⌧ )e jw⌧ d⌧ = H(jw), 8w 2 R.
0
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ZSR for single-pole systems using eigenresponse
• We will see that yZS will be continuous at time 0 when H = P/Q is strictly proper
(i.e., bn = 0 so that m := the degree of P < n := degree of Q); thus, h has no bn
term and there is no “direct coupling” between input and output in the system.
• So given a strictly proper system with a single pole p (characteristic value), i.e., n = 1,
the ZSR to a sinusoidal input f = Aesot+j will be of the form
yZS (t) = H(so )f (t) + ce
pt
= H(so )Aeso t+j + ce
pt
, 8t
0.
• So, if H is strictly proper, we can solve for the coefficient c of the system mode e
yZS (0) = 0, i.e.,
0 = H(so )Aej + c ) c =
pt
using
H(so )Aej .
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Remark: time origin at
1 for an asymptotically stable system
• If we take the origin of time to be 1 for an asymptotically stable LTI system, then the
system is in steady-state in “finite time” when the characteristic-modes component of the
response will be negligible.
• So, for frequency w 2 R, the response to the persistent sinusoidal input f (t) = Aej(wt+
is, 8t 2 R,
✓Z 1
◆
Z 1
jw(t ⌧ )+j
jw⌧
h(⌧ )Ae
d⌧ =
h(⌧ )e
d⌧ Aej(wt+ ) = H(jw)Aej(wt+ ),
1
)
1
where the frequency w 2 R and the integral is finite since h is assumed Dirichlet.
• Thus, H is the bilateral Fourier transform of the impulse response h,
Z 1
h(⌧ )e jw⌧ d⌧ = H(jw), 8w 2 R.
1
• For a causal system (and as in the previous slide),
Z 1
h(⌧ )e jw⌧ d⌧ = H(jw), 8w 2 R.
0
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Resonance of the marginally stable series RLC circuit
• Recall that the series RLC circuit with capacitor-voltage output is marginally stable when
R =
p 0 (damping factor ⇣ = 0) with two purely imaginary characteristic values at
±j/ LC =: ±jwo .
• All sinusoidal inputs are, of course, bounded.
• Consider an input of the form
f (t) = ejwt = cos(wt) + j sin(wt).
• When w 6= wo , the steady-state total response to this complex-valued, bounded sinusoidal input, consisting of the eigenresponse H(jw)f (t) plus a linear combination of the
(persistent) characteristic modes e±jwot, is bounded
• However, if w = wo then Q(jwo ) = 0 and P (jwo ) = wo2 6= 0; so |H(jw)| diverges
as w ! wo.
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Resonance of the marginally stable series RLC circuit (cont)
• That is, though f (t) = ejwo t is a bounded input, the output (of the marginally stable
circuit with resonant frequency wo) is not bounded.
• This phenomenon is called resonance.
• When we study transient response of resonant circuits in the (complex) frequency domain,
we will see that the total response of this system to f (t) = ejwot has an unbounded output
component tejwot.
• In the following, we will refer to the characteristic values of the system (roots of the
characteristic polynomial Q, asymptotes of transfer function H ), as system poles.
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Zero-state step response of causal LTI systems
• The zero-state step response of a causal LTI system with impulse response h is yu = u ⇤ h.
• That is, 8t < 0, yu (t) = 0, and 8t 0,
Z 1
yu (t) =
h(⌧ )u(t
1
Z t
=
h(⌧ )d⌧
⌧ )d ⌧
0
= H(0) · 1 + linear combo. of char. modes
where the (steady-state, zero-frequency/DC) eigenresponse is
Z 1
H(0) =
h(⌧ )d⌧.
0
• So, if the system is asymptotically stable, then
lim yu (t) = H(0).
t!1
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134
Step response of a first-order circuit - time constant, 90% rise time
• Recall that this circuit has system equation D y + (RC)
1
response h(t) = 2RC
e t/(RC)u(t).
1y
1f
= (2RC)
and impulse
• Also, H(0) = P (0)/Q(0) = 0.5 (consistent with the infinite DC impedance of a
capacitor so that the circuit is an even voltage divider in DC steady-state).
• The (ZS) step response is yu (t) = (h ⇤ u)(t) =
t 0.
Rt
0
h(⌧ )d⌧ = 0.5(1
t/(RC) )u(t),
e
• The 90% rise-time of a system is the first time t90 at which it attains 90% of the terminal
(steady-state) value of its step response,
yu (t90 ) = 0.9 · 0.5 ) t90 = RC ln(10)
where RC is the time constant of this circuit; i.e., t90 is ln(10) ⇡ 2.3 time-constants.
135
Step response of series RLC circuit - preliminaries
• In this series RLC circuit, the capacitor voltage has been designated the output signal.
• Thus, the system equation is
• Again, P (s) = (LC)
1,
D2 y + RL
1Dy
+ (LC)
1y
= (LC)
1f .
but the characteristic polynomial remains
Q(s) = s2 + RL
1
s + (LC)
1
= s2 + 2wo ⇣s + wo2 ,
where again the resonant frequency and damping factor are, respectively,
p
p
wo = 1/ LC and ⇣ = (R/2) C/L.
p
• So the characteristic values / poles again are wo (⇣ ± ⇣ 2 1).
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Step response of series RLC circuit - overdamped case, ⇣ > 1
• In the overdamped case (⇣ > 1), there are two di↵erent, negative real poles, r± .
• Again, we’ll see later how the impulse response is
wo
p
h(t) =
e r+ t
2
2 ⇣
1
er
t
u(t), t 2 R.
• Thus, the ZS step response is yu = h ⇤ u,
✓
◆
wo2
wo2
yu (t) =
1+
e r+ t +
er t u(t), t 2 R.
r+ (r+ r )
r (r
r+ )
• Again, note how the zero-state step response consists of
– the eigenresponse H(0)u(t) = u(t), with H(s) = P (s)/Q(s) = (LC)
RL 1 s + (LC) 1 ) so that H(0) = 1,
1 /(s2 +
– and a linear combination of characteristic modes, whose coefficients are such that the
initial conditions of yu are zero consistent with a zero-state step response, yu(0) = 0
and (D y)(0) = 0.
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Step response of series RLC circuit - overdamped case, ⇣ > 1 (cont)
• One can directly check that the step response monotonically increases:
yu (0) = 0
(D yu )(t)
0 8t 0
lim yu (t) = 1 = H(0) = P (0)/Q(0).
t!1
• To find an approximate 90% rise time in the special case when ⇣
exp(r t90 ) ⌧ exp(r+ t90 ) < 1),
wo2
er+t90
r+ (r+ r )
✓
◆
1
10wo2
ln
r+
r+ (r+ r )
0.9 = 0.9yu (1) = yu (t90 ) ⇡ 1 +
) t90 ⇡
1 (so that
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Step response of series RLC circuit - overdamped case (cont)
• This is the step response yu (t) for the series RLC circuit example with R = 700⌦,
L = 0.1H, and C = 1µF, so that
wo = 3162rad/s and ⇣ = 1.107.
• Since ⇣ > 1, this is the overdamped case.
• From the plot, t90 ⇡ 0.0015s.
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Step response of series RLC circuit - critically damped case, ⇣ = 1
• In the critically damped case (⇣ = 1), there is a repeated (double), negative real pole at
wo .
• The impulse response is now
h(t) = wo2 te
wo t
u(t), t 2 R.
• Thus, the ZS step response h ⇤ u is (after integration by parts),
yu (t) =
1
noting the characteristic modes e
wo t
wo t e
wo t
and te
wo t .
e
wo t
u(t), t 2 R,
• Again, the step response monotonically increases:
yu (0) = 0
(D yu )(t)
0 8t 0
lim yu (t) = 1 = H(0).
t!1
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Step response of series RLC circuit - underdamped case, 0 < ⇣ < 1
• In the underdamped case (0 < ⇣ < 1), there are two complex-conjugate poles of magnitude
wo and negative real part,
q
wo(⇣ ± j 1 ⇣ 2 ).
• Here the impulse response is
h(t) =
wo
p
1
⇣2
e
wo ⇣t
q
sin(wo 1
⇣ 2 t)u(t), 8t 2 R.
• Thus, the ZS step response h ⇤ u is (after integration by parts): 8t 2 R,
yu (t) =
1
e
wo ⇣t
cos(wo
q
1
2
⇣ t)
p
⇣
1
⇣2
e
wo ⇣t
q
sin(wo 1
2
⇣ t)
!
u(t).
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Step response of series RLC circuit - underdamped case (cont)
• One can directly check that
yu (0) = 0 and
lim yu (t) = 1,
t!1
but yu is not monotonically increasing in the underdamped case.
• Indeed, yu overshoots its terminal value yu (1) = 1.
• By solving thep
first-order condition, D yu = 0, we get that local extrema of yu occur at
times k⇡/(wo 1 ⇣ 2) for integers k > 0.
• In particular, k = 1 corresponds to the maximum overshoot,
!
p
⇡
p
yu
yu (1) = e ⇡⇣/ 1
wo 1 ⇣ 2
⇣2
> 0,
see the following figure.
• Note that the overshoot decreases with damping factor ⇣ .
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Step response of series RLC circuit - underdamped case (cont)
• This is the previous numerical example but with the resistance reduced to R = 100⌦ so
that ⇣ = 0.16 < 1, i.e., the underdamped case, and unchanged w0 = 3162rad/s.
p
• Here, we
see
maximum
overshoot
of
about
0.6
⇡
exp(
⇡⇣/
1
p
⇡/(wo 1 ⇣ 2 ) seconds.
⇣ 2 ) at about 0.001 ⇡
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Toward frequency domain methods
• We now begin our journey into the frequency domain.
• We first study Fourier-series representation of periodic signals in the time domain.
• Our origin of time will be 1 so that a steady-state eigenresponse of an asymptotically
stable system is reached in finite time.
• Subsequently, we will study the full Zero-State Response (ZSR, including modes) using the
bilateral Fourier transform on the real frequency domain, w 2 R.
• The Fourier transform can accommodate aperiodic input signals and not asymptotically
stable and noncausal systems.
• Finally, as we have just done in the time domain, we will study the total system response
(ZSR + ZIR) using the unilateral Laplace transform in the complex frequency domain,
s 2 C.
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Fourier-series representation of periodic signals - Outline
• Approximating Dirichlet signals f : I ! R by minimizing square error over a finite timeinterval (period) I ⇢ R
– Linear vector space of signals
– Norm, inner product, and LLSE estimation
– Orthogonal signal-sets
– LLSE given (projection onto the span of) orthogonal signal-sets
• Fourier series representation of periodic, Dirichlet signals f : R ! R
– Trigonometric Fourier series: DC, fundamental, and harmonic components.
– Compact trigonometric Fourier series.
– Exponential Fourier series.
• Parseval’s theorem for signal power.
• Eigenresponse to a periodic input, i.e., steady-state response of an asymptotically stable
LTI system.
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Dirichlet signals on a finite time-interval
• We begin our study of Fourier series by reviewing basics of linear algebra
– for continuous-time signals f := {f (t) | t 2 I} on a finite time-interval I ⇢ R (later
a period of the assumed-periodic f ),
– instead of vectors v 2 Rk , which can be thought of as discrete-time signals on a finite
time-interval, i.e., v := {vl | l 2 {1, 2, ..., k}} for positive integer k 1.
• Again, we will restrict consideration to Dirichlet signals f : I ! R, i.e., satisfying
R
– the weak Dirichlet condition of absolute integrability, I |f (t)|dt < 1,
– the strong Dirichlet conditions: over any finite interval ⇢ I : (i) f has a finite number
of jump discontinuities, and (ii) f has a finite number of minima and maxima.
• The set of all such signals forms a linear vector space, e.g., it is closed under scalar multiplication and vector addition: 8 such signals f, g and scalars ↵, 2 R, ↵f + g : I ! R
is also Dirichlet.
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Square error between a signal and its estimator
• Suppose the signal fˆ is an approximation of f over the time-interval I .
• The square error between f and fˆ is
||f
fˆ||2 :=
Z
I
|f (t)
i.e., the squared L2 norm of the error signal " := f
fˆ(t)|2 dt,
fˆ.
• This definition of norm has the properties of the Euclidean norm of vectors of a finitedimensional vector space, i.e.,
qR
– 8 (Dirichlet) signals x, ||x|| :=
|x(t)|2 dt = 0 , x = 0, i.e., x is the zero
I
signal (x(t) = 0 for “almost all” t 2 I ),
– 8 signals x and scalars c 2 R, and ||cx|| = |c| · ||x||.
– 8 signals x, y , ||x + y||  ||x|| + ||y|| (triangle inequality).
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LLSE estimator - projection and inner product
• Consider signals f, g : I ! R with g 6= 0.
• We wish to approximate f “given g ”, i.e., using fˆ = ↵g by minimizing over the scalar
↵ 2 R the square error
||f fˆ||2 = ||f ↵g||2 =: E(↵).
• By definition of norm and linearity of the integral,
Z
E(↵) =
|f (t) ↵g(t)|2 dt
I
Z
Z
Z
=
|f (t)|2 dt 2↵ f (t)g(t)dt + ↵2 |g(t)|2 dt,
i.e., E is quadratic.
I
I
I
• Since E is convex, it is minimized at ↵ satisfying the first-order optimality condition,
Z
Z
0 = E 0 (↵) =
2 f (t)g(t)dt + 2↵ |g(t)|2 dt
I
I
R
f (t)g(t)dt
I
) ↵ = R
|g(t)|2 dt
I
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LLSE estimator - projection and inner product (cont)
• That is, the Linear, Least Square-Error (LLSE) estimator, of f given g 6= 0 is
✓R
◆
f (t)g(t)dt
< f, g >
I
R
g =
g,
2
|g(t)| dt
||g||2
I
where we have identified the inner product operator mapping two signals to a scalar,
Z
< f, g > :=
f (t)g(t)dt 2 R.
I
• The LLSE estimator of the signal f given the signal g 6= 0 is the projection of f onto the
one-dimensional vector space (of signals)
{↵g | ↵ 2 R} = span{g}.
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LLSE estimator - LLSE signal and given signals are orthogonal
• The error signal for this LLSE estimator is
" := f
< f, g >
g = f
||g||2
fˆ.
• Note that the inner product of real-valued signals is obviously linear in both arguments and
commutative.
• Also, for all signals g ,
< g, g > = ||g||2 .
• Thus, if g 6= 0 then
< ", g > = < f
= 0,
i.e., the error " = f
< f, g >
g, g > = < f, g >
||g||2
< f, g >
< g, g >
||g||2
↵g of the LLSE estimator of f given g is orthogonal to g .
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LLSE estimator given orthogonal signal-sets
• Consider a group of k
1 non-zero signals
g1 , g2 , ..., gk : I
! R
that are orthogonal to each other, i.e.,
Z
8p 6= q 2 {1, 2, ..., k},
gp (⌧ )gq (⌧ )d⌧ =: < gp , gq > = 0.
I
• For any signal f : I ! R, we can easily generalize the LLSE estimator given one signal
to find the LLSE that is a linear combination of these k
1 given signals: for scalars
↵1 , ↵2 , ...↵k 2 R,
fˆ = ↵1 g1 + ↵2 g2 + ... + ↵k gk ,
i.e., the LLSE estimator.
• To this end, define the square-error function,
E(↵1 , ↵2 , ...↵k ) = ||f
fˆ||2 = ||"||2 = < ", " > .
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LLSE estimator given orthogonal signal-sets (cont)
• By the linearity and commutativity property of inner product,
Z
E(↵1 , ↵2 , ...↵k ) =
|f (⌧ ) ↵1 g1 (⌧ ) ↵2 g2 (⌧ ) ...
Iˆ
= <f
k
X
l=1
= ||f ||2 + 2
k
X
↵l g l , f
X
↵l g l >
l=1
↵p ↵q < g p , g q > +
p<q
↵k gk (⌧ )|2 d⌧
X
l
2↵l < f, gl > +↵2l ||gl ||2
• By the orthogonality assumption, i.e., 8p 6= q , < gp , gq >= 0,
X
E(↵1 , ↵2 , ...↵k ) = ||f ||2 +
2↵l < f, gl > +↵2l ||gl ||2
l
• Again, the coefficients ↵l that minimize E are obtained by the k first-order conditions
0 =
@E
@↵l
=
2↵l < f, gl > +2↵l ||gl ||2 , l 2 {1, 2, ..., k}.
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LLSE estimator given orthogonal signal-sets (cont)
• Thus, the LLSE coefficients are
↵l
=
< f, gl >
=
||gl ||2
R
IR
f (⌧ )gl (⌧ )d⌧
I
|gl (⌧ )|2 d⌧
• So for an orthogonal signal-set, the LLSE coefficients are the same as those of LLSE
estimators given just the individual signals.
• Note: If the given signal-set {g1 , g2 , ..., gk } is not orthogonal, the Gram-Schmidt procedure
can be used to find an orthogonal signal-set that spans the linear vector space of the given
one, where
span{g1, g2, ..., gk } = {↵1g1 + ↵2g2 + ... + ↵k gk | ↵1, ↵2, ..., ↵k 2 R}.
• Actually, Gram-Schmidt will work on a given signal-set that is not even linearly independent
to produce an orthonormal basis of its span.
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LLSE estimators with signals illustrated as vectors
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LLSE estimator on a finite time-interval - example
• Consider the exponential signal f = 5e2t for t 2 I = [ 2, 2].
• To find the LLSE estimator of f given the signals g1 (t) = 1 and g2 (t) = t, t 2 I ,
fˆ = ↵1 g1 + ↵2 g2 ,
first note that
< g1 , g2 > =
Z
2
g1 (t)g2 (t)dt =
2
= 0.
Z
2
2
1 2
t
2
1 · t dt =
2
2
• Thus, g1 and g2 are orthogonal.
• Recall that the product of an even and odd signal is odd, and that zero is the integral (or
average value) of an odd signal over an interval of time I that equally straddles the origin.
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LLSE estimator on a finite time-interval - example (cont)
• Since g1 and g2 are orthogonal, the LLSE coefficients are
R2
R2
(5/2)e2t
f (t)g1 (t)dt
5e2t · 1dt
< f, g1 >
2
2
a1 =
=
=
=
R
R
2
2
||g1 ||2
4
|g1 (t)|2 dt
12 dt
2
a2 =
< f, g2 >
||g2 ||2
2
2
2
= (5/8)(e4 e 4 ) = (5/4) cosh(4).
R2
R2
(5/2)te2t (5/4)e2t
f (t)g2 (t)dt
5e2t tdt
2
= R2
= R22
=
(1/3)t3 |2 2
|g (t)|2 dt
t2 d t
2 2
2
= (45/64)e4 + (75/64)e
• So, the LLSE of f given {g1 , g2 } is
4
fˆ = (5/4) cosh(4)g1 + ((45/64)e4 + (75/64)e
4
)g2 .
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2
2
Fourier series representation - Basic theorem
• Now consider periodic signals on R.
• Recall that if the ratio of the periods of g1 and g2 is rational, or if one is periodic and the
other is constant, then all linear combinations ↵1g1 + ↵2g2 are periodic, i.e., 8 scalars
↵ 1 , ↵ 2 2 R.
• The following theorem is the basic to representation periodic signals by Fourier series.
• Theorem: The span of the set of signals
{cos(kwo t), t 2 R}, k 2 {0, 1, 2, 3, ...} and {sin(kwo t), t 2 R}, k 2 {1, 2, 3, ...}
includes all piecewise-continuous, periodic Dirichlet signals with period T := 2⇡/wo > 0
that are valued at the midpoint of their jump discontinuities.
• Note that the constant signal is included in this signal set:
if k = 0 then 8t 2 R, 1 = cos(kwot).
• Also for k > 0, the period of cos(kwo t) and sin(kwo t) is 2⇡/(kwo ) = T /k.
• So, linear combinations of these signals will be periodic with period T /k for some
k 2 {1, 2, ...} or constant (corresponding to k = 0).
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Fourier series coefficients (coordinates w.r.t. trigonometric basis)
• Suppose f : R ! R is Dirichlet and periodic with period
2⇡
> 0,
wo
where wo is the fundamental frequency of f .
T
=:
• Define the infinitely large, trigonometric signal-set:
gc,k := {cos(kwo t), t 2 R}, k 2 Z
0
and gs,k := {sin(kwot), t 2 R}, k 2 Z
1
• Note that all of these g have the property that 8t 2 R, g(t) = g(t + T ), consistent with
g either being constant (k = 0), or having period T /k for some integer k > 0 so that
8t 2 R, g(t) =
=
=
=
g(t + T /k)
g(t + T /k + T /k) = g(t + 2T /k) ...
g(t + (k 1)T /k + T /k)
g(t + T ).
• The LLSE of f given this signal set over an arbitrary interval I of length T is
fˆ =
1
X
k=0
↵k gc,k +
1
X
k=1
k gs,k
= ↵0 +
1
X
(↵k gc,k +
k gs,k ),
k=1
where by the previous theorem, the {↵k , k } are the coordinates of f w.r.t. the trig basis,
i.e., the LLSE has zero square error, ||"||2 = ||f fˆ||2 = 0.
158
Orthogonality of the given trigonometric signals - di↵erent cosines
• Again, wo := 2⇡/T is the fundamental frequency of T -periodic f .
• Let
R
T
...dt represents integration over any interval of length T (i.e., any period of f ).
• 8 integers k 6= l
0,
< gc,k , gc,l > =
=
Z
Z
cos(kwo t) cos(lwo t)dt
T
T
0.5 cos((k + l)wo t) + 0.5 cos((k
l)wo t) dt
0
T
0.5
0.5
sin((k + l)wo t) +
sin((k l)wo t)
(k + l)wo
(k l)wo
0
0.5
0.5
=
sin((k + l)wo T ) +
sin((k l)wo T )
(k + l)wo
(k l)wo
0.5
0.5
=
sin((k + l)2⇡) +
sin((k l)2⇡)
(k + l)wo
(k l)wo
= 0.
=
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Orthogonality of the given trigonometric signals - di↵erent sines
• 8 integers k 6= l
1,
< gs,k , gs,l > =
Z
sin(kwo t) sin(lwo t)dt
ZT T
=
0.5 cos((k l)wo t)
0.5 cos((k + l)wo t) dt
0
0.5
sin((k
(k l)wo
= 0
=
l)wo t)
0.5
sin((k + l)wo t)
(k + l)wo
T
0
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Orthogonality of the given trigonometric signals - a sine and cosine
• 8 integers k
0, l
< gc,k , gs,l > =
=
Z
Z
1,
cos(kwo t) sin(lwo t)dt
T
T
0.5 sin((l + k)wo t) + 0.5 sin((l
k)wo t) dt
0
=
=
= 0
0.5
cos((l + k)wo t)
(l + k)wo
0.5
(cos((l + k)2⇡)
(l + k)wo
T
0.5
cos((l k)wo t)
(l k)wo
0
0.5
1)
(cos((l k)2⇡)
(l k)wo
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Fourier coefficients/coordinates
• By the orthogonality of the given trigonometric signal-set, the LLSE of f over a period
T = 2⇡/wo has coefficients
R
f (t) cos(kwo t)dt
< f, gc,k >
8k 0, ↵k =
= RT
2
||gc,k ||
| cos(kwo t)|2 dt
T
8 Z
2
>
>
f (t) cos(kwo t)dt if k > 0
<
T T
Z
=
1
>
>
f (t)dt if k = 0
:
T T
R
f (t) sin(kwo t)dt
< f, gs,k >
8k 1, k =
= RT
2
||gs,k ||
| sin(kwo t)|2 dt
T
Z
2
=
f (t) sin(kwo t)dt.
T T
• Note that 8k > 0, T 1 ||gc,k ||2 = T 1 ||gs,k ||2 = 1/2, consistent with our previous
calculation of the power of a sinusoid with unit amplitude.
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1)
Fourier coefficients/coordinates (cont)
• So, at all points t of continuity of f (when f = fˆ),
f (t) = ↵0 +
1
X
(↵k cos(kwo t) +
k
sin(kwo t)) =: fˆ(t).
k=1
• Note that ↵0 = T
1
R
T
f (t)dt is just the mean/average value (DC component) of f .
• The T -periodic (k = 1) component is the fundamental of T -periodic f .
• For k
2, the T /k-periodic component is the (k
1)th harmonic of f .
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Trigonometric Fourier series - square-wave example
• The Fourier coefficients of this square wave f of period 7 and duty cycle 4 are
Z
Z
2
1 T /2
1 2
5
20
↵0 =
f (t)dt =
5dt =
t
=
.
T
7
7
7
T /2
2
2
Z
Z
2 T /2
2 2
8k 1, ↵k =
f (t) cos(kwo t)dt =
5 cos(k(2⇡/7)t)dt
T
7 2
T /2
2
=
8k
1,
k
=
5
10
sin(k(2⇡/7)t)
=
sin(4k⇡/7).
k⇡
k⇡
2
Z
2 T /2
f (t) sin(kwo t)dt = 0.
T
T /2
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Trigonometric Fourier series - square-wave example (cont)
• So,
f (t) =
1
X 10
20
+
sin(4k⇡/7) cos(2k⇡t/7) =: fˆ(t), 8t 2 R,
7
k⇡
k=1
except possibly at points t of jump discontinuity of f where fˆ(t) = 5/2, e.g., at t = ±2.
• Note k = 0 8k
1 simply because the integrand is odd on the interval of integration
[ 3.5, 3.5] - recall that the product of an even (f ) and odd (sine) function is odd.
• Also note the interesting identity that ensues at time t = 0 (a point of continuity of f ):
2⇡
7
1
X
sin(4k⇡/7)
=
k
k=1
.
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Trigonometric Fourier series - triangle-wave example
↵0 =
8k
1, ↵k
8k
1,
k
=
=
=
=
1
T
2
T
2
T
3
5
Z
Z
Z
✓
T /2
1
f (t)dt =
10
T /2
T /2
Z
5
3t dt =
5
3 2
t
20
5
= 0.
5
f (t) cos(kwo t)dt = 0.
T /2
T /2
f (t) sin(kwo t)dt =
T /2
1
5
Z
5
3t sin(k⇡t/5)dt
5
5
52
t cos(k⇡t/5) +
sin(k⇡t/5)
k⇡
(k⇡)2
30
30
cos(k⇡) =
( 1)k+1
k⇡
k⇡
◆
5
after integration by parts
5
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Trigonometric Fourier series - even and odd signals
• Thus, for this triangle wave f ,
f (t) =
1
X
30
k=1
k⇡
( 1)k+1 sin(k⇡t/5)
8t 2 R except possibly at t = 5k for odd k 2 Z (i.e., points of jump discontinuity in f
where the RHS fˆ(5k) = 0).
• Note ↵k = 0 8k 0 in the previous triangle-wave example simply because the integrand
is odd on the interval of integration [ 5, 5] - recall that the product of an odd (f ) and
even (cos) function is odd.
• In general, if T -periodic f (t) is even on [ T /2, T /2], then 8k
a cosine (even) trigonometric Fourier series.
• Alternatively, if f (t) is odd on [ T /2, T /2] then 8k
(odd) trigonometric Fourier series.
1,
k
= 0, i.e., f has
0, ↵k = 0, i.e., f has a sine
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Partial Fourier series approximation
• Consider the partial Fourier series fˆK of T -periodic, Dirichlet f up to the (K
harmonic, i.e., 8K 0, t 2 R,
fˆK (t) := ↵0 +
K
X
↵k cos(kwo t) +
k
1)th
sin(kwo t),
k=1
• For Dirichlet f (in L2 ),
Z
lim
|fˆK (t)
K!1
f (t)|2 dt =
T
lim ||fˆK (t)
K!1
f (t)||2 = 0.
• So, at points t of continuity of f ,
lim fˆK (t) := fˆ(t) = f (t).
K!1
• In the following we will no longer point out possible di↵erences between f and its (complete)
Fourier series fˆ at points of jump discontinuity of f .
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Partial Fourier series approximation - square-wave example
• In the following, we plot fˆ0 = 20/7, fˆ1 , fˆ10 , and fˆ40 for the previous square-wave example.
• The non-uniform convergence near the jump discontinuity is called the Gibbs phenomenon.
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Parseval’s theorem for signal power
• Recall that the average power of a sinusoid A cos(wt + ), t 2 R, is A2 /2.
• Discrepancies only at points of jump discontinuity will not a↵ect mean power calculations.
• So, the Fourier series representation of a T -periodic, Dirichlet signal f can be used to
compute its mean power, i.e.,
Z
1
Pf = T
|fˆ(t)|2 dt
= T
= T
=
↵20
1
1
Z
Z
+
T
↵0 +
T
↵20 +
T
1
X
k=1
k sin(kwo t)
↵2k cos2 (kwo t) +
2
2
k sin (kwo t)
k=1
1
X
k=1
1 ✓ 2
X
↵k
2
+
2
(↵k cos(kwo t) +
2
k
2
dt
dt by orthogonality
◆
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Parseval’s theorem for signal power (cont)
• Suppose we want to compute the fraction ' of T -periodic f ’s power in the frequency
band [w1, w2] where w2 > w1 0 are frequencies in radians/s and wo = 2⇡/T is the
fundamental of f .
• First identify which harmonics of f lie in the band, i.e., for what integers k
[w1 , w2 ], equivalently
0 is kwo 2
k1 := dw1 /wo e = dw1 T /(2⇡)e  k  k2 := bw2 /wo c = bw2 T /(2⇡)c ,
where for z 2 R, dze is the smallest integer
• So,
' =
z and bzc is the greatest integer  z .
k2
1 X ↵2k +
Pf
2
2
k
,
k=k1
where we’ve defined b0 = a0 to simplify this expression when k1 = 0,
P2
and kk=k
... := 0 when k1 > k2 .
1
• A complementary problem is to find the smallest frequency interval [w1 , w2 ] for which
' = 0.95.
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Parseval’s theorem for signal power - example
• To find the fraction of the power of the previous square wave (with period T = 7) in the
frequency band [20.5Hz, 100.5Hz], recall its Fourier series:
1
X 10
20
+
sin(4k⇡/7) cos(2k⇡t/7).
7
k⇡
k=1
• Thus, its total mean power is
Pf
:=
✓
20
7
◆2
+
✓
1
X
1 10
k=1
2
k⇡
sin(4k⇡/7)
◆2
.
• Since the frequency band in radians/s is [41⇡, 201⇡] and the signal’s period is 7,
– the lower harmonic index in the band is k1 = d41⇡ · 7/(2⇡)e = d143.5e = 144,
– the upper one is k2 = b201⇡ · 7/(2⇡)c = b703.5c = 703.
• So, the fraction of f ’s mean power in [20.5Hz, 100.5Hz] is
✓
◆2
703
X
1 10
1
Pf
sin(4k⇡/7) .
2 k⇡
k=144
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Parseval’s theorem for signal power - example (cont)
• Note that Pf is easily computed directly as
Z
1
1
100
Pf =
|f (t)|2 dt = 52 · 4 =
⇡ 14.
T T
7
7
• Also note that the f ’s DC power is
✓ Z
◆2
✓ ◆2
1
20
2
f (t)dt
= a0 =
⇡ 8.
T T
7
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The compact trigonometric Fourier series
• Using the trigonometric identity
8↵, , w 2 R, ↵ cos(wt) +
sin(wt) =
q
↵2 +
2
cos(wt
tan
1
( /↵)).
we obtain the compact trigonometric Fourier series (referenced to cosine) of any T -periodic,
Dirichlet f from its trigonometric one:
fˆ(t) = A0 cos(
Ak
=
k
=
q
↵2k +
tan
1
0)
+
1
X
Ak cos(kwo t +
k=1
2
k
(
k ),
t 2 R, where
k /↵k )
with b0 := 0.
• Note that the amplitudes Ak
0, 8k
• Also, if a0 < 0 then A0 = |a0 | and
0.
0
= ⇡.
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Trigonometric and compact trigonometric Fourier series
• Note that f (t) = 5 cos(6t + 3) + 7 sin(13t + 5) is periodic since the ratio, 13/6 or
6/13, of the periods, 2⇡/6 and 2⇡/13, of its components is rational.
• The period of f is the least common integer multiple of its components’ periods, i.e.,
T = 2⇡ = 6(2⇡/6) = 13(2⇡/13) seconds (so f ’s fundamental is wo = 1 radian/s).
• The above f is given in compact trigonometric form and has only 5th and 12th harmonics
(no DC or fundamental components in particular):
f (t) = 5 cos(6t + 3) + 7 cos(13t + 5 ⇡/2)
• We can write the above f as a trigonometric Fourier series,
f (t) = [5 cos(3) cos(6t)
5 sin(3) sin(6t)] + [7 sin(5) cos(13t) + 7 cos(5) sin(13t)]
• The derived trigonometric Fourier series of the previous: (a) even square-wave happened to
be in compact form, and (b) odd triangle-wave can be transformed to compact form again
by using the identity 8v 2 R, sin(v) = cos(v ⇡/2).
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Compact trigonometric Fourier series - square-wave example
• For this (not even) square-wave example with period T = 7, the DC trigonometric Fourier
coefficient is
Z 2.5+T
Z
3
1
1 3
5
20
↵0 =
f (t)dt =
5dt =
t
=
.
T
7 1
7 1
7
2.5
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Compact trigonometric Fourier series - square-wave example (cont)
• The other trigonometric Fourier coefficients are, 8k 1:
Z 2.5+T
Z
2
2 3
↵k =
f (t) cos(kwo t)dt =
5 cos(2k⇡t/7)dt
T
7 1
2.5
3
=
=
k
=
5
5
sin(2k⇡t/7)
=
(sin(6k⇡/7) sin( 2k⇡/7))
k⇡
k⇡
1
10
cos(2k⇡/7) sin(4k⇡/7).
k⇡
Z 2.5+T
Z
2
2 3
f (t) sin(kwo t)dt =
5 sin(2k⇡t/7)dt
T
7 1
2.5
3
=
=
5
5
cos(2k⇡t/7)
=
(cos(6k⇡/7)
k⇡
k⇡
1
10
sin(2k⇡/7) sin(4k⇡/7).
k⇡
cos( 2k⇡/7))
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Compact trigonometric Fourier series - square-wave example (cont)
• So, the compact trigonometric Fourier coefficients are A0 = 20/7, 0 = 0, and 8k 1,
r
q
10
4k⇡
2k⇡
2k⇡
10
4k⇡
2
2
Ak =
↵k + k =
| sin(
)| cos2 (
) + sin2 (
) =
| sin(
)|
k⇡
7
7
7 ◆
k⇡
7
✓
sin(4k⇡/7) sin(2k⇡/7)
tan 1 ( k /↵k ) =
tan 1
k =
sin(4k⇡/7) cos(2k⇡/7)
⇢
tan 1 (tan(2k⇡/7)) =
2k⇡/7 if sin(4k⇡/7) > 0
=
1
tan (tan(2k⇡/7)) + ⇡ =
2k⇡/7 + ⇡ if sin(4k⇡/7) < 0
1
X
20
10
) fˆ(t) =
+
| sin(4k⇡/7)| cos(2k⇡t/7 2k⇡/7 + ⇡ 1{sin(4k⇡/7) < 0})
7
k⇡
=
k=1
1
X 10
20
+
sin(4k⇡/7) cos(2k⇡(t
7
k⇡
1)/7),
k=1
consistent with this square wave being the previous example even square-wave delayed by
one second, where
• the indicator function 1B = 1 if B is true, else 1B = 0.
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The complex-exponential Fourier series
• Recall the Euler-De Moivre identities: 8v 2 R,
ejv = cos(v) + j sin(v)
ejv e jv
j ejv + j e
sin(v) =
=
2j
2
ejv + e jv
cos(v) =
.
2
jv
• We obtain the complex-exponential Fourier series of any T -periodic, Dirichlet signal f by
substituting these identities into its trigonometric one:
1
X
fˆ(t) =
Dk ejkwo t ,
k= 1
where, again, wo = 2⇡/T ,
↵0 = D 0
and 8k
1,
Dk =
↵k
j
k
and D
2
k
=
↵k + j
2
k
.
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The complex-exponential Fourier series (cont)
• By substituting the expressions for the trigonometric Fourier coefficients, we get that
Z
1
8k 2 Z, Dk =
f (t)e jkwo t dt.
T T
0, a k , b k 2 R
• Note that if f is real valued, then 8k
) 8k 2 Z, D
• In turn, 8k
k
= Dk ,
1, Dk ejkwo t = D
k
the complex conjugate of Dk 2 C.
jkwo t ,
e
consistent with f being real valued.
• Similarly, we can transform the complex-exponential to the trigonometric Fourier coefficients: 8k 1,
↵k = D k + D
k
and
k
= j(Dk
D
k ).
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Complex-exponential Fourier series - example
• Define 8t 2 R,
f (t) =
1
X
2(t 5k)
7e
(u(t
5k)
u(t
3
5k)),
k= 1
i.e., a Dirichlet signal with period T = 5 and duty cycle 3 (within each period).
• The complex-exponential Fourier coefficients are, 8k 2 Z,
Z
Z
1
1 T
Dk =
f (t)e jkwo t dt =
f (t)e jkwo t dt
T T
T 0
Z
1 3
1
7
=
7e 2t e jk2⇡t/5 dt =
·
e
5 0
5
(2 + jk2⇡/5)
⇣
⌘
7
=
1 e 3(2+jk2⇡/5) .
10 + jk2⇡
3
(2+jk2⇡/5)t
0
181
Complex-exponential Fourier series - orthogonality
• For any period T > 0, the C-valued signal-set
n
o
ge,k (t) := ejk2⇡t/T = ejkwo t , t 2 R
k2Z
is orthogonal (and a basis for the set of all periodic Dirichlet functions with fundamental
frequency wo or some whole multiple, kwo).
• Orthogonality is easily verified: 8k 6= l 2 Z
Z
Z
< ge,k , ge,l > :=
ge,k (t)ge,l (t)dt =
ejkwote
T
• Moreover, 8k 2 Z,
||ge,k ||2 =< ge,k , ge,k > =
j(k
Z
T
l)wo
ej(k
l)wo t
ge,k (t)ge,k (t)dt =
T
dt
T
1
=
jlwo t
= 0
0
Z
T
|ge,k (t)|2 dt =
Z
1dt = T.
T
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Complex-exponential Fourier series - orthogonality (cont)
• Note that inner product for C-valued signals is linear in the first argument and conjugate
linear in the second argument.
• So, for inner product thus defined, we have again that the complex-exponential Fourier
coefficients for T -periodic, Dirichlet f are, 8k 2 Z,
Dk
=
=
=
< f, ge,k >
||ge,k ||2
R
f (t)ge,k (t)dt
T
1
T
Z
T
f (t)e
jkwo t
dt.
T
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Spectrum (frequency-domain representation) of a signal
• Recall the previous 10-periodic, odd, triangle-wave example signal, f .
• The trigonometric Fourier coefficients of f are ↵k = 0 for k
for k 1.
0 and
k
= ( 1)k 30/(k⇡)
• Thus, the exponential Fourier coefficients are
8
if k = 0
< 0
( 1)k+1 j15/(k⇡) if k > 0
Dk =
: ( 1)k j15/(k⇡)
if k < 0
• We can plot magnitude and phase spectrum of f , respectively:
15
for k 6= 0, and
k⇡
8
< don’t care if k = 0
⇡/2
if k > 0 and odd or k < 0 and even
=
: ⇡/2
if k < 0 and odd or k > 0 and even
|Dk | =
\Dk
versus kwo = k⇡/5 for k 2 Z.
• If w := kwo 2 R, then 8k 6= 0, Dk =
15wo
k⇡wo
=
15wo
⇡w
=
3
.
w
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Spectrum (frequency-domain representation) of a signal (cont)
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Complex-exponential Fourier series - Parseval’s theorem
• We can argue as before that the mean power of T -periodic, Dirichlet f is
Z
Z
Pf = T 1
|f (t)|2 dt = T 1
f (t)f (t)dt
= T
= T
= T
=
1
Z
1
X
k= 1
=
k= 1
|Dk | T
|Dk |
2
1
Dl ejlwo t dt
l= 1
Dk Dl
Dk Dk
2
1
X
Dk ejkwo t
k= 1 l= 1
1
X
k= 1
1
X
T
1
X
T k= 1
1
1
X
X
1
1
T
Z
Z
Z
ejkwot ejlwot dt
T
ejkwot ejkwotdt
by orthogonality
T
1dt
T
2
= |D0 | + 2
1
X
k=1
|Dk |2 .
where the last equality is because f is R-valued ) 8k 2 Z, D
k
= Dk ) |Dk | = |D
k |.
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Complex-exponential Fourier series - Parseval’s theorem - example
P
2(t 5k) (u(t
• Recall that 5-periodic f (t) = 1
5k) u(t 3 5k)), t 2 R,
k= 1 7e
had complex-exponential Fourier series coefficients: 8k 2 Z,
⇣
⌘
7
Dk =
1 e 3(2+jk2⇡/5)
10 + jk2⇡
7
=
(1 e 6 cos(6k⇡/5) + j e 6 sin(6k⇡/5)).
10 + jk2⇡
• By Parseval’s theorem, the mean power of f is
Pf
=
1
X
k= 1
1
X
⇣
7
1
10 + jk2⇡
e
3(2+jk2⇡/5)
⌘
2
q
7
p
(1 e 6 cos(6k⇡/5))2 + (e
2
100
+
(k2⇡)
k= 1
Z
Z
1
1 3
49
2
=
|f (t)| dt =
49e 4t dt =
(1 e 12 ).
T T
5 0
20
=
6
sin(6k⇡/5))2
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Complex-exponential Fourier series - power in a frequency band
• Consider the frequency band band [w1 , w2 ] with w2 > w1
0 radians/s.
• To compute the amount of T -periodic f ’s mean power in this band, recall that we need to
identify which of its harmonics k lie in it:
k1 := dw1 /wo e  k  bw2 /wo c =: k2 ,
where again the fundamental frequency of f is wo = 2⇡/T .
• Using the complex-exponential Fourier series of f , note that we need to account for both
positive (kwo) and negative ( kwo) frequencies for this index set k.
• That is, if {Dk }k2Z are the complex-exponential Fourier coefficients of f and k1 > 0,
then the amount of f ’s power in this band is
k2
X
k=k1
(|D
k|
2
+ |Dk |2 ) =
k1
X
k= k2
|Dk |2 +
k2
X
k=k1
|Dk |2 = 2
k2
X
k=k1
|Dk |2 .
• Note that if k1 = 0 (i.e., w1 = 0), then we should include the term |D0 |2 (representing
f ’s DC power) only once in the expression above.
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Eigenresponse
• Recall that for an asymptotically stable system with transfer function H , the steady-state
response to the (persistent) sinusoidal input f (t) = Aej(wt+ ), t 2 R, with amplitude
A > 0, frequency w, and phase , is the eigenresponse,
y(t) = H(jw)Aej(wt+
)
= |H(jw)|Aej(wt+
+\H(jw))
, t 2 R.
• That is,
– the magnitude response or system gain of the system is |H(jw)|, or 20 log10 |H(jw)|
in decibels (dB):
input magn. 20 log10 A dB ! output magn. 20 log10 A + 20 log10 |H(jw)| dB
– the phase response of the system is \H(jw):
input phase
radians ! output phase
+ \H(jw) radians
• So, the eigenresponse to real-valued f (t) = A cos(wt + ), t 2 R, is real-valued
+ \H(jw)), t 2 R.
y(t) = A|H(jw)| cos(wt +
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Eigenresponse and Fourier series
• Given a complex-exponential Fourier series of an arbitrary (Dirichlet) periodic input,
f (t) =
1
X
Dk ejkwo t
k= 1
the eigenresponse of an asymptotically stable, LTI system H is, by linearity,
y(t) =
1
X
k= 1
Dk H(jkwo )ejkwo t , t 2 R.
• Equivalently, the eigenresponse to f with the compact Fourier-series representation
f (t) =
1
X
Ak cos(kwo t +
k)
k=0
is
y(t) =
1
X
k=0
|H(jkwo )|Ak cos(kwo t +
k
+ \H(jkwo )),
t 2 R.
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Eigenresponse and Fourier series - circuit example
• For this example LTIC circuit, the transfer function is
R
R
=
R + (jwL||R)
R + ((jwL) 1 + R
jw/2 + R/(2L)
jw + R/(2L)
P (jw)
Q(jw)
H(jw) =
=
=
1) 1
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Eigenresponse and Fourier series - circuit example (cont)
• The transfer function can be verified by the system equation Q(D)y = P (D)f and
H = P/Q, where by KCL
0
) 0
) Q(D)y
=
f
y
R
Df
+
0
+ i where L D i = f
y
Dy
Df Dy
0 Dy
f y
+ Di =
+
+
R
R
R
L
R
1
R
:= D y +
y =
Df +
f =: P (D)f.
2L
2
2L
=
Dy
y
R
R
+
0
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Eigenresponse and Fourier series - circuit example (cont)
• So, the magnitude response in dB at frequency w is
p
(w/2)2 + (R/(2L))2
(w/2)2 + (R/(2L))2
20 log10 |H(jw)| = 20 log10 p
= 10 log10
2
2
w2 + (R/(2L))2
w + (R/(2L))
and the phase response is
\H(jw) = tan
1
✓
w/2
R/(2L)
◆
tan
1
✓
w
R/(2L)
◆
.
• Note that for very low frequencies limw!0 |H(jw)| = H(0) = 1 (= 0 dB), and for
high frequencies limw!1 |H(jw)| = 0.5 (⇡ 6 dB).
• So, higher frequencies are attenuated by about 6 dB, while low frequencies are not significantly attenuated, i.e., they pass through the system (because the DC impedance of the
inductor is zero).
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Eigenresponse and Fourier series - circuit example (cont)
P
2(t 5k) (u(t
• Again recall that f (t) = 1
5k) u(t 3
k= 1 7e
complex-exponential Fourier series coefficients,
⇣
⌘
7
Dk =
1 e 3(2+jk2⇡/5) .
10 + jk2⇡
5k)), t 2 R, had
• So, if R = 10k⌦ and L = 0.1H (i.e., R/L = 105 ), then the eigenresponse of this
circuit to this input is, 8t 2 R,
y(t) =
=
1
X
k= 1
1
X
k= 1
H(jk2⇡/5)Dk ejk2⇡t/5
⇣
jk⇡/5 + 105 /2
7
·
1
jk2⇡/5 + 105 /2 10 + jk2⇡
e
3(2+jk2⇡/5)
⌘
ejk2⇡t/5.
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LTIC system magnitude and phase responses
• Recall that a SISO LTIC system relating input signal f to output y by P (D)f = Q(D)y
has (zero state) transfer function H(s) = P (s)/Q(s).
• The roots of Q are called system characteristic values or poles, while the roots of P are
system zeros.
• Recall that for an asymptotically stable system with transfer function H , the steady-state
(eigen-) response to real-valued, persistent, sinusoidal input
f (t) = A cos(wt + ), t 2 R
is real-valued sinusoidal output at the same frequency
y(t) = A|H(jw)| cos(wt + + \H(jw)), t 2 R,
with input parameters: amplitude A > 0, frequency w 0, and phase .
• the magnitude response or system gain of the system is |H(jw)|, or 20 log10 |H(jw)|
in decibels (dB):
input magnitude 20 log10 A dB ! output magnitude 20 log10 A + 20 log10 |H(jw)| dB
• the phase response of the system is \H(jw):
input phase
radians ! output phase
+ \H(jw) radians
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Bode plots of the system
• The magnitude Bode plot of the system depicts the amplitude gain in decibels 20 log10 |H(jw)|
as a function of the log-frequency log10 w for w 0, i.e., a “log-log” plot.
• The phase Bode plot of the system depicts the phase change \H(jw) as a function of
log10 w for w 0, i.e., a semi-log plot.
• In the following discussion of Bode plots, logarithm is assumed base 10.
• Note that DC (w = 0) gain/phase-change is depicted at log w = 1, i.e., the origin
on a Bode plot is a low frequency log w = 0 ) w = 1 radians/s, but not DC.
• A change in frequency by a factor of 10 is called a decade (by a factor of 8 is called an
octave).
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Example of a second-order transfer function
• Recall that for this second-order, series RLC circuit, the transfer function is
H(s) =
LC 1
s2 + RL 1 s + (LC)
1
.
• Consider the overdamped case where there are two real poles r± < 0, and define the
positive frequencies w1 := r+ < w2 := r .
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Example of a second-order transfer function (cont)
• Thus, the transfer function can be written as
H(jw) =
H(0)
=
(1 + jw/w1 )(1 + jw/w2 )
(1
H(0)
jw/r+ )(1
jw/r )
,
where here the DC gain is
H(0) =
LC 1
LC
=
w1 w2
r+ r
1
= 1.
• For example, if R = 1010⌦, L = 10mH and C = 1µF, then w1 = 103 radians/s and
w2 = 105 radians/s.
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“Standard form” transfer function to explain Bode plots
• Consider a transfer function that is a proper, rational polynomial
H(jw) =
P (jw)
, for frequencies w
Q(jw)
0,
where by “proper” we mean that the degree of P  that of Q.
• In polar form (not indicating dependence on real frequency w
|H|ej\H
• Thus,
|H| =
|P |
|Q|
=
0),
|P |ej\P
|Q|ej\Q
and \H = \P
\Q.
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“Standard form” transfer function to explain Bode plots (cont)
• So, for the above second-order RLC circuit, we can write
20 log |H(jw)| = 20 log |H(0)| 20 log |1 + jw/w1 | 20 log |1 + jw/w2 |
\H(jw) = \H(0) \(1 + jw/w1 ) \(1 + jw/w2 )
• More generally, assume that P and Q have only real roots and no common roots, i.e., 8k, l,
wz,k 6= wp,l .
• Factoring P and Q we get for the case of no DC poles or zeroes (i.e., 8k,
0, so that DC gain in decibels is finite):
Qm
(1 + jw/wz,k )
H(jw) = H(0) Qnk=1
(i)
(1
+ jw/wp,k )
k=1
wz,k ,
wp,k 6=
• If H has r  m DC zeroes (i.e., 0 DC gain or 1 in decibels) or r  n DC poles (i.e., 1
DC gain), then for some scalar ⌘ 2 R, respectively:
Q r
(jw)r m
(1 + jw/wz,k )
Qn k=1
H(jw) = ⌘
(ii)
(1
+
jw/wp,k )
k=1
Qm
k=1 (1 + jw/wz,k )
or H(jw) = ⌘
(iii)
Qn r
r
(jw)
k=1 (1 + jw/wp,k )
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“Standard form” transfer function to explain Bode plots (cont)
• Since 8w > 0, |jw| = w and \jw = ⇡/2, the phase and magnitude for the above three
cases (i)-(iii) are, for w 0:
20 log |H(jw)|
=
20 log |H(0)| +
m
X
20 log |1 + jw/wz,k |
k=1
\H(jw)
=
\H(0) +
m
X
=
\(1 + jw/wz,k )
20 log |⌘| + r20 log w +
20 log |1 + jw/wp,k |
n
X
\(1 + jw/wp,k )
m r
X
20 log |1 + jw/wz,k |
\⌘ + r
k=1
20 log |H(jw)|
=
20 log |⌘|
r20 log w +
m
X
n
X
\(1 + jw/wp,k )
20 log |1 + jw/wz,k |
X
⇡
+
\(1 + jw/wz,k )
2
m
=
\⌘
r
k=1
20 log |1 + jw/wp,k |
(ii)
(ii)
k=1
k=1
\H(jw)
n
X
k=1
X
⇡
+
\(1 + jw/wz,k )
2
m r
=
(i)
k=1
k=1
\H(jw)
(i)
k=1
k=1
20 log |H(jw)|
n
X
n r
X
20 log |1 + jw/wp,k |
k=1
n r
X
\(1 + jw/wp,k )
(iii)
k=1
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Magnitude Bode plots with DC zeros or poles
• If there is a DC pole or zero, the magnitude Bode plot in decibels is obviously not finite at
DC.
• Note the terms
20 log |⌘| ± r20 log w
for the above magnitude Bode plots with DC poles or zeros (of degree r).
• As a function of log w, these terms have
– slope ±20r and
– y -intercept 20 log |⌘|, where
– the y -axis is log w = 0, i.e., w = 1 radians/s (not DC where log w =
w = 0).
1 or
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(iii)
Magnitude Bode plot: log |1 + jw/w1| vs. log w with w1 > 0
• So, to construct the magnitude Bode plot, we first need to construct that of its component
terms of the form
20 log |1 + jw/w1 | = 10 log(1 + (w/w1 )2 ) versus log w, w
0,
for an arbitrary frequency w1 > 0.
• Case 1: If w < w1 /10 (i.e., log w <
that |1 + jw/w1| ⇡ 1 and
1 + log w1 ), then 1.1 > |1 + jw/w1 | > 1 so
20 log |1 + jw/w1 | ⇡ 0.
• Case 2: If w = w1 , then
20 log |1 + jw/w1 | = 20 log |1 + j| = 20 log
p
2 ⇡ 3.
• Case 3: If w > 10w1 (i.e., log w > 1 + log w1 ), then |1 + jw/w1 | ⇡ w/w1 and
20 log |1 + jw/w1 | ⇡ 20 log w
20 log w1 ;
as a function of log w, this line is of slope 20dB/decade with x-intercept log w1.
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Magnitude Bode plot: log |1 + jw/w1| vs. log w with w1 > 0 (cont)
• These three cases are summarized in the following figure.
• Note that if w1 is real but negative, the magnitude Bode plot above does not change; the
above analysis applies to w1 = |w1|.
• If w1 is complex-valued,
– the same three cases above apply using its modulus |w1| (instead of w1) leading to
the same asymptotic magnitude Bode plot,
– however at log |w1| in Case 2, the true magnitude Bode plot will not di↵er by 3dB
from (the corner of) the asymptotic plot, cf. the resonant circuit example.
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Magnitude Bode plot: log |1 + jw/w1| vs. log w (cont)
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Magnitude Bode plots - rules of thumb
• So, to sketch the asymptotic magnitude Bode plot of a transfer function H :
1. Mark o↵ on the x-axis the modulus of the non-zero zeroes and poles of H .
2. Each zero, respectively pole, contributes an additional 20 dB/decade slope only for
frequencies larger than (i.e., to the right of) its modulus.
3. If the DC (from
1 = log 0) gain is finite, then mark it o↵ on the y-axis; else
4. If there are r DC zeroes (Case (ii)), then the initial (again from
is 20r dB/decade.
1 = log 0) slope
5. If there are r DC poles (Case (iii)) then the initial slope is -20r dB/decade.
• If the kth repeated zero (resp., pole) is real, then the true Bode plot will be 3k dB larger
(resp., smaller) than the asymptotic Bode plot at the zero (resp., pole).
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Magnitude Bode plot - second-order system example
• Suppose two poles at w1 = 103 radians/s and w2 = 105 radians/s and DC gain
H(0) = 1 (20 log 1 = 0 dB), consider again the example second-order transfer function
H(0)
(1 + jw/w1 )(1 + jw/w2 )
) 20 log |H(jw)| = 0 20 log |1 + jw/103 | 20 log |1 + jw/105 |.
H(jw) =
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Phase Bode plot - \(1 + jw/w1) vs. log w for w1 > 0
• To plot
\(1 + jw/w1 ) = tan
1
(w/w1 ), for w1 > 0,
we again take three cases.
• Case 1: If w < w1 /10 (i.e., log w <
0  tan
• Case 2: If w = w1 , then tan
1
1 + log w1 ), then
(w/w1 ) < tan
1
(w/w1 ) = tan
1
(1/10) ⇡ 0.
1
(1) = ⇡/4 radians (=45 degrees).
• Case 3: If w > 10w1 (i.e., log w > 1 + log w1 ), then
⇡/2
tan
1
(w/w1 ) > tan
1
(10) ⇡ ⇡/2 radians (=90 degrees).
• These three cases are summarized in the following figure.
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Phase Bode plot - \(1 + jw/w1) vs. log w for w1 > 0
• So, the change in phase due to the term
\(1 + jw/w1 ) = tan
1
(w/w1 )
is ⇡/2 radians over a two decade interval [w1/10, 10w1], i.e., a slope of 45 degrees/decade.
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Phase Bode plot - discussion
• Unlike the magnitude response, for the phase response:
– the pole or zero has an e↵ect on the one decade before w1,
– the e↵ect of the pole or zero “saturates” to a total change in phase of ±⇡/2 one
decade after w1, and
– we do not annotate deviations between the true and asymptotic plots (as 3 dB gaps).
• If
1 < H(0) < 0, then the initial phase \H(0) = ±⇡ .
• Note that if w1 is real but negative, the phase change is instead
at log( w1) = log |w1|, i.e., -45 degrees/decade.
⇡/2 over two decades
• Also, if w1 is complex-valued, the phase change remains ⇡/2 over two decades, but the
phase Bode plot departs from the asymptotic and is steeper at log |w1|, cf. the resonant
circuit example.
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Phase Bode plot - second-order system example
• For two poles at w1 = 103 rad/s and w2 = 105 rad/s and DC gain H(0) = 1,
consider again the example second-order transfer function
H(0)
(1 + jw/w1 )(1 + jw/w2 )
) \H(jw) = 0 tan 1 (w/103 ) tan 1 (w/105 )
H(jw) =
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Bode plots - second-order system example 2
H(s) =
10s + 107
10(s + 106 )
10(1 + s/106 )
=
=
s2 + 10100s + 106
(s + 104 )(s + 102 )
(1 + s/104 )(1 + s/102 )
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Bode plots - example with DC zero
H(s) =
10s(s + 100)
10 · 100
) |H(j1)| ⇡
= 10
4
s + 10
104
1
=
20dB.
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Op-amp circuit example - noninverting configuration
• KCL at the negative input terminal (with node voltage f ) gives
0 =
0
f
R1
+
y
f
+ C D(y
RF
f)
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Op-amp circuit example (cont)
• Thus, the system equation and transfer function are:
1
Q(D)y = D y +
y
CRF
=
) H(s) =
=
the last equality when C = 0.01µF, R1
✓
◆
1
1
1
Df +
+
f = P (D)f
C RF
R1
s + C 1 (R1 1 + RF 1 )
P (s)
=
Q(s)
s + (CRF ) 1
s + 1010
,
s + 10
= 100k⌦, RF = 10M⌦.
• So, this system has
– a pole at
10,
– a zero at
1010 ⇡
103 , and
– DC gain H(0) ⇡ 102 = 40dB.
• The zero being larger than the pole will restore the slope to 0 dB/decade in the magnitude
plot, and the phase to 0 degrees in the phase plot.
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Op-amp circuit example - asymptotic Bode plots
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Op-amp circuit example 2 - inverting configuration
• Here, y = vc so KCL at the negative input terminal (with node voltage 0) gives
0 =
0
f
R1
+
y
0
RF
+ C D(y
0) ) H(s) =
(CR1 ) 1
s + (CRF ) 1
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Op-amp circuit example 2 - inverting configuration (cont)
• So when C = 0.01µF, R1 = 100k⌦, and RF = 10M⌦,
H(s) =
103
s + 10
• The DC gain is again 20dB, but the DC phase is ±⇡ , i.e., H(0) =
• At high-frequencies, |H(jw)| ⇠ 103 /w ! 0(=
⇡/2.
100 < 0.
1 dB) and \H(jw) ⇠ \( 1/j) =
• Note that the previous noninverting op-amp at high frequencies, |H(jw)| ⇠ 1(= 0 dB)
and \H(jw) ⇠ \1 = 0.
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Op-amp circuit example 2 - asymptotic Bode plots
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Resonant/tuned circuits - complex-conjugate poles (0  ⇣ < 1)
• Recall the series RLC circuit transfer function with output signal the capacitor voltage,
H(s) =
s2
(LC) 1
+ RL 1 s + (LC)
1
=
s2
wo2
+ 2⇣wo s + wo2
• H has a complex-conjugate pair of poles when damping factor ⇣ < 1 each with magnitude
wo :
q
wo (⇣ ± j 1 ⇣ 2 )
• H is marginally stable (divergent eigenresponse at resonant frequencies ±wo , i.e., poles at
s = ±jwo ) when ⇣ = 0 (R = 0).
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Resonant/tuned circuits - complex-conjugate poles (cont)
• Consider H(jwo ) =
j/(2⇣) as a function of ⇣  1.
• For the magnitude response:
– If ⇣ = 1, then |H(jwo)| = 1/2 = 6 dB consistent with a double pole at
and a DC gain |H(0)| = 1(= 0 dB).
wo
– If ⇣ = 0.5, then |H(jwo)| = 1(= 0 dB), i.e., the 6dB gap disappears and the true
Bode plot meets the asymptotic one at its corner.
– lim⇣!0 |H(jwo)| = 1; this resonance e↵ect is why this system is called a “tuned
circuit”: it dramatically amplifies only in a narrow band around wo.
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Resonant/tuned circuits - complex-conjugate poles (cont)
• The phase
\H(jw) =
tan
1
✓
2⇣
wo /w
w/wo
◆
.
• The change in phase due to the pair of poles (each with magnitude wo ) is ⇡ as frequency
w increases past wo ; note that the denominator is positive for w < wo and negative for
w > wo .
• As ⇣ ! 0, this change in phase becomes more abrupt, not gradual at
over two decades.
90 degrees/decade
• Recall that for L = 0.1H, and C = 1µF, the resonant frequency is wo = (LC)
3162 radians/s.
• Furthermore, if R = 100⌦ then ⇣ = (R/2)
case).
p
0.5
=
C/L = 0.16 < 1 (the underdamped
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Resonant/tuned circuits - complex-conjugate poles (cont)
• The following Bode plots are for the underdamped cases corresponding to R = 100⌦ (⇣ =
0.16) and R = 10⌦ (⇣ = 0.016), as well as the critically damped case (⇣ = 1) with a
repeated (double) pole at wo (note the -6dB gap and more gradual -90 degrees/decade
slope at log wo).
• The overdamped case (⇣ > 1) would have two real poles straddling log wo and Bode plots
similar to the previous two-pole example of a series RLC transfer function.
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Resonant/tuned circuits - magnitude and radian-phase Bode plots
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Resonant/tuned circuits - DC zero
• Recall that if the output of the RLC circuit y is instead the voltage across the resistor, then
H has a DC zero:
H(s) =
RL 1 s
s2 + RL 1 s + (LC)
1
=
s2
2⇣wo s
.
+ 2⇣wo s + wo2
• The following asymptotic Bode plots are for the underdamped case case when wo = 3162
radians/s.
• The DC gain is 0(= 1 dB) and that the gain increases from DC at 20 dB/decade,
consistent with a DC zero.
• The DC phase is ⇡/2 radians since \H(jw) ⇡ \j for very small frequencies w # 0.
• At frequency wo the system response is H(jwo ) = 1, i.e., 0dB and 0 phase.
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Resonant/tuned circuits - DC zero - asymptotic plots for 0  ⇣  1
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Bilateral Fourier transform
• We now study in the frequency domain the ZSR of a LTI system to a not necessarily periodic
input.
• We will employ the bilateral Fourier transform spanning positive and negative frequencies
(as the complex-exponential Fourier series for periodic signals).
• Outline of topics to be covered:
– Derivation of bilateral Fourier transform (FT) from Fourier series of periodically extended pulse of finite support.
– Basic FT pairs and properties.
– Parseval’s theorem for signal energy.
– Ideal filters and causality issues.
– Modulation and demodulation
– Poisson’s identity and Nyquist sampling.
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Periodic extensions of aperiodic signals of finite support
• Recall periodic extensions of pulses x of finite support.
• In this example, the pulse x has support of duration 4 and for T
extension of x depicted in this figure is
xT (t) :=
1
X
x(t
kT ) =
k= 1
1
X
(
kT
k= 1
4, the T -periodic
x)(t), t 2 R,
i.e., superposition of time-shifted pulses x by all integer (n) multiples of T .
• Consider the Fourier series of xT : 8t 2 R,
xT (t) =
1
X
k= 1
Dk,T ejk2⇡t/T , where Dk,T =
1
T
Z
T /2
x(⌧ )e
jk2⇡⌧ /T
d⌧.
T /2
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From Fourier series to Fourier transform - periodic extension (cont)
• Define the function,
X(k2⇡/T ) :=
• Thus, for large T , Dk,T ⇡ T
Z
1
x(⌧ )e
jk2⇡⌧ /T
d⌧.
1
1 X(k2⇡/T ).
• Also, 8t 2 R, x(t) = limT !1 xT (t).
• So, substituting this approximation of the Fourier coefficients D into the complex-exponential
Fourier series of xT gives, by Riemann integration, 8t 2 R,
x(t) =
lim
T !1
=
where w = 2⇡/T so that T
1
Z
1
1
X
X(k2⇡/T )ejk2⇡t/T T
1
k= 1
X(w)ejwt dw/(2⇡)
w= 1
converges to the di↵erential dw/(2⇡) as T ! 1.
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Fourier transform and inverse Fourier transform
• In summary, we defined the Fourier transform of time-domain signal x : R ! R as
Z 1
X(w) :=
x(t)e jwt dt 8 frequencies w 2 R.
1
• Using the complex-exponential Fourier series of the periodic extension of x, we heuristically
derived the inverse Fourier transform of the frequency-domain signal X : R ! R as
Z 1
1
x(t) :=
X(w)ejwt dw 8 time t 2 R.
2⇡ 1
• We write X = F x (or F {x}) and x = F
mation.
1X ,
again emphasizing functional transfor-
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Fourier transform properties - even magnitude and odd phase
• If x : R ! R, i.e., x is a real-valued signal, then X = F x has
– even magnitude, i.e., |X( w)| = |X(w)| 8w 2 R, and
– odd phase, i.e., \X(w) =
\X( w) 8w 2 R,
• That is, X( w) = X(w) 8w 2 R:
Z 1
X( w) =
x(t)e j( w)t dt
1
Z 1
=
x(t)ejwt dt
=
Z
1
1
x(t)e
jwt dt
1
(since x is R-valued)
= X(w) 8w 2 R.
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Fourier transform properties - linearity & di↵erentiation
• The Fourier transform is a linear mapping simply because of the linearity of integration.
• For all di↵erentiable (Dirichlet) signals x, the Fourier transform of a time derivative is
8w 2 R, (F D x)(w) = jw(F x)(w).
• Recalling D := d , this property is proved by simple integration by parts: 8w 2 R,
dt
Z 1
(F D x)(w) =
(D x)(t)e jwt dt
1
Z 1
=
e jwtdx(t)
1
Z 1
1
= x(t)e jwt 1
x(t)de jwt
1
Z 1
= 0 + jw
x(t)e jwt dt (⇤)
1
= jwX(w),
where (*) is by the (weak) Dirichlet condition of absolute integrability which implies that
limt!±1 x(t) = 0.
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Fourier transforms applied to SISO systems given by ODE
• We can use the Fourier transform (FT) to analyze the zero-state response of the system
P (D)f
= Q(D)y.
• Let the FT of the input and output are, respectively, F = F f and Y = F y .
• By the linearity and di↵erentiation properties of the FT: 8w 2 R,
P (jw)F (w) = Q(jw)Y (w).
• Thus, the transfer function,
H(jw) =
P (jw)
Y (w)
=
, 8w 2 R.
Q(jw)
F (w)
• Change of notation: In the following discussion of Fourier transforms, we will use
“H(w)” instead of “H(jw)” for the transfer function, i.e.,
H(w) =
P (jw)
Y (w)
=
, 8w 2 R.
Q(jw)
F (w)
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Fourier transform of the unit impulse, , and H = F h
• By the ideal sampling property, the Fourier transform of the impulse is, 8w 2 R,
Z 1
(F )(w) =
(t)e jwt dt = e jwt t=0 = 1.
1
• So, F ⌘ 1.
• Note that by inverse Fourier transform, this means
Z 1
ejwtdw 8t 2 R!
2⇡ (t) =
1
• Thus, the LTI SISO system with input f =
in the frequency domain
Y = HF
=
has ZSR y = h (impulse response), so that
= H1 = H,
and so,
F h = H,
i.e., the Fourier transform of the impulse response is the transfer function.
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Transfer function is Fourier transform of (ZS) impulse response
• In summary, given the input f , to solve for the ZSR y the LTI SISO system described as
P (D)f = Q(D)y (or as y = f ⇤ h), we will employ the Fourier transform:
y
where H = F h and F = F f .
= F
1
{HF },
• From this, we can conclude the convolution property:
F {f ⇤ h} = F {f }F {h}.
• Sometimes the inverse Fourier transform is easily computed by integration.
• Otherwise, certain Fourier-transform properties and known Fourier-transform pairs can simplify computing F 1.
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Direct proof of the convolution property, F {f ⇤ h} = F {f }F {h}.
F {f ⇤ h}(w) =
=
=
=
=
Z
Z
Z
Z
Z
1
(f ⇤ h)(t)e jwt dt
1
1 Z 1
f (⌧ )h(t ⌧ )d⌧ e jwt dt
1
1
1 Z 1
f (⌧ )h(t ⌧ )e jw(t ⌧ ) jw⌧ d⌧ dt
1
1
1 Z 1
0
f (⌧ )h(t0 )e jwt jw⌧ d⌧ dt0 (t0 = t
1
1
Z 1
1
0
f (⌧ )e jw⌧ d⌧
h(t0 )e jwt dt0
1
⌧)
1
= F {f }(w)F {h}(w)
= F (w)H(w) 8w 2 R.
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Fourier transform pairs - exponential functions of time
• Again, let X = F x.
• If the causal signal x(t) = eso t u(t), t 2 R, with so 2 C and Re{so } < 0, then
Z 1
Z 1
1
X(w) =
x(t)e jwt dt =
esote jwtdt =
.
jw
so
1
0
• Note that if so 2 R (so that x is R-valued), then |X(w)| = (w2 + s2o ) 0.5 is even,
\X(w) = tan 1 (w/so ) is odd, and, again, X is the frequency domain representation
of an aperiodic signal x.
• Also note that the “anti-causal” signal { eso t u( t), t 2 R} has the same Fourier transform expression but requires Re{so} > 0 for convergence.
• For example, 8w 2 R,
F {e
3t
u(t)}(w) =
1
jw + 3
and F {e2tu( t)}(w) =
• Neither the causal signal e3t u(t) nor the anti-causal signal e
they do not have a finite integral.
2t u(
1
2
jw
.
t) are Dirichlet because
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Fourier analysis of RL circuit with exponential input
• Recall that for this circuit, the transfer function is
H(w) =
jw/2 + R/(2L)
1
R/(4L)
=
+
8w 2 R.
jw + R/(2L)
2
jw + R/(2L)
• So, by linearity and the Fourier-transform pairs we have already discussed, the impulse
response of this system is
h(t) =
1
R
(t) +
e
2
4L
Rt/(2L)
u(t) 8t 2 R.
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Inverse Fourier transforms by known pairs
• Again, note that if we try to use the integral definition of F 1 on the frequency domain
signal c (constant) or b/(jw+a) we do not get an integrand corresponding to an indefinite
integral, i.e., we get
Z 1
Z 1
1
1
b
cejwt dw or
ejwtdw.
2⇡ 1
2⇡ 1 jw + a
• We found the inverse Fourier transforms by remembering the easily computed Fourier transforms F {c } for constant c 2 R and F {e atu(t)} for constant a > 0.
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Fourier analysis of RL circuit with exponential input (cont)
• Let R = 2⌦ and L = 1H so that H(w) = (jw/2 + 1)/(jw + 1).
• So, if the input of this circuit is f (t) = 4e
the ZSR y is, 8w 2 R,
Y (w) = H(w)F (w) =
=
3t u(t),
t 2 R, then the Fourier transform of
jw/2 + 1
4
2jw + 4
·
=
jw + 1 jw + 3
(jw + 1)(jw + 3)
1
1
+
jw + 1
jw + 3
where the last equality is a partial-fraction expansion (PFE) of Y .
• Thus, the ZSR y(t) = (e
t
+e
3t )u(t).
• Note that y is the sum of a system characteristic mode e
H|jw= 3 f (t) = (1/4)4e 3t = e 3t .
t
and the forced response,
• We will subsequently study in detail the PFE of rational polynomials in s = jw for purposes
of computing the inverse Laplace transform.
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Fourier transform properties - time shifts
• The time-shift property is: 8to , w 2 R,
Z 1
(F { to x})(w) =
( to x)(t)e jwt dt
1
Z 1
=
x(t to )e jwt dt
1
Z 1
0
=
x(t0 )e jw(t +to ) dt0 (t0 := t
1
Z 1
0
jwto
= e
x(t0 )e jwt dt0
= e
jwto
to )
1
X(w)
• So a time shift by to corresponds to multiplication in the frequency domain by a complexexponential sinusoid with period 2⇡/to.
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Fourier transform properties - frequency shifts
• The frequency-shift (modulation) property is: 8wo , t 2 R,
Z 1
1
(F 1 wo X)(t) =
( wo X)(w)ejwt dw
2⇡ 1
Z 1
1
=
X(w wo )ejwt dw
2⇡ 1
Z 1
1
0
=
X(w0 )ejt(w +wo ) dw0
2⇡ 1
Z 1
1
0
= ejtwo
X(w0 )ejtw dw0
2⇡ 1
= ejtwo x(t).
(w0 := w
wo )
• A baseband signal x multiplied by a carrier signal at frequency wo radians/s leads to a
frequency shifted (modulated) signal ejtwo x(t).
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Fourier transform properties - time and frequency scaling
• If X = F x and scalar ↵ 6= 0, the time/frequency scaling property is easily shown by
change-of-variables:
⇣w ⌘
1
(F S1/↵ x)(w) := (F {x(↵t)})(w) =
X
.
|↵|
↵
• For example,
jw
+2
j w6
where x(t) = e
F
1
!
2t u(t)
D 6x(6t) = D 6e
u(t) =
12t
72e
u(t) + 6 (t)
and u(6t) = u(t); equivalently,
jw
jw
=6
+2
jw + 12
j w6
jw
1
=
36
·
·
j w6 + 2
6 j w6 + 2
F
1
F
1
!
j w6
!
12t
6De
72
+6
jw + 12
F
1
!
u(t) =
72e
12t
u(t) + 6 (t)
2t
36(S1/6 D x)(t) = 36 S1/6 ( 2e
=
jw
=
w
j6 +2
12t
72e
72e
12t
12t
u(6t) + 36 (6t) =
u(t) + (t))
72e
12t
u(t) + 6 (t)
u(t) + 6 (t)
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Fourier transform properties - composition order
• The order of composed frequency-shift and time-shift operations matters, i.e., these operations are not commutative.
• Given X = F x and scalars to , wo .
• If frequency shift (modulation) before time shift:
x(t
to )ejwo (t
to )
=
to {x(t)e
jwo t
F
} ! X(w
w o )e
jwto
• On the other hand, if time shift before frequency shift:
x(t
to )ejwo t = (
to x)(t)e
jwo t
F
! X(w
w o )e
j(w wo )to
• Similarly, other operation pairs (e.g., involving di↵erentiation, time/frequency scaling) are
not commutative.
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Fourier transform pairs - complex-exponentials and impulses
• Note that by the time shift property, the Fourier transform of an impulse delayed by to is a
complex-exponential sinusoid in the frequency domain with period 2⇡/to:
F{
to
}(w) = e
jto w
w 2 R.
• Similarly, an impulse located at wo in the frequency domain corresponds to a sinusoid in
the time-domain with frequency wo:
F
1
{
wo
}(t) =
1 jwo t
e
t 2 R.
2⇡
• The presence of impulses in the frequency domain indicates discrete frequency components,
e.g., of periodic signals.
• Again note that a complex exponential (over a time or frequency domain) cannot be directly
integrated.
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Fourier transform pairs - sine and cosine
• By direct integration we can verify that
F 1 {⇡
F 1 { j⇡
wo
wo
+⇡
+ j⇡
wo
wo
}(t) = cos(wo t) t 2 R
}(t) = sin(wo t) t 2 R
• By the Euler-De Moivre identity, the aperiodic, one-sided functions (real ↵ > 0)
e
e
↵t
cos(wo t)u(t) =
↵t
sin(wo t)u(t) =
1 ( ↵+jwo )t
e
u(t) +
2
1 ( ↵+jwo )t
e
u(t)
2j
1 ( ↵
e
2
1 (
e
2j
jwo )t
u(t),
↵ jwo )t
u(t)
• Recalling the FT for such one-sided exponential functions, we can show by linearity that for
↵ > 0, w 2 R,
1
1
1
1
↵ + jw
·
+ ·
=
2 ↵ + j(w wo )
2 ↵ + j(w + wo )
(↵ + jw)2 + wo2
1
1
1
1
wo
F {e ↵t sin(wo t)}(w) =
·
·
=
2j ↵ + j(w wo ) 2j ↵ + j(w + wo )
(↵ + jw)2 + wo2
F {e
↵t
cos(wo t)}(w) =
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Fourier transform of f (t) = cos(wot), t 2 R
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Fourier transform of tk esotu(t) is k!/(jw
so)k+1
• Recall that SISO LTI systems have characteristic modes of the form tk eso t u(t) where k
is an integer and so 2 C a characteristic value of the system.
0
• As we are considering only asymptotically stable systems in our discussion of Fourier transforms, we can assume Re{so} < 0.
• To evaluate the Fourier transform of such a characteristic mode, we need to integrate by
parts: for k > 0 and Re{so} < 0:
Z 1
Z 1
1
F {tk eso t u(t)}(w) =
tk eso t e jwt dt =
tk de(so jw)t
so jw 0
0
Z 1
1
1
1
k (so jw)t
=
t e
e(so jw)tdtk
so jw
so jw 0
0
Z 1
k
=
tk 1 e(so jw)t dt
(jw so ) 0
k
=
F {tk 1 eso t u(t)}(w) w 2 R.
jw so
• So by induction, we can show that for integers k > 0 and Re{so } < 0:
F {tk eso t u(t)}(w) =
(jw
k!
so )k+1
8w 2 R.
248
Fourier transform pairs - sinc and rectangle pulse
• Consider the even rectangular pulse of height 1 and width 2wo in the frequency domain:
R(w) = u(w + wo )
• Its inverse Fourier transform is
r(t) = F
1
{R}(w) =
=
=
=
=
u(w
wo )
Z 1
1
R(w)ejwt dw
2⇡ 1
Z wo
1
ejwtdw
2⇡ wo
1 1 jwo t
· (e
e jwot)
2⇡ jt
wo ejwo t e jwo t
·
⇡
2jwo t
wo
· sinc(wo t).
⇡
where
sinc(v) := sin(v)/v.
• Note that by L’Hopital’s rule, limv!0 sinc(v) = 1.
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Fourier transform pairs - sinc and rectangle pulse (cont)
• The rectangular pulse has finite support but its FT pair, the sinc, has infinite support.
• Note that as the frequency wo increases,
– the rectangle pulse R in the frequency domain is broader, but
– the sinc function in the time domain is narrower; indeed, the width between the zero
crossings of the central lobe is ⇡/wo ( ⇡/wo) = 2⇡/wo.
• The two plotted sinc functions have wo 2 {1, 5}.
• In the limit as wo ! 1 this approaches the
$ 1 Fourier-transform pair.
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Duality
• We have noted several “dual” FT properties and pairs, e.g., time and frequency shifts,
sinusoids and impulses.
• Duality is a consequence of the similarity of the Fourier and inverse Fourier transforms.
• If X = F x, i.e.,
X(w) =
Z
1
1
x(t)e
jwt
dt 8w 2 R,
then we can consider the Fourier transform (not inverse FT) of X :
Z 1
Z 1
1
(F X)(w) =
X(t)e jwt dt = 2⇡ ·
X(t)ej( w)t dt
2⇡
1
1
= 2⇡x( w) 8w 2 R,
noting how we have used the definition of inverse FT for the last equality and have switched
the roles of frequency (w) and time (t).
• Duality of Fourier transforms can be stated more compactly as:
recalling S
1
X = F x , F X = 2⇡ S 1 x,
is the domain-inversion (reflection in y-axis) operation.
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Duality - example
• For example, recall that
wo
r(t) =
sinc(wot), t 2 R !F
⇡
R(w) = u(w
wo )
u(w + wo ), w 2 R.
• By duality, we expect (replacing parameter wo with to now representing a time value):
to
2⇡ sinc( wto ) = 2to sinc(wto ), w 2 R,
⇡
where we note that the sinc function is also even.
u(t
to )
u(t + to ), t 2 R !F
• Exercise: Verify this by directly integrating to compute F R.
R1
• Exercise: Suppose the “procedure” or “function” FT(⇣, v) returns
⇣(a)e jav da
1
when passed a signal ⇣ : R ! R and real number v . Show how to use this procedure to
compute (i) (F x)(w) and (ii) (F 1X)(t).
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Signal energy by Parseval’s theorem
• Consider an energy signal x : R ! R with Fourier transform X = F x,
Z 1
Z 1
Ex =
|x(t)|2 dt =
x(t)x(t)dt.
1
1
• Substituting x = F 1 X , using Fubini’s theorem, and then F , we get Parseval’s theorem:
Z 1
Z 1
Z 1
1
1
X(v)ejvt dv
X(w)ejwt dw dt
Ex =
2⇡ 1
1 2⇡
1
Z 1
Z 1
Z 1
1
1
jvt
=
X(v)e dv
X(w)e jwt dwdt
2⇡ 1
1 2⇡
1
✓Z 1 jvt
◆
Z 1
Z 1
1
e
jwt
=
X(w)
X(v)
e
dt dv dw
2⇡ 1
1
1 2⇡
Z 1
Z 1
Z 1
1
1
=
X(w)
X(v) (w v)dv dw =
X(w)X(w)dw
2⇡ 1
2⇡ 1
1
Z 1
1
=
|X(w)|2 dw
2⇡ 1
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Parseval’s theorem - example
• Recall that if X(w) = u(w + wo ) u(w
then its inverse Fourier transform is x(t) =
wo ), w 2 R (i.e., an
wo
sinc(wot), t 2 R.
⇡
even rectangle pulse),
• The energy of this signal in the time domain cannot be directly computed, but by using
Parseval’s theorem,
Z 1
1
Ex =
|X(w)|2 dw
2⇡ 1
Z wo
1
=
dw
2⇡ wo
wo
=
⇡
• Equating to
R1
1
|x(t)|2 dt and substituting ⌧ = wo t, we get an interesting result:
Z 1
sinc2(⌧ )d⌧ = ⇡.
1
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Parseval’s theorem - energy in a frequency band
• Suppose we want to find the energy of the response y of an ideal band-pass filter to a given
input signal f , with filter pass-band [w1, w2] such that 0 < w1 < w2 radians/s.
• Y = F y satisfies
|Y (w)| =
• So, by Parseval’s theorem,
Ey
=
=
⇢
|F (w)| if w 2 [w1 , w2 ]
0
else
Z w1
Z w2
1
1
2
|F (w)| dw +
|F (w)|2 dw
2⇡ w2
2⇡ w1
Z
1 w2
|F (w)|2 dw,
⇡ w1
where the last equality is because f is R-valued (and hence F = F f is an even function).
• Exercise: Find the transfer function H and impulse response h = F
filter.
1H
of this band-pass
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Ideal Amplifier
• An ideal amplifier has impulse response h = A
function H(w) = Aej⌧ w , w 2 R.
⌧
for delay ⌧ > 0, and hence transfer
• Thus, the ZSR of an ideal amplifier to any (Dirichlet) input signal f is, for some ⌧ 2 R,
y(t) = (f ⇤ h)(t) = Af (t
t 2 R,
⌧ ),
i.e., an amplified (by A) and undistorted (though potentially delayed by ⌧ ) version of the
input signal; equivalently,
Y (w) = F (w)H(w) = AF (w)e
jw⌧
,
w 2 R.
• Such a constant response (A = |H(w)| 8w 2 R) across the whole frequency band is not
feasible.
• Typically, only an approximately distortionless amplification is possible over a limited frequency band.
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Ideal Filters
• Similarly, we may be interested in “filtering out” all frequencies w (H(w) = 0) of an input
signal f except for a certain frequency band, i.e., except for the pass-band of frequencies
w where H(w) = 1 (or H(w) = e jw⌧ for delay ⌧ ).
• Thus, the ZSR
Y (w) = F (w)H(w) =
⇢
F (w) if w is in the pass-band
0
else
• If the pass-band is of the form:
– [0, w0] for w0 > 0 (more precisely [ w0, w0] in the Fourier domain), then the filter
is a low-pass filter, and w0 (or sometimes 2w0) is called the bandwidth of the filter;
S
– ( 1, w0] [w0, 1) for w0 > 0, then the filter is a high-pass filter;
S
– [ w1, w0] [w0, w1] for w1 > w0 > 0, then the filter is a band-pass filter.
• Feasibility also requires causality, equivalently the Paley-Wiener criterion in the frequency
domain which precludes H(w) = 0 on whole intervals of the real line.
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Low-pass, band-pass and high-pass filters
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Feasible low-pass filter (LPF) or amplifier
• We therefore see that an ideal low-pass filter (without delay) will have transfer function in
the form of an even rectangle pulse,
H(w) = A(u(w + w0 )
w0 )) 8w 2 R,
u(w
where A > 1 if it’s an ideal amplifier over the “low” frequency band [ w0, w0].
• Thus, the impulse response is
h(t) =
Awo
sinc(wot), t 2 R.
⇡
• Note that a non-causal impulse response h that is not anti-causal can be delayed to a causal
impulse response ⌧ h, where h(t) = 0 8t < ⌧ ;
• thus, the ZSR to f of the causal LTI system is f ⇤
⌧h
=
⌧ (f
⇤ h).
• But the impulse response of an ideal LPF is anti-causal (meaning that the support of h
extends to 1).
• Hence this ideal filter is anticausal and infeasible.
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Approximating an ideal LPF H with a causal LPF H̃
• Consider again the ideal LPF H(w) = u(w + wo )
u(w
• Suppose the anticausal (sinc) impulse response h = F
then truncated to yield a causal impulse response
h̃ = u
⌧h
) H̃ = (2⇡)
1
U ⇤ {H(w)e
1H
jw⌧
w o ), w 2 R .
is delayed by ⌧ > ⇡/wo > 0
} where U = F u.
• The idea is that ⌧ > 0 is sufficiently large so that the square error
Z 1
Z ⌧
|h(t ⌧ ) h̃(t)|2 dt =
|h(t)|2 dt
is small.
1
1
• If this is the case, the transfer function of the causal LPF H̃ = F h̃ will be a slightly distorted
version of H (both in the pass and non-pass bands) with a roughly linear phase of negative
slope due to the time delay.
• Note that just the delayed version of the impulse response ⌧ h still corresponds to an
ideal, anticausal LPF with transfer function H(w)e j⌧ w , w 2 R.
• In a subsequent course on signal processing, you will see how to feasibly trade-o↵ lower
distortion for complexity (e.g., degree n of the causal filter’s characteristic polynomial)
through the use of special polynomials, e.g., from the Butterworth or Chebyshev families.
260
Example causal LPF h̃ = u
⌧h
with wo = 2 and delay ⌧ = 20
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Modulation of a baseband signal
• A signal f = F 1 F is said to be a baseband signal if it is band limited, i.e., if there is
frequency w0 > 0 such that F (w) = 0 8|w| > w0.
• For example, w0 ⇡ 20kHz = 20000 · 2⇡ radians/s for an audio signal discernible to the
human ear.
• Suppose the audio signal is to be (wireless) transmitted over the air in a frequency band
assigned to a radio station,
[
[ wc w0 ,
wc + w0 ] [wc w0 , wc + w0 ]
where typically wc
w0 .
• So, the audio signal needs to be frequency shifted to the carrier frequency wc prior to
transmission over the air.
• Again, frequency shifting is accomplished by multiplication by a sinusoid in the time domain,
f (t) cos(wc t), t 2 R.
which is the modulated signal traveling over the air.
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Modulation and demodulation of a baseband signal - example
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Demodulation of a baseband signal
• The radio receiver needs to demodulate the signal f (t) cos(wc t) after it has been received
over the air.
• Demodulation is accomplished by first applying the same frequency shift as modulation,
f (t) cos(wc t) · cos(wc t) = f (t) cos2 (wc t)
✓
◆
1
1
= f (t)
+ cos(2wc t) ,
2
2
t 2 R.
and then applying a low-pass filter to eliminate the high-frequency components (at ±2wc)
to recover the baseband signal,
1
f.
2
• In later courses on communications, you will learn of di↵erent modulation techniques, including those that involve first sampling and quantizing the continuous-time baseband signal.
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Sampling - preliminaries
• Sampling a continuous-time, band-limited signal f is a first step to processing it digitally
(i.e., applying a digital filter to its sample sequence).
• Recall the ideal sampling property of the unit impulse: if f is continuous at ⌧ then
Z 1
(t ⌧ )f (t)dt = f (⌧ ),
1
i.e., the sample at time ⌧ , f (⌧ ), is extracted from the signal f by multiplying the signal
with an impulse at ⌧ and integrating.
• Suppose the continuous-time signal f is being sampled every Ts = 2⇡/ws seconds to
obtain the sequence {f (kTs)}k2Z.
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Sampling - picket-fence function
• To capture such periodic sampling at frequency ws , we multiply f by the “picket-fence”
function,
pTs (t) =
1
X
(t
k= 1
to obtain
pTs (t)f (t) =
1
X
k= 1
i.e., the impulse (t
f (t) (t
kTs ) =
kTs ) 8t 2 R,
1
X
f (kTs ) (t
k= 1
kTs ) 8t 2 R,
kTs ) is weighted by the f -sample f (kTs ).
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The picket-fence function
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Impact of sampling in the frequency domain - Poisson’s identity
• To visualize the impact of sampling on the signal, we now derive F {pTs f }.
• Obviously, pTs is Ts -periodic.
• The complex-exponential Fourier series of pTs has coefficients Dk = Ts
leads to Poisson’s identity:
pTs (t) :=
1
X
k= 1
(t
kTs ) =
1
1 X jkws t
e
,
Ts
k= 1
1
8k 2 Z; this
8t 2 R!
• Thus, if F = F f then by the linearity and frequency-shift properties,
F {pTs f }(w) =
1
1 X
F f (t)ejkws t (w) =
Ts
k= 1
1
1 X
F (w
Ts
k= 1
kws ),
8w 2 R.
• So, sampling has the e↵ect of superposing copies of F frequency-shifted by all integer
multiples of the sampling frequency, ws = 2⇡/Ts.
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Nyquist sampling rate for band-limited signals
• Suppose the signal f is band-limited, i.e., there is a finite wb > 0 such that F (w) = 0
8|w| > wb .
• The frequency-shifted copies of F that constitute F pTs f will not overlap if
ws wb > w b
) ws > 2wb .
• If this is the case, note that we can recover the original signal f from pTs f by applying a
low-pass filter (with filter bandwidth between wb and wc wb).
• Thus, if ws > 2wb , then the complete “information” in f is preserved in its samplesequence pTs f , equivalently (in terms of information) the samples {f (kTs)}1
k= 1 .
• 2wb is called the Nyquist sampling frequency of f .
• Sampling at less than the Nyquist sampling frequency may prevent recovery of f from pTs f ,
e.g., aliasing.
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Sampling band-limited signals - example and exercise
• Consider the signal x(t) = Asinc(wo t) with
• X(w) := (F x)(w) = (⇡A/wo )(u(w + wo )
u(w
wo )) (check F
1X
= x).
• So, the Nyquist simpling frequency of x is 2wo radians/s, i.e., the Nyquist sampling period
is 2⇡/(2wo) = ⇡/wo seconds.
• Consider again the ideal sampling (picket fence) function p(t) =
sampling period T > 0.
• Recall that (F xp)(w) =
P1
r= 1 X(w
P1
k= 1
(t
kT ) for
k2⇡/T )/T .
• Plot F xp and show that, in particular, if T = ⇡/wo (the Nyquist sampling period) then
(F xp)(w) = (⇡A/wo )/(⇡/wo ) = A (a constant).
• So by inverse Fourier transform, x(t)p(t) = A (t).
• Does this make sense in light of x(t)p(t) =
P1
k= 1 x(kT )
(t
kT )?
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Nyquist sampling (ws > 2wb) and aliasing
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Unilateral Laplace transform for transient response - Outline
• Laplace transform definition and region of convergence.
• Basic LT pairs and properties.
• Inverse LT of rational polynomials by partial fraction expansion.
• Total response of SISO LTIC systems Q(D)y = P (D)f .
• The steady-state eigenresponse revisited.
• Complex-frequency(s)-domain equivalent circuits.
• System composition and canonical realizations.
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Unilateral Laplace transform
• We define the unilateral Laplace transform (LT) of time-domain signal x : [0, 1) ! C as
X = Lx = L{x} where
Z 1
X(s) :=
x(t)e st dt
0
and:
– the LT is defined for complex frequencies s 2 C for which this integral converges,
– 0
is just before the chosen origin of time, 0. .
• The lower limit of integration is taken as “0 ” so that:
– the LT of the impulse is well defined, i.e., we’re not “splitting hairs”,
1 = (L )(s), 8s 2 C, and
– the LT can accommodate “initial conditions.”
• Input functions of the form f = gu (u the unit step), will have zero initial conditions by
definition, i.e.,
(Dk f )(0 ) = 0, for all integers k
0.
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Region of convergence for Laplace transform
• Suppose that the signal x : [0, 1) ! C is bounded by an exponential:
9 A > 0, ↵ 2 R such that 8t 0 , |x(t)|  Ae↵t ,
i.e., 8t 0 ,
Ae↵t  |x(t)|  Ae↵t .
• If x is bounded by an exponential Ae↵t , t 0 , then the region of convergence (RoC) of
Lx contains a portion of C that is an open right-half plane.
• To see why, recall absolute integrability ) integrability; and:
Z 1
Z 1
|x(t)e st |dt =
|x(t)|e Re{s}t dt
0
Z0 1

Ae(↵ Re{s})t dt by exp. bdd.
0
=
• Thus, if ↵
• Note that ↵
Ae(↵ Re{s})t
↵ Re{s}
1
0
=
A
Re{s}
↵
< 1.
Re{s} < 0 then X(s) exists and |X(s)| < 1 by the triangle inequality.
Re{s} < 0 , Re{s} > ↵ 2 R, i.e., s is an element of an open RHP ⇢ C.
274
Laplace transforms of exponential functions of time and their RoCs
• Consider the exponential signal x(t) = eso t u(t), t 2 R, with a particular so 2 C (which
is bounded by 1eRe{so}t, i.e., ↵ = Re{so} ).
• The Laplace transform of the above exponential signal is
Z 1
Z 1
X(s) =
x(t)e st dt =
e (s so)tdt =
0
=
0
1
s
so
1
s
so
e
(s so )t
1
0
, for Re{s}>Re{so }.
• Note: x(t) has no (t) term, so it doesn’t matter whether we integrate from 0
or 0.
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Laplace transforms pairs - (Lu)(s) = 1/s when Re{s} > 0.
• The Laplace transform of the unit step is just a special case of that of an exponential
function.
(Lu)(s) =
• If x(t) = 1 for t
0
1
s
when Re{s} > 0.
then x 6= u since u(0 ) = 0 but
(Lx)(s) =
1
s
when Re{s} > 0,
i.e., Lx = Lu.
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Ambiguity of inverse Laplace transform, L
• For Dirichlet signals x that do not have an impulse
component at the origin 0,
= Lx = Lxu
X
• For example, if x(t) = eso t for t
Re{s}>Re{so}, but note that
1
0 , then (Lx)(s) = (Lxu)(s) = 1/(s
so ) for
x(0 ) = 1 while (xu)(0 ) = 0,
i.e., x and xu di↵er in their initial values.
• So, there is some ambiguity when inverting L
signal.
1X
to find the corresponding time-domain
• This ambiguity is important as, generally, we want
– zero initial conditions when inverting s-domain terms X of the ZSR (i.e., L
xu), and
1X
– possibly non-zero initial conditions when inverting s-domain terms of the ZIR (i.e., L
x without the step u factor so that (Dk x)(0 ) 6= 0 is possible).
• As a default, we will use the form L
1X
=
1X
= xu in the following.
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Laplace transform properties - linearity
• Just as the Fourier transform, the Laplace transform is linear.
• With the Laplace transform, we must be sure to stipulate the RoC for the linear combination
as the intersection (smallest open RHP) of those of the component signals.
• For example, if x(t) = (↵0 es0 t + ↵1 es1 t )u(t) for ↵0 , ↵1 , s0 , s1 2 C and t
X = Lx is
X(s) = ↵0 ·
1
s
s0
+ ↵1 ·
1
s
s1
0 , then
, when Re{s} > max{Re{s0 }, Re{s1 }}.
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=
Laplace transform properties - time shifting
• Again, as the Fourier transform, the Laplace transform has a time-shift property.
• If we do not delay the negative-time support of a signal x into positive time, or there is no
negative-time support of x (x is causal), then we can simply write
(L
⌧ x)(s)
= e
⌧s
(Lx)(s) if ⌧ > 0,
where the RoC is unchanged by time shifting.
• For example, if x(t) = eso t u(t) for t 0 and fixed ⌧ > 0 then
Z 1
Z 1
st
(L ⌧ xu)(s) =
x(t ⌧ )e dt =
eso(t ⌧ )u(t ⌧ )e stdt
0
0
Z 1
0
0
=
esot u(t0)e s(t +⌧ )dt0 (t0 := t ⌧ )
⌧
Z 1
0
s⌧
= e
e (s so)t dt0 (u and lower limit of integration)
0
e s⌧
=
, when Re{s}>Re{so }.
s so
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Laplace transform properties - time-shifting (cont)
• However, if we delay the negative-time support of a signal to positive time, then we must
subtract it out in order to relate to (unilateral) Lx = X .
• That is, if ⌧ > 0 then
(L
⌧ x)(s)
Z
1
= (L{x(t ⌧ ); t 0 })(s) =
x(t ⌧ )e
0
Z 1
0
= e ⌧s
x(t0 )e st dt0 (no unit-step u factor)
Z
⌧
1
Z 0
dt0 + e ⌧ s
x(t0 )e
0
⌧
Z 0
0
= e ⌧ s X(s) + e ⌧ s
x(t0 )e st dt0 .
= e
⌧s
x(t0 )e
st0
st0
st
dt
d t0
⌧
• E.g., if x(t) = eso t (no unit-step u factor) for t 2 R and ⌧ > 0, then
(L
⌧ x)(s)
=
e ⌧s
s so
e
⌧s
s
e ⌧ so
e
=
so
s
⌧ so
so
, when Re{s}>Re{so }.
• Similarly, if we advance (⌧ < 0) a signal x’s positive time support into negative time, we
need to add it back to relate the result to X = Lx.
280
Laplace transform pairs - sinusoids
• In the special case of an exponential function that is a sinusoid, x(t) = ejwo t u(t) for
w o 2 R,
X(s) =
1
, when Re{s}> 0.
s jwo
• By Euler-De Moivre identity and linearity, we get that
s
L{cos(wo t)u(t)}(s) =
, when Re{s}> 0
s2 + wo2
wo
L{sin(wo t)u(t)}(s) =
, when Re{s}> 0
s2 + wo2
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Laplace transform properties - frequency shift
• As the for the Fourier transform, multiplication by an exponential function in time results
in a frequency shift.
• That is, if X = Lx and so 2 C, then
for all s 2 C such that s
L{eso t x(t)}(s) = X(s
so ),
so is in the RoC of X .
• For example, for all ↵, wo 2 R,
L{e↵t cos(wo t)u(t)}(s) =
(s
↵t
L{e sin(wo t)u(t)}(s) =
(s
s ↵
, when Re{s}> ↵
↵)2 + wo2
wo
, when Re{s}> ↵
↵)2 + wo2
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LT pairs - L{tk esotu(t)}(s) = k!/(s
so)k+1 when Re{s}> Re{so}
• Again recall that SISO LTI systems have characteristic modes of the form tk eso t u(t) where
k 0 is an integer and so 2 C a characteristic value of the system.
• As for the Fourier transform, to evaluate the Laplace transform of such a characteristic
mode, we need to integrate by parts: for k > 0 and Re{s} > Re{so}:
Z 1
Z 1
1
L{tk eso t u(t)}(s) =
tk eso t e st dt =
tk de(so s)t
so s 0
0
Z 1
1
1 k (so s)t
1
=
t e
e(so s)tdtk
so s
so s 0
0
Z 1
k
=
tk 1 e(so s)t dt
s so 0
k
=
L{tk 1 eso t u(t)}(s)
s so
• So by induction, we can show that for integers k > 0 and Re{s} > Re{so }:
L{tk eso t u(t)}(s) =
(s
k!
.
so )k+1
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Laplace transform properties - di↵erentiation
• For all di↵erentiable (Dirichlet) signals x, with X := L{x}, the Laplace transform of a
time derivative is (recall D := d ),
dt
(L D x)(s) = sX(s)
x(0 ).
• This property is proved by simple integration by parts: 8s in the RoC of X = Lx,
Z 1
(L D x)(s) =
(D x)(t)e st dt
0
Z 1
=
e stdx(t)
0
Z 1
1
= x(t)e st 0
x(t)de st
0
Z 1
=
x(0 ) + s
x(t)e st dt (⇤)
0
=
x(0 ) + sX(s),
where (*) because limt!1 x(t)e st = 0 is a necessary condition for the convergence of
X (i.e., for s to be in the RoC of x).
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Laplace transform properties - di↵erentiation (cont)
• For two derivatives:
(L D2 x)(s) =
=
=
=
(L D(D x))(s)
s(L D x)(s) (D x)(0 )
s(sX(s) x(0 )) (D x)(0 )
s2 X(s) sx(0 ) (D x)(0 ).
• So, the derivative property can be inductively generalized to: 8s 2 RoC of X = Lx and
integers k 1,
(L Dk x)(s) = sk X(s)
k 1
X
sl (Dk
1 l
x)(0 )
l=0
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Laplace transform properties - Initial Value Theorem
• Suppose Y = Ly where y is Dirichlet and continuous at the time 0.
• The initial value theorem states that
y(0) =
• To see why:
sY (s) =
Z
1
0
y(t)se
lim sY (s).
s!1
st
dt and
lim se
s!1
st
=
.
• So, the IVT follows from the ideal sampling property of the impulse.
• If Y = P̃ /Q is a rational polynomial with the degree of P̃ <
1 + degree of Q,
– then sY (s) is still a strictly proper rational polynomial with degree of sP̃ (s) < degree
of Q(s),
– so y(0) = 0 by L’Hopital’s rule on sY (s) and the IVT.
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Laplace transform properties - Initial Value Theorem - example
• For example, if y(t) = 2e
2t
Y (s) =
i.e., 0 = the degree of P̃ <
2e
3t
then y(0) = 0 and
2
s+2
2
2
=
,
s+3
(s + 2)(s + 3)
1+ degree of Q =
1 + 2 so that
lim sY (s) = 0 = y(0).
s!1
• For another example, if y(t) = (3 + e
Y (s) = (4s + 6)/(s2 + 2s); thus
lim sY (s) =
s!1
2t )u(t)
lim
s!1
then y(0) = 4 and
4s + 6
= 4 = y(0).
s+2
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Laplace transform properties - Final Value Theorem
• The final value theorem states that if limt!1 y(t) exists, then
lim y(t) = lim sY (s).
t!1
• To see why, recall that
sY (s)
) lim sY (s)
s!0
s!0
Z
1
y(0 ) = (L D y)(s) =
(D y)(t)e
0
Z 1
y(0 ) =
dy(t) = y(1) y(0 )
st
dt =
Z
1
e
st
dy(t)
0
0
• For example, if y(t) = (3 + e 2t )u(t) then y(1) = 3 and
✓
◆
3
1
s
lim sY (s) = lim s
+
= lim 3 +
= 3 = y(1).
s!0
s!0
s!0
s
s+2
s+2
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Laplace transforms applied to SISO systems given by ODE
• We can use the Laplace transform to analyze the total transient response of the SISO, LTIC
system with input and output signals, respectively, f : [0, 1) ! R, y : [0 , 1) ! R:
P (D)f
= Q(D)y.
• We assume Dirichlet input functions of the form f = gu, u the unit step, with zero initial
conditions:
(Dk f )(0 ) = 0, for all integers k
so, by the derivative property,
• Recall Q(D) =
a n = 1.
Pn
k=0 ak
(LD k f )(s)
=
Dk and P (D) =
0;
sk (Lf )(s).
Pm
k=0 bm D
m
, where 8k, ak , bk 2 R and
• Again, let Y = Ly and F = Lf .
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Laplace transforms applied to SISO systems given by ODE (cont)
• By the derivative property for the Laplace transform of the ODE,
n
X
P (s)F (s) = Q(s)Y (s)
ak
k=1
k 1
X
sl (Dk
1 l
y)(0 ).
l=0
for s in the intersection of the RoC of f and the system characteristic modes.
• Thus, the total response,
Y (s) =
P (s)
F (s) +
Q(s)
Pn
k=1 ak
Pk
1 l
k 1 l
y)(0
l=0 s (D
)
Q(s)
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Laplace transforms applied to SISO systems given by ODE (cont)
• Thus, the total response
Y (s) = YZS (s) + YZI (s)
where
YZS (s) =
P (s)
F (s) = H(s)F (s),
Q(s)
H(s) = P (s)/Q(s) the transfer function of the system, and
Pn
Pk 1 l k 1 l
y)(0 )
k=1 ak
l=0 s (D
YZI (s) =
Q(s)
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Laplace analysis of RL circuit with exponential input
• The nodal equation at y for this circuit is, by KCL,
Z t
f (t) y(t)
f (⌧ ) y(⌧ )
0 y(t)
+ i(0 ) +
d⌧ +
R
L
R
0
= 0, t
0.
• Di↵erentiating this equation, multiplying by R/2, and rearranging gives the standard form:
Dy +
R
y
2L
=
1
R
Df +
f.
2
2L
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Laplace analysis of RL circuit with exponential input (cont)
• Taking Laplace transform gives (with f (0 ) = 0):
1
R
sF (s) +
F (s)
2
2L
y(0 )
) Y (s) = H(s)F (s) +
, where
Q(s)
YZS (s)
P (s)
s/2 + R/(2L)
1
R/(4L)
H(s) =
=
=
=
+
.
F (s)
Q(s)
s + R/(2L)
2
s + R/(2L)
sY (s)
y(0 ) +
R
Y (s) =
2L
• Note that by linearity, the impulse response of this system is
h(t) = (L
1
1
R
(t) +
e
2
4L
H)(t) =
Rt/(2L)
u(t) 8t
0
.
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Laplace analysis of RL circuit with exponential input (cont)
• Suppose R = 4⌦, L = 1H, the initial condition y(0 ) = 2V , and the input signal
f (t) = 2e 3t u(t) for t 0 , i.e., F (s) = 2/(s + 3) for Re{s} > 3.
• Thus, the total response
y(0 )
Q(s)
s/2 + 2
2
2
s+4
2
=
·
+
=
+
s+2 s+3
s+2
(s + 2)(s + 3)
s+2
✓
◆
1
2
2
=
+
+
s+3
s+2
s+2
Y (s) = H(s)F (s) +
where we have “expanded the fraction” for the last equality, and the RoC for Y is Re{s} >
max{ 3, 2} = 2.
• Thus, by linearity, the total response
y(t) = ( e
yZS (t) = ( e
3t
3t
+ 2e
+ 2e
2t
2t
)u(t) + 2e 2t 8t 0 ,
)u(t) and yZI (t) = 2e
2t
8t
0
.
• Note that yZS (0 ) = 0 while yZI (0 ) = 2 = y(0 ), and the forced/eigen-response
is H( 3)f (t) = e 3tu(t).
294
Laplace analysis of a circuit - approach
• In the following, we will solve transient total response of circuits by
– first finding a complex-frequency(s)-domain equivalent circuit that includes its initial
conditions,
– then solving the circuit for the total transient response Y (s),
– then performing partial fraction expansion separately on the ZIR and ZSR,
– finally inverting the Laplace transform separately for the ZSR and ZIR using known
Laplace-transform pairs.
• Regarding Laplace-transform pairs, we will
– use the unit-step u factor in the time-domain for ZSR terms (so that the ZSR has zero
initial conditions at 0 ),
– otherwise not for ZIR terms (so that the ZIR is continuous at time 0 and meets the
given initial conditions at 0 ).
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Partial Fraction Expansion (PFE)
• In the previous example we inverted the Laplace transform of a strictly proper rational
polynomial of the form M (s)/N (s) by using a PFE.
• A rational polynomial M (s)/N (s) is strictly proper if the degree of M < that of N .
• Use long division to obtain a strictly proper rational polynomial from one that is not.
• For example,
3s + 2
23
= 3+
s◆ 7
s 7
✓
1 3s + 2
) L
(t) = 3 (t) + 23e7t u(t), t
s 7
• Here, (3s + 2)/(s
0.
7) is not a strictly proper rational polynomial while 23/(s
7) is.
• If H is proper but not strictly so, then: h = L 1 H will have an impulse component (bn ),
the ZSR may not be continuous at the origin, and there is “direct coupling” between input
and output.
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PFE (cont)
• In the following, we will consider the strictly proper rational polynomial M (s)/N (s) with
all coefficients 2 R and K poles (roots of the polynomial N ).
• That is, K := degree of N > degree of M , and so
K
Y
N (s) =
(s
pk ).
k=1
• We also assume M and N have no common roots, i.e., no “pole-zero cancellation” issue
to consider.
• Note that the RoC for M (s)/N (s) is the open RHP {s 2 C | Re{s} > maxk Re{pk }}.
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PFE - the case of no repeated poles
• Suppose there are no repeated “poles” for M/N (roots of N ), i.e.,
8k 6= l, pk 6= pl .
• In this case, we can write the PFE of M/N as
M (s)
N (s)
=
K
X
l=1
cl
s
pl
where the scalars (Heaviside coefficients) cl 2 C are
cl
=
Q
M (s)
pk )
k6=l (s
= lim
s=pl
s!pl
M (s)
(s
N (s)
pl ) =
M (s)
(s
N (s)
pl )
.
s=pl
• That is, to find the Heaviside coefficient ck over the term s pk in the PFE, we have
removed (covered up) the term s pk from the denominator N (s) and evaluated the
remaining rational polynomial at s = pk .
• This approach, called the Heaviside cover-up method, works even when p is C-valued.
• Given the PFE of M/N , (L
1 M/N )(t)
=
PK
l=1 cl e
pl t u(t).
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PFE - proof of Heaviside cover-up method
• To prove that the above formula for the Heaviside coefficient cl is correct, note that the
claimed PFE of M/N is
PK
Q
K
X
pk )
cl
l=1 cl
k6=l (s
=
s pl
N (s)
l=1
• Thus, the PFE equals M/N if and only if the numerator polynomials are equal, i.e., i↵
K
X
M (s) =
l=1
cl
Y
(s
pk ) =: M̃ (s).
k6=l
• Again, two polynomials are equal if their degrees, L, are equal and either:
– their coefficients are the same, or
– they agree at L + 1 (or more) di↵erent points, e.g., two lines (L = 1) are equal if
they agree at 2 (= L + 1) points.
• Since M is a polynomial of degree < K , it suffices to check that whether M = M̃ for
all s = pk , k 2 {1, 2, ..., K}, i.e., this would create K equations in < K unknowns (the
coefficients of M ).
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PFE - proof of Heaviside cover-up method (cont)
• But note that any pole pr of M/N is a root of all but the r th term in M̃ , so that
Y
M̃ (pr ) = cr
(pr pk )
0
k6=r
M (s0 )
= @Q
0
pk 0 )
k0 6=r (s
=
Q
s0 =pr
1
A
Y
M (pr )
(pr
pk 0 )
k0 6=r (pr
Y
(pr
pk )
k6=r
pk )
k6=r
= M (pr ).
• Q.E.D.
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PFE - the case of no repeated poles - example
• To find the inverse Laplace transform of a strictly proper rational polynomial X = M/N ,
first factor its denominator N , e.g.,
s2 + 5s
s2 + 5s
=
, for Re{s} >
s3 + 9s2 + 26s + 24
(s + 4)(s + 3)(s + 2)
X(s) =
2.
• So, by PFE
c4
c3
c2
c4 (s + 3)(s + 2) + c3 (s + 4)(s + 2) + c2 (s + 4)(s + 3)
+
+
=
s+4
s+3
s+2
(s + 4)(s + 3)(s + 2)
) M (s) = 1s2 + 5s + 0 = c4 (s + 3)(s + 2) + c3 (s + 4)(s + 2) + c2 (s + 4)(s + 3) = M̃ (s).
X(s) =
• We can solve for the 3 constants ck by comparing the 3 coefficients of quadratic M and
M̃ .
• The Heaviside cover-up method suggests we try s =
c4 =
s2
+ 5s
(s + 3)(s + 2)
• Thus, x(t) = (L
=
2, c3 =
s= 4
1 X)(t)
= ( 2e
2,
s2
+ 5s
(s + 4)(s + 2)
4t
+ 6e
3t
3,
4 to solve for c2 , c3 , c4 :
= 6, c2 =
s= 3
3e
s2 + 5s
(s + 4)(s + 3)
=
3
s= 2
2t )u(t).
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PFE - the case of a non-repeated, complex-conjugate pair of poles
• Again, recall that for polynomials with all coefficients 2 R, all complex poles will come in
complex-conjugate pairs, p1 = p2.
p
• The case of non-repeated poles p1 , p2 = ↵±j (↵, 2 R, j :=
1) that are complexconjugate pairs can be handled as above, leading to corresponding complex-conjugate Heaviside coefficients c1, c2, i.e., c1 = c2.
• In the PFE, we can alternatively combine the terms
c1
c2
r1 s + r0
+
=
s p1
s p2
(s ↵)2 +
where by equating the two numerator polynomials’ coefficients,
r0 =
c1 p 2
c2 p 1 =
• Note that
⇢
r1 s + r0
1
L
(s ↵)2 +
2
2
2Re{c1 p2 } 2 R and r1 = c1 + c2 = 2Re{c1 } 2 R.
(t) = L
1
= r1 e
⇢
r1
↵t
cos( t)u(t) +
(s
s ↵
↵)2 +
2
+
r0 + r1 ↵
r0 + r1 ↵
e
↵t
·
(s
↵)2 +
2
sin( t)u(t), t
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(t)
0
.
PFE - non-repeated complex-conjugate pair of poles - example
• To find the inverse Laplace transform of
3s + 2
,
s3 + 5s2 + 10s + 12
X(s) =
first factor the denominator to get
3s + 2
.
(s2 + 2s + 4)(s + 3)
X(s) =
• Note that the poles of X are
3 and
• So, we can expand X to
X(s) =
p
1 ± j 3 (so X ’s RoC is Re{s} >
1).
r1 s + r0
c3
+
,
s2 + 2s + 4
s+3
where by the Heaviside cover-up method,
c3 =
s2
3s + 2
+ 2s + 4
=
1.
s= 3
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PFE - non-repeated complex-conjugate pair of poles - example (cont)
• To find r1 , r0 , we will compare coefficients of the numerator polynomials of X and its PFE,
i.e.,
0s2 + 3s + 2 = (r1 s + r0 )(s + 3) + c3 (s2 + 2s + 4) (⇤)
= (r1 1)s2 + (3r1 + r0 2)s + 3r0 4.
• Thus, by comparing coefficients
0 = r1
1, 3 = 3r1 + r0
2, 2 = 3r0
4
we get
r0 = 2 and r1 = 1.
• Note how s =
3 in (*) gives c3 =
1 as Heaviside cover-up did.
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PFE - non-repeated complex-conjugate pair of poles - example (cont)
• Thus, by substituting, then completing the square in the denominator and rearranging the
numerator,
X(s) =
=
s+2
1
+
+ 2s + 4
s+3
p
s+1
1
3
p
p
+p ·
2
2
2
(s + 1) + ( 3)
3 (s + 1) + ( 3)2
s2
• So,
x(t) = (L
✓
=
e
1
.
s+3
1
X)(t)
p
1
t
cos( 3t) + p e
3
t
p
sin( 3t)
e
3t
◆
u(t).
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PFE - the general case of repeated poles
• If a particular pole p of M (s)/N (s) is of order r
then the PFE of M (s)/N (s) has the terms
c1
s
p
+
c2
cr
+ ... +
(s p)2
(s p)r
with ck 2 C 8k 2 {1, 2, ..., r}, where
(i.e., corresponding to poles 6= p).
r
X
=
k=1
(s
ck
M (s)
=
p)k
N (s)
)
M (s)
(s
N (s)
p)r
(s)
(s) represents the other PFE terms of M (s)/N (s)
• Note that equating M/N to its PFE and multiplying by (s
M (s)
(s
N (s)
p)r ,
1, i.e., N (s) has a factor (s
= cr +
r 1
X
ck (s
p)r
p)r gives
k
+
(s)(s
p)r
k=1
p)r
= cr ,
s=p
i.e., Heaviside cover-up (of the entire term (s
p)r ) works for cr .
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PFE - the general case of repeated poles (cont)
• To find cr
1,
we di↵erentiate the previous display to get
d M (s)
(s
ds N (s)
p)r
=
r 1
X
ck (r
1 k
k)(s
p)r
ck (r
k)(s
+
k=1
= cr
) cr
1
=
✓
1
+
r 2
X
k=1
d M (s)
(s
ds N (s)
p)
r
◆
d
(s)(s
ds
p)r
1 k
+
p)r
d
(s)(s
ds
p)r
s=p
• If we di↵erentiate the original display k 2 {0, 1, 2, ..., r 1} times and then substitute
s = p, we get (with 0! := 1)
✓ k
◆
d M (s)
r
(s
p)
= k!cr k
dsk N (s)
s=p
✓ k
◆
1
d M (s)
r
) cr k =
(s p)
.
k! dsk N (s)
s=p
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PFE - the general case of repeated poles - example
• To find the inverse Laplace transform of
3s + 2
,
(s + 1)(s + 2)3
X(s) =
write the PFE of X as
X(s) =
c1
c2,1
c2,2
c2,3
+
+
+
,
2
s+1
s+2
(s + 2)
(s + 2)3
so clearly the RoC of X is Re{s} >
1.
• By Heaviside cover-up
c1 =
3s + 2
(s + 2)3
=
s= 1
1 and c2,3 =
3s + 2
s+1
= 4.
s= 2
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PFE - the general case of repeated poles - example (cont)
• Also,
1
1!
c2,2 =
1
2!
c2,1 =
✓
✓
d 3s + 2
ds s + 1
d2 3s + 2
d s2 s + 1
• Thus,
X(s) =
) x(t) = (L
1
◆
X)(t) =
=
s= 2
◆
=
s= 2
1
1
1! (s + 1)2
= 1
s= 2
1
2
2! (s + 1)3
= 1
s= 2
1
1
1
4
+
+
+
8Re{s} >
s+1
s+2
(s + 2)2
(s + 2)3
e t + e 2t + te 2t + 2t2e 2t u(t).
• Exercise: Find the ZSR y to input f (t) = 2ej3t u(t) of the marginally stable system
H(s) = 4/(s2 + 9).
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PFE and eigenresponse for asymptotically stable systems
• The total response of a SISO LTI system to input f is of the form
Y (s) = H(s)F (s) +
P1 (s)
P (s)
P1 (s)
=
F (s) +
= YZS (s) + YZI (s).
Q(s)
Q(s)
Q(s)
where P1 depends on the initial conditions and the RoC is the intersection of that of input
F = Lf and the system characteristic modes.
• Suppose the system is asymptotically stable (equivalently BIBO stable) and the input is a
sinusoid at frequency wo, f (t) = Aej(wot+ )u(t) with A > 0 ) F (s) = Aej /(s
jwo ).
• Since jwo cannot be a system pole (owing to asymptotic stability all poles have strictly
negative real part), we can use Heaviside cover-up on
YZS (s) = H(s)F (s) =
P (s)
Aej
Q(s)(s jwo )
to get that the total response is of the form
YZS (s) =
H(jwo )Aej
+ modes = H(jwo )F (s) + modes,
s jwo
where Aej is the phasor of f .
310
1
PFE and eigenresponse for asymptotically stable systems (cont)
• Thus, the total response of an asymptotically stable system to a sinusoidal input f at
frequency wo 0 is
y(t) = H(jwo )f (t) + linear combination of characteristic modes.
• So by asymptotic stability, the steady-state response is the eigenresponse, i.e., as t ! 1,
y(t) ⇠ H(jwo )f (t) = H(jwo )Aej(wo t+ ) . = |H(jwo )|Aej(wo t+
+\H(jwo ))
.
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Complex-frequency(s)-domain equivalent circuits - introduction
• To solve for the transient total response of SISO LTIC circuits by Laplace transform, we
can use Kirchho↵’s laws and current-voltage characteristics of the individual devices to find
– the ODE P (D)f = Q(D)y relating input f to output y ,
– and the initial conditions (Dk y)(0 ) from those given -typically the initial inductor
currents/fluxes and capacitor voltages/charges, i.e., initial conditions of circuit “state”
variables.
• Alternatively, we can find the s-domain equivalent circuit from the s-domain equivalents of
the individual devices, and apply KVL/KCL to solve for transient total response.
• Independent and dependent sources in the s-domain simply involve the Laplace transform
of the associated signals.
• Similarly, an ideal op-amp has the same characteristics in the s-domain (infinite input
impedance and zero di↵erential input voltage).
• By linearity of the Laplace transform, a resistor with resistance R⌦ relating a current i to
voltage v by Ohm’s law v = Ri, in the s-domain is simply V = RI where V = Lv and
I = Li.
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s-domain equivalent capacitors and inductors
• The current-voltage characteristic of an inductor is v = L D i, so in the s-domain,
V (s) = sLI(s)
Li(0 ) or I(s) =
1
1
V (s) + i(0 ).
sL
s
• The current-voltage characteristic of a capacitor is i = C D v , so in the s-domain,
I(s) = sCV (s)
Cv(0 ) or V (s) =
• Recall
L
1
⇢
1
s
1
1
I(s) + v(0 ).
sC
s
= u, the unit step,
but when considering initial conditions (ZIR terms), we take
⇢
1 1
L
(t) = 1 8t 0 .
s
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s-domain equivalent capacitors and inductors (cont)
• The Thevinin or Norton equivalent may be more convenient depending on the circuit.
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s-domain equivalent circuits - example
• The s-domain equivalent of a first-order op-amp circuit is depicted above for vC (0 ) = 2V
and the Norton equivalent s-domain capacitor - so that CvC (0 ) = 0.5 · 2 = 1.
• By KCL at the negative terminal of the op-amp and at vc :
0+
F
Vc
2
Y
F
1
+ 0.5s(0
+
Vc
Vc )
F
2
= 0
( 1) = 0 (impedance of cap. is 1/(0.5s))
• Eliminating Vc = (F + 2)/(s + 1) we get
Y (s) =
3s + 2
F (s)
2(s + 1)
1
= H(s)F (s) + YZI (s).
s+1
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s-domain equivalent circuits - example (cont)
• So, if the input f (t) = 2e
response,
3t u(t),
t
3s + 2
(s + 1)(s + 3)
c1
c3
=
+
s+1
s+3
where by Heaviside cover-up for the PFE,
Y (s) =
c1 = [(3s + 2)/(s + 3)]s=
• Thus,
1
=
0 , then F (s) = 2/(s + 3) and the total
1
= YZS (s) + YZI (s)
s+1
1
s+1
0.5 and c3 = [(3s + 2)/(s + 1)]s=
y(t) = ( 0.5e t + 3.5e 3t )u(t) e
= ( 0.5e t u(t) + H( 3)f (t))
= yZS (t) + yZI (t), t 0
3
= 3.5.
t
e
t
• Regarding initial conditions, note that
– by KCL at the negative terminal in the time-domain,
0 = y(0 )
f (0 ) + (vc (0 )
f (0 ))/2 = y(0 ) + vc (0 )/2 = y(0 ) + 1,
so that, consistent with the above solution,
y(0 ) =
1 = yZI (0 ).
316
Circuit example with switch instantaneously thrown at time t = 0
• The switch is thrown at time t = 0 after a long time (i.e., when t < 0).
• The input is DC so that at time 0 , the circuit (consisting of just the battery, the 2H
inductor and 1⌦ resistor) is in DC steady-state; check that for t < 0, this first-order
circuit is asymptotically stable with transfer function 1/(2s + 1), i.e., with system pole
(characteristic value) 1/2.
• An inductor has zero DC impedance, so the output y(0 ) = 10V/1⌦ = 10A by Ohm’s
law.
• For time t
0 , the input f (t) = 10u(t) so that f (0 ) = 0; the negative-time
evolution of the circuit is accounted for by the initial system states, e.g., y(0 ) = 10.
• Because the circuit is closed for time t < 0, the initial (time 0 ) current through the 1H
inductor is zero and the initial voltage of the 1F capacity is zero.
317
Circuit example with switch thrown at time t = 0 (cont)
• Writing KVL in the time domain for time t
f
y
0, we get
= 10u = 2 D y + vc + D y + y, where
= D vc and vc is the capacitor voltage.
• Instead of simply di↵erentiating KVL to eliminate vc (and using D f = 10 ), we will take
Laplace transform of these two equations taking care to including the initial values of the
2H inductor but to use zero initial values for the 1H inductor and 1F capacitor, i.e.,
F (s) =
10
s
= 2(sY (s)
) Y (s) =
YZS (s) =
YZI (s) =
y(0 )) + Vc (s) + sY (s) + Y (s) and Y (s) = sVc (s)
10
20s
+ 2
= YZS (s) + YZI (s) where
+s+1
3s + s + 1
10
s
10
= 2
·
= H(s)F (s) and
2
3s + s + 1
3s + s + 1 s
20s
2y(0 )s
=
.
3s2 + s + 1
3s2 + s + 1
3s2
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Circuit example involving a switch thrown at time t = 0 - ZSR
• Note that the poles of YZS (roots of 3s2 + s + 1) are complex:
p
p
1 + j 11
1 j 11
p=
and p =
.
6
6
• To obtain the total response in the time domain, we can either complete the square (3s2 +
s + 1 = 3((s + 1/6)2 + 35/108)) and then use the known Laplace transforms of sine
and cosine, or just use Heaviside cover-up on the complex-conjugate poles:
YZS (s) =
c =
10
10/3
c
c
=
=
+
, where
+s+1
(s p)(s p)
s p
s p
10/3
10/3
10
=
= j p .
s p s=p
p p
3 11
3s2
p
• Thus, c = j10/(3 11) and
yZS (t) = cept u(t) + cept u(t), 8t
0
.
• Note that there is no forced response (eigenresponse) component H(0)f (t) for the DC
input f since the transfer function H has a DC zero, i.e., H(0) = 0.
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Circuit example with switch thrown at time t = 0 - ZIR and total resp.
• Similarly, we can perform PFE on YZI to get
YZI (s) =
a =
20s
(20/3)s
a
a
=
=
+
, where
+s+1
(s p)(s p)
s p
s p
(20/3)s
20p/3
=
.
s p s=p
p p
3s2
• So, again noting no step-function u factors,
yZI (t) = aept + aept , 8t
0
.
• The total response is clearly
y(t) = yZS (t) + yZI (t) = cept u(t) + cept u(t) + aept + aept , 8t
0
.
• Exercise: check the computed y(0 ) = yZI (0 ) = 2Re{a} = 10.
• This example is Lathi Problem 6.4-5, p. 465.
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Circuit example with switch thrown at time t = 0 - s-domain equiv.
• The s-domain equivalent circuit would have input voltage F (s) = 10/s, output current
Y (s), inductive impedences 2s and s, capacitive impedance 1/s, and resistance 1.
• To preserve the single loop, we use Thevinin equivalent s-domain (reactive) devices for time
t 0.
• Since the initial conditions are zero for the 1H inductor and 1F capacitor, their s-domain
representations (of s and 1/s impedences, respectively) will not have (i.e., have zero)
associated independent voltage sources.
• The 2H inductor will have a
wise.
Ly(0 ) =
20V battery referenced positive counterclock-
• Thus, by KVL in the s-domain
F (s) =
10
s
= 2sY (s)
20 + s
1
Y (s) + sY (s) + Y (s)
10
20s
+ 2
= YZS (s) + YZI (s),
+s+1
3s + s + 1
as computed above by applying Laplace transform to time-domain KVL.
) Y (s) =
3s2
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s-domain equivalent circuits - RLC circuit example (w/o switch)
• For the series RLC circuit, we’ve used Thevinin equivalent devices for the s-domain equivalent circuit so that there remains a single loop.
• By KVL (voltage increases counterclockwise),
vc (0 )
1
+
I(s) Li(0 ) + sLI(s) + Y (s) F (s)
s
sC
I(s) = Y (s)/R
vc (0 )
1
sL
) 0 =
+
Y (s) Li(0 ) +
Y (s) + Y (s) F (s) (eliminating I )
s
sRC
R
0 =
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s-domain equivalent circuits - RLC circuit example (cont)
• Thus, the total response is
Y (s) =
=
=
YZS (s) =
YZI (s) =
F (s) + Li(0 ) vc (0 )s 1
(sRC) 1 + sLR 1 + 1
sRL 1
sRi(0 ) vc (0 )RL 1
F (s) + 2
2
1
1
s + sRL + (LC)
s + sRL 1 + (LC) 1
YZS (s) + YZI (s), where
sRL 1
P (s)
F (s) =
F (s) = H(s)F (s), and
2
1
1
s + sRL + (LC)
Q(s)
sRi(0 ) vc (0 )RL 1
P1 (s)
=
.
2
1
1
s + sRL + (LC)
Q(s)
• In the following, we will take L = 1H, C = 0.25F, vc (0 ) = 2V, i(0 ) = 1A.
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s-domain equivalent circuits - RLC circuit example (cont)
• For an overdamped case, take R = 5⌦, so that Q(s) = s2 +5s+4 = (s+1)(s+4).
• If the input f (t) = 3e2t u(t) () F (s) = 3/(s
2)), then the total response is
15s
5s 10
+
(s + 1)(s + 4)(s 2)
(s + 1)(s + 4)
✓
◆ ✓
◆
5/3
5/3
10/3
5
10
=
+
+
+
by PFE
s 2
s+1 s+4
s+1
s+4
✓
◆
5 2t
5
10 4t
e + e t
e
u(t) + 5e t + 10e 4t
) y(t) =
3
3
3
= yZS (t) + yZI (t), 8t 0 .
Y (s) =
• To check this solution is satisfies the initial conditions:
– By Ohm’s law, y(0 ) = Ri(0 ) = 5 · 1 = 5 = yZI(0 ).
– By KVL in the time-domain,
0 = vc (0 ) + LR
1
(D y)(0 ) + y(0 ) f (0 ) = 2 + 0.2(D y)(0 ) + 5
) (D y)(0 ) =
35 = (D yZI )(0 ).
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0
s-domain equivalent circuits - RLC circuit example (cont)
• Now consider the marginally stable case with R = 0 and the output redefined to be
Y := Vc so that KVL is
0 = Y + sLI Li(0 ) F where I = sCY
(LC) 1
sy(0 ) + i(0 )/C
) Y (s) =
F (s) +
s2 + (LC) 1
s2 + (LC) 1
4
2s + 4
=
F (s) + 2
s2 + 4
s +4
Cy(0 )
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s-domain equivalent circuits - RLC circuit example (cont)
• So, the total response to the sinusoidal input f (t) = 3e2jt u(t) is
12
2s + 4
+ 2
2
(s + 2j)(s 2j)
s +4
✓
◆ ✓
◆
3/4
3/4
3j
s
2
=
+
+
+
2
+
2
s + 2j
s 2j
(s 2j)2
s2 + 4
s2 + 4
✓
◆
3 2jt
3
) y(t) =
e
+ e2jt 3jte2jt u(t) + 2 cos(2t) + 2 sin(2t)
4
4
= yZS (t) + yZI (t), t 0 .
Y (s) =
• Note that resonance phenomenon for the input f is a sinusoid at the system resonant
frequency of 2 = (LC) 0.5 radians/s, leading to an unbounded response (te2jt) to a
bounded input f .
• Also note that yZI (0 ) = 2 = y(0 ) and i(0 ) = 1 = C(D y)(0 ) = 0.25(D y)(0 )
) (D y)(0 ) = 4 = yZI (0 ).
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System composition - for the ZSR of a LTIC system
• SISO system with input f and ZSR y , where F = Lf , Y = Ly :
• An ideal amplifier for scalar a (recall inverting and noninverting op-amp circuit configurations):
• An integrator system - recall (Lu)(s) = s
1
and (f ⇤ u)(t) =
Rt
0
f (⌧ )d⌧, t
0:
• Exercise: show how a integrator can be implemented using a simple op-amp circuit with a
output-feedback capacitor and an input-side resistor.
327
System composition - summer
• One can realize a summer with inverting and/or non-inverting op-amp circuit configurations.
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System composition - non-inverting summer implementation
• The voltage at the positive input terminal is v+ = v =
Rb
y.
Rb +Rf
• With system inputs v1 , v2 , KCL at the positive input terminal is
0 =
0
v+
Ra
+
2
X
vk
k=1
2
X
R̃
) y =
vk ,
R
k=1 k
v+
Rk
where R̃ =
Rb
Rb + Rf
2
X 1
1
+
Ra
Rk
k=1
!!
1
.
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Canonical system realizations - single pole system
• Note that for the system with transfer function
H(s) =
a
Y (s)
=
,
s+b
F (s)
we can write
sY (s) = aF (s)
bY (s).
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Canonical system realizations - by PFE using single-pole systems
• Given n distinct real poles, we have by PFE of strictly proper H = P/Q,
H(s) =
n
X
k=1
s
ck
.
pk
• So, one can realize this system by summing the outputs of n single-pole systems.
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System composition - tandem systems
• Recall how an op-amp has a separate power supply, decoupling input from output so that
the transfer function H1 does not depend on its load (system H2).
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System composition - tandem systems - realizing repeated poles
• So, one can realize part of a transfer function H with, say, twice-repeated (triple) pole p
leading to PFE terms
c1
c2
c3
+
+
,
s p
(s p)2
(s p)3
by using three tandem systems:
• Letting the input of this subsystem be F and the output be Y , note that
Y (s) = c1 Fs (s)
+ c2 (sF (s)
+ c3 (sF (s)
.
p
p)2
p)3
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Canonical system-realizations - direct form
• Consider the proper transfer function H = P/Q, i.e., m  n.
• The direct-form realization employs the interior system state X := F/Q,
i.e., F = QX and Y = P X where the former implies
F (s) =
n
X
ak sk X(s) with an = 1
k=0
) sn X(s) = F (s)
n
X1
ak sk X(s).
k=0
• For n = 2, there are two “system states” (outputs of integrators), X and sX (x and
D x):
334
Canonical system-realizations - direct form (cont)
• Now adding Y = P X , we final get the direct-form canonical system-realization of H :
• Again, state variables taken as outputs of integrators, here: x, D x, ... Dn
1
x.
• If bn = b2 =
6 0, there is direct coupling of input and output, H is proper but not strictly
so, h = L 1H has a impulse component b2 , and the ZSR may be discontinuous at time
0.
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Canonical system realizations - direct form (cont)
• Note that this n = 2 example above can be used to implement a pair of complex-conjugate
poles as part of a larger PFE-based implementation (with otherwise di↵erent states); e.g., for
n = 2, H(s) = P (s)/Q(s) where
Q(s) = s2 + a1 s + a0 = (s
for ↵,
↵)2 +
2
2 R, so the poles are ↵ ± j .
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Canonical system realizations - by PFE
• In the general case of a proper transfer function, we can use partial-fraction expansion (after
long division to get a strictly proper rational polynomial)
– grouping the terms corresponding to a complex-conjugate pair of poles, i.e., a secondorder denominator, and
– using a direct-form realization for these terms.
• Besides the PFE-based and direct-form realizations, there are other (zero-state) system
realizations, e.g., “observer” canonical.
• For proper rational-polynomial transfer functions H = P/Q, all of these realizations involve
n (= degree of Q) integrators, the output of each being a di↵erent interior state variable
of the system.
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Signal tracking by a Proportional-Integral (PI) system
• “Open-loop” system is PI: G(s) = Kp + Ki /s.
• The goal is that the error signal e = f
the input f .
• Y = GE = G(Y
y ! 0, i.e., the “closed-loop” output y tracks
F ) ) Y = F G/(1 + G).
• Thus, the closed-loop transfer function (F to Y , with negative feedback) is
✓
◆
G(s)
1
↵
Ki
H(s) =
=
Kp +
, where ↵ :=
.
1 + G(s)
1 + Kp
s+↵
Ki + Kp
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Signal tracking by a PI system (cont)
• Now assume that ↵ > 0 (H is stable).
• Suppose we want to see how the PI parameters Kp , Ki a↵ect tracking of a step f = Au,
i.e., we want the (ZS) step response:
✓
◆
A
A
Kp
↵
Y (s) = H(s)F (s) = H(s) =
+
s
1 + Kp
s
s(s + ↵)
✓
◆
A
Kp + 1
1
=
by PFE
1 + Kp
s
s+↵
A
) y(t) = Au(t)
e ↵tu(t)
1 + Kp
• Exercise: Find an expression for the time t > 0 at which y(t) = 0.9A.
• Exercise: Derive how this closed-loop PI system would track a ramp, f (t) = Atu(t).
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Feedback control for stabilization by pole placement
• Consider this simple feedback control system where the fixed open-loop system (plant)
G = P/Qo and constant-gain (proportional) K output-feedback combine to give the
closed-loop system H :
X=F
KY and Y = GX ) Y = GF
KGY =
G
F = HF.
1 + KG
• Thus, the closed-loop characteristic polynomial Q = Qo + KP because
H
=
G
P
P
=
=
.
1 + KG
Qo + KP
Q
• E.g., if P (s) = 1, Qo (s) = s 5 and K > 5, then G is unstable, but Q(s) = s 5+K
has a root at 5 K < 0 and so the closed-loop system is stable.
• So, proportional feedback control may be used to stabilize a system by pole placement.
340