Signals and Systems
Lecture 15 Wednesday 12th December 2017
DR TANIA STATHAKI READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING IMPERIAL COLLEGE LONDON
β’ Consider a discrete-time signal π₯(π‘) sampled every π seconds.
π₯ π‘ = π₯0πΏ π‘ + π₯1πΏ π‘ β π + π₯2πΏ π‘ β 2π + π₯3πΏ π‘ β 3π + β―
β’ Recall that in the Laplace domain we have:
β πΏ π‘ = 1 β πΏ π‘ β π = πβπ π
β’ Therefore, the Laplace transform of π₯ π‘ is:
π π = π₯0 + π₯1πβπ π + π₯2π
βπ 2π + π₯3πβπ 3π + β―
β’ Now define π§ = ππ π = π(π+ππ)π = πππcosππ + ππππsinππ.
β’ Finally, define
π[π§] = π₯0 + π₯1π§β1 + π₯2π§
β2 + π₯3π§β3 + β―
The z-transform derived from the Laplace transform.
β’ From the Laplace time-shift property, we know that π§ = ππ π is time
advance by π seconds (π is the sampling period).
β’ Therefore, π§β1 = πβπ π corresponds to one sampling period delay.
β’ As a result, all sampled data (and discrete-time systems) can be
expressed in terms of the variable π§.
β’ More formally, the unilateral π β transform of a causal sampled
sequence:
π₯ π = {π₯ 0 , π₯ 1 , π₯ 2 , π₯ 3 , β¦ }
is given by:
π[π§] = π₯0 + π₯1π§β1 + π₯2π§
β2 + π₯3π§β3 + β― = π₯[π]π§βπβ
π=0 , π₯π = π₯[π]
β’ The bilateral π βtransform for any sampled sequence is:
π[π§] = π₯[π]π§βπ
β
π=ββ
πβπ: the sampled period delay operator
Definition Purpose Suitable for
Laplace
transform πΏ π = π(π)πβπππ π
β
ββ
Converts integral-
differential
equations to
algebraic equations.
Continuous-time signal
and systems analysis.
Stable or unstable.
Fourier
transform πΏ(π) = π(π)πβππππ π
β
ββ
Converts finite
energy signals to
frequency domain
representation.
Continuous-time,
stable systems.
Convergent signals
only. Best for steady-
state.
Discrete
Fourier
transform
πΏ[πππ] =
π»π[ππ»]πβππππππ΅πβππ=ββ
π sampling period
Ξ©0 = π0 π = 2π/π0
Converts discrete-
time signals to
discrete frequency
domain.
Discrete time signals.
π§ β
transform πΏ[π] = π[π]πβπ
β
π=ββ
Converts difference
equations into
algebraic equations.
Discrete-time system
and signal analysis;
stable or unstable.
Laplace, Fourier and π β transforms
β’ Find the π§ βtransform of the causal signal πΎππ’[π], where πΎ is a constant.
β’ By definition:
π[π§] = πΎππ’ π π§βπ =
β
π=ββ
πΎππ§βπ
β
π=0
= πΎ
π§
πβ
π=0
= 1 +πΎ
π§+
πΎ
π§
2
+πΎ
π§
3
+ β―
β’ We apply the geometric progression formula:
1 + π₯ + π₯2 + π₯3 + β― =1
1 β π₯, π₯ < 1
β’ Therefore,
π[π§] =1
1βπΎ
π§
, πΎ
π§< 1
=π§
π§βπΎ, π§ > πΎ
β’ We notice that the π§ βtransform exists for certain values of π§. These values
form the so called Region-Of-Convergence (ROC) of the transform.
Example: Find the π βtransform of π π = πΈππ[π]
β’ Observe that a simple equation in π§-domain results in an infinite sequence
of samples.
β’ The figures below depict the signal in time (left) and the ROC, shown with
the shaded area, within the π§ βplane.
Example: Find the π βtransform of π π = πΈππ[π]cont.
β’ Find the π§ βtransform of the anticausal signal βπΎππ’[βπ β 1], where πΎ is a
constant.
β’ By definition:
π[π§] = βπΎππ’ βπ β 1 π§βπ =
β
π=ββ
βπΎππ§βπ
β1
π=ββ
= β πΎβππ§π
β
π=1
= β π§
πΎ
πβ
π=1
= βπ§
πΎ
π§
πΎ
πβ
π=0
= βπ§
πΎ1 +
π§
πΎ+
π§
πΎ
2
+π§
πΎ
3
+ β―
β’ Therefore,
π[π§] = βπ§
πΎ
1
1βπ§
πΎ
, π§
πΎ< 1
=π§
π§βπΎ, π§ < πΎ
β’ We notice that the π§ βtransform exists for certain values of π§, which consist
the complement of the ROC of the function πΎππ’[π] with respect to the
π§ βplane.
Example: Find the π βtransform of π π = βπΈππ[βπ β π]
β’ We proved that the following two functions:
The causal function πΎππ’ π and
the anti-causal function βπΎππ’[βπ β 1] have:
The same analytical expression for their π§ βtransforms.
Complementary ROCs. More specifically, the union of their ROCS
forms the entire π§ βplane.
β’ Observe that the ROC of πΎππ’ π is π§ > πΎ .
β’ In case that πΎππ’ π is part of a causal systemβs impulse response, we see
that the condition πΎ < 1 must hold. This is because, since limπββ
πΎ π = β, for
πΎ > 1, the system will be unstable in that case.
β’ Therefore, in causal systems, stability requires that the ROC of the systemβs
transfer function includes the circle with radius 1 centred at origin within the
π§ βplane. This is the so called unit circle.
Summary of previous examples
β’ By definition πΏ 0 = 1 and πΏ π = 0 for π β 0.
π[π§] = πΏ π π§βπ = πΏ 0 π§β0 = 1
β
π=ββ
β’ By definition π’ π = 1 for π β₯ 0.
π[π§] = π’ π π§βπ =βπ=ββ π§βπ =β
π=0 1
1β1
π§
, 1
π§< 1
=π§
π§β1, π§ > 1
Example: Find the π βtransform of πΉ[π] and π[π]
β’ We write cosπ½π =1
2πππ½π + πβππ½π .
β’ From previous analysis we showed that:
πΎππ’[π] β π§
π§βπΎ, π§ > πΎ
β’ Hence,
πΒ±ππ½ππ’[π] β π§
π§βπΒ±ππ½, π§ > πΒ±ππ½ = 1
β’ Therefore,
π π§ =1
2
π§
π§βπππ½ +π§
π§βπβππ½ =π§(π§βcosπ½)
π§2β2π§cosπ½+1, π§ > 1
Example: Find the π βtransform of ππ¨π¬π·ππ[π]
β’ Find the π§ βtransform of the signal depicted in the figure.
β’ By definition:
π π§ = 1 +1
π§+
1`
π§2+
1
π§3+
1
π§4= π§β1 π
4
π=0
=1 β π§β1 5
1 β π§β1=
π§
π§ β 11 β π§β5
π βtransform of 5 impulses
π βtransform Table
No. π[π] πΏ[π]
π]
π βtransform Table
No. π[π] πΏ[π]
Inverse π βtransform
β’ As with other transforms, inverse π§ βtransform is used to derive π₯[π] from π[π§], and is formally defined as:
π₯ π =1
2ππ π[π§]π§πβ1ππ§
β’ Here the symbol indicates an integration in counter-clockwise
direction around a closed path within the complex π§-plane (known as
contour integral).
β’ Such contour integral is difficult to evaluate (but could be done using
Cauchyβs residue theorem), therefore we often use other techniques to
obtain the inverse π§ βtransform.
β’ One such technique is to use the π§ βtransform pair table shown in the
last two slides with partial fraction.
Find the inverse π βtransform in the case of real unique poles
β’ Find the inverse π§ βtransform of π π§ =8π§β19
(π§β2)(π§β3)
Solution
π[π§]
π§=
8π§ β 19
π§(π§ β 2)(π§ β 3)=
(β196
)
π§+
3/2
π§ β 2+
5/3
π§ β 3
π π§ = β19
6+
3
2
π§
π§β2+
5
3
π§
π§β3
By using the simple transforms that we derived previously we get:
π₯ π = β19
6πΏ[π] +
3
22π +
5
33π π’[π]
Find the inverse π βtransform in the case of real repeated poles
β’ Find the inverse π§ βtransform of π π§ =π§(2π§2β11π§+12)
(π§β1)(π§β2)3
Solution
π[π§]
π§=
(2π§2β11π§+12)
(π§β1)(π§β2)3=
π
π§β1+
π0
(π§β2)3+
π1
(π§β2)2+
π2
(π§β2)
We use the so called covering method to find π and π0
π =(2π§2 β 11π§ + 12)
(π§ β 1)(π§ β 2)3 π§=1
= β3
π0 =(2π§2 β 11π§ + 12)
(π§ β 1)(π§ β 2)3 π§=2
= β2
The shaded areas above indicate that they are excluded from the entire
function when the specific value of π§ is applied.
Find the inverse π βtransform in the case of real repeated poles cont.
β’ Find the inverse π§ βtransform of π π§ =π§(2π§2β11π§+12)
(π§β1)(π§β2)3
Solution
π[π§]
π§=
(2π§2β11π§+12)
(π§β1)(π§β2)3=
β3
π§β1+
β2
(π§β2)3+
π1
(π§β2)2+
π2
(π§β2)
To find π2 we multiply both sides of the above equation with π§ and let
π§ β β.
0 = β3 β 0 + 0 + π2 β π2 = 3
To find π1 let π§ β 0.
12
8= 3 +
1
4+
π1
4β
3
2β π1 = β1
π[π§]
π§=
(2π§2β11π§+12)
(π§β1)(π§β2)3=
β3
π§β1β
2
π§β2 3 β1
(π§β2)2+
3
(π§β2) β
π π§ = β3π§
π§β1β
2π§
π§β2 3 βπ§
(π§β2)2+
3π§
(π§β2)
Find the inverse π βtransform in the case of real repeated poles cont.
π π§ = β3π§
π§β1β
2π§
π§β2 3 βπ§
(π§β2)2+
3π§
(π§β2)
β’ We use the following properties:
πΎππ’[π] βπ§
π§βπΎ
π(πβ1)(πβ2)β¦(πβπ+1)
πΎππ! πΎππ’ π =
π§
(π§βπΎ)π+1
β’ Therefore,
π₯ π = [β3 β 1π β 2π(πβ1)
8 β 2π β
π
2 β 2π + 3 β 2π]π’[π]
= β 3 +1
4π2 + π β 12 2π π’[π]
Find the inverse π βtransform in the case of complex poles
β’ Find the inverse π§ βtransform of π π§ =2π§(3π§+17)
(π§β1)(π§2β6π§+25)
Solution
π π§ =2π§(3π§ + 17)
(π§ β 1)(π§ β 3 β π4)(π§ β 3 + π4)
π[π§]
π§=
(2π§2β11π§+12)
(π§β1)(π§β2)3=
π
π§β1+
π0
(π§β2)3+
π1
(π§β2)2+
π2
(π§β2)
Whenever we encounter a complex pole we need to use a special partial
fraction method called quadratic factors method.
π[π§]
π§=
2(3π§+17)
(π§β1)(π§2β6π§+25)=
2
π§β1+
π΄π§+π΅
π§2β6π§+25
We multiply both sides with π§ and let π§ β β:
0 = 2 + π΄ β π΄ = β2
Therefore,
2(3π§+17)
(π§β1)(π§2β6π§+25)=
2
π§β1+
β2π§+π΅
π§2β6π§+25
Find the inverse π βtransform in the case of complex poles cont.
2(3π§+17)
(π§β1)(π§2β6π§+25)=
2
π§β1+
β2π§+π΅
π§2β6π§+25
To find π΅ we let π§ = 0:
β34
25= β2 +
π΅
25 β π΅ = 16
π[π§]
π§=
2
π§β1+
β2π§+16
π§2β6π§+25 β π π§ =
2π§
π§β1+
π§(β2π§+16)
π§2β6π§+25
β’ We use the following property:
π πΎ π cos π½π + π π’[π] β π§(π΄π§+π΅)
π§2+2ππ§+ πΎ 2 with π΄ = β2, π΅ = 16, π = β3, πΎ = 5.
π =π΄2 πΎ 2+π΅2β2π΄ππ΅
πΎ 2βπ2 = 4β25+256β2β(β2)β(β3)β16
25β9= 3.2, π½ = cosβ1 βπ
πΎ= 0.927πππ,
π = tanβ1 π΄πβπ΅
π΄ πΎ 2βπ2= β2.246πππ.
Therefore, π₯ π = [2 + 3.2 cos 0.927π β 2.246 ]π’[π]
β’ Since π§ = ππ π = π π+ππ π = πππππππ where π =2π
ππ , we can map the
π βplane to the π§ βplane as below.
For π = 0, π = ππ and π§ = ππππ. Therefore, the imaginary axis of the
π βplane is mapped to the unit circle on the π§ βplane.
= Im(s)
Mapping from π βplane to π§ βplane
Mapping from π βplane to π§ βplane cont.
β’ For π < 0, π§ = πππ < 1. Therefore, the left half of the π βplane is mapped
to the inner part of the unit circle on the π§ βplane (turquoise areas).
β’ Note that we normally use Cartesian coordinates for the π β plane
π = π + ππ and polar coordinates for the π§ βplane (π§ = ππππ).
Mapping from π βplane to π§ βplane cont.
β’ For π > 0 , π§ = πππ > 1 . Therefore, the right half of the π βplane is
mapped to the outer part of the unit circle on the π§ βplane (pink areas).
Find the inverse π βtransform in the case of complex poles
β’ Using the results of todayβs Lecture and also Lecture 9 on stability of causal continuous-time systems and the mapping from the π βplane to the π§ βplane, we can easily conclude that:
A discrete-time LTI system is stable if and only if the ROC of its system function π»(π§) includes the unit circle, π§ = 1.
A causal discrete-time LTI system with rational π§ βtransform π»(π§) is stable if and only if all of the poles of π»(π§) lie inside the unit circle β
i.e., they must all have magnitude smaller than 1. This statement is based on the result of Slide 5.