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Signals and Systems
Lecture 15 Wednesday 12th December 2017
DR TANIA STATHAKI READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING IMPERIAL COLLEGE LONDON
โข Consider a discrete-time signal ๐ฅ(๐ก) sampled every ๐ seconds.
๐ฅ ๐ก = ๐ฅ0๐ฟ ๐ก + ๐ฅ1๐ฟ ๐ก โ ๐ + ๐ฅ2๐ฟ ๐ก โ 2๐ + ๐ฅ3๐ฟ ๐ก โ 3๐ + โฏ
โข Recall that in the Laplace domain we have:
โ ๐ฟ ๐ก = 1 โ ๐ฟ ๐ก โ ๐ = ๐โ๐ ๐
โข Therefore, the Laplace transform of ๐ฅ ๐ก is:
๐ ๐ = ๐ฅ0 + ๐ฅ1๐โ๐ ๐ + ๐ฅ2๐
โ๐ 2๐ + ๐ฅ3๐โ๐ 3๐ + โฏ
โข Now define ๐ง = ๐๐ ๐ = ๐(๐+๐๐)๐ = ๐๐๐cos๐๐ + ๐๐๐๐sin๐๐.
โข Finally, define
๐[๐ง] = ๐ฅ0 + ๐ฅ1๐งโ1 + ๐ฅ2๐ง
โ2 + ๐ฅ3๐งโ3 + โฏ
The z-transform derived from the Laplace transform.
โข From the Laplace time-shift property, we know that ๐ง = ๐๐ ๐ is time
advance by ๐ seconds (๐ is the sampling period).
โข Therefore, ๐งโ1 = ๐โ๐ ๐ corresponds to one sampling period delay.
โข As a result, all sampled data (and discrete-time systems) can be
expressed in terms of the variable ๐ง.
โข More formally, the unilateral ๐ โ transform of a causal sampled
sequence:
๐ฅ ๐ = {๐ฅ 0 , ๐ฅ 1 , ๐ฅ 2 , ๐ฅ 3 , โฆ }
is given by:
๐[๐ง] = ๐ฅ0 + ๐ฅ1๐งโ1 + ๐ฅ2๐ง
โ2 + ๐ฅ3๐งโ3 + โฏ = ๐ฅ[๐]๐งโ๐โ
๐=0 , ๐ฅ๐ = ๐ฅ[๐]
โข The bilateral ๐ โtransform for any sampled sequence is:
๐[๐ง] = ๐ฅ[๐]๐งโ๐
โ
๐=โโ
๐โ๐: the sampled period delay operator
Definition Purpose Suitable for
Laplace
transform ๐ฟ ๐ = ๐(๐)๐โ๐๐๐ ๐
โ
โโ
Converts integral-
differential
equations to
algebraic equations.
Continuous-time signal
and systems analysis.
Stable or unstable.
Fourier
transform ๐ฟ(๐) = ๐(๐)๐โ๐๐๐๐ ๐
โ
โโ
Converts finite
energy signals to
frequency domain
representation.
Continuous-time,
stable systems.
Convergent signals
only. Best for steady-
state.
Discrete
Fourier
transform
๐ฟ[๐๐๐] =
๐ป๐[๐๐ป]๐โ๐๐๐๐๐๐ต๐โ๐๐=โโ
๐ sampling period
ฮฉ0 = ๐0 ๐ = 2๐/๐0
Converts discrete-
time signals to
discrete frequency
domain.
Discrete time signals.
๐ง โ
transform ๐ฟ[๐] = ๐[๐]๐โ๐
โ
๐=โโ
Converts difference
equations into
algebraic equations.
Discrete-time system
and signal analysis;
stable or unstable.
Laplace, Fourier and ๐ โ transforms
โข Find the ๐ง โtransform of the causal signal ๐พ๐๐ข[๐], where ๐พ is a constant.
โข By definition:
๐[๐ง] = ๐พ๐๐ข ๐ ๐งโ๐ =
โ
๐=โโ
๐พ๐๐งโ๐
โ
๐=0
= ๐พ
๐ง
๐โ
๐=0
= 1 +๐พ
๐ง+
๐พ
๐ง
2
+๐พ
๐ง
3
+ โฏ
โข We apply the geometric progression formula:
1 + ๐ฅ + ๐ฅ2 + ๐ฅ3 + โฏ =1
1 โ ๐ฅ, ๐ฅ < 1
โข Therefore,
๐[๐ง] =1
1โ๐พ
๐ง
, ๐พ
๐ง< 1
=๐ง
๐งโ๐พ, ๐ง > ๐พ
โข We notice that the ๐ง โtransform exists for certain values of ๐ง. These values
form the so called Region-Of-Convergence (ROC) of the transform.
Example: Find the ๐ โtransform of ๐ ๐ = ๐ธ๐๐[๐]
โข Observe that a simple equation in ๐ง-domain results in an infinite sequence
of samples.
โข The figures below depict the signal in time (left) and the ROC, shown with
the shaded area, within the ๐ง โplane.
Example: Find the ๐ โtransform of ๐ ๐ = ๐ธ๐๐[๐]cont.
โข Find the ๐ง โtransform of the anticausal signal โ๐พ๐๐ข[โ๐ โ 1], where ๐พ is a
constant.
โข By definition:
๐[๐ง] = โ๐พ๐๐ข โ๐ โ 1 ๐งโ๐ =
โ
๐=โโ
โ๐พ๐๐งโ๐
โ1
๐=โโ
= โ ๐พโ๐๐ง๐
โ
๐=1
= โ ๐ง
๐พ
๐โ
๐=1
= โ๐ง
๐พ
๐ง
๐พ
๐โ
๐=0
= โ๐ง
๐พ1 +
๐ง
๐พ+
๐ง
๐พ
2
+๐ง
๐พ
3
+ โฏ
โข Therefore,
๐[๐ง] = โ๐ง
๐พ
1
1โ๐ง
๐พ
, ๐ง
๐พ< 1
=๐ง
๐งโ๐พ, ๐ง < ๐พ
โข We notice that the ๐ง โtransform exists for certain values of ๐ง, which consist
the complement of the ROC of the function ๐พ๐๐ข[๐] with respect to the
๐ง โplane.
Example: Find the ๐ โtransform of ๐ ๐ = โ๐ธ๐๐[โ๐ โ ๐]
โข We proved that the following two functions:
The causal function ๐พ๐๐ข ๐ and
the anti-causal function โ๐พ๐๐ข[โ๐ โ 1] have:
The same analytical expression for their ๐ง โtransforms.
Complementary ROCs. More specifically, the union of their ROCS
forms the entire ๐ง โplane.
โข Observe that the ROC of ๐พ๐๐ข ๐ is ๐ง > ๐พ .
โข In case that ๐พ๐๐ข ๐ is part of a causal systemโs impulse response, we see
that the condition ๐พ < 1 must hold. This is because, since lim๐โโ
๐พ ๐ = โ, for
๐พ > 1, the system will be unstable in that case.
โข Therefore, in causal systems, stability requires that the ROC of the systemโs
transfer function includes the circle with radius 1 centred at origin within the
๐ง โplane. This is the so called unit circle.
Summary of previous examples
โข By definition ๐ฟ 0 = 1 and ๐ฟ ๐ = 0 for ๐ โ 0.
๐[๐ง] = ๐ฟ ๐ ๐งโ๐ = ๐ฟ 0 ๐งโ0 = 1
โ
๐=โโ
โข By definition ๐ข ๐ = 1 for ๐ โฅ 0.
๐[๐ง] = ๐ข ๐ ๐งโ๐ =โ๐=โโ ๐งโ๐ =โ
๐=0 1
1โ1
๐ง
, 1
๐ง< 1
=๐ง
๐งโ1, ๐ง > 1
Example: Find the ๐ โtransform of ๐น[๐] and ๐[๐]
โข We write cos๐ฝ๐ =1
2๐๐๐ฝ๐ + ๐โ๐๐ฝ๐ .
โข From previous analysis we showed that:
๐พ๐๐ข[๐] โ ๐ง
๐งโ๐พ, ๐ง > ๐พ
โข Hence,
๐ยฑ๐๐ฝ๐๐ข[๐] โ ๐ง
๐งโ๐ยฑ๐๐ฝ, ๐ง > ๐ยฑ๐๐ฝ = 1
โข Therefore,
๐ ๐ง =1
2
๐ง
๐งโ๐๐๐ฝ +๐ง
๐งโ๐โ๐๐ฝ =๐ง(๐งโcos๐ฝ)
๐ง2โ2๐งcos๐ฝ+1, ๐ง > 1
Example: Find the ๐ โtransform of ๐๐จ๐ฌ๐ท๐๐[๐]
โข Find the ๐ง โtransform of the signal depicted in the figure.
โข By definition:
๐ ๐ง = 1 +1
๐ง+
1`
๐ง2+
1
๐ง3+
1
๐ง4= ๐งโ1 ๐
4
๐=0
=1 โ ๐งโ1 5
1 โ ๐งโ1=
๐ง
๐ง โ 11 โ ๐งโ5
๐ โtransform of 5 impulses
๐ โtransform Table
No. ๐[๐] ๐ฟ[๐]
๐]
๐ โtransform Table
No. ๐[๐] ๐ฟ[๐]
Inverse ๐ โtransform
โข As with other transforms, inverse ๐ง โtransform is used to derive ๐ฅ[๐] from ๐[๐ง], and is formally defined as:
๐ฅ ๐ =1
2๐๐ ๐[๐ง]๐ง๐โ1๐๐ง
โข Here the symbol indicates an integration in counter-clockwise
direction around a closed path within the complex ๐ง-plane (known as
contour integral).
โข Such contour integral is difficult to evaluate (but could be done using
Cauchyโs residue theorem), therefore we often use other techniques to
obtain the inverse ๐ง โtransform.
โข One such technique is to use the ๐ง โtransform pair table shown in the
last two slides with partial fraction.
Find the inverse ๐ โtransform in the case of real unique poles
โข Find the inverse ๐ง โtransform of ๐ ๐ง =8๐งโ19
(๐งโ2)(๐งโ3)
Solution
๐[๐ง]
๐ง=
8๐ง โ 19
๐ง(๐ง โ 2)(๐ง โ 3)=
(โ196
)
๐ง+
3/2
๐ง โ 2+
5/3
๐ง โ 3
๐ ๐ง = โ19
6+
3
2
๐ง
๐งโ2+
5
3
๐ง
๐งโ3
By using the simple transforms that we derived previously we get:
๐ฅ ๐ = โ19
6๐ฟ[๐] +
3
22๐ +
5
33๐ ๐ข[๐]
Find the inverse ๐ โtransform in the case of real repeated poles
โข Find the inverse ๐ง โtransform of ๐ ๐ง =๐ง(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3
Solution
๐[๐ง]
๐ง=
(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3=
๐
๐งโ1+
๐0
(๐งโ2)3+
๐1
(๐งโ2)2+
๐2
(๐งโ2)
We use the so called covering method to find ๐ and ๐0
๐ =(2๐ง2 โ 11๐ง + 12)
(๐ง โ 1)(๐ง โ 2)3 ๐ง=1
= โ3
๐0 =(2๐ง2 โ 11๐ง + 12)
(๐ง โ 1)(๐ง โ 2)3 ๐ง=2
= โ2
The shaded areas above indicate that they are excluded from the entire
function when the specific value of ๐ง is applied.
Find the inverse ๐ โtransform in the case of real repeated poles cont.
โข Find the inverse ๐ง โtransform of ๐ ๐ง =๐ง(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3
Solution
๐[๐ง]
๐ง=
(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3=
โ3
๐งโ1+
โ2
(๐งโ2)3+
๐1
(๐งโ2)2+
๐2
(๐งโ2)
To find ๐2 we multiply both sides of the above equation with ๐ง and let
๐ง โ โ.
0 = โ3 โ 0 + 0 + ๐2 โ ๐2 = 3
To find ๐1 let ๐ง โ 0.
12
8= 3 +
1
4+
๐1
4โ
3
2โ ๐1 = โ1
๐[๐ง]
๐ง=
(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3=
โ3
๐งโ1โ
2
๐งโ2 3 โ1
(๐งโ2)2+
3
(๐งโ2) โ
๐ ๐ง = โ3๐ง
๐งโ1โ
2๐ง
๐งโ2 3 โ๐ง
(๐งโ2)2+
3๐ง
(๐งโ2)
Find the inverse ๐ โtransform in the case of real repeated poles cont.
๐ ๐ง = โ3๐ง
๐งโ1โ
2๐ง
๐งโ2 3 โ๐ง
(๐งโ2)2+
3๐ง
(๐งโ2)
โข We use the following properties:
๐พ๐๐ข[๐] โ๐ง
๐งโ๐พ
๐(๐โ1)(๐โ2)โฆ(๐โ๐+1)
๐พ๐๐! ๐พ๐๐ข ๐ =
๐ง
(๐งโ๐พ)๐+1
โข Therefore,
๐ฅ ๐ = [โ3 โ 1๐ โ 2๐(๐โ1)
8 โ 2๐ โ
๐
2 โ 2๐ + 3 โ 2๐]๐ข[๐]
= โ 3 +1
4๐2 + ๐ โ 12 2๐ ๐ข[๐]
Find the inverse ๐ โtransform in the case of complex poles
โข Find the inverse ๐ง โtransform of ๐ ๐ง =2๐ง(3๐ง+17)
(๐งโ1)(๐ง2โ6๐ง+25)
Solution
๐ ๐ง =2๐ง(3๐ง + 17)
(๐ง โ 1)(๐ง โ 3 โ ๐4)(๐ง โ 3 + ๐4)
๐[๐ง]
๐ง=
(2๐ง2โ11๐ง+12)
(๐งโ1)(๐งโ2)3=
๐
๐งโ1+
๐0
(๐งโ2)3+
๐1
(๐งโ2)2+
๐2
(๐งโ2)
Whenever we encounter a complex pole we need to use a special partial
fraction method called quadratic factors method.
๐[๐ง]
๐ง=
2(3๐ง+17)
(๐งโ1)(๐ง2โ6๐ง+25)=
2
๐งโ1+
๐ด๐ง+๐ต
๐ง2โ6๐ง+25
We multiply both sides with ๐ง and let ๐ง โ โ:
0 = 2 + ๐ด โ ๐ด = โ2
Therefore,
2(3๐ง+17)
(๐งโ1)(๐ง2โ6๐ง+25)=
2
๐งโ1+
โ2๐ง+๐ต
๐ง2โ6๐ง+25
Find the inverse ๐ โtransform in the case of complex poles cont.
2(3๐ง+17)
(๐งโ1)(๐ง2โ6๐ง+25)=
2
๐งโ1+
โ2๐ง+๐ต
๐ง2โ6๐ง+25
To find ๐ต we let ๐ง = 0:
โ34
25= โ2 +
๐ต
25 โ ๐ต = 16
๐[๐ง]
๐ง=
2
๐งโ1+
โ2๐ง+16
๐ง2โ6๐ง+25 โ ๐ ๐ง =
2๐ง
๐งโ1+
๐ง(โ2๐ง+16)
๐ง2โ6๐ง+25
โข We use the following property:
๐ ๐พ ๐ cos ๐ฝ๐ + ๐ ๐ข[๐] โ ๐ง(๐ด๐ง+๐ต)
๐ง2+2๐๐ง+ ๐พ 2 with ๐ด = โ2, ๐ต = 16, ๐ = โ3, ๐พ = 5.
๐ =๐ด2 ๐พ 2+๐ต2โ2๐ด๐๐ต
๐พ 2โ๐2 = 4โ25+256โ2โ(โ2)โ(โ3)โ16
25โ9= 3.2, ๐ฝ = cosโ1 โ๐
๐พ= 0.927๐๐๐,
๐ = tanโ1 ๐ด๐โ๐ต
๐ด ๐พ 2โ๐2= โ2.246๐๐๐.
Therefore, ๐ฅ ๐ = [2 + 3.2 cos 0.927๐ โ 2.246 ]๐ข[๐]
โข Since ๐ง = ๐๐ ๐ = ๐ ๐+๐๐ ๐ = ๐๐๐๐๐๐๐ where ๐ =2๐
๐๐ , we can map the
๐ โplane to the ๐ง โplane as below.
For ๐ = 0, ๐ = ๐๐ and ๐ง = ๐๐๐๐. Therefore, the imaginary axis of the
๐ โplane is mapped to the unit circle on the ๐ง โplane.
= Im(s)
Mapping from ๐ โplane to ๐ง โplane
Mapping from ๐ โplane to ๐ง โplane cont.
โข For ๐ < 0, ๐ง = ๐๐๐ < 1. Therefore, the left half of the ๐ โplane is mapped
to the inner part of the unit circle on the ๐ง โplane (turquoise areas).
โข Note that we normally use Cartesian coordinates for the ๐ โ plane
๐ = ๐ + ๐๐ and polar coordinates for the ๐ง โplane (๐ง = ๐๐๐๐).
Mapping from ๐ โplane to ๐ง โplane cont.
โข For ๐ > 0 , ๐ง = ๐๐๐ > 1 . Therefore, the right half of the ๐ โplane is
mapped to the outer part of the unit circle on the ๐ง โplane (pink areas).
Find the inverse ๐ โtransform in the case of complex poles
โข Using the results of todayโs Lecture and also Lecture 9 on stability of causal continuous-time systems and the mapping from the ๐ โplane to the ๐ง โplane, we can easily conclude that:
A discrete-time LTI system is stable if and only if the ROC of its system function ๐ป(๐ง) includes the unit circle, ๐ง = 1.
A causal discrete-time LTI system with rational ๐ง โtransform ๐ป(๐ง) is stable if and only if all of the poles of ๐ป(๐ง) lie inside the unit circle โ
i.e., they must all have magnitude smaller than 1. This statement is based on the result of Slide 5.
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