DSP & Digital Filters
Transcript of DSP & Digital Filters
DSP & Digital Filters
Lecture 1 z-Transform
DR TANIA STATHAKIREADER (ASSOCIATE PROFESSOR) IN SIGNAL PROCESSINGIMPERIAL COLLEGE LONDON
β’ Recall that in order to describe a continuous-time signal π₯(π‘) in frequency
domain we use:
β The Continuous-Time Fourier Transform (or Fourier Transform):
π(π) = ΰΆ±
ββ
β
π₯(π‘)πβπππ‘ππ‘
β The Laplace Transform
π π = ΰΆ±
ββ
β
π₯(π‘)πβπ π‘ππ‘
β’ The above transforms and their basic properties are considered known in
this course.
β’ If you have doubts please consult any book on Signals and Systems.
Continuous-time signals
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‘ = 1β 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―
Discrete-time signals
The z-transform derived from the Laplace transform
β’ From the Laplace time-shift property, we know that an additional term
π§ = ππ π in the Laplace domain, corresponds to time-advance by πseconds (π is the sampling period) of the original function in time.
β’ Accordingly, π§β1 = πβπ π corresponds to a time-delay of one sampling
period.
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― = Οπ=0
β π₯[π]π§βπ, π₯π = π₯[π]
β’ The bilateral π βtransform for any sampled sequence is:
π[π§] =
π=ββ
β
π₯[π]π§βπ
πβπ: the sampling period delay operator
β’ Find the π§ βtransform of the causal signal πΎππ’[π], where πΎ is a constant.
β’ By definition:
π[π§] =
π=ββ
β
πΎππ’ π π§βπ =
π=0
β
πΎππ§βπ =
π=0
βπΎ
π§
π
= 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 rational equation in π§-domain corresponds to an
infinite sequence of samples in time-domain.
β’ The figures below depict the signal in time (left) for πΎ < 1 and the ROC,
shown with the shaded area, within the π§ βplane.
Example: Find the π βtransform of π π = πΈππ π cont.
β’ Consider the causal signal π₯ π = Οπ=1πΎ πΎπ
ππ’[π] with π(π§) = Οπ=1πΎ π§
π§βπΎπ.
β’ In that case the ROC is the intersection of the ROCs of the individual
terms, i.e., the intersection of the sets π§ > πΎπ i.e., ROC: π§ > πΎmax
β’ In case that π₯ π is the impulse response of a system, the transfer function
of the system is the rational function π(π§) = Οπ=1πΎ π§
π§βπΎπwith poles πΎπ.
β’ The above analysis yields the following properties regarding the ROC:
PROPERTY:
If π π is a causal signal, the ROC of its π βtransform is π > πΈπ¦ππ±
with πΈπ¦ππ± the maximum magnitude pole of the π βtransform.
β In the general case of π π being a right-sided signal (RSS) the
ROC is as above but might not include β (think why).
PROPERTY:
No pole can exist in ROC.
Generic form of a causal signal
β’ The signal π₯ π = Οπ=1πΎ πΎπ
ππ’[π] is bounded only if πΎπ < 1 βπ or πΎmax < 1.
β’ In that case the ROC includes a circle with radius equal to 1. This is known
as the unit circle.
β’ The above observation yields the following property:
PROPERTY:
If the ROC of πΏ(π) includes the unit circle in π βplane, then the signal
in time is bounded and its Discrete Time Fourier Transform exists.
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 unit circle.
Generic form of a causal signal cont.
β’ Find the π§ βtransform of the anti-causal signal βπΎππ’[βπ β 1], where πΎ is a
constant.
β’ By definition:
π[π§] =
π=ββ
β
βπΎππ’ βπ β 1 π§βπ =
π=ββ
β1
βπΎππ§βπ = β
π=1
β
πΎβππ§π = β
π=1
βπ§
πΎ
π
= βπ§
πΎ
π=0
βπ§
πΎ
π
= 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’ Consider the anti-causal signal π₯ π = Οπ=1πΎ βπΎπ
ππ’[βπ β 1] with
π§ βtransform π(π§) = Οπ=1πΎ π§
π§βπΎπ.
β’ In that case the ROC is the intersection of the sets π§ < πΎπ , i.e., ROC:
π§ < πΎmin
β’ In case that π₯ π is the impulse response of a system, the transfer function
of the system is the rational function π(π§) = Οπ=1πΎ π§
π§βπΎπwith poles πΎπ.
β’ The above analysis yield the following property regarding ROCs:
PROPERTY:
If π π is an anti-causal signal, the ROC of its π βtransform is π <πΎmin with πΎmin the minimum magnitude pole of the π βtransform.
β In the general case of π π being a left-sided signal (LSS) the ROC
is as above but might not include π (think why).
Generic form of an anti-causal signal
β’ We proved that the following two functions:
βͺ The causal function πΎππ’ π and
βͺ the anti-causal function βπΎππ’[βπ β 1] have:
β The same analytical expression for their π§ βtransforms.
β Complementary ROCs. More specifically, the union of their ROCS
forms the entire π§ βplane.
β’ The above observations verify that the analytical expression alone is not
sufficient to define the π§ βtransform of a signal. The ROC is also required.
Summary of previous examples
β’ Example: Find the π§ βtransform of the two-sided signal:
π₯ π = 2ππ’[π] β 4ππ’[βπ β 1]
Based on the previous analysis we have:
π π§ =π§
π§β2+
π§
π§β4, ROC: π§ > 2 β© π§ < 4 or ROC: 2 < π§ < 4
β’ Example: Find the π§ βtransform of the two-sided signal:
π₯ π = 4ππ’[π] β 2ππ’[βπ β 1]
Based on the previous analysis we have:
π π§ =π§
π§β2+
π§
π§β4, ROC: π§ > 4 β© π§ < 2 or ROC: β
PROPERTY:
If π π is two-sided signal then the ROC of its π βtransform is of the
form:
β πΎ1 < π < πΎ2 with πΎ1, πΎ2 poles of the system or
β β
Two-sided signals
β’ By definition πΏ 0 = 1 and πΏ π = 0 for π β 0.
π[π§] =
π=ββ
β
πΏ π π§βπ = πΏ 0 π§β0 = 1
β’ By definition π’ π = 1 for π β₯ 0.
π[π§] = Οπ=βββ π’ π π§βπ = Οπ=0
β π§βπ =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] βπ§
π§βπΎ, π§ > πΎ
β’ Hence,
πΒ±ππ½ππ’[π] βπ§
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=0
4
π§β1 π =1 β π§β1 5
1 β π§β1=
π§
π§ β 11 β π§β5
π βtransform of 5 impulses
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
2ππΰΆ»π[π§]π§πβ1ππ§
β’ Here the symbol Χ― indicates an integration in counter-clockwise
direction around a circle within the ROC and π§ = π πππ.
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 a π§ βtransform pairs Table shown in the
last two slides with partial fraction expansion.
Inverse π βtransform: Proof
Proof:
1
2ππΰΆ»π[π§]π§πβ1ππ§ =
1
2ππΰΆ»
π=ββ
β
π₯[π]π§βπ π§πβ1ππ§
= Οπ=βββ π₯[π]
1
2ππΧ― π§πβπβ1ππ§ = Οπ=ββ
β π₯[π] πΏ π βπ = π₯[π]
β For the above we used the Cauchyβs theorem:
1
2ππΧ― π§πβ1ππ§ = πΏ π for π§ = π πππ anti-clockwise.
ππ§
ππ= ππ πππ β
1
2ππΧ― π§πβ1ππ§ =
1
2πππ=02π
π πβ1ππ(πβ1)π ππ πππππ =
π π
2ππ=02π
ππππ ππ = π π πΏ π
[ π π
2ππ=02π
ππππ ππ = α0 π β 0
π π
2π2π = π π π = 0
]
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)(πβ3)
Solution
π[π§]
π§=
8π§ β 19
π§(π§ β 2)(π β 3)=(β
196)
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)
βͺ We use the so called covering method to find π and π0
π = ΰΈ(2π§2 β 11π§ + 12)
(π§ β 1)(π§ β 2)3π§=1
= β3
π0 = ΰΈ(2π§2 β 11π§ + 12)
(π§ β 1)(π§ β 2)3π§=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)
βͺ 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
βͺ 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:
βͺ πΎππ’[π] βπ§
π§βπΎ
βͺπ(πβ1)(πβ2)β¦(πβπ+1)
πΎππ!πΎππ’ π β
π§
(π§βπΎ)π+1
[β2π§
π§β2 3 = (β2)π§
π§β2 2+1 β (β2)π(πβ1)
222!πΎππ’ π = β2
π(πβ1)
8β 2ππ’[π]
β’ Therefore,
π₯ π = [β3 β 1π β 2π(πβ1)
8β 2π β
π
2β 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΅
25β π΅ = 16
π[π§]
π§=
2
π§β1+
β2π§+16
π§2β6π§+25β 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] βπ§(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 = tanβ1π΄πβπ΅
π΄ πΎ 2βπ2= β2.246πππ.
Therefore, π₯ π = [2 + 3.2 cos 0.927π β 2.246 ]π’[π]