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8/10/2019 Retenedor de orden cero
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Chapter 3
Impulse Sampling
If a continuous-time signal x(t) is sampled in a periodic manner, mathe-matically the sampled signal may be represented by
x(t) =
k=
x(t) (t kT) =x(t)
k=
(t kT) (1)
orx(t) =
k=
x(kT) (t kT) (2)
x*(t) is the sampled version of the continuous-time signal x(t).Low limits may be changed to k= 0, if x(t) = 0, t
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Data Hold
Data hold is a process of generating a continuous-time signal h(t) from adiscrete-time sequence x(kT). The signal h(t) during the time intervalkT t (k+ 1)Tmay be approximated by a polynomial in as follows:
h(kT+) =ann +an1
n1 + +a1+a0
where 0 Tnote: h(kT) =x(kT)
h(kT+) =ann +an1
n1 + +a1+x(kT)
If the data hold circuit is annth-order polynomial extrapolator, it is calledan nth-order hold. It uses the past n + 1 discrete data x((k n)T),x((k n + 1)T), , x(kT) to generate h(kT+).
ZERO-ORDER HOLD
Ifn= 0 in the above equation, we have a zero order hold so that
h(kT+) =x(kT) 0 < T, k= 0, 1, 2,
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Transfer Function of ZOH
h(t) = x(0) [u(t) u(t T)] +x(T) [u(t T)] u(t 2T)] +
x(2T)[u(t 2T) u(t 3T)] +
=k=0
x(kT) [u(t kT)] u(t (k+ 1)T)]
Now,L[u(t kT)] = ekTs
s
thus, L[h(t)] = H(s) =k=0
x(kT) ekTs e(k+1)Ts
s
= 1 eTs
s Gh0(s)
k=0
x(kT)ekTs
X(s)
Thus, transfer function of ZOH
= H(s)
X
(s) =
1 eTs
s
FIRST-ORDER HOLD
h(kT+) =a1+x(kT), 0 < T, k= 0, 1, 2,
nowh((k 1)T) =x((k 1)T)
so thath((k 1)T) = a1T+x(kT) =x((k 1)T)
or
a1 =x(kT) x((k 1)T)
T
h(kT+) =x(kT) +x(kT) x((k 1)T)
T , 0 < T
kT4T3T2TT
x (kT)
h(t)
Input to FOH
Output from FOH
-T
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TRANSFER FUNCTION
Consider a unit step function input
x(t) =k=0
u(kT) (t kT) =k=0
(t kT)
h(t) = (1 +
t
T) u(t)
(t T)
T u(t T) u(t T)
H(s) = (1
s+
1
T s2)
1
T s2 eTs
1
s eTs
= 1 eTs
s +
1 eTs
T s2
= (1 eTs) T s + 1
T s2
The Laplace transform of unit step
X(s) = L[u(t)] = 11 eTs
thus
Gh1(s) = H(s)
X(s)= (1 eTs)2
T s + 1
T s2
=
1 eTs
s
2T s + 1
T
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Obtaining z transform of functions involving the term 1eTs
s
Suppose the transfer function G(s) follows a zero-order hold
ZOH G(s)
X(s) = 1 eTs
s G(s)
X(z) = Z[X(s)] = (1 z1)Z
G(s)
s
Suppose a transform function G(s) follows a first-order hold
FOH G(s)
X(s) = (1 eTs
s )2
T s + 1
T G(s)
X(z) = (1 z1)2 Z[T s + 1
T s2 G(s)]
Reconstructing Original Signals from Sampled Signals
Sampling Theorem:Defines=
2T
where T is the sampling period, then if
s >21
where 1 is the highest-frequency component present in the continuous-timesignal x(t), then the signal x(t) can be reconstructed completely from thesampled signal x(t).
Intuitive proof
The frequency spectrum of a sampled signal x(t) is given by
X(j) = 1
T
k=X(j +jsk)
* See next page
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Spectrum of Sampled Signal
Fourier series representation of a train of impulses
k=
(t kT) =
n=
Cn ej( 2n
T )t (3)
where
Cn = 1
T
T2
T2
k=(t kT)ejn(
2tT ) dt
= 1
T
T2
T2
(t)ejn(2tT ) dt
(t) is the only value in the integral limit range. So
Cn= 1
T (4)
note: by sifting property,
(t) g(t) dt= g(0)
Thus, (3) & (4)
k= (t kT) = 1T
n= ej(2n
T )t
Fourier series representation of sum of impulses
x(t) =
k=
x(t) (t kT)
L {x(t)} =
x(t)
1
T
n=
ejnst
esT dt
where s= 2
T
X(s) = 1
T
n=
x(t) ejnst est dt
= 1
T
n=
x(t) e(sjns)t dt
Laplace transform with a change of variable
X
(s) =
1
T
n= X(s jns)
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Aliasing
Ifs
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Ifs 21 , the original signal can be recovered by using an ideal lowpass filter.
GI(j) =
1 12s
12s
0 elsewhere
the inverse Fourier transform gives
gI(t) =
1
T
sin(st2)
s t2 unit impulse response
Use a ZOH to approximate ideal low pass filter.
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Frequency response of ZOH
Gh0 =1 eTs
s
Since1 eTs
s =
eTs2 (eT
s2 eT
s2 )
s
s= j e
j
2 T(eT
j
2 eTj
2 )
j
sin T2
ej2 T
2
sin = ej ej
2j
Gh0(j) = T sin( T
2)
T2
ejT2
The ZOH does not approximate an ideal low pass filter very well. Higherorder holds do a better job but are more complex and have more time delay,
which reduces stability margin. So ZOHs get used a lot.
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The Pulse Transfer Function
Convolution Summation
y(kT) =k=0
g(kT hT) x(kT)
=k=0
x(kT hT) g(kT)
x(kT) g(kT)
where g(kT) is the systems weighting sequence. i.e. g(k) = Z1 {G(z)}
Starred Laplace Transfrom
Y(s) = G(s) X(s)
Note: X(s) is periodic with period 2s
since X(s) =X(s jsk), fork =0, 1, 2, . G(s) is not periodic.
Taking the starred transform {See notes on next page }
Y(s) = [G(s)X(s)] = [G(s)]X(s) = G(s)X(s)
Note: X(s) = 1T
k= X(s +jsk) + 12x(0
+)
Y(z) =G(z) X(z)
Since the z transform can be seen to be the starred Laplace transform witheTs replaced by z; i.e. X(s) =X(z)
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Obtaining The Pulse Transfer Function
The presence or absence of an inputsampler is crucial in determining thepulse transfer function of a system.For figure a,
Y(s) = G(s) X(s)
Y(s) =G(s) X(s)
Y(z) =G(z) X(z)
For figure b,Y(s) =G(s) X(s)
Y(s) = [G(s) X(s)] = [GX(s)] =GX(z) =G(z)X(z)
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Methods for obtaining the z transform
1. X(z) = Z[X(s) expanded into partial fractions Xi(s)] =
i {Z[Xi(s)]}
2. X(z) =
k=0 x(kT)zk
3. X(z) =
residue of X(s)zzeTs
at pole of X(s)
Pulse transfer function of cascaded elements
For figure (a)
U(s) =G(s) X(s) U(s) =G(s) X(s)
Y(s) =H(s) U(s) Y(s) =H(s) U(s)
Y (s) = H(s) G(s) X(s)
= G(s) H(s) X(s)
Y(z) =G(z) H(z) X(z) Y(z)X(z)
=G(z) H(z)
For figure (b)
Y(s) =G(s) H(s) X(s) =GH(s) X(s)
Y (s) = [GH(s)] X(s)
Y(z) =GH(z) X(z) Y(z)
X(z)=GH(z) = Z[GH(s)]
N ote that G(
z)
H(
z) =
GH(
z) = Z[
GH(
s)]
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Pulse transfer function of closed-loop systems
C(s) =G2(s) U
(s) (5) C(s) =G2(s) U
(s) (6)
E(s) =R(s) H(s) C(s) (7)
U(s) =G1(s) E(s) = G1(s) R(s) G1(s) H(s) C(s)
= G1(s) R(s) G1(s) H(s) G2(s) U(s)
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U(s) = [G1(s) R(s)] [G1(s) Gs(s) H(s) U
(s)]
= [G1R(s)] [G1G2H(s)]
U(s)
{1 + [G1G2H(s)]} U(s) = [G1R(s)]
U(s) = [G1R(s)]
1 + [G1G2H(s)](8)
(6)&(8)
C(s) = G2(s) [G1R(s)]
1 + [G1G2H(s)] C(z) = G
2(z) G1R(z)1 +G1G2H(z)
Example
Find C(z)R(z)
C(z)
R(z)=
G(z)
1 +G(z), for H(s) = 1
now
G(z) = Z[G(s)] = Z
(1 eTs)
K
s(s + 1)
= (1 z1)Z
K
s
K
s + 1
= (1 z1)
K
1 z1
K
1 eTz1
= K(1 eT)z1
1 eTz1
closed-loop
C(z)
R(z)=
K(1 eT)z1
1 + [K (K+ 1)eT]z1
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Obtaining response between consecutive sampling instants
The z transform does not give the response between sampling instants.Can use the following methods to do so.
1. The Laplace transform method
2. The modified z transform method
3. The state space method
Will look only at first method for now.
1. The Laplace transform method
Example: Consider the following system
We know that
C(s) = G(s) E(s) =G(s) R(s)
1 +GH
(s)
Thus
c(t) = L1[C(s)] = L1
G(s) R(s)
1 +GH(s)
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Mapping between the s plane and the z plane
z= eTs
lets= + j
z= eT(+j) =eT ejT =eTej(T+2k)
now|z| =eT
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Constant-attenuation loci
Constant-frequency loci
Constant-damping ratio line
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Stability analysis of closed-loop systems in the z domain
Consider the following pulse transfer function
C(z)
R(z) =
G(z)
1 +GH(z)
Stability may be accessed by looking at the roots of the the characteristicequation.P(z) = 1 + GH(z) = 0
1. For stability, require that the roots |zi| 0
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General Form of the Jury stability table
Row z0 z
1 z
2 z
n-k z
n-1 z
n
1 a0 a
1 a
2 a
n-k a
n-1 a
n
2 an a
n-1 a
n-2 a
k a
1 a
0
3 b0 b
1 b
2 b
n-k b
n-1
4 bn-1
bn-2
bn-3
bk b
0
5 c0 c
1 c
2 c
n-2
6 cn-2
cn-3
cn-4
c0
::
::
::
::
::
2n-5 p0 p
1 p
2 p
3
2n-4 p3 p
2 p
1 p
0
2n-3 q0 q
1 q
2
where
bk =
a0 ankan ak
ck = b0 bn1kbn1 bk
dk =
c0 cn2kcn2 ck
...
q0 =
p0 p3p3 p0
q2 =
p0 p1p3 p2
The necessary and sufficient condition for the F(z) to have no roots onand outside the unit circle are
F(1)> 0
F(1)
>0 n even |bn1|
|c0| > |cn2||d0| > |dn3||q0| > |q2|
(n 1) constraints
For a second order system, n=2, Jurys table contains only one row
F(1)> 0
F(1)> 0
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|a0| < an
Bilinear transformation and Routh stability criterion
Transform the z plane to the w plane by
z=w + 1
w 1
w =
z+ 1
z 1
which maps the inside of the unit circle in the zplane into the left half ofthe w plane. The unit circle in zplane maps into the imaginary axis in thew plane and the outside of the unit circle in zplane maps into RHP ofplane.
The w plane is similar to s plane (but not quantitatively) canuse Routh test.
LetP(z) =a0z
n +a1zn1 + +an1z+an= 0
a0w + 1
w 1
n+a1
w + 1w 1
n1+ +an1
w + 1w 1
+a0 = 0
Q(w) =b0wn +b1w
n1 + +bn1w +bn= 0
Requires a lot of computation but can now use Routh test.
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