retenedor de orden cero

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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