|
Impulse
response of a sampled data system |
Characteristics
of a digital controller (in the difference equations form) to give a
specified response of the system to a
unit step function. |
Position
servo operated by a sampler and a zero order hold |
To obtain the
state equation of a discrete data system
described by its difference equation |
Solution of
the discrete time model of a system |
|
To obtain the state equation of a
discrete data system described by its
difference equation |
Discrete model of the
tape-drive system
when a digital controller is used
|
Examples of discrete and continuous
signals |
Stability of a discrete time system using bilinear
transformation |
Stability from the characteristic polynomial corresponding to the pulse
transfer function of a sampled data system
|
|
Position servo operated by a sampler and
a zero order hold |
Examples of discrete and continuous
signals |
Drawing the simulation diagram of a system
represented by the transfer function
G(z |
To obtain the equivalent discrete time
equation from the state space description of a continuous time plant |
The z-transform |
|
Pulse transfer function of a discrete time system |
|
|
|
OBJECTIVE TYPE QUESTIONS:
|
Problem 9.1: Impulse response
of a sampled data system
1/[s(s+1)] Zero-order hold
Fig.1
Solution:
From Fig.
G(s)=(1-exp(-st))/[s2(s+1)] =(1-exp(-st))[1/s2 –1/s +1/(s+1)]
G(z)= Z{G(s)}(1-z2)Z{1/s2 –1/s +1/(s+1)}
For T=1, we therefore have
G(z)= z exp(-1)+1-2exp(-1) = 0.3678z+0.2644
(z-1)(z-exp(-1)) z2-1.3678+.3678
For unit impulse R(z) = 1 so that C9z) =G(z)
Dividing denominator into the numerator, we get
C (z) = .3678z-1 + .7675 z-2 + .9145 z-3 +…
The response is
C(0)=0
C(T)= .3678
C(2T)= .7675
C(3T)= .9145 etc.
Problem 9.2: Characteristics
of a digital controller (in the difference equations form) to give a
specified response of the system to a
unit step function.
Fig. 2 shows a digital control
system.
Digital controller
Hold Plant
-
Fig. 2
(a)
Show that the z
transform, D(z), of the digital controller
is given by
D(z) =
[1/G(z)].[C(z)/R(z)]/[1 – {C(z)/R(z)}]
where G (z) is the z
transform of zero order hold and plant.
(b)The
transfer function of the plant is
(s+2)/[s(s+1)].
Determine the characteristics of
a digital controller (in the difference equations form) such that the response
of the system to a unit step function will be c (t) = 5(1-e-2t). The
sampling period is 1 s.
Use the following table of
Time function -
u (t)-1/s-z/(z-1)
t-1/s2-zT/(z-1)2
e-at-1/(s+a)-z/(z-e-aT)
Solution:
For the plant and zero order hold G(s) = G1(s)G2(s) = (1- e-Ts) (s+2)/[s2(s+1)]
G1(s) = 1-e-Ts.
G2(s) = (s+2)/s2(s+1) =(2/s2)-(1/s)+(1/(s+1))
G(z) =[(z-1)/z]{ [2z/(z-1)2] –[z/(z-1)]+[z/(z-.368)]
=(1.368z-0.104)/(z2-1.368z+0.368)
For
C(s) =5[(1/s)-(1/(s+2))],
Then C(z) =5[((z/(z-1)) –(z/(z-0.135))] = 4.32z/[(z-1)(z-0.135)]
C(z)/R(z) =4.32/(z-.135) and 1-(C(z)/R(z)) =(z-4.45)/(z-.135)
D(z) =[z2-1.368z+.368]/[1.368z-.104].[4.32/(z-4.45)]
= [4.32z2-5.91z+1.59]/[1.368z2-6.202z+.46]
Dividing numerator and denominator by 1.368z2 and cross-multiplying gives
M(z) -4.53z-1M(z)+.34z-2M(z) =3.16E(9z)-4.32z-1E(z) +1.16z-2E(z)
Inverting gives the controller characteristics
m(k)
=4.53m(k-1)-.34m(k-2)+3.16e(k)-4.32e(k-1)+1.16e(k-2)
Problem 9.3: Position servo
operated by a sampler and a zero order hold
(a)
Derive a state
variable equation for a system governed
by the second –order difference equation
c(k+2)+3 c(k+1)
+2 c(k)
where u (k) is the input and c (k) is the output.
(b)
Fig.4 shows a
position servo operated by an input via a sampler and a zero order hold with
the position c and velocity dc/dt represented by x1and
x2 respectively. If the states at the sampling instants alone are required ,determine the A and B matrices in the following equation
x1(k+1) = [A
] x1(k) + [B]
u(k)
x2
(k+1) x2(k)
Process 2/(s2+3s+2) Zero-order hold
T
Fig.4
Solution:
(a)
Let x1(k) =c(k)
X2(k) = x1(k+10 =c(k+1)
Then x1(k+1) = x2(k)
X2(k+1) =-2x1(k)-3x2(k0+u(k)
Then in matrix form:
-2 -3 1
(b)
X(k+1) = j(T) x(k) + Y(T) u(k)
j(T) =L-1(sI-A)-1=
s -1
-1
2 s+3
= 2e-t -e-2t e-t -e-2t
-2e-t +e-2t -e-t +2e-2t
-2e-T +e-2T -e-T +2e-2T
T
Also Y(T) = ò j(ג)
b. d ג
0
T
= 2e-ג -e-2ג e-ג -e-2ג
0 d
ג
-2e-ג +e-2ג -e-ג +2e-2ג 1
0
= 0.5 -e-T +
0.5 e-2T
e-T -
e-2T
Setting T =1,
j(T) = 0.6005 0.2326
-.4652 -0.0973
Y(T) = 0.1998
0.2326
x(k+1) = 0.6005 0.2326 x(k) + 0.1998 u(k)
-.4652 -0.0973 0.2326
Problem 9.4: To obtain the
state equation of a discrete data system described by its difference equation
(a)
Write a short note
on the existence and uniqueness of the solution of state equations.
(b) A
discrete –data system is described by the difference equation
c (k+2)
+5c(k+1) + 3c(k) =u (k+1) +2u(k)
where u (k) is
the reference input and c (k) is the output. Show that the state equation of
the system is
x
(k+1) = 0
1 x (k) +
1 u (k)
-3 -5 -3
|
Assume zero initial conditions and describe the
technique used to arrive at the above equation. |
Solution:
(b)
Taking z-transform of the difference equation, and assuming zero initial conditions yields
Z2 c (z)+5z c (z) +3c(z)=zu (z) +2u(z)
c (z)/u (z) =(z+2)/(z2+5z+3)
Characteristic eqn;
Z2+5z+3 =0
Define the state variables as x1 (k)= c (k), x2 (k)= x2 (k+1)- u (k),
Substitution of the state variables into the original difference eqn. gives
x1 (k+1)=x2 (k) + u (k),
x2 (k+1)= -3 x1 (k)-5 x2 (k)-3 u (k),
Problem 9.5: Solution of the
discrete time model of a system
G2 H G1
+
|
R(s) +
+
C(s) |
-
x1
(k+1) 0.1 0
0 0 x1 (k) 1
u (k+1)
x2
(k+1) = 0.7
0 .1 0 0 x2 (k) +
0
x3
(k+1) 0 0.82
0.08 0 x3 (k) 0
x4
(k+1) 0 0 0.85 0.02 x4 (k) 0
y (k) = 0
0
0
0.97
In the above, xi
(k) is the number of students in the ith year of the
four-year degree program in academic year k, u (k) represents the input (that
is, the enrollment in the first-year class in year k) and y (k) is the output
(that is, the number of students graduating in year k).
In July 1992, the college had
188 students in the first year class, 126, in the second year class, 103 in the
third year class, and 72 in the fourth year class.
The table below gives the
history of admissions to the first year class in the subsequent years.
Year-1993-1994-1995-1996-1997-1998-1999
Admissions-275-320-350-400-450-450-430
Use this model to determine the number of students graduating in May of each year from 1994 to 2002.
Solution:
(a) There are two
forward paths from R(s) to C(s), with transmittance G1 and G2.
There is on loop with transmittance –G1H, and this is touching both
the forward paths. Hence,
C(s)/R(s) = (G1+G2)/(1+G1H)
(b) The discrete –time model may be written as
x (k+1) = F.x
(k) +G.u (k)
y (k+1) = C.x (k+1)
From the given data,
x (1992)=
126
103
72
u (1993)=275, u (1994)=320,u (1995)= 350,u (1996)=400, u (1997)=450, u (1998)=450,
u (1999)=430.
Y (1994)=C [(Fx
(1992)+G.u ((1993)] = 86 (Rounding off as integer)
Y (1995)= C [(Fx
(1993)+G.u ((1994)] = 94
Y (1996)= C [(Fx
(1994)+G.u ((1995)] = 107
Y (1997)= C [(Fx
(1995)+G.u ((1996)] = 159
Y (1998)= C [(Fx
(1996)+G.u ((1997)]= 196
Y (1999)= C [(Fx
(1997)+G.u ((1998)] = 220
Y (2000)= C [(Fx
(1998)+G.u ((1998)] =249
Y (2001)= C [(Fx
(1999)] =281
These are not affected by u (2000) and u (2001)
Y (2002)= C [(Fx
(2000)] =290
Problem 9.6: To obtain the
state equation of a discrete data system described by its difference equation
(a) Derive a state variable equation for a
system governed by the second-order difference equation:
c (k+2) +3c(k+1)+4c(k) = u (k)
where u (k) and
c (k) are the input and output signals respectively. Write the values of the A,
B, C, D matrices.
Draw
the block diagram of the system, showing the A, B, C, D matrices on it.
3 -4 1
C= [1 0 ] D=0
If the input is zero, and x (0) = 0,
1
solve for x (k) and y (k), k= 1,2,3,4 etc.
Solution:
(a) Set x1 (k) =c (k)
and x2 (k) =x1 (k+1)=c (k+1). Then we have
X1 (k+1) = x2 (k)
X2 (k+1) =-4x1(k)-3x2(k)+u (k)
In matrix-vector form, x (k+1) =Ax (k)+Bu (k)
4 -3 1
Unit delay T=1
C
B
|
|
A
(b)
As u (k) = 0, we have
x (k+1) =Ax (k) with x (o)= 0
x (1) = 1
4
13
40
121 and y = (0, 1, -4, 13, -40 etc)
(a)Fluid flowing through an orifice can be
represented by the non-linear equation
Fig.6 shows the variables,
and K is a constant. Determine a linear approximation to the fluid flow
equation.
(b) A slack loop is
maintained between two sets of drive rolls, Fig.7, in a certain computer tape
deck.
O O
O O
L
(t)
Fig.7
All rolls have diameter D.
Under normal operating conditions drive speeds are W1(t)= W2(t)= W’ and slack loop length
is L (t) = L’. Starting at time t0, there are small
independent variations in the drive speeds so that W1(t)= W’ +w1(t) and W2(t)= W’ +w2(t). Derive an
expression for the resulting change in the loop length l (t).
(i) assuming that wi =(wi) des, i = 1, 2.Neglect roller dynamics.
(ii) assuming that wi and (wi) des are related by the differential equation
tdwi/dt =(wi) des-wi , i = 1,2, where t is the time constant of the drive motors.
Solution:
(a) Q = KDP1/2, where DP= P1-P2
¶Q=K/(2DP01/2) ¶P
(b) dL/dt = pD (W1-W2)/2=pD (w1-w2)/2
L = L’ +l
Therefore, dL/dt = dl/dt
Hence, dl/dt = pD (w1-w2)/2, l (t0) =0
©
(i) The incremental model is dl/dt = (D/2)[w1(t)-w2(t)]
Due to the assumed absence of roller dynamics, w1and
w2
are constant over the interval kT< t
<
(k+1) T.
Therefore, l (kT+T)=l (kT) +(DT/2)[w1(kT)-w2(kT)]
(ii) Assuming first-order dynamics,
kT+T
l (kT+T) = l (KT) +(D/2)ò[w1(t)-w2(t)] dt
kT
w1(t)-w2(t) =[w1(kT) des-w2(kT) des](1-e- (t-kT)/t)
Therefore,
L (kT+T) =l (kT) + =[w1(kT) des-w2(kT) des](DT/2)[1-(t/T)(1-e-T/t)]
Problem 9.8: Examples of
discrete and continuous signals
State whether the following signals are discrete or
continuous.
(i) elevation contours on a map
(ii) the score of basketball game
(iii) signals leaving or entering the CPU of a
computer.
Solution:
(i) continuous (ii)discrete (iii)discrete
Problem 9.9: Stability of
a discrete time system
using bilinear transformation
(a)
How would you modify
the Routh-Hurwitz criterion to study the stability of
a closed-loop discrete-time system?
(b)
The characteristic
polynomial of a closed-loop discrete-time system is given by
z3
+ 5z2 + 3z +2 =0.
Determine the stability
of the system using the bilinear transformation.
Solution:
(b)
f(z)= z3+5z2+3z+2=0
Substitute z= (w+1)/(w-1) and obtain
F(w) =11 w3-w2+w-3 =0
Construct Routh table. From Routh table,there is one sign change in the first column.
Hence UNSTABLE system.
(a)
Discuss the use of
the bilinear transformation
z = (r+1)/(r+2)
to determine the stability of a sampled data control
system.
f(z) = z3 +5.94z2 + 7.2 z –0.368
Investigate the stability of the system.
Solution:
(b)
Using the bilinear transformation, the system characteristic equation becomes
((r+1)/(r-1))3+5,94((r+1)/(r-1))2+7.7(r+1)/(r-1)-.368=0
or
f( r)=14.27r3+2.3 r2-11.47r+3.13=0
r3 14.27 -11.47
r2 2.3 3.13
r1 -31.1
r0 3.13
Two sign changes. Two zeroes of he characteristic polynomial f(z) will remain outside the unit circle. System is UNSTABLE.
Problem 9.11: Position servo
operated by a sampler and a zero order hold
(a)
What is the state
transition matrix? Explain its use.
(b)
Fig.9 shows a
sampled data position servo system with error-sampling and zero-order hold
(ZOH). Fig. 10 shows its state-variable diagram. Determine qo and dqo/dt at
sampling instants. Find the solution of the state vector x(k) for a step input qi (t) at sampling instants T= 1 and 2 seconds. Assume zero
initial conditions. Use state transition matrix.
ZOH
ZOH 1/s 1/s
·
Fig.10
Solution:
(b) Double vertical bars indicate matrices.
A= 0 1
0 -1
1 t=T
= Laplace-1 s -1
0 s+1 t=T
0 1/s t=T
= 1 1-e-T = 1 .632
0 e-T 0
.368
T
y(T)= òf(l)Bdl
0
y(T)= ò 1 1-e-l 0 dl
0 0 e-l 1
T + e-T-1
= -e-T+1
T=1
= .368
.632
State variable equations become
X(k+1) = 1 .632 X(k) + .368 [r(k)-X(k)]
0 .368 .652
Problem 9.12: Examples of
discrete and continuous signals
-
(a)
State whether the
following signals are discrete or continuous.
i.
Temperature in a
room
ii.
Digital clock
display
iii.
The output of a
loudspeaker
Solution:
ii. Discrete
iii.
Continuous
Problem 9.13: Drawing
the simulation diagram of a system represented by the
transfer function G(z)
(a)
State whether the
following systems are discrete or continuous:
i.
Elevation contours
in a map
ii.
Temperature in a
room
iii.
Digital clock
display
iv.
The score of a basketball game
v.
The output of a loudspeaker.
(b)
A discrete –time system can be simulated on a digital computer in the same way
as a continuous-time system on an analogue computer. The only difference is
that integrators are replaced by delays. This implies that the blocks
containing s-1 are replaced by blocks containing z-1. Use
this approach to
draw the simulation diagram for a system represented by the transfer function
3z2
+ 4z
z3 – 1.2z2
+0.45 z – 0.05
Solution:
1- 1.2z-1 +.45z-2-.05z-3
3 4 +
y
|
+ z-1 z-1 z-1 |
1.2 -.45 .05
(a) Describe a numerical method for the solution
of continuous time state equations.
(b)The
state space description for a
continuous-time plant is
A = 0 1 B = 0
0 -1 1
Obtain
the equivalent discrete-time equation. Take the sampling instant, T = 1s.
Solution:
(b) x(t) = exp(A(t-t0))x(t0) +¦ exp(A(t-t0))Bu(t)dt
Setting t0=kT and t= (k+1)T and assuming u(t)=u(kT) for kT less than/equal to t less than(k+1)T,
we get
X(k+1) =f(T)x(k)+q(T)u(k)
Where f(T) =exp(At)
q(T)=¦exp(Al)Bdl
0 -exp(-t)
Selecting T=1s,we get f(T) = 1 .632
0 ,368
|
|
.632
Therefore X(k+1) = 1 .632 x1(k) + .368 u(k)
0 ,368 x2(k) .632
Problem 9.15: The
z-transform
F(z) = z sin wt
z2-2z cos wt +1
(b)
Find f(kT) if
F(z) = 10z
(z-1)(z-2)
(c)
If G*(s) =å f(nT) e-nTs, show that the closed form solution for
G(s) = 1
(s+p)
is
G*(s) = 1
1-e T(s+p).
Solution:
(a)g(t)=e-pt.
= å¥n=0 e-nT (p+s)
= å¥n=0 Xn where X= e-T (p+s)
å¥n=0 Xn =1/(1-X)
G*(s) = 1
1-e-T (p+s)
(b) sinwt=
exp(jwt)-exp(-jwt)
2j
F(z) =(1/2j){z - z }
z-exp(jwt) z-exp(jwt)
= zsinwt
z2-2zcoswt+1
|
(c ) X(z)/z = 10 |
(z-1)(z-2)
X(z)
= -10z + 10z
(z-1) (z-2)
x(kT) = -10 +10.2k
=10(-1 +2k) (k=0,1,2…)
or, x(0)=0
x(T)= 10
x(2T)= 30
x(3T)= 70
x(4T)= 150
Problem 9.16: Pulse transfer
function of a discrete time system
G(s) = 10
s(s+1)
G(s)
-
Show that the system is
unstable.
(b)
Obtain the pulse transfer function of the closed-loop system shown in Fig.13.
G(s)
-
H(s)
Solution:
G(z)
= 10(1- exp(-1))z
(z-1) (z-exp(-1))
Characteristic equation: 1+G(z) =0
(z-1) (z-exp(-1)) +10 (1-exp(-1))z=0
exp(-1)=).368
therefore, z2 +4.952z +.368=0
z= =0.076,-4.876
Magnitude of one root >1
Therefore, unstable.
(b)
R G(s)
H(s)
E(s) =R(s) – B(s)
E(z)= R(z) – B(z)
C(z) = G(z) E(z)
B(z) = H(z)C(z)
Therefore, C(z) =[ R(z) –H(z)C(z)]G(z)
Or, C(z)/R(z) = G(z)
1+H(z) G(z)
Answer the following:
(i)
In multiple- rate
sampling
(a)
the sampling
instants are random
(b)
two concurrent sampling operations occur at tk = pT1and qT2 where T1and
T2 are constants and p,q are integers
(c)
the sampling instants are equally spaced, or tk = kT (k = 0, 1,
2….)
(d)
the pattern of the tk is repeated
periodically
Ans: ( b )
(ii)
Match List E with
List F given below.
|
List E |
List F |
|
||
|
A |
Analogue controller |
I |
Are
high performance controllers and are combinations of analogue & digital
controllers. |
|
|
B |
Digital controller |
II |
Represent
the variables in the equations by continuous physical quantities and can be
designed that will serve as nondecision making controllers |
|
|
C |
Hybrid controller |
III |
Operate
only on numbers and are currently being used for the solution of optimal
operation of industrial plants. |
|
The correct matching is
(a)AII BIII CI
(b) AI BII CIII
( c ) AIII BII CI
(d)AI BIII CII
Ans: (a)
( iii) The
z-transform of f(t) = et is
(a)
z/(z-1)
(b)
z/(z-eT)
(c)
z/(z- e-jT)
(d)
Tz/(z-1)2
Ans: ( b)
(iv ) The initial
and final values of the time function corresponding to the z-transform
4z3-5z2+8z
are
(a)
Zero and
indeterminate
(b)
2 and 14
(c)
4 and 28
(d)
8 and 56
respectively.
|
Ans: ( c) |
( v )
Match List E with List F in the following Table of z transforms.
List E, List F
A, x(t)=d(t), I, x(z)= z
/(z-1)
B, x(t)=u(t), II, x(z)=Tz/(z-1)2
C, x(t)=t, III, x(z)=z/(z-e-T)
D, x(t)=e-t, IV, x(z)=1
The correct matching is
(a)AIII BII CIV DI
(b)AIV BI CII DIII
©AII BIII CI DIV
(d)AI BIVCIII DII
( vi ) Match List E with List F in the following Table
of transducer types.
A, Sampled data
transducer, I, A
transducer in which the input signal is a quantized signal and the output
signal is a smoothed continuous function of time.
B, Digital transducer, II,
A transducer in which the input signal is a continuous function of time and the
output signal is a uantized signal which can assume
only certain discrete levels
C, Analogue to Digital
transducer, III, A transducer in which the input and output signals occur only
at discrete instants of time, but the magnitudes of the signals are unquantized
D, Digital to Analogue
transducer, IV, A transducer in which the input and output signals occur only
at discrete instants of time, and the signal magnitudes are quantized
The correct matching is
(a)AIII BII CIV DI
(b)AIV BI CII DIII
©AIII BIV CII DI
(d)AI BIVCIII DII
Ans: ( c) TOP