PTFE 4761 Final Practice Problems
(problems from Spring 2007 Final Exam)
1)
(note that here I leave T to be whatever value;
it’s also okay to take a value of T, e.g. T = 1 to
make the notation simpler)
a)
The difference equation for a discrete time step dt = T:
)
(
)
(
)
(
)
(
2
)
(
)
(
nT
c
T
nT
c
T
nT
c
nT
y
T
nT
y
T
nT
y
+

=
+

.
Multiplying with T and collecting terms:
(1 + 2T) y(nT+T) – y(nT) = c(nT+T) + (T – 1) c(nT)
(note that we have used a “forward” difference equation;
a “backward” one, i.e. using y(nT) –
y(nTT) will also work)
b) Taking the z transform of the equation above:
(
using the relationship that if y(nT)
→
Y(z)
then
y(nT+T)
→
z Y(z),
with
y(0) = 0)
(1+2T) z Y(z) – Y(z) = z C(z) + (T1) C(z)
[ (1+2T) z – 1 ] Y(z) = [ z + T – 1 ] C(z)
)
(
1
)
2
1
(
1
)
(
z
C
z
T
T
z
z
Y

+

=
then the transfer function
1
)
2
1
(
1
)
(
)
(
)
(

+

=
=
z
T
T
z
z
C
z
Y
z
G
2)
b
z
z
a
z
z
G
=
2
)
(
2
and we apply a unit step input:
c(nT) = 1 for n > 0,
C(z) =
1

z
z
Then Y(z) = G(z) C(z) =
1
2
2

z
z
b
z
z
a
z
I assume that “zero steadystate error” means that the output will be
y(nT) = c(nT) = 1 for n
→∞
.
Using the final value theorem in the zdomain:
y
ss
=
b
a
b
a
b
z
z
a
z
z
z
b
z
z
a
z
z
z
z
Y
z
z
nT
y
z
z
z
n
+
=
+
+
=
=


=

=
→
→
→
∞
→
3
1
2
1
1
2
lim
1
2
1
lim
)
(
1
lim
)
(
lim
2
1
2
1
1
Since we want y
ss
to be 1:
1 + a = 3 + b
a = 2 + b
b)
For dead beat process, the pole(s) of G(z) have to be 0.
One way is for the denominator to be:
(z
2
+ 2z + b) = (z0)(z0) = z
2
.
But this is not possible.
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 Fall '09
 Unknown
 Limit superior and limit inferior

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