Solution to 316 Final Examination Su06
1.
Using the direct-substitution method and a sampling rate of fs = 10 , find the
transfer function of a discrete-time filter which approximates a continuous-time
system with a transfer function
3s ( s + 1)
H (s) = 2
.
s + 5s + 6
z 2 + b1z + b0
The transfer function can be written in the form H ( z ) = A 2
. Find the
z + a1z + a0
numerical values of A, a1 , a0 , b1 and b0 .
A = 3 , a1 = _-1.5595 , a0 = 0.6065 , b1 = -1.905 , b0 = 0.905
fs = 10 ⇒ Ts = 1 / 10
(
)
3s ( s + 1)
( s − 0 ) ( s + 1) ⇒ H z = 3 ( z − 1) z − e
H (s) = 2
=3
( )
s + 5s + 6
( s + 3) ( s + 2 )
z − e−3 /10 z − e−1/ 5
(
H (z) = 3
−1 /10
)(
( z − 1) ( z − 0.905 ) = 3 z 2 − 1.905 z + 0.905
( z − 0.7408 ) ( z − 0.8187 ) z 2 − 1.5595 z + 0.6065
)
2.
Draw a parallel realization of the transfer function
z 2 − 1.8 z + 0.7
H (z) = 5 2
.
z + 0.3z − 0.1
z 2 − 1.8 z + 0.7
2.1z − 0.8 

= 5 1 − 2

2

z + 0.3z − 0.1
z + 0.3z − 0.1 
1
z 2 + 0.3z − 0.1 z 2 − 1.8 z + 0.7
H (z) = 5
)
z 2 + 0.3z − 0.1
− 2.1z + 0.8


0.543 
13.215 2.715
2.1z − 0.8
2.6443

=5−
+
+
H (z) = 5 1 −
= 5 1 −


z + 0.5 z − 0.2
( z + 0.5 ) ( z − 0.2 )   z + 0.5 z − 0.2 

5
X(z)
1
z
2.715
-0.2
1
z
0.5
13.215
Y(z)
3.
Indicate for each of the following transfer functions whether the system is
Stable (S)
Marginally stable (therefore also unstable) (MS)
or
Unstable without being marginally stable (US).
(a)
3s 2 + s + 8
H (s) = 2
2 s + 7 s + 10
S
MS
US
Poles at −1.75 ± j1.3919 . In the open left half-plane. Therefore stable.
(b)
H (z) =
3z 2 + z + 8
2 z 2 + 7 z + 10
S
MS
US
Poles at −1.75 ± j1.3919 or 2.236 e± j 2.4697 . Outside the unit circle. Therefore
unstable.
(c)
A continuous-time feedback system with forward-path transfer
s 2 + 5 s + 10
and feedback-path transfer function H 2 ( s ) = −1 .
function H1 ( s ) = 2
s + 2s + 3
S
MS
US
s 2 + 5 s + 10
2
s 2 + 5 s + 10
s 2 + 5 s + 10
s
+
s
+
2
3
=
H (s) =
=−
⇒ One pole at s = −7 / 3 ⇒ Stable
s 2 + 5 s + 10
−3s − 7
3s + 7
1− 2
s + 2s + 3
(d)
A discrete-time feedback system with forward-path transfer
z 2 + 5 z + 10
function H1 ( z ) = 2
and feedback-path transfer function H 2 ( z ) = −1 .
z + 2z + 3
S
MS
US
z 2 + 5 z + 10
2
z 2 + 5 z + 10
z 2 + 5 z + 10
H ( z ) = z 2+ 2 z + 3 =
=−
⇒ One pole at z = −7 / 3 ⇒ Unstable
z + 5 z + 10
−3z − 7
3z + 7
1− 2
z + 2z + 3
4.
An FIR filter is designed with an impulse response h [ n ] = 8 ( u [ n ] − u [ n − 3]) .
(a)
How many poles and zeros does its transfer function have and where are
they located?
2 Zeros at −0.5 ± j 0.866
2 Poles at 0
From the definition of the z transform,
∞
(
H ( z ) = 8∑ ( u [ n ] − u [ n − 3]) z − n = 8 1 + z −1 + z −2
n=0
(
)
H ( z) = 8 1 + 1 / z + 1 / z2 = 8
)
z2 + z + 1
z2
Zeros at z 2 + z + 1 = 0 ⇒ z = −0.5 ± j 0.866 (which are on the unit circle)
Poles at z 2 = 0 ⇒ Double pole at z = 0
Alternate Solution:
3
z
z
z −2
Z
−2 z − 1
u [ n ] ←→
⇒ u [ n ] − u [ n − 3] ←→
−
=z
z −1
z −1 z −1
z −1
Z
z2 + z + 1
z − 1 z3 − 1
)
z3 − z2
z2 − 1
z2 − z
z −1
z −1
0
z2 + z + 1
u [ n ] ←→
,
z2
Z
(b)
The rest of the solution is as above.
At what values of Ω would the filter response be absolutely zero?
e jΩ = −0.5 ± j 0.866 ⇒ Ω = ±2π / 3