EG7016 - Design of Discrete Systems
Draft Exam Solutions– June 2003
1.

a
i
The Z transform of the sequence xk is defined as X(z) =
x
k 0
k
z k .
Consider the continuous signal x(t) from which xk is derived.
A time function xs(t) can be defined such that xs(t) = x(t).d(t) where

d(t) =
  (t  kT ) (A “Dirac comb”).
k 0
X(z) is the Laplace transform of xs(t) (with esT replaced by the
complex variable z).
b
i
z 1
1  0.2 z 1  0.15 z 2
z 1
=
(1  0.3z 1 )(1  0.5 z 1 )
The transfer function =
=
z
( z  0.3)( z  0.5)
There are poles are at z = 0.3 and z = -0.5 and there is a zero at z = 0.
ii
The unit sample response is the inverse transform of the transfer function.
1.25
1.25
z 1
The transfer function is
=
1
1
1
(1  0.3z )(1  0.5 z ) (1  0.3z ) (1  0.5 z 1 )
Therefore, using Z transform tables supplied, xk = 1.25{(0.3)k – (-0.5)k}.
e  jT
1  0.2e  jT  0.15e  2 jT
iii
The frequency response is
iv
When f= 0.5 (ω = π) and T = 0.5 then z = ejωT = ejπ/2 and is at the point P
(0,j) in the complex plane. The poles are at P1 and P2 and the zero at Z (the
origin).
Unit
circle
Imaginary part of z
P
Real part of z
P1
Z
P2
Gain = PZ/(PP1.PP2) and phase = π/2 – tan-1PZ/P1Z – (π - tan-1PZ/P2Z)
But PZ = 1, PP1 = 12  0.52 = 1.118 and PP2 = 12  0.32 = 1.044
Therefore at ω, gain = 0.86 and phase = -1.4 rad.
2.
i
The two filters required can most appropriately be designed respectively by;
1. the bi-linear transformation with pre-distortion and band transformation, starting
from a low-pass analogue 4th order Butterworth standard prototype using a
sampling interval of 0.05 seconds or less.
2. the Fourier series (windows) method starting with the ideal band-pass
specifications and a preferred sampling interval of 0.05 seconds.
Candidates should choose (with justification) the correct formulae from those
supplied and put in the appropriate figures in the right places for the band edges and
the sampling interval. A correct choice of sampling interval is important. The linear
phase filter also needs an appropriate data window to be chosen and, again, this
should be justified and the correct figures should be inserted into the formula for the
chosen window.
ii Candidates are expected to bring out most of the following points.
1. The first design has a better cut-off rate than the second although it is of lower
order. This is because it is recursive.
2. The band edges are exact in the first design but not the second (unless some
special measures are taken).
3. The first design is more efficient (execution time and/or chip area) because it is
of lower order.
4. The second design avoids phase distortion because it is a linear phase filter.
5. The second design has a long start up transient because of the pure time delay.
6. The first design is useful in applications where power-density information is of
most interest (e.g. – speech processing) the second is appropriate where it is
important to retain relative timing information of either components in the passband or between two channels (e.g. – instrumentation for event detection).
3.
The standard third order, continuous ITAE filter has a transfer function is
1
1
=
, by partial fractions,
2
3
1  2.15s  1.75s  s
(0.708  s)(1.412  1.042s  s 2 )
0.85s  0.284
0.85
=
0.708  s (1.412  1.042s  s 2 )
s
1
= 0.85
- 0.85
0.708  s
( s  0.521) 2  (1.068) 2
1
- 0.284
( s  0.521) 2  (1.068) 2
But before applying the Impulse Invariance transformations, the given step response needs
to be speeded up by a factor of 4 so that the step response of the discrete filter will reach its
final value (1) for the first time in 1 second. Thus we need to replace s by s/4 in the above
transfer function and find the z-transform equivalent of:
s
1
1
3.4
- 13.6
- 4.54
.
2
2
2.832  s
( s  2.084)  (4.272)
( s  2.084) 2  (4.272) 2
Applying to this the Impulse Invariance transformations from tables provided gives;
2.084
1  e  2.084x 0.1 [cos( 4.272 x0.1) 
sin( 4.272 x0.1)] z 1
1
4.272
H(z) = 3.4
- 13.6
1  2e  2.084x 0.1 cos 4.272 x0.1z 1  e  2 x 2.084x 0.1 z  2
1  e  2.832x 0.1 z 1
1
sin( 4.272 x0.1) z 1
4
.
272
- 4.45
1  2e  2.084x 0.1 cos 4.272 x0.1z 1  e  2 x 2.084x 0.1 z  2
1
0.1z 1
1  0.9 z 1
= 4.7
13.6
4.45
1  1.48 z 1  0.66 z 2
1  0.812 z 1
1  1.48 z 1  0.66 z 2
 8.9  10.8 z 1  6.48 z 2
=
1  0.668 z 1  1.86 z 2  0.536 z 3
[Note: marks will be given for correct process and only minimal deduction will
be made for minor errors in manipulation or arithmetic]
4.
a)
What we want to see here is both (i) if the student did the designed exercise and (ii) if he/she took
the right conclusions about it. The answer should mention that the best resolution was found for
the widest possible rectangular window, but that was also the one which produced the highest side
lobes and the worst variance. With smoother windows the side lobes died off more quickly and the
variance was reduced but the spectral resolution was poorer.
[15 marks]
b)
Pm = 1
Qm = 1+j
1
We know that W8 
Pm+1 = ?
W81
Qm+1 = ?
2
2

j = 0.707 – 0.707j, and the radix-2 Decimation-in-time butterfly is
2
2
given by
Pm+1 = Pm + WNk Qm
Qm+1= Pm - WNk Qm
So, using the input values given, we have
Pm+1 = Pm + WNk Qm = 1 + (0.707 – 0.707j)(1+j) =
= 1+0.707-0.707j+0.707j+0.707 = 2.4140
Qm+1= Pm - WNk Qm = 1 – (0.707 – 0.707j)(1+j) =
= 1-0.707+0.707j-0.707j-0.707 = - 0.4140
[5 marks]
5.
a) Solution: Given that the Fourier Transform of x(t) is

X(f)   x(t)e  j2ft dt

For the function given in figure 1 this reduces to
10
X(f)   x(t)e  j2ft dt
0
=
1
=
 j 2f

10
0
e - j2ft d(-j2ft ) =

1
e  j 2ft
 j2π2

10
0
=
1
1
e  jf 10
e  j 2f 10  e 0 =
e  j 2f 10  e 0   jf 10 =
 j 2f
 j 2f
e




1
1 e  jf 10  e  jf 10  jf 10
sin( 10f )  e  jf 10 =

e
=
f
f
2j
= 10
sin( 10f )  jf 10
e
= 10 sinc(10f)  e  jf 10
10f
[8 marks]
b) Solution: The sketch is
10
9
8
7
6
5
4
3
2
1
0
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
the shape, the amplitude and the location of the zeroes are all important.
[6 marks]
c) Solution: The consequences are:
-
There will be leakage due to the fact that H(f) will be convolved with X(f) and X(f) is not
an impulse: the width of the main lobe is not zero and there are side lobes.
The amplitude of the first side lobe is 2.2475, or -12.97 dB.
The spectral resolution of the DFT will be limited to approximately 0.1 Hz.
[6 marks]