DESIGN OF FIR DIGITAL FILTERS
Solution 17.1
Rectangular window:
N-1
WR (ej)
e-jwn
n=-N+l
2N-2
=
e-j(n-N+l)
n=O
2N-2
e -jw(-N+l)
n=0
jw(-N+l)
e-
1-ejw(2N-l)
1-e~jW
sin ((2N-1),,)
2
WRW
WR ( e W)
sin
4
The width of the main lobe is
which, for N>> 1 is approximately
2N-1
27. N
Bartlett window:
0 < n
Define w R(n) = 1
(n -
B
N +
1) =
-j
)=e (N-i)
sin
si
sin
Therefore
(e)
WB W
(e W
wN
_K (with N = M + 1).
-N
­
2
2
2
sin
1
N) 1
1~
w (n) * w R (n)
R
N
From Eq. 7.76 of the text
WR(e
-
otherwise
= 0
Then w
< 'N
__
(Sin -T
In this case the width of the main lobe is N
which is twice that for
the rectangular window.
S17.1
Raised cosine window:
N-l
e-jon
a + Cos (Tr)n
)
WH(e
-(N-l)
N-1
N-1
jwn +
e c~2
e-jn(w
-(N-l)
sin
N-1
e -jn
) +
(w-
T
N-1
-(N-l)
2N-l
(o (2N-1
sin
sin[
sin
2
2N-l
(-iTV
(w---)
1
Assuming that N >> 1, this can be rewritten as
sin (Nw +T)
sinwN
sin (Nw-TV)
+
)=a_ sin
WH(e
WH~ec
2
+ 2
2N
Lsin-(Nw+7)
sin
(Nw-Tr)
This is the superposition of three terms as sketched in
Figure S17.1-l.
sin W N
sin -­
2
(NO+-T)
sin
s sin (NW-
2N1
sin
TV
-N
7T)
(Nw-T)
2TV
N
Figure S17.1-1
which
From this figure we observe that the first values of w for
Consequently for this
the superposition will be zero are w = ±
-.
window also, the width of the main lobe isN.
S17.2
)
Solution 17.2
(a)
Since h1 (n) and h2 (n) are related by a circular shift, their
DFTs are related by
H2 (k)2
=
(k)
W 4kH
8 H1 (
(-l)kH1 (k)
k
Thus, their magnitudes are equal.
Since the DFT corresponds to samples of the Fourier transform,
the values of H1 (k) are the samples of H1 (ejW) indicated in
(b)
Figure S17.2-1
H (e
)W
DFT values
-7T
7T
7
Figure S17.2-1
From (a),
H2 (k)
=
(-1)kH (k).
Thus the values H2 (k) are as indicated
Since these alternate in polarity, the continuous
frequency response of which these are samples, must go through zero
in between these samples as indicated.
in Figure S17.2-2.
S17.3
H2 (ejW
,
'
I
\
I
-T TI
Figure S17.2-2
Thus h2 (n) obviously does not correspond to a good low pass filter,
even though the magnitude of its DFT values are identical to those
of the low pass filter h (n).
Solution 17.3
(a) Since E(eJW)
H (eJW)
d
transform of E(e W),
e(n) = hd(n)
-
=
H(e 3), and e(n) is the inverse Fourier
-
h(n).
Tr
(b)
e2
JE(ejw)1 2 dw
1
-1?
7T +00
1
f[
.i
n=-o
+00
E e(n)e(k)ej wnejwk
dw
k=-0
Interchanging the order of integration and summation
e 2= 1
S17.4
+00
+00
E
X
1
7
e (n) e (k) J
ejw (k-n) dw
=0
k
=27
k = n
n
Thus,
c
2
(c)
2
e
n=-co
=
E:=
2
(n)
From the results of parts (a) and (b),
E+o [hd(n)-h(n)]
2
n=-o
or, since h(n) = 0, n < 0 and n > N,
2
2 N-1
E2 = EQ [hd(n)-h(n)]2
n=0
+
-l
*O
2
E h d(n) +
n=N
n=-o
h
hd(n).
Clearly, the choice of h(n) cannot affect the last summations.
The
first is non-negative and consequently its minimum value is zero
which is achieved for h(n)
=
hd(n).
It should be stressed that although a rectangular window minimizes
the mean square error, it does not generally result in the best
frequency characteristic in terms, for example of passband or
stopband ripple.
SL7. 5
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Resource: Digital Signal Processing
Prof. Alan V. Oppenheim
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