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Added on - 13 Dec 2021

Showing pages 1 to 4 of 14 pages
Solution 1
a)
As per the theory the filtering will be ok here as we can easily remove overlapping noise through
filter processing. In case if we apply filtering then two popular filtering method of adaptive
manner, are perfect techniques for processing this.
b)Now we can have
Design:
Transfer function of filter is:
H(z)=Y(z)
X(z)=

k=0
M
bkzk
1+
k=0
N
akzk
So here M is the fir filter length. Order of filter is given by N=M-1.
Response of filter is
y(n)=
k=0
M1
bkx(nk)
y(n)=
k=0
M1
h(k)x(nk)
Coefficients:
Without Window
b =0.01690.25280.56670.25280.0169
Rectangular Window
b =0.01690.25280.56670.25280.0169
At 95% unity**, the poles are:
p1,2=.95[.5878±j.8090
p1,2=.5584±j.7686
H(z)=(z.5878+j.8090)(z.5878j.8090)
(z.5584+j.7686)(z.5584j.7686)
With the reduced canonical form:
H(z)=11.175z1+z2
11.117z1+.903z2
The recursive algorithm becomes:
yn=xn1.175xn1+xn2+1.117yn1.903yn2
Output:
Without Window
Rectangular Window
c)
Zeroes of the functions are:
z1,2=cos(54°)±jsin(54°)
z1,2=.5878±j.8090  