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Chapter 6 Answers 6.6 (b) the impulse response h1[n] is as shown in figure s6.6,as was increase ,it is clear that the significant central lobe of h1[n] becomes more concentrated around the origin. consequently. h[n]=h1[n](-1)^n also becomes more concentrated about the origin. 6.7 the frequency response magnitude |H(jw)| is as shown in figure s6.7.the frequency response of the bandpass filter G(jw) will be given by
G( jω) = FT{2h(t) cos(4000π t)}
= H ( j(ω ? 4000π )) + H ( j(ω + 4000π ))
This is as shown in figure s6.7
H(j
-4000
-2000
-1000
π π1000
2000
π4000
G(jω )
-6000π -4000π -2000π 0 2000π 4000π 6000π
Figure S6.7
(a) from the figure ,it is obvious that the passband edges are at 2000∏rad/sec and 6000∏rad/sec. this
translates to 1000HZ and 3000Hz,respectively.
(b) (b)from the figure ,it is obvious that the stopband edges are at 1600∏ rad/sec.this translates to 800Hz and
3200 Hz, respectively.
6.8 taking the Fourier transform of both sides of the first difference equation and simplifying, we obtain the
frequency response H(e^jw)of the first filter.
M
∑∑H (e jω )
=
Y (e jω ) X (e jω )
bk e? jωk = k=0
N
1? ak e? jωk
.
k =1
Taking the Fourier transform of both sides of the second difference equation and simplifying ,we obtain the
frequency response H1(e^jw) of the second filter.
M
∑∑H (e jω )
=
Y (e jω ) X (e jω )
(?1)k bk e? jωk = k=0
N
1? (?1)k ak e? jωk
.
k =1
This may also be written as
M
∑∑H (e jω ) =
Y (e jω ) X (e jω )
=
b e? j(ω ?π )k k k =0 N
1 ? ak e? j(ω ?π
)k
= H (e j(ω ?π ) ).
k =1
Therefore .the frequency response of the second filter is obtained bu shifting the frequency response of the first
filter by ∏.although the first fitter has its passband between-wp and wp. Therefore, the second filter will have
its passband between ∏-wp and ∏+wp.
6.9 taking the Fourier transform of the given differential equation and simplifying .we obtain the frequency of
the LTI system to be
H (e jω
)
=
Y (e jω ) X (e jω )
=
2 5 + jω
Taking the inverse Fourier transform, we o
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