( b ) By the definition of h[nl [Eq. (2.3011 and Eq. (6.145) we obtain in VS .NET

Encode QR Code in VS .NET ( b ) By the definition of h[nl [Eq. (2.3011 and Eq. (6.145) we obtain

( b ) By the definition of h[nl [Eq. (2.3011 and Eq. (6.145) we obtain
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h [ n ] = 6 [ n ] + 6 [ n - 11 h [ n ]=
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From Eq. (6.146)
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Fig. 6-18
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CHAP. 61
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FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
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which are sketched in Fig. 6-19.
Fig. 6-19
Let R,
, be the 3-dB bandwidth of the system. Then by definition (Sec. 5.7) ,
we obtain
i13dB =
We see that the system is a discrete-time wideband low-pass finite impulse response (FIR) filter (Sec. 2 . 9 0 .
6.35. Consider the discrete-time LTI system shown in Fig. 6-20. where a is a constant and
Fig. 6-20
FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
( a ) Find the frequency response H(S1) of the system. ( b ) Find the impulse response h [ n ]of the system.
( c ) Sketch the magnitude response ( H ( a ) ( the system for a of
( a ) From Fig. 6-20 we have y [ n ] - a y [ n - 11 = x [ n ]
= 0.9
and a
= 0.5.
(6.147)
Taking the Fourier transform of Eq. (6.147)and by Eq. (6.771,we have
( b ) Using Eq. (6.371, we obtain
h [ n ]= a n u [ n ]
From Eq. (6.148)
which is sketched in Fig. 6-21 for a = 0.9 and a = 0.5. We see that the system is a discrete-time low-pass infinite impulse response (IIR) filter (Sec. 2 . 9 0
-n 2
Fig. 6-21
6.36. Let h L p F [ nbe the impulse response of a discrete-time 10~4-pass ] filter with frequency . response H L p F ( R )Show that a discrete-time filter whose impulse response h [ n ] is given by
h[nl = ( l)"hLPF[nl
is a high-pass filter with the frequency response
H ( S 1 )= H L P F ( a
CHAP. 61 FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS
Since - 1 = el", we can write
( 6.152) h [ n ] = ( - l ) " h L P F [ n ] eJ""hLPF[n] = Taking the Fourier transform of Eq. (6.152) and using the frequency-shifting property (6.44), we obtain
H(R)= H L P F ( R - ~ ) which represents the frequency response of a high-pass filter. This is illustrated in Fig. 6-22.
+ -a,
n-R,
Fig. 6-22 Transformation of a low-pass filter to a high-pass filter.
6.37. Show that if a discrete-time low-pass filter is described by the difference equation
then the discrete-time filter described by
is a high-pass filter.
Taking the Fourier transform of Eq. (6.153),we obtain the frequency response H L p F ( Rof ) the low-pass filter as
If we replace R by ( R - a ) in Eq. (6.155), then we have
which corresponds to the difference equation
FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
6.38. Convert the discrete-time low-pass filter shown in Fig. 6-18 (Prob. 6.34) to a high-pass filter.
From Prob. 6.34 the discrete-time low-pass filter shown in Fig. 6-18 is described by [Eq.
( 6.145
Using Eq. (6.154), the converted high-pass filter is described by which leads to the circuit diagram in Fig. 6-23. Taking the Fourier transform of Eq. (6.157) and by Eq. (6.77), we have
From Eq. (6.158)
and which are sketched in Fig. 6-24. We see that the system is a discrete-time high-pass FIR filter.
Fig. 6-23
6.39. The system function H ( z ) of a causal discrete-time LTI system is given by
where a is real and la1 < 1. Find the value of b so that the frequency response H ( R ) of the system satisfies the condition IH(n)l= 1 Such a system is called an all-pass filter.
By Eq. (6.34) the frequency response of the system is
all R
(6.160)
CHAP. 6 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS 1
Fig. 6-24
Then, by E . (6.160) q
which leads to
J b+ e-]'I= 11 - ae-jnl
Ib+cosn-jsinRI=Il -acosR+jasinRl
1 + b 2 + 2bcosR= 1 + a 2 - 2acosO
(6.162)
and we see that if b = -a, Eq. (6.162)holds for all R and Eq. (6.160)is satisfied.
6.40. Let h [ n ] be the impulse response of an FIR filter so that
h [ n ]=0 n<O,nrN
Assume that h [ n ] is real and let the frequency response H ( R ) be expressed as
H ( I 2 ) = 1H ( f l ) ) e ~ ~ ( ~ )
( a ) Find the phase response 8 ( R ) when h [ n ] satisfies the condition [Fig. 6-25(a)]
h[n]= h [ N - 1 - n ]
(6.163)
( b ) Find the phase response B ( R ) when h [ n ] satisfies the condition [Fig. 6-25(b)]
h [ n ]= - h [ N - 1 - n ] (6.164)
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
N even
N- I
I I I I I I
N odd
Fig. 6-25
Taking the Fourier transform of Eq. (6.163) and using Eqs. (6.431, (6.461, and (6.62).we obtain
H ( R ) = H * ( R ) e-j(N-')R
or Thus,
IH(f))le1flfl)= ) H ( n ) ( e - i o ( ~ ~ e - ~ ( N - I ) n e(n)= -e(n)
- ( N -i ) n
e ( n )= -+(N- 1 ) ~
which indicates that the phase response is linear. ( b ) Similarly, taking the Fourier transform of Eq. (6.164, we get ~ ( n= )- H * ( R ) e - ~ ( " - ' ) f l or Thus,
I H ( f l ) l e i 0 ( n ) , IH(n)(e~ne-l@(fl)e-~(N-l)fl
e(n)= T - q n )
- ( N - i p
which indicates that the phase response is also linear.
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
6.41. Consider a three-point moving-average discrete-time filter described by the difference
equation
Find and sketch the impulse response h [ n ]of the filter. ( b ) Find the frequency response H(IR) of the filter. (c) Sketch the magnitude response IH(IR)I and the phase response 8(IR) of the filter.
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