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then, from Probs. (6.35) and (6.38) the discrete-time system can be constructed as a cascade connection of two systems as shown in Fig. 6-3Ma). From Fig. 6-3Ma) it is seen that we can replace two unit-delay elements by one unit-delay element as shown in Fig. 6-30( 6).
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CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
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Fig. 6-30 Simulation of a differentiator.
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( b ) By Eq. (6.184) the frequency response Hd(R) of the discrete-time system is given by
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Note that when R -C 1, we have 2 R R Hd(R) =j-tan=j- =jw T, 2 T s if R = oT, (Fig. 6-31).
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FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
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6.46. Consider designing a discrete-time LTI system with system function H J z ) obtained by applying the bilinear transformation to a continuous-time LTI system with rational system function H,(s). That is,
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Show that a stable, causal continuous-time system will always lead to a stable, causal discrete-time system.
Consider the bilinear transformation of Eq. (6.183)
Solving Eq. (6.188) for z, we obtain
Setting s =jw in Eq. (6.1891, we get
Thus, we see that the jw-axis of the s-plane is transformed into the unit circle of the z-plane. Let
z =re'"
s = a +j o
Then from Eq. (6.188)
r2- 1 2 r sin $ 2 1+r2+2rcosR+'1+r2+2rcos~ Hence,
2 2 r sin R T, 1 + r 2 + 2 r c o s R
From Eq. (6.191a)we see that if r < 1, then a < 0, and if r > 1, then cr > 0. Consequently, the left-hand plane (LHP) in s maps into the inside of the unit circle in the z-plane, and the right-hand plane (RHP) in s maps into the outside of the unit circle (Fig. 6-32). Thus, we conclude that a stable, causal continuous-time system will lead to a stable, causal discrete-time system with a bilinear transformation (see Sec. 3.6B and Sec. 4.6B). When r = 1, then o = 0 and
sin 0
T, I
+ cos R
2 R -tanT, 2
CHAP. 61
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
s-plane
z-plane
Fig. 6-32 Bilinear transformation.
U ~ circle I
From Eq. (6.193) we see that the entire range
< w <a, is mapped only into the range
-Trsn
6.47. Consider the low-pass RC filter in Fig. 6-28(a). Design a low-pass discrete-time filter by the bilinear transformation method such that its 3-dB bandwidth is ~ / 4 .
Using Eq. (6.192), R,
, = 7r/4 corresponds to ,
2 7r 0.828 -tan=Ts 2 Ts 8 Ts = l / R C . Thus, the system function H,(s) of the RC filter is given by
w,,,
= -tan-
R,,,
From Prob. 5.55(a), w ,
Let H J z ) be the system function of the desired discrete-time filter. Applying the bilinear transformation (6.183)to Eq. (6.1951, we get
from which the system in Fig. 6-33 results. The frequency response of the discrete-time filter is
At R = 0, Hd(0)= 1 , and at R sponse.
/ 4J,H d ( r / 4 )= 0.707 = I / (
fi, which
is the desired re-
FOURIER ANALYSIS O F DISCRETE-TIME SIGNALS AND SYSTEMS
[CHAP. 6
Fig. 6-33 Simulation of an RC filter by the bilinear transformation method.
6.48. Let h [ n ] denote the impulse response of a desired IIR filter with frequency response H ( R ) and let h , [ n ] denote the impulse response of an FIR filter of length N with frequency response H,(R). Show that when
h o [ n ]=
(:["I
OsnsN-1 otherwise
(6.198)
the mean-square error e 2 defined by
is minimized.
By definition (6.27)
H(R)=
n = -m
h[n]e-J'"
H,(R)=
m -m
ho[n]e-JRn
where e [ n ] = h [ n ]- h , [ n ] . By Parseval's theorem (6.66) we have
The last two terms in Eq. (6.201) are two positive constants. Thus, that is, Note that Eq. (6.198) can be expressed as where w [ n ] is known as a rectangular window function given by
w [ n ]= OsnsN-1 otherwise
is minimized when
CHAP. 61
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
DISCRETE FOURIER TRANSFORM
6.49. Find the N-point DFT of the following sequences x [ n ] :
From definitions (6.92) and (1.45), we have
Figure 6-34 shows x [ n ] and its N-point DFT X [ k ] .
Fig. 6-34
( b ) Again from definitions (6.92) and (1.44) and using Eq. (1.90), we obtain
Figure 6-35 shows x [ n ] and its N-point DFT X [ k ] .
Fig. 6-35
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
6.50. Consider two sequences x [ n ] and h [ n ] of length 4 given by
( a > Calculate y [ n ] = x [ n ] 8 h [ n ] by doing the circular convolution directly.
( b ) Calculate y [ n ] by DFT.
The sequences x [ n ] and h [ n ] can be expressed as
x[n]= {l,O,- 1,O)
B Eq. (6.108) y
h [ n ] = ( l I , 1 ~1 , ~ , ~ )
The sequences x [ i ] and h [ n - iImod4for n Eq. (6.108) we get
0 , 1 , 2 , 3 are plotted in Fig. 6-36(a).Thus, by
which is plotted in Fig. 6-36(b). ( b ) B Eq. (6.92) y
Then by Eq. (6.107) the DFT of y [ n ] is
Y [ k ]= X [ k ] H [ k ]= ( 1 - w,Zk)(l + i~qk ;wqZk + twak) +
- 1 +i w k 2 4
Since W:k
-lw2k4 4
1 ~ 3 -Lw4k k
-Lw5k
(w:)~ l k and wdjk= W4( 4 + ' ) = wqk, we obtain = k
y [ k ] = $ + $ ~ q k - f ~ : ~ - i wk~ 0 , ~ , 2 , 3 =~ 1
Thus, by the definition of DFT [Eq. (6.9211 we get
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
6.51. Consider the finite-length complex exponential sequence
~ [ n=]
( a ) Find the Fourier transform
OsnsN-1 otherwise
X(fl) x[n]. of ( b ) Find the N-point DFT X[k] x[n]. of
FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6
From Eq. (6.27)and usingEq.(1.90), we have
- ,j(R-RuXN( b ) Note from Eq. (6.98) that
I)/Z
sin [( R - R,) ~ / 2 ] sin[(R - R , ) / 2 ]
we obtain
6.52. Show that if x [ n ] is real, then its DFT X [ k ] satisfies the relation
where
denotes the complex conjugate.
From Eq. (6.92)
Hence, if x [ n ] is real, then x * [ n ] = x [ n ] and
6.53. Show that
where * denotes the complex conjugate and
X [ k ] = DFT{x[n])
We can write Eq. (6.94) as
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
Noting that the term in brackets in the last term is the DFT of X * [ k ] ,we get
which shows that the same algorithm used to evaluate the DFT can be used to evaluate the IDFT.
6.54. T h e DFT definition in Eq. (6.92) can be expressed in a matrix operation form as X=WNx (6.206)
where
The N x N matrix WN is known as the D F T matrix. Note that WN is symmetric; that is, W z = WN, where W is the transpose of WN. :
(a) Show that
; where W is the inverse of WN and W,* is the complex conjugate of WN. (6) Find W, and W;' explicitly.
If we assume that the inverse of W, exists, then multiplying both sides of Eq. (6.206) by W i we obtain
which is just an expression for the IDFT. The IDFT as given by Eq. (6.94) can be expressed in matrix form as
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