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FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS in .NET
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10, T, = 2, and w , 2n/To = n in Eq. (5.107) (Prob. 5 . 9 , we get
rad, the filter passes all harmonic components Since the cutoff frequency o, of the filter is 4 7 ~ of x ( t ) whose angular frequencies are less than 4 n rad and rejects all harmonic components of x ( t ) whose angular frequencies are greater than 477 rad. Therefore, 20 y ( t ) = 5 + sinnt x + s i n 3 ~ t 3n
5  5 4 Consider an ideal lowpass filter with frequency response
The input to this filter is Find the value of o, such that this filter passes exactly onehalf of the normalized energy of the input signal x ( t ). CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
From Eq. (5.155) Then
The normalized energy of x ( t ) is
Using Parseval's identity (5.641, the normalized energy of y ( t ) is
1 d w, =/  o0 2  17 tan' 4+ 2 2
E = 2 " 8 from which we obtain
'"c = tan  = 1 o, = 2 rad/s
5.55. T h e equivalent bandwidth of a filter with frequency response H ( o ) is defined by
where IH(w)lm, denotes the maximum value of the magnitude spectrum. Consider the lowpass RC filter shown in Fig. 5  6 ( a ) . ( a ) Find its 3dB bandwidth W,. , ( b ) Find its equivalent bandwidth We,. From Eq. (5.91) the frequency response H ( w ) of the RC filter is given by
H(o)= l+joRC
l+j(o/o,) where o, = 1 /RC. Now
The amplitude spectrum lH(w)l is plotted in Fig. 56(b). When w IH(o,)l = I / & . Thus, the 3dB bandwidth of the RC filter is given by = o, = 1/RC, FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
[CHAP. 5
( 6 ) From Fig. 56(b) we see that IH(O)I Rewriting H(w) as
1 is the maximum of the magnitude spectrum.
and using Eq. (5.179), the equivalent bandwidth of the RC filter is given by (Fig. 532) Fig. 532 Filter bandwidth.
5.56. The risetime t , of the lowpass RC filter in Fig. 56(a) is defined as the time required for a unit step response to go from 10 to 90 percent of its final value. Show that where f, W,,,/2.rr
1/2.rrRC is the 3dB bandwidth (in hertz) of the filter.
From the frequency response H(w) of the RC filter, the impulse response is
Then, from Eq. (2.12) the unit step response d t ) is found to be
Dividing the first equation by the second equation on the righthand side, we obtain e ( r ~  r d / R C= 9 CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Fig. 533 which indicates the inverse relationship between bandwidth and risetime.
5.57. Another definition of bandwidth for a signal x ( t ) is the 90 percent energy containment bandwidth W,, defined by where Ex is the normalized energy content of signal x ( t ) . Find the W , for the following signals: ( a ) x ( t ) = e"'u(t), a > 0 sin at ( b ) x(t)= (a) rr l From Eq. (5.155) ~ ( t=)e  " u ( t ) X( W ) = a +jo
From Eq. (1.14) Now, by Eq. (5.180) FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
from which we get
Thus.
From Eq. (5.137) sin at ~ ( t = X(w) ) 7~t Using Parseval's identity (5.64), we have =pu(w) = Iwl < a
Iwl > a
Then, by Eq. (5.180) from which we get
Ww = 0.9a rad/s
Note that the absolute bandwidth of x(t) is a (radians/second). 5.58. Let x ( t ) be a realvalued bandlimited signal specified by [Fig. 534(b)] Let x,(t be defined by
( a ) Sketch x $ t ) for
T, < r/o, and for T, > r/oM.
( b ) Find and sketch the Fourier spectrum X $ o ) of x J r ) for T, < r/oM and for T, > n/w,. (a) Using Eq. (I.26), we have
The sampled signal x , ( r ) is sketched in Fig. 534(c) for Tq< r/w,, and in Fig. 534(i) for T, > T / w ~ . The signal x,(t) is called the ideal sampled signal, T, is referred to as the sampling interr.al (or period), and f , = 1/T, is referred to as the sampling rate (or frequency ).

