Fig. 5-26 Gaussian pulse and its Fourier transform. in .NET

Creator QR Code JIS X 0510 in .NET Fig. 5-26 Gaussian pulse and its Fourier transform.

Fig. 5-26 Gaussian pulse and its Fourier transform.
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5.44. Using t h e Fourier transform, redo Prob. 2.25.
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The system is described by y'(t)
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+2y(t)=x(t) +xf(t)
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Taking the Fourier transforms of the above equation, we get jwY(w)
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+ 2Y(w)=X ( w ) +jwX(w)
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CHAP. 51
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FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
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+ 2)Y(w) = (1 + jw) X(w )
Hence, by Eq. (5.67) the frequency response H(w) is
Taking the inverse Fourier transform of H(w), the impulse response h(t) is
Note that the procedure is identical to that of the Laplace transform method with s replaced by j w (Prob. 3.29).
5.45. Consider a continuous-time LTI system described by
Using the Fourier transform, find the output y(t) to each of the following input signals:
( a ) x(t) = eP'u(t)
(b) x(r)=u(t)
Taking the Fourier transforms of Eq. (5.1661, we have jwY(w) Hence,
+ 2Y(w) = X(w)
From Eq. (5.155)
Therefore,
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
( b ) From Eq. (5.154)
Thus, by Eq. (5.66) and using the partial-fraction expansion technique, we have
1 1 w )+ 2 + jw jw(2 + j w )
where we used the fact that f ( w ) 6 ( w ) = f(O)6(o) [Eq. (1.2511. Thus,
We observe that the Laplace transform method is easier in this case because of the Fourier transform of d t ) .
5.46. Consider the LTI system in Prob. 5.45. If the input x ( t ) is the periodic square waveform shown in Fig. 5-27, find the amplitude of the first and third harmonics in the output y ( t ) .
Note that x ( t ) is the same x ( t ) shown in Fig. 5-8 [Prob. 5.51. Thus, setting A and w , = 2rr/T0 = rr in Eq. (5.1061, we have
10, To = 2,
Next, from Prob. 5.45
Fig. 5-27
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
Thus, by Eq. (5.74) we obtain
The harmonic form of y(r) is given by [Eq. (5.15)]
where D k is the amplitude of the kth harmonic component of y(t ). By Eqs. (5.11) and (5.16), D, and d , are related by
Thus, from Eq. (5.167), with m
= 0,
we obtain
D , = 21d,I
j ( ja) a2
I=1.71
With m = 1, we obtain
5.47. The most widely used graphical representation of the frequency response H(w) is the Bode plot in which the quantities 2010glo~H(w)l 8,(0) are plotted versus w, with and w plotted on a logarithmic scale. The quantity 2010glolH(o)l is referred to as the magnitude expressed in decibel! (dB), denoted by (H(o)l,,. Sketch the Bode plots for the following frequency responses:
( c ) H(w) =
104(1 j o ) (10 jw)(lOO + jw)
For o < 10, <
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
For o s 10,
On a log frequency scale, 2010glo(w/10) is a straight line with a slope of 20 dB/decade (a decade is a 10-to-1 change in frequency). This straight line intersects the 0-dB axis at o = 10 [Fig. 5-28(a)]. (This value of o is called the comer frequency.) At the corner frequency w = 10 ~(10)1,, = 2010glo11+ jll
= 2010glofi
-- 3 dB
The plot of (H(w)ldB sketched in Fig. 5-28(a). Next, is
Then O,(o)
= tan-'
- --, 0 10
aso-0
radian (rad). The plot of BH(o) is sketched in At w = 10, OH(lO)= tan-' 1 = ~ / 4 Fig. 5-28(b). Note that the dotted lines represent the straight-line approximation of the Bode plots.
For o
100,
> For w > 100,
On a log frequency scale -2010glo(w/100) is a straight line with a slope of -20 dB/decade. This straight line intersects the 0-dB axis at the corner frequency o = 100 [Fig. 5-29(a)]. At the corner frequency o = 100 H(100)1,,
-2010glofi = -3 dB
The plot of IH(w)ldBis sketched in Fig. 5-29(a). Next
Then
At o = 100, 8,(100) Fig. 6).
-tanp' 1 = - ~ / 4 rad. The plot of OH(w) is sketched in
CHAP. 51
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
First, we rewrite H(w) in standard form as H(o) Then
lO(1 +jw) ( 1 +jw/lO)(l +jo/100)
Note tha~t there are three corner frequencies, o = 1, w = 10, and w = 100. A,t corner frequency w = 1 H ( ~ ) I , , = 20 + 20loglO& - 2010g,,m6At corner frequency w = 10 ~ ( 1 0 ) ~ ~ , = 2 0 + 2 0 l o ~ , , ~ - 2 0 1 o ~ ,0, 1 0 - l o m b 37dB 2 &g At corner frequency w = 100 The Bode amplitude plot is sketched in Fig. 5-30(a). Each term contributing to the overall amplitude is also indicated. Next, OH(w) = tan-' w - tan-' - - t a n - ' 10 100 Then
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