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- tan-'(0.1) eH(lO) = tan-'(10) - tan-'(1)
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e H ( l ) = tan-'(1) The plot of eH(w) is sketched in Fig. 5-30(b).
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- tan-'(0.01) - tan-'(0.1)
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8,(100) = tan-'(100) - tan-'(10) - tan-'(1)
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5.48. An ideal ( -7r/2) radian (or -90") phase shifter (Fig. 5-31) is defined by the frequency response
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( a ) Find the impulse response h(t ) of this phase shifter. (6) Find the output y ( t ) of this phase shifter due to an arbitrary input x(t).
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( c ) Find the output y(t) when x ( t )
(a) Since
e - j " l 2 = -j
= cos o o t .
e J " / 2 = j,
H(w) can be rewritten as H(w) = -jsgn(w) (5.170)
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
[CHAP. 5
Fig. 5-30 Bode plots.
CHAP. 51
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
Fig. 5-31 - n / 2 rad phase shifter.
sgn(t)
where
Now from Eq. (5.153)
and by the duality property (5.54) we have
- jsgn(w)
(5.172)
since sgn(w) is an odd function of w. Thus, the impulse response h(t) is given by h(t)
( b ) By Eq. (2.6)
F - ' [ ~ ( w ) ]= F - ' [ - j s g n ( w ) ]
(5.173)
The signal y(t) defined by Eq. (5.174) is called the Hilbert transform of x ( t ) and is usually denoted by f (l). ( c ) From Eq. (5.144) cos wet H T[S(W - w,,) Then Y(w) = X ( w ) H ( w ) = T [ s ( ~- w,)
= - j sgn(w,)S(w ~ =
+ a ( w + wO)]
- w,,)
+ S(w + w,)][-jsgn(w)] - jT sgn( -w,,)S(w + w,)
- w,) + j.rrS(w + w,) - 1. Thus, from Eq. (5.145) we get since sgn(oo) = 1 and sgn( -w,)
-jd(w y ( t ) =sin w,t Note that cos(w,t
- 7/21 = sin wot.
H ( o ) = A ( w ) +j B ( o )
5.49. Consider a causal continuous-time LTI system with frequency response Show that the impulse response h ( t ) of the system can be obtained in terms of A(w) or B(w) alone.
FOURIER ANALYSIS O F TIME SIGNALS AND SYSTEMS
[CHAP. 5
Since the system is causal, by definition
h(t)=0
Accordingly, Let
where he(r) and h,(t) are the even and odd components of h(r), respectively. Then from Eqs. ( 1 . 5 ) and ( 1 . 6 ) we can write
h ( r ) = 2he(r)= 2h,(r)
From Eqs. (5.616) and ( 5 . 6 1 ~ ) have we
h,(t) -A(w)
Thus, by Eq. (5.175)
h,(t) -jB(w)
Equations ( 5 . 1 7 6 ~ ) (5.176b) indicate that h ( t ) can be obtained in terms of A ( w ) or B ( w ) and alone.
5.50. Consider a causal continuous-time LTI system with frequency response
H ( o ) = A ( o ) +j B ( o )
If the impulse response h ( t ) of the system contains no impulses at the origin, then show that A ( w ) and B ( w ) satisfy the following equations:
As in Prob. 5.49, let
h ( r ) = h e ( r )+ h o ( t ) Since h ( t ) is causal, that is, h(r) = 0 for t < 0, we have h e ( t )= - h O ( t )
Also from Eq. (5.175) we have
t <O
h,(t) =h,(t) Thus, using Eq. (5.1521, we can write
h , ( t ) = h , ( r ) sgn(r) h o ( r ) = h e ( r ) sgn(t)
Now, from Eqs. (5.6161, (5.61~1, (5.153) we have and
CHAP. 51
FOURIER ANALYSIS OF TIME SIGNALS A N D SYSTEMS
Thus, by the frequency convolution theorem (5.59) we obtain
Note that A ( w ) is the Hilbert transform of B ( w )[Eq. (5.17411 and that B ( w )is the negative of the Hilbert transform of A(w).
5.51. The real part of the frequency response H ( w ) of a causal LTI system is known to be r S ( w ) . Find the frequency response H ( o ) and the impulse function h ( t ) of the system.
H ( w )=A ( w )
+jB( w )
Using Eq. (5.177b1, with A ( w ) = 7 ~ 8 ( w ) , obtain we
Hence,
and by Eq. (5.154)
h(t) =u(t)
FILTERING
Consider an ideal low-pass filter with frequency response
The input to this filter is sin at x(r) = 7T t Find the output y ( r ) for a < w,. ( b ) Find the output y( t ) for a > w,. ( c ) In which case does the output suffer distortion
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