barcode lib ssrs With h l ( n ) being in cascade with g(n), the overall frequency response becomes in Software

Encoder Code 128C in Software With h l ( n ) being in cascade with g(n), the overall frequency response becomes

With h l ( n ) being in cascade with g(n), the overall frequency response becomes
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~ ( e j " = H I(eiW)[Hz(ei") )
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+ H 3 ( e j wH4(eJW)1 )
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FOURIER ANALYSIS
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(b) The frequency responses of the systems in this interconnection are
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[CHAP. 2
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Suppose that the frequency response of a linear shift-invariant system is piecewise constant as shown in the following figure:
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Describe how this filter may be implemented as a parallel connection of low-pass filters.
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This filter may be viewed as a summation of a low-pass filter, a bandpass filter, and a high-pass filter. Because both a bandpass filter and a high-pass filter may be synthesized using a parallel connection of low-pass filters, we may proceed as follows. First, we put an allpass filter H3(eJW) A3 in parallel with a low-pass filter with a cutoff = frequency 02 and a gain of Az - A3. This parallel network has a frequency response
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To produce the correct magnitude over the lower band, Iwl iwl, we add a third low-pass filter in parallel with the other two. This filter has a cutoff frequency of wl and a gain of A I - A2.
Two linear shift-invariant systems are connected in a feedback network as illustrated in the figure below.
Assuming that the overall system is stable, so that H(ej") exists, show that the frequency response of this feedback network is . Y (ej") F (ei") J" H(e ) - X(ejm) 1 - F(ej")G(ej")
To analyze this network, we begin by noting that
CHAP. 21
FOURIER ANALYSIS
which, in the frequency domain, becomes
W (el") = X ( e J W ) G ( e J " ) Y( e l W )
Because
y ( e j w )= F ( e J w ) w ( e l " )
then Solving for ~ ( r i " yields )
Y ( e l w )= F ( e j w ) [ x ( e J w ) G ( e I w ) Y( e l w ) ]
Therefore, the frequency response is
The Discrete-Time Fourier Transform
A linear shift-invariant system is described by the LCCDE
Find the value of b so that IH(eJm)l is equal to I at w = 0, and find the hay-power point (i.e., the is frequency at which I H(ejW)I2 equal to one-half of its peak value, which occurs at w = 0).
The frequency response of the system described by this difference equation is
Because
l~(ej")= l~
- 0.5e-jw)( 1 - 0 . 5 e j w )
1.25 - cos w
I H ( e J w ) lwill be equal to 1 at o = 0 if
h2 --
1.25 - 1
This will be true when b = f0.5. To find the half-power point, we want to find the frequency for which
I H ( ~ ' " ) =~ I 0.25 = 0.5 1 .25 - cos o
This occurs when cos o = 0.75 or o = 0.2317.
Consider the system defined by the difference equation
where a and b are real, and la I < 1. Find the relationship between a and h that must exist if the frequency response is to have a constant magnitude for all w , that is, I H ( ~ ~ =) 1 * I
FOURIER ANALYSIS
[CHAP. 2
Assuming that this relationship is satisfied, find the output of the system when a = and
x(n) = (;)"u(n)
The frequency response of the LSI system described by this difference equation is
The squared magnitude is
1 + h2 2b cos w ( b + e-jo)(h + eJ") I H ( ~ ' "=I ~ ) I + a2 - 2a cos w ( 1 - ae-JU)(l- aeju)
Therefore, it follows that I H(e'")12 = 1 if and only if b = -a. if x(n) = ( i ) " u ( n )Y ( e j U is given by , ) With a = and h =
Using the DTFT pair
given in Table 2- 1, and using the linearity and delay properties of the DTFT, we have
= What we observe from this example is that although I H (eJU)l 1, the nonlinear phase has a significant effect on the values of the input sequence.
Show that the group delay of a linear shift-invariant system with a frequency response H(ejw)may be expressed as H R ( ~ ~ , ) GR ( d w ) H I (ejW)G(ejw)
t h ( ~= )
I H(ejw)I2
where HR(eJw) Hl(ejw)are the real and imaginary parts of H(ejw),respectively, and G R ( e j Wand and ) G r ( e j w ) the real and imaginary parts of the DTFT of nh(n). are
In terms of magnitude and phase, the frequency response is
Note that if we take the logarithm of H(eJW), have an explicit expression for the phase we
Differentiating with respect tow, we have
Equating the imaginary parts of both sides of this equation yields
CHAP. 21
If we define
FOURIER ANALYSIS
d -H(eJ") = Hk(ei") dw
+j~;(ej")
where HA(eJW) the derivative of the real part of H(eiw) and H;(ejW)is the derivative of the imaginary part, the is group delay may be written as
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