barcode generator for ssrs Therefore, the frequency range 5 kHz 5 f 5 10 kHz corresponds to a digital frequency range in Software

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Therefore, the frequency range 5 kHz 5 f 5 10 kHz corresponds to a digital frequency range
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and the desired digital filter is a bandstop filter that has a frequency response as illustrated in the figure below.
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A major problem in the recording of electrocardiograms (ECGs) is the appearance of unwanted 60-Hzinterference in the output. The causes of this power line interference include magnetic induction, displacement currents in the leads on the body of the patient, and equipment interconnections. Assume that the bandwidth of the signal of interest is 1 kHz, that is,
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CHAP. 31
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SAMPLING
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The analog signal is converted into a discrete-time signal with an ideal A/D converter operating using a sampling frequency tY. The resulting signal x(n) = x,(nTs) is then processed with a discrete-time system that is described by the difference equation
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The filtered signal, y(n), is then converted back into an analog signal using an ideal D/A converter. Design a system for removing the 60-Hz interference by specifying values for f,,a, and b so that a 60-Hz signal of the form w,(t) = A sin(l20nt) will not appear in the output of the D/A converter.
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The signal that is to have the 60-Hz noise removed is bandlimited to 1000 Hz. Therefore, in order to avoid aliasing when the signal is sampled, we require a sampling frequency
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Using the minimum rate of 2000 Hz, note that a 60-Hz signal wo(t) = sin(l20nt) becomes
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where = 0.06n. Recall that complex exponentials are eigenfunctions of linear shift-invariant systems. Therefore, if the input to an LSI system is x(n) = eJn"O,the output is
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Because
wo. w(n) will be removed from x ( n ) if we design a filter so that H(ejw)is equal to zero at w = f Because H(eJw)is a second-order filter with a frequency response
it may be factored as follows: H(eJ") = (1 - ~ e - ~ " ) (- fie-]") l Therefore, H (ejw)will be zero for w = fwo if a! = eJw and B = e--1". In this case, we have
Thus, our requirements are that
a = -2 cos wo = -2 cos(0.06n)
and f, = 2000.
The following system is used to process an analog signal with a discrete-time system.
xo(r)
.Discrete-time
System
YO@)
SAMPLING
[CHAP. 3
Suppose that x a ( t ) is bandlimited with X,( f ) = 0 for If 1 > 5 kHz as shown in the figure below,
+Xdf)
and that the discrete-time system is an ideal low-pass filter with a cutoff frequency of n/2. Find the Fourier transform of y a ( t ) if the sampling frequencies are f l = f2 = 10 kHz. Repeat for f l = 20 kHz and f2 = 10 kHz. Repeat for f l = 10 kHz and f2 = 20 kHz.
When the sampling frequencies of the C/D and D/C converters are the same, and x , ( t ) is bandlimited with X,(jS2) = 0 for If21 > n / T l , this system is equivalent to an analog filter with a frequency response
) Therefore, if ~ ( e j "is a low-pass filter with a cutoff frequency 71/2. the cutoff frequency of H a ( j f 2 ) ,denoted by n o , is given by
Thus,
fo =
= 2500 Hz
When the sampling frequencies of the C/D and D/C are different, it is best to plot the spectrum of the signals as they progress through the system. Wilh X,( f ) as shown above, the discrete-time Fourier transform of x ( n ) is
Because the cutoff frequency of the discrete-time low-pass filter is n/2, y ( n ) = x ( n ) , and the output of the D/C converter is as plotted below.
With
= 10 kHz, we are sampling x u ( [ )at the Nyquist rate, and the spectrum of x ( n ) is
+ XkJ")
CHAP. 31
SAMPLING
and the output of the low-pass filter is as shown below.
+ y(el")
Therefore, the spectrum of y o ( ( )is as follows:
+Ya(f)
Consider the system in Fig. 3-9 for implementing a continuous-time system in terms of a discrete-time system. Assume that the input to the C/D converter is bandlimited to Q0 = Q,/2 and that the unit sample response of the discrete-time system is
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