barcode generator for ssrs (a) xa(t) is real with Xu(f ) nonzero only for 9 kHz < ( f1 < 12 kHz. in Software

Encoder ANSI/AIM Code 128 in Software (a) xa(t) is real with Xu(f ) nonzero only for 9 kHz < ( f1 < 12 kHz.

(a) xa(t) is real with Xu(f ) nonzero only for 9 kHz < ( f1 < 12 kHz.
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(b) xa(t) is real with X u (f ) nonzero only for 18 kHz < 1 f 1 < 22 kHz.
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(c) xa(t) is complex with X,( f ) nonzero only for 30 kHz < f < 35 kHz.
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(a) For this signal, the bandwidth is B = f2
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- f l = 3 kHz, and f2 = 12 = 4B is an integer multiple of B . Therefore, the minimum sampling frequency is f, = 2B = 6 kHz.
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( b ) For this signal, B = 4 kHz and f2 = 22, which is not an integer multiple of B . With Lfi/B1 = 5, if we let B' = f 2 / 5 = 4.4, f2 is an integer multiple of B', and .ra(t) may be sampled with a sampling frequency of f, = 2 8 ' = 8.8 kHz.
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(c) For a complex bandpass signal with a spectrum that is nonzero for fi < f < f 2 , the minimum sampling frequency is f, = f2 - f l . Thus, for this signal, f, = 5 kHz.
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How many bits are needed in a n A/D converter if w e want a signal-to-quantization noise ratio of at least a 9 0 d B Assume that xa(t) is gaussian with a variance ,: and that the range of the quantizer extends from -3a, to 3ax; that is, X,,, = 30; (with this value for X,,,, only about one out of every 1000 samples will exceed the quantizer range).
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CHAP. 31
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SAMPLING
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For a (B + I)-bit quantizer, the signal-to-quantization noise ratio is SQNR = 6.02B With X , = 30, this becomes ,, SQNR = 6.02B
x + 10.8 1 - 2010g m ,,
+ 10.81 - 20l0g3 = 6.02B + 10.81 - 9.54 = 6.02B + 1.27
If we want a signal-to-quantization noise ratio of 90 dB, we require
An image is to be sampled with a signal-to-quantization noise ratio of at least 80 dB. Unlike many other signals, the image samples are nonnegative. Assume that the sampling device is calibrated so that the sampled image intensities fall within the range from 0 to 1. How many bits are needed to achieve the desired signal-to-quantization noise ratio
For a bipolar signal with amplitudes that fall within the range [-X,,,,, is
X,,,], the signal-to-quantization noise ratio
SQNR = 6.02B + 10.81 - 201og&
For a nonnegative signal that is confined to the interval [0, I], the signal-to-quantization noise ratio is equivalent to the bipolar case if we set X,,, = 0.5. If we assume that the intensities of the image are uniformly distributed over the interval [0, I], u2 = I
Therefore,
SQNR = 6.02B
rn + 10.81 - 2010g - = 6.02B + 6.03 2
and for a signal-to-quantization noise ratio of 80 dB, we require
or B
+ 1 = 14 bits.
Suppose that we have a set of unquantized samples, x(n), that are nonnegative for all n. A method for quantizing x ( n ) that is often used in speech processing is as follows. First, we form the sequence y(n) = log[x(n)l Then y(n) is quantized with a (B
+ 1)-bit uniform quantizer, 9(n) = C [y(n)l = y(n) + e(n)
W ) = exp(9(n))
The quantized signal samples are then obtained by exponentiating J(n),
Show that if' e(n) is small, the signal-to-quantization noise ratio is independent of the signal power.
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we have, for R(n), R(n) = exp{log[x(n)l < If e(n) < 1, we may use the expansion to write &n) = x(n)l I
+ e(n)} = x(n) . exp(e(n))
exp{e(n)) x 1
+ e(n)
+ e(n)] = x(n) + f (n)
where f (n) = x(n)e(n) is a (signal-dependent) quantization noise. If we assume that the quantization noise e(n) is statistically independent of x(n), Elf2(n)) = ~ { x ~ ( n E{e2(n)) )}. and the signal-to-quantization noise ratio is E{x2(n)l SQNR = 10 log - -10 log E (e2(n)) = E { f '(n)) which is independent of the signal power.
Discrete-Time Processing of Analog Signals
A continuous-time signal xa(t) is to be filtered to remove frequency components in the range 5 kHz 5 f i 10 kHz. The maximum frequency present in xa(t) is 20 kHz. The filtering is to be done by sampling x,(t), filtering the sampled signal, and reconstructing an analog signal using an ideal D/C converter. Find the minimum sampling frequency that may be used to avoid aliasing, and for this minimum sampling rate, find the frequency response of the ideal digital filter H ( e J m )that will remove the desired frequencies from xa (t).
Because the highest frequency in x,(t) is 20 kHz, the minimum sampling frequency to avoid aliasing is f, = 40 kHz. The relationship between the continuous frequency variable R and the discrete frequency variable o is given by
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