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Sources of corruption include multipath interference, changing channel gains, interference from other users, broadband noise, and narrowband interferences.
Sources of corruption include multipath interference, changing channel gains, interference fromother users, broadband noise, and narrowband interferences.

Broadband noise

When the signal arrives at the receiver, it is small and must be amplified. While it is possible to build high-gainamplifiers, the noises and interferences will also be amplified along with the signal. In particular, any noise in the amplifier itself will be increased. This is often called “thermal noise” and is usually modelled as white(independent) broadband noise. Thermal noise is inherent in any electronic components and is caused by smallrandom motions of electrons, like the Brownian motion of small particles suspended in water.

Such broadband noise is another reason that a bandpass filter is applied at the front end of the receiver. Byapplying a suitable filter, the total power in the noise (compared to the total power in the signal) can often be improved. [link] shows the spectrum of the signal as a pair of triangles centered at the carrier frequency ± f c with bandwidth 2 B . The total power in the signal is the area under the triangles. The spectrum of the noiseis the flat region, and its power is the shaded area. After applying the bandpass filter, the power in the signalremains (more or less) unchanged, while the power in the noise is greatly reduced. Thus, the signal-to-noise ratio (SNR) improves.

Narrowband noise

Noises are not always white; that is, the spectrum may not always be flat. Stray sine waves (and othersignals with narrow spectra) may also impinge on the receiver. These may be caused by errant transmitters that accidentlybroadcast in the frequency range of the signal, or they may be harmonics of a lower frequency waveas it experiences nonlinear distortion. If these narrowband disturbances occur out of band, they willautomatically be attenuated by the bandpass filter just as if they were a component of the wideband noise.However, if they occur in the frequency region of the signal, they decrease the SNR in proportion to theirpower. Judicious use of a “notch” filter (one designed to remove just the offending frequency) can be an effectivetool.

[link] shows the spectrum of the signal as the pair of triangles, along with three narrowband interferersrepresented by the three pairs of spikes. After the bandpass filter, the two pairs of out-of-band spikes are removed,but the in-band pair remains.

The signal-to-noise ratio is depicted graphically as the ratio of the power of the signal (the area under the triangles) to the power in the noise (the shaded area). After the bandpass filter, the power in the noise decreases, and so the SNR increases.
The signal-to-noise ratio is depicted graphically as the ratio of the power of the signal (the area under thetriangles) to the power in the noise (the shaded area). After the bandpass filter, the power in the noise decreases,and so the SNR increases.
Three narrowband interferers are shown in the top figure (the three pairs of spikes). The BPF cannot remove the in-band interferer, though a narrow notch filter can, at the expense of changing the signal in the region where the narrow band noise occurred.
Three narrowband interferers are shown in the top figure (the three pairs of spikes). The BPF cannot remove the in-band interferer,though a narrow notch filter can, at the expense of changing the signal in the region where the narrow band noise occurred.

Applying a narrow notch filter tuned to the frequency of the interferer allows its removal,although this cannot be done without also affecting the signal somewhat.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
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Dominic
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Lukman Reply
_3_2_1
felecia
⅗ ⅔½
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_½+⅔-¾
felecia
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
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Ifeanyi
on number 2 question How did you got 2x +2
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combine like terms. x + x + 2 is same as 2x + 2
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x*x=2
felecia
2+2x=
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×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Naagmenkoma
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
Shedrak
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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What is the expressiin for seven less than four times the number of nickels
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Software receiver design. OpenStax CNX. Aug 13, 2013 Download for free at http://cnx.org/content/col11510/1.3
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