Dbu 1: Fill & Download for Free

Decibels are not a unit, but rather a logarithmic scale. Any time decibels are referenced, it is meaningless without a level or starting reference.For example, for sound volume, most often the scale is dBA SPL. This means sound pressure level, A weighted. There's a predefined level of what 0 dBA SPL is, and the measured level is in relation to that. You can be louder (positive dB SPL) or quieter (negative dB SPL) than 0. Typically it is rare to find negative dB SPL as it would be below the threshold of hearing.Similarly in the electrical portion (mixers, CD players, etc.) there are predefined levels. dBu (decibel unloaded) – Used for professional audio equipment, like mixers, etc. and dBV (decibel volts) – Used for consumer audio equipment, like CD players, etc.So, what are the dBV and dBu versions of the reference level? 0 dBV for consumer equipment = 1 Volt. 0 dBu for professional equipment = 0.7746 Volts. So on your mixer, when your meters are reading 0, and your fader is at unity you are (providing your equipment and gain structure are set properly) outputting 0.7746 volts to your amplifier / recorder /etc. you can lower the fader and still hear sound. So it is common to have levels of -5, -10, -40 dBu, etc all the way down to -infinity where the signal is fully attenuated.

If you read the math in plain English it makes sense.The right side of that equation translates to: "The probability that the attacker gets the original plaintext through a random guess".The left side is conditional probability, and translates to: "The probability that the attacker gets the original plaintext through a random guess, given that he knows what that plaintext encrypts to."So in those terms, a perfect cipher is one where if you intercept the ciphertext of a message and try to figure out what the plaintext is, your chances of success are no better than if you just randomly guessed without any ciphertext. A cipher that can conceal information this well is "perfect".Here's an example of an imperfect cipher: The old-fashioned shift cipher using the English alphabet. Let's suppose you intercepted the ciphertext DBU (that's the "c" in the equation) and you know that it was encrypted with a shift cipher. The plaintext was three characters because the ciphertext is three characters. There are [math]26^3[/math] three-letter sequences available and so the probability on the right side of the equation -- randomly guessing the plaintext correctly -- is [math]1/26^3[/math]. (Remember, you don't know "m" here.) On the other hand, knowing the encryption method and having the ciphertext in hand, there are only 26 possible plaintexts -- one for each value of the shifting -- and so your chances of guessing the plaintext given the ciphertext are now [math]1/26[/math]. This is the probability on the left side above. In other words, an "imperfect" cipher is one where knowledge of the ciphertext improves your chances of actually guessing the right plaintext.By contrast, the prototype example of a perfect cipher is a one-time pad, which we can build using the Vigenere cipher using a key that is the same length as the message. For example, suppose you received the ciphertext DBU that was encrypted with a three-letter keyword. If you just randomly guess what the three-letter plaintext was, you again have a [math]1/26^3[/math] chance of being right. But this is without using the ciphertext. Does having the ciphertext DBU help? In this case, no because all three-letter plaintext possibilities are equally likely. In the shift cipher example, not all plaintexts are equally likely -- for example we could not have "COW" as the plaintext because that would have required shifting "C" forward by 1 but "O" forward by 13. However, with the Vigenere cipher we are precisely shifting each character by a different amount. And since the keyword is the same length as the message, each character can have its own independent shift amount. Therefore I can come up with a key that would cause DBU to decrypt to any three-letter sequence I choose. Hence all plaintexts are equally likely and knowledge of the ciphertext doesn't change that fact, which means that the probability of guessing "m" with the ciphertext is the same as the probability of guessing "m" without it. That's what makes the one-time pad "perfect".

Cat (c+1=D,a+1=b & t+1=u)next series is Dbu,like thatDbu (D+1=E,b+1=c & u+1=v)Cat,Dbu,Ecv,_next upcoming series isE+1=fC+1=dV+1=uFdu is next upcoming seriesCat,Dbu,Ecv,Fdu answer