a given signal has a bit rate of 3000. if each signal element carries four bits, what is the baud rate?
475880
baud rate = bit rate/N\nbaud rate = 3000/4\nbaud rate = 750baud/s
955578
About 410 kilobits per second. Kilobits per second (Kbps) is the standard unit for measuring download speeds. Don't confuse it with kilobytes per second (KB/s) which is seldom used. There are eight bits in a byte.\n\n3 MB = 3 * 2^10 = 3,072 kilobytes\n3,072 kilobytes / 60 seconds = 51.2 KB/s\n51.2 KB/s * 8 bits in a byte = 409.6 Kbps (about 410)\n\nPlease note: the 2-to-the-power-of-10 in the first line reflects there being 1,024 kilobytes in a megabyte, rather than 1,000. Although sometimes 1,000 is used to simplify explanations because people can easily work with multiples of 10.\n\nEnjoy your physics homework or whatever it is!
1033626
1&=0.55$\nso for 30000& you get 54545,5$\n54545,5=240x+720 =>240x=53825.5 => x=224,27 \nfor 30000& you get 224 gadgets. \nCareful, you don't get 224,27gadgets!
835943
Bit rate refers to the number of bits of data in a certain time period.  The higher the bit rate, the more information per second a song or video file holds, and therefore the higher the quality of playback.\n\nMP3s can be encoded one of 2 ways.  Constant bitrate, or variable bitrate.  Constant 192kbps for example means that the song has been encoded at a constant rate of 192kbps, so EVERY second of audio will be 192kb in size.\n\nWith variable bit rate things get a bit more clever, and the codecs can pick out which bits need more attention.  If you pick variable 192kbps then the average bit rate of any section is 192kbps, but parts with little sound, or with uncomplicated sound, will be at a lower bit rate, whereas more complex sections (for example with more drums going on) will be at a higher bit rate.\n\nIt doesn't do much to reduce the file size, because it still averages out at the same bit rate, but it allows some sections to sound a lot better (with less compression) than having one flat bit rate.
510406
315mHz \n\nLove,\nKenneth Horowitz
569807
What bitrate do you have it compressed to? I would say the minimum should be 128kbts/sec
826399
123 a big big news for you!...\n:)! http://www.osoq.com/funstuff/extra/extra04.asp?strName=123
358234
1 bit (b)\n\n1 byte (B) = 8 bits\n\n1 Kilobyte (K / KB) = 2^10 bytes = 1,024 bytes \n\n1 Megabyte (M / MB) = 2^20 bytes = 1,048,576 bytes \n\n1 Gigabyte (G / GB) = 2^30 bytes = 1,073,741,824 bytes \n\n1 Terabyte (T / TB) = 2^40 bytes = 1,099,511,627,776 bytes\n\n1 Petabyte (P / PB) = 2^50 bytes = 1,125,899,906,842,624 bytes\n\n1 Exabyte (E / EB) = 2^60 bytes = 1,152,921,504,606.846,976 bytes\n\n1 Zettabyte (Z / ZB) = 2^70 bytes = 1,180,591,620,717,411,303,424 bytes\n\n1 Yottabyte (Y / YB) = 2^80 bytes = 1,208,925,819,614,629,174,706,176 bytes
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
761907
560 = N / (1- u/ 340) --------------- [1]\n\n480 = N *(1+ u/ 340) --------------- [2]\n\n[2] Divided by [1] gives \n\n480 / 560 = (1 + u/ 340) *(1 - u/ 340)\n\n 0.8571 = 1- (u/ 340) ^2.\n\n(u/ 340) ^2 = 0.1429\n\nu = 128.5 m/s.
354973
One bit corresponds to one transistor in memory.  One byte consists of 8 bits.  Every letter or character occupies one byte (or 8 bits) of memory.  The standard used to assign those 8-digit binary numbers to every letter, numeral, and other character is called ASCII.
215179
No loss at all, ever?  Not really.  To get a BER  (bit error rate) of zero, you would need an infinite SNR (signal to noise ratio), or an infinite amount of bandwidth, or an infinite amount of time.  None, of course are practical, or for that matter, possible.\n\nExtremely, extremely low BER's are possible, so low that when accompanied with error detection or correction schemes, can be assumed for all practical purposes to be zero.
449888
Actually, SuperShoey is correct.  Sorry for the incorrect answer I gave, I wasn't thinking.  When talking in BITS, there are 1,000 kilobits in one megabit.  However, there are 1,024 Kilobytes in a MegaByte.  There are 8 bits in a byte.\n\nThis is where people are getting confused.  They believe that because there are actually 1,024 bytes in a KB, that there will also be 1,024 bits in a kilobit.  This is not so.\n\n1 Mbps = 1,000Kbps
133260
.68181818181818 etc. \n(a little more than 1/2 a foot per second)
398218
Successive approximation A/D's work similar to a binary search algorithm.  The result can be very accurate if the A/D is designed correctly and the rate of the analog signal is relatively slow compared to the clock of the A/D, or a sample and hold block is used.  \nhere is an example of the operation (for this example a 4bit and that the input voltage is 1.2V):\n1.  using a DAC apply 1000 as a reference the  "1" is the test bit (say 1000 =1V)\n2.  Compare analog signal to this reference (result true) therefore the test bit is kept\n3.  Now apply the next test bit and the kept bit 1100 (the one in the 4's place is the new test bit and the one in the 8's place is the kept bit 1100= 1.5V)\n4.  Compare analog signal to this reference (result is false 1.2V <1.5V) therefore test bit is not kept\n5.  Now apply the next test bit and the kept bit 1010 (the one in the 2's place is the new test bit and the "10" in the 8,4's place is the kept bits 1010= 1.25V)\n6. Compare analog signal to this reference (result is false 1.2V <1.25V) therefore test bit is not kept.\n7. Now apply the next test bit and the kept bit 1001 (the one in the 1's place is the new test bit and the "100" in the 8,4,2's place is the kept bits 1001= 1.125V)\n8. Compare analog signal to this reference (result is true 1.2V >1.125V) therefore test bit is kept.\nResult is 1001\n\nHope this helps
213974
You should be able to figure this out.\n\nKbps is thousands of bits each second\n\nKB is thousands of bytes.  A byte is usually 8 bits.\n\nConsider 80x8 Kb (640 thousand)\n\nto see how many seconds, divide the data size by speed.\n\n640 thousand bits / 32 thousand bits each second\n\n640,000/ 32,000 is definitely more than 3 seconds and far less than 10 years.
1168573
To answer the question properly, need to know more about the nature of the signal.\n\nWhat is the signal bandwidth (what is the fastest that the signal is changing?)\n\nWhat is the dynamic range of the signal? (What is the ratio of the are the maximum-to-minimum values expected)\n\nWhat amount (%) of error can you tolerate in each measurement?\n\nFlash conversion is very fast, but is power hungry and limited in its number of bits. I'd probably choose this one only if there were no other alternatives. Considering that seismic events are relatively slow with respect to the typical speeds of electronics, I might expect that a flash converter may not be necessary.\n\nThere are several very fast successive-approximation ADCs that could probably do the job. And they are cheap!\n\nSigma-delta converters are relatively slow, but have a very wide dynamic range. Useful for high precision across a wide range.\n\nCheck out www.analog.com. They make a lot of converters, you can probably find some application notes on the topic.