DECIBEL SCALE PROBLEMS HOW TO
DECIBEL SCALE PROBLEMS FREE

While I’m on the subject of light, how far does light travel through space in one year? The answer is about 5.879 trillion miles, which we call a light year. Eighty-seven percent of 186,282.397 miles per second is 162,065.685 miles per second still really fast! It takes a mere 0.00000000117 second (1.17 nanosecond) for RF to travel through a foot of coax.

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How long does it take for RF signals to travel through 1 foot of hardline feeder coax? Assume the cable’s velocity of propagation is 87 percent, which means that RF signals travel through the coax at 87 percent of the free space (vacuum) value of the speed of light. In case you were wondering, Earth’s equatorial circumference is about 24,900 miles, or 256,948,534 dollar bills laid end to end. In a blink of an eye, then, our light beam will have traversed 32,599 miles, or roughly one-and-a-third trips around Earth’s equator. How far will light travel in the blink of an eye? Normal eyeblink duration is typically 150 to 200 milliseconds (ms), so for this example let’s call it 175 ms. In one second, that beam of light will go around the Earth nearly 7-1/2 times! For this example, I’m ignoring terrain obstructions, reflection, atmospheric refraction, the atmosphere’s slightly different velocity of propagation compared to a vacuum, and the fact that light prefers to travel in a straight line. How fast? Let’s imagine a beam of light that could somehow travel unimpeded around the Earth’s equator. In a vacuum, light or any electromagnetic signal, for that matter zips along at 299,792,458 meters per second, or 186,282.397 miles per second. One measure that we deal with, often indirectly, is the speed of light. Nice mathematical shorthand! Speed of light Can you imagine dealing with numbers like these on a daily basis? The decibel allows us to express these same levels as 0 dBmV, +20 dBmV and +48 dBmV respectively, using the formula dBmV = 20log (signal level in mV/1mV). 0.001 volt isn’t too cumbersome as far as the number of zeros to the right of the decimal point, but how about the same signal level expressed as power, or 0.00000001333 watt (13.33 nanowatts)? A line extender’s typical perchannel input level is 0.010 volt (10 millivolts) or 0.000001333 watt (1.33 microwatt), and its per-channel output is 0.251189 volt (251.189 mV), or 0.000841276 watt (841.276 µW). Think about the minimum per-channel input to a subscriber terminal, as defined in Part 76 of the FCC’s Rules: 1 millivolt (75 ohms), or 0.001 volt. Too many zeroes Have you ever wondered why we use the decibel to express signal levels?Have you ever wondered why we use the decibel to express signal levels? Why not just use the signal’s actual voltage or power? One reason is the very small numbers that make up typical cable network signal levels. Our 1,000,000 Hz frequency becomes 1 megahertz (MHz), and 1,000,000 bits per second is 1 megabit per second (Mbps). It’s generally inconvenient to deal with large numbers like that, which is where the International System of Units (SI) – specifically SI prefixes () – can help. From that perspective, a million is a pretty big number. If you were to count to 1,000,000, how long would it take? If we assume the average person needs second per number, and you count nonstop (no sleeping, eating, drinking, or …) to 1,000,000, it will take a million seconds. It also might represent data rate – say, 1,000,000 bits per second. When discussing an RF signal, 1,000,000 might be the signal’s frequency, as in 1,000,000 Hertz (Hz). In reality, it’s hard to grasp just how big or small numbers can be.

Fortunately, there are several ways to simplify how we describe those numbers, which I’ll get to shortly. (I'll leave this step up to you.Ours is a world of numbers, and in cable we tend to use a lot of really big and really small numbers. Now we can find 'I' by using our calculators to raise 10 to the -0.9 power. If not, then we can rewrite this in exponential form:

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If you know how to do inverse logarithms on your calculator you can find 'I' now. And since by definition this becomes -12. Use the property of logarithms,, to rewrite as.We can start by finding the / If we understand logarithms we will know that this must be -12. Solving for I means we have to get an equation we I is by itself on one side. (I know this is where I got confused when I first looked at the problem.)Īnd, as your teacher says, we still need to solve for I. And the zero in the logarithm is a subscript, not the digit of a number. It took me a while to understand what was going on. You can put this solution on YOUR website!

Decibel scale problems