Sometimes, technical specifications report the maximum frequency of the product instead of the bytes per second, which is inconvenient. This article will explain how to apply the proper conversion factor to translate frequency into bytes per second.

In order to perform the conversion, other important information is needed. It is required that the bus width, the number of bits transfered in one clock tick, to be known. The bus width makes a difference.

A product that operates at 4 bits is slower than a product that operates at 8 bits if both frequencies are the same. There are other advanced concepts to recognize as well such as double data rate vs single data rate but those topics are beyond this article. For the time being, let us consider the simple basic cases.

As an example, let us consider a specification that specifies a frequency of **X** such as 1000Hz over a 8 bit bus. In one second, there is exactly 1000 clock ticks, each tick is 1ms apart. At each tick, 8 bits are sent over the bus. Then in one second, 8 bits are sent 1000 thousand times, for a total of 8000 bits. This gives us an effective transfer rate of 8000bit per second, or 1000bytes per second.

The conversion is simple, take the number of bits of the bus line and multiply it by the number of clock ticks in one second to get the effective bytes per second.

In the previous example, one may consider taking a shortcut and noticed that replacing the *hz* with *bit per second* produced the same result.

While this is true for a bus size of 8 bits, it is not for other sizes. Consider a bus width of 4 bits, then in the same example with the frequency is 1000hz, the effective bit per second becomes 4 bits multiplied with 1000hz is only 4000bit per second. This is only half of speed.

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