Difference between revisions of "Why is dB defined as 20log 10?"

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X_dB = 10log_10 (X/X0)
 
X_dB = 10log_10 (X/X0)
  
This way, if X is 10 times X0, we read X is 10dB greater than X0. On the usefulness of the log_10, it allows to express a very large range of ratios in a range of moderate size, allowing one to clearly visualize huge changes of some quantity.
+
This way, if X is 10 times X0, we read X is 10dB greater than X0. On the usefulness of utilizing the log_10: it allows to express a very large range of ratios in a range of moderate size, allowing one to clearly visualize huge changes of some quantity.
  
 
Many contributions to Control Theory have come from the area of Electrical Engineering. Since in a circuit with constant resistance, the power developed is proportional to the square of the applied voltage, then if V and V0 are two voltages of interest (I/O)
 
Many contributions to Control Theory have come from the area of Electrical Engineering. Since in a circuit with constant resistance, the power developed is proportional to the square of the applied voltage, then if V and V0 are two voltages of interest (I/O)

Latest revision as of 03:00, 3 October 2006

Q Why is dB defined as 20log 10? What is so special about 20?


A This convention is in place due to the original definition of Decibel (dB) as a measure of a ratio between powers. If X, X0 are power measurements( X0 could be the input or reference power, X the output ), then the dB measurement was defined as

X_dB = 10log_10 (X/X0)

This way, if X is 10 times X0, we read X is 10dB greater than X0. On the usefulness of utilizing the log_10: it allows to express a very large range of ratios in a range of moderate size, allowing one to clearly visualize huge changes of some quantity.

Many contributions to Control Theory have come from the area of Electrical Engineering. Since in a circuit with constant resistance, the power developed is proportional to the square of the applied voltage, then if V and V0 are two voltages of interest (I/O) one finds that:

V_dB= 10log_10(V^2/V0^2) = 20log_10(V/V0)

which is the reason why Electrical engineers use the dB defined as 20log_10(V/V0), and so Control theorists do!


--Franco