# why derivative and logarithm operators amplify the noise effects?

It is said, do not use derivative operator in a the control system. because it will amplify the noise effect. Can someone explain it and give some mathematical reference about it please?
Also I heard, we shouldn't use logarithm operator too, for the same reason. How come?
Thanks in advance.

• A pure derivative's gain increases with frequency, reaching infinite gain at infinite frequency. – endolith Sep 4 '14 at 21:21

## 1 Answer

The derivative operator is linear and time-invariant and can be described by its frequency response

$$X(j\omega)=j\omega\tag{1}$$

From (1) it is obvious that its gain increases linearly with frequency, and high frequencies (where the SNR is usually bad) are greatly amplified (as already pointed out in endolith's comment).

The logarithm is a memoryless nonlinearity and it compresses the input signal. This operation is used in many signal processing algorithms, such as feature extraction for speech recognition. In these algorithms this operation is essential to the performance and it definitely does not deteriorate the signal. What is important, however, is to deal with the singularity of the logarithm as its argument approaches zero because: $$\lim_{x\rightarrow 0+}\log(x)=-\infty$$ If nothing is done about it, the logarithm will produce very high negative peaks during periods where the signal is almost zero (i.e. when there's only noise). This problem is easily handled by defining a threshold, either on the value of the logarithm or on the input signal. Note that the logarithm can only be applied to non-negative signals (usually magnitudes or squared magnitudes of frequency domain data).

• Thanks for your complete answer ^^. specially for logarithm @matt-l – SAH Sep 5 '14 at 7:45
• Interesting question and answer - may I ask why the "SNR is usually bad" at high frequencies? – FreshAir Nov 25 '15 at 2:46