# Digital Signal Processing - Static Systems

Some systems have feedback and some do not. Those, which do not have feedback systems, their output depends only upon the present values of the input. Past value of the data is not present at that time. These types of systems are known as static systems. It does not depend upon future values too.

Since these systems do not have any past record, so they do not have any memory also. Therefore, we say all static systems are memory-less systems. Let us take an example to understand this concept much better.

## Example

Let us verify whether the following systems are static systems or not.

• $y(t) = x(t)+x(t-1)$
• $y(t) = x(2t)$
• $y(t) = x = \sin [x(t)]$

### a) $y(t) = x(t)+x(t-1)$

Here, x(t) is the present value. It has no relation with the past values of the time. So, it is a static system. However, in case of x(t-1), if we put t = 0, it will reduce to x(-1) which is a past value dependent. So, it is not static. Therefore here y(t) is not a static system.

### b) $y(t) = x(2t)$

If we substitute t = 2, the result will be y(t) = x(4). Again, it is future value dependent. So, it is also not a static system.

### c) $y(t) = x = \sin [x(t)]$

In this expression, we are dealing with sine function. The range of sine function lies within -1 to +1. So, whatever the values we substitute for x(t), we will get in between -1 to +1. Therefore, we can say it is not dependent upon any past or future values. Hence, it is a static system.

From the above examples, we can draw the following conclusions −

• Any system having time shifting is not static.
• Any system having amplitude shifting is also not static.
• Integration and differentiation cases are also not static.