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A linear system follows the laws of superposition. This law is necessary and sufficient condition to prove the linearity of the system. Apart from this, the system is a combination of two types of laws −
Both, the law of homogeneity and the law of additivity are shown in the above figures. However, there are some other conditions to check whether the system is linear or not.
The conditions are −
Examples of nonlinear operators −
(a) Trigonometric operators Sin, Cos, Tan, Cot, Sec, Cosec etc.
(b) Exponential, logarithmic, modulus, square, Cube etc.
(c) sa(i/p) , Sinc (i/p) , Sqn (i/p) etc.
Either input x or output y should not have these nonlinear operators.
Examples
Let us find out whether the following systems are linear.
a) y(t) = x(t)+3
This system is not a linear system because it violates the first condition. If we put input as zero, making x(t) = 0, then the output is not zero.
b) y(t) = sintx(t)
In this system, if we give input as zero, the output will become zero. Hence, the first condition is clearly satisfied. Again, there is no nonlinear operator that has been applied on x(t). Hence, second condition is also satisfied. Therefore, the system is a linear system.
c) y(t) = sin(x(t))
In the above system, first condition is satisfied because if we put x(t) = 0, the output will also be sin(0) = 0. However, the second condition is not satisfied, as there is a nonlinear operator which operates x(t). Hence, the system is not linear.
If we want to define this system, we can say that the systems, which are not linear are nonlinear systems. Clearly, all the conditions, which are being violated in the linear systems, should be satisfied in this case.
Conditions
The output should not be zero when input applied is zero.
Any nonlinear operator can be applied on the either input or on the output to make the system nonlinear.
Examples
To find out whether the given systems are linear or nonlinear.
a) y(t)=e^{x(t)}
In the above system, the first condition is satisfied because if we make the input zero, the output is 1. In addition, exponential nonlinear operator is applied to the input. Clearly, it is a case of NonLinear system.
b) y(t) = x(t+1) +x(t−1)
The above type of system deals with both past and future values. However, if we will make its input zero, then none of its values exists. Therefore, we can say if the input is zero, then the time scaled and time shifted version of input will also be zero, which violates our first condition. Again, there is no nonlinear operator present. Therefore, second condition is also violated. Clearly, this system is not a nonlinear system; rather it is a linear system.
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