Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Signals and Systems: Linear and Non-Linear Systems
Linear System
A system is said to be linear if it obeys the principle of homogeneity and principle of superposition.
Principle of Homogeneity
The principle of homogeneity says that a system which generates an output y(t) for an input x(t) must produce an output ay(t) for an input ax(t).
Superposition Principle
According to the principle of superposition, a system which gives an output 𝑦1(𝑡) for an input 𝑥1(𝑡) and an output 𝑦2(𝑡) for an input 𝑥2(𝑡) must produce an output [𝑦1(𝑡) + 𝑦2(𝑡)] for an input [𝑥1(𝑡) + 𝑥2(𝑡)].
Therefore, for a continuous-time linear system,
[𝑎𝑦1(𝑡) + 𝑏𝑦2(𝑡)] = 𝑇[𝑎𝑥1(𝑡) + 𝑏𝑥2(𝑡)] = 𝑎𝑇[𝑥1(𝑡)] + 𝑏𝑇[𝑥2(𝑡)]
Also, for a discrete-time linear system,
[𝑎𝑦1(𝑛) + 𝑏𝑦2(𝑛)] = 𝑇[𝑎𝑥1(𝑛) + 𝑏𝑥2(𝑛)] = 𝑎𝑇[𝑥1(𝑛)] + 𝑏𝑇[𝑥2(𝑛)]
Hence, we can say that a system is linear if the output of the system due to weighted sum of inputs is equal to the weighted sum of outputs.
Filter circuits, communications channels, etc. are few examples of linear systems.
Non-Linear System
A system is said to be a non-linear system if it does not obey the principle of homogeneity and principle of superposition.
Generally, if the equation describing the system contains square or higher order terms of input/output or product of input/output and its derivatives or a constant, the system will be a non-linear system. Triangulation of GPS signals is an example of non-linear system.
Numerical Example
Check whether the given systems are linear or non-linear systems −
$\mathrm{\frac{\mathrm{d^{2}}y(t) }{\mathrm{d} t^{2}}+3ty(t)=t^{2}x(t)} $
$\mathrm{3\frac{\mathrm{d}y(t) }{\mathrm{d} t}+4y(t)=x^{2}(t)} $
Solution
The given system is,
$$\mathrm{\frac{\mathrm{d^{2}}y(t) }{\mathrm{d} t^{2}}+3ty(t)=t^{2}x(t)} $$
Consider an input 𝑥1(𝑡) which produces an output 𝑦1(𝑡), then,
$$\mathrm{\mathrm{\frac{\mathrm{d^{2}}y_{1}(t) }{\mathrm{d} t^{2}}+3ty_{1}(t)=t^{2}x_{1}(t)}\: \: \cdot \cdot \cdot (1)} $$
And an input 𝑥2(𝑡) which produces an output 𝑦2(𝑡), then,
$$\mathrm{\mathrm{\frac{\mathrm{d^{2}}y_{2}(t) }{\mathrm{d} t^{2}}+3ty_{2}(t)=t^{2}x_{2}(t)}\: \: \cdot \cdot \cdot (2)} $$
Now, the linear combination of the equations (1) and (2) gives,
$$\mathrm{ \begin{Bmatrix} a\mathrm{\frac{\mathrm{d^{2}}y_{1}(t) }{\mathrm{d} t^{2}}+a3ty_{1}(t)}\ \end{Bmatrix}+\begin{Bmatrix} b\mathrm{\frac{\mathrm{d^{2}}y_{2}(t) }{\mathrm{d} t^{2}}+b3ty_{2}(t)}\ \end{Bmatrix}=at^{2}x_{1}(t)+bt^{2}x_{2}(t) } $$
$$\mathrm{\Rightarrow \frac{\mathrm{d} ^{2}}{\mathrm{d} t^{2}}\left [ ay_{1}(t)+by_{2}(t) \right ]+3t\left [ ay_{1}(t)+by_{2}(t) \right ]}$$
$$\mathrm{= t^{2}\left [ ax_{1}(t)+bx_{2}(t) \right ]\: \: \cdot \cdot \cdot (3)}$$
Where, [𝑎𝑦1(𝑡) + 𝑏𝑦2(𝑡)] is the weighted sum of outputs and [𝑎𝑥1(𝑡) + 𝑏𝑥2(𝑡)] is the weighted sum of inputs.
Hence, the equation (3) shows that the weighted sum of inputs to the given system generates an output that is equal to the weighted sum of outputs to each of the individual inputs. Therefore, the given system is a linear system.
The given system is,
$$\mathrm{3\frac{\mathrm{d} y(t)}{\mathrm{d} t}+4y(t)=x^{2}(t)}$$
Let the output 𝑦1(𝑡) is corresponding to the input 𝑥1(𝑡), then,
$$\mathrm{3\frac{\mathrm{d} y_{1}(t)}{\mathrm{d} t}+4y_{1}(t)=x_{1}^{2}(t)\: \: \cdot \cdot \cdot (1)}$$
Similarly, the output 𝑦2(𝑡) is corresponding to the input 𝑥2(𝑡), then,
$$\mathrm{3\frac{\mathrm{d} y_{2}(t)}{\mathrm{d} t}+4y_{2}(t)=x_{2}^{2}(t)\: \: \cdot \cdot \cdot (2)}$$
Then, the linear combination (i.e., homogeneity and superposition) of the equations (1) and (2) can be written as,
$$\mathrm{\begin{Bmatrix} a3\frac{\mathrm{d} y_{1}(t)}{\mathrm{d} t}+a4y_{1}(t)\ \end{Bmatrix}+\begin{Bmatrix} b3\frac{\mathrm{d} y_{2}(t)}{\mathrm{d} t}+b4y_{2}(t)\ \end{Bmatrix}=ax_{1}^{2}(t)+bx_{2}^{2}(t)}$$
$$\mathrm{\Rightarrow 3\frac{\mathrm{d} }{\mathrm{d} t}\left [ ay_{1}(t)+by_{2}(t) \right ]+4\left [ ay_{1}(t)+by_{2}(t) \right ]=ax_{1}^{2}(t)+bx_{2}^{2}(t)}$$
Where, $\mathrm{\left [ ay_{1}(t)+by_{2}(t) \right ]}$ is the weighted sum of outputs but $\mathrm{\left [ ax_{1}^{2}(t)+bx_{2}^{2}(t) \right ]} $ is not the weighted sum of inputs. Here, the principle of superposition is not satisfied. Therefore, the given system is a non-linear system.
30K+ Views
