Bithi Mitra
Lecturer, Dept. of EEE, NUB
Linear & Non-Linear system
Linear System
The system which follows the principle of superposition is known as linear system.
Non-Linear System
The system which does not follow the principle of superposition is known as non-linear system.
Principle of Superposition
Superposition simply implies that response resulting from several input signals can be computed as the sum
of the responses resulting from each input acting alone.
Mathematically
Principle of superposition is the combination of two different laws:
a. Law of additivity.
b. Law of Homogeneity.
Law of Additivity
Step-1
In order to understand the law of additivity let us take a system where input is 𝑥1(𝑡). After passing through
the system the output becomes 𝑦1(𝑡)
𝑥1(𝑡).
System
Input
𝑦1(𝑡)
Output
Now we will change the input. Instead of 𝑥1(𝑡) let us assume that the new input is 𝑥2(𝑡). After passing
through the system the new output will be 𝑦2(𝑡).
𝑥2(𝑡).
Input
System
𝑦2(𝑡)
Output
Now we will add two output
Lt Col K M Mustafizur Rahman, EECE Dept, MIST
Step-1
𝑦1(𝑡)
𝑥1(𝑡)
System
Output
Input
Σ
𝑦1(𝑡) + 𝑦2(𝑡)
𝑦2(𝑡)
𝑥2(𝑡)
System
Output
Input
Step-2
𝑥1(𝑡) + 𝑥2(𝑡)
Σ
System
𝑦/(𝑡)
Step-2
In second step, instead of adding output, we will add two inputs 𝑥1(𝑡) + 𝑥2(𝑡) and feed it through thesystem.
After going through the system new output generated is 𝑦/(𝑡)
Now there are two possibilities
(1) 𝑦/(𝑡) = 𝑦1(𝑡) + 𝑦2(𝑡)
(2) 𝑦/(𝑡) ≠ 𝑦1(𝑡) + 𝑦2(𝑡)
If the two outputs are same then the system is said to follow the law of additivity.
Law of Homogeneity
Say, we have same system as in case of law of additivity.
𝑥(𝑡).
𝑦(𝑡)
System
Output
Input
Step-1
This time we will not perform any addition but multiplication. We will multiply a constant k to the obtained
output (𝑡)
𝑥(𝑡)
Input
System
𝑦(𝑡)
𝑘
𝑘𝑦(𝑡)
Constant
Lt Col K M Mustafizur Rahman, EECE Dept, MIST
Step-2
In this step we will multiply same constant to the input (𝑡) and multiplied result will act as the input to the
system. Then the generated output will be compared with 𝑘𝑦(𝑡)
𝑥(𝑡)
𝑦(𝑡)
System
𝑘
Input
𝑘𝑦(𝑡)
Constant
𝑘
𝑘𝑥(𝑡)
System
𝑦/(𝑡)
Constant
Now there are two possibilities
(1)𝑦/(𝑡) = 𝑘𝑦(𝑡)
(2)𝑦/(𝑡) ≠ 𝑘𝑦(𝑡)
If the two outputs are same then the system is said to follow the law of homogeneity.
Mathematical Problems on Linear and Non-Linear system
Example-1
Test the linearity y(𝑡) = 𝑥(𝑡2)
Output
𝑥(𝑡)
Output
System
(𝑡) = 𝑥(𝑡2)
Output
Input
Functionality/Nature of the system
It is very important to understand the functionality of the system. Above system’s functionality is that it
replaces 𝑡 to 𝑡2
Law of Additivity
Step-1
𝑥1(𝑡)
Sys
𝑦1(𝑡) = 𝑥1(𝑡2)
𝑥2(𝑡)
Sys
𝑦2(𝑡) = 𝑥2(𝑡2)
∴ 𝑦1(𝑡) + 𝑦2(𝑡) = 𝑥1(𝑡2) + 𝑥2(𝑡2)
Step-2
Sys
𝑥1(𝑡) + 𝑥2(𝑡)
𝑦/(𝑡) = 𝑥1(𝑡2) + 𝑥2(𝑡2)
New Input
Note: Nature of the system is that “ It replaces 𝑡 to 𝑡2 of the input signal. Whatever may be the input, the
system replaces 𝑡 to 𝑡2
By comparing step-1 and step-2 we can conclude that the system follows law of additivity.
Law of Homogeneity
Step-1
𝑥(𝑡)
System
𝑦(𝑡)
𝑘
𝑘
𝑘(𝑡)
System
𝑘(𝑡) = 𝑘𝑥(𝑡2)
Step-2
𝑥(𝑡)
𝑦/(𝑡) = 𝑘𝑥(𝑡2)
Comparing two output we can say that both are same and equal. Thus system follows the law of
homogeneity.
As the system follows both law of homogeneity and law of additivity, therefore, given system is a linear
system.
Example-2: Test the linearity y(𝑡) = 𝑥2(𝑡)
solution:
Law of additivity:
Consider that 𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
𝒚𝟏 (𝒕) = 𝒙𝟐𝟏 (𝒕)
𝒚𝟐 (𝒕) = 𝒙𝟐𝟐 (𝒕)
𝒚𝟏 (𝒕) + 𝒚𝟐 (𝒕) = 𝒙𝟐𝟏 (𝒕) + 𝒙𝟐𝟐 (𝒕) … … … … (𝟏)
𝒚(𝒕) = 𝒙𝟐 (𝒕) = {𝒙𝟏 (𝒕) + 𝒙𝟐 (𝒕)}𝟐 … … … … (𝟐)
By comparing equation-1 and equation-2 it can be concluded that the system does not follow law
of additivity. As the system does not follow law of additivity, therefore, given system is nonlinear system.
Note: Once it is confirmed that that system is not following law of additivity then proving law of
homogeneity is not required.
Example 03:
Determine whether the following system is Linear or Non-linear
y(𝑡) = a
𝒅𝒙(𝒕)
𝒅𝒕
Solution:
Law of additivity:
Law of homogeneity:
Consider that 𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
𝑘𝑦(𝑡) = 𝑘 × a
So, 𝑦1 (𝑡) = 𝑎
𝑑𝑥1 (𝑡)
𝑑𝑡
𝑘𝑥(𝑡) → 𝑦 ′ (𝑡) = 𝑎
𝑑𝑥2 (𝑡)
𝑑𝑡
𝑑𝑥1 (𝑡)
𝑑𝑥2 (𝑡)
𝑦1 (𝑡) + 𝑦2 (𝑡) = 𝑎 ( 𝑑𝑡
+ 𝑑𝑡
)
𝑑𝑥(𝑡)
𝑑𝑡
𝑑(𝑘𝑥(𝑡))
𝑑𝑥(𝑡)
=𝑘×𝑎
𝑑𝑡
𝑑𝑡
𝑦2 (𝑡) = 𝑎
Comparing two output we can say that both are same and
equal. Thus system follows the law of homogeneity.
Again,
𝑑
y(𝑡) = a {𝑥1 (𝑡) + 𝑥2 (𝑡)}
𝑑𝑡
𝑑𝑥 (𝑡)
1
= 𝑎 ( 𝑑𝑡
+
𝑑𝑥2 (𝑡)
𝑑𝑡
)
= 𝑦1 (𝑡) + 𝑦2 (𝑡)
Comparing two output we can say that both are same
and equal. Thus system follows the law of additivity.
As the given system follows the law of additivity and homogeneity, the system is linear.
Example 04: Determine whether the following system is Linear or Non-linear
𝑦(𝑡) = 𝒆𝒙(𝒕)
Solution:
Law of homogeneity:
Law of additivity:
𝑘𝑦(𝑡) = 𝒌 𝒆𝒙(𝒕)
Consider that 𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
𝑘𝑥(𝑡) → 𝑦 ′ (𝑡) = 𝒆𝒌 𝒙(𝒕)
So, 𝑦1 (𝑡) = 𝑒 𝑥1 (𝑡)
𝑦1 (𝑡) = 𝑒
𝑥2 (𝑡)
Comparing two output we can say that both are not
same and equal. Thus system does not follow the law
of homogeneity.
𝑦1 (𝑡) + 𝑦2 (𝑡) = 𝑒 𝑥1 (𝑡) + 𝑒 𝑥2 (𝑡)
Again,
𝑦(𝑡) = 𝑒 (𝑥2 (𝑡)+𝑥2 (𝑡))
= 𝑒 𝑥1 (𝑡) × 𝑒 𝑥2 (𝑡)
≠ 𝑦1 (𝑡) + 𝑦2 (𝑡)
Comparing two output we can say that both are not same and equal. Thus system does not follow
the law of additivity.
As the given system does not follow the law of additivity and homogeneity, the system is non - linear.
Example 05: Determine whether the following system is Linear or Non-linear
y(𝑡) = t x(t)
Solution:
Law of additivity:
Law of homogeneity:
Consider that 𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
𝑘𝑦(𝑡) = 𝑘 t x(t)
So, 𝑦1 (𝑡) = 𝑡 𝑥1 (𝑡)
𝑘𝑥(𝑡) → 𝑦 ′ (𝑡) = 𝑘 𝑡 𝑥(𝑡)
𝑦2 (𝑡) = 𝑡 𝑥2 (𝑡)
Comparing two output we can say that both are
same and equal. Thus system follows the law of
homogeneity.
𝑦1 (𝑡) + 𝑦2 (𝑡) = 𝑡 𝑥1 (𝑡) + 𝑡 𝑥2 (𝑡)
= 𝑡{𝑥1 (𝑡) + 𝑥2 (𝑡)}
Again,
𝑦(𝑡) = 𝑡 𝑥(𝑡)
= 𝑡 {𝑥1 (𝑡) + 𝑥2 (𝑡)}
= 𝑦1 (𝑡) + 𝑦2 (𝑡)
Comparing two output we can say that both are same
and equal. Thus system follows the law of additivity.
As the given system follows the law of additivity and homogeneity, the system is linear.
Example 06:
Determine whether the following system is Linear or Non-linear
y(𝑡) = 10 x(t) + 5
Solution:
Law of additivity:
Law of homogeneity:
𝑘𝑦(𝑡) = 𝑘 × {10 𝑥(𝑡) + 5}
Consider that 𝑥(𝑡) = 𝑥1 (𝑡) + 𝑥2 (𝑡)
So, 𝑦1 (𝑡) = 10 𝑥1 (𝑡) + 5
𝑦2 (𝑡) = 10 𝑥2 (𝑡) + 5
𝑦1 (𝑡) + 𝑦2 (𝑡) =10 𝑥1 (𝑡) + 5 + 10 𝑥2 (𝑡) + 5
𝑘𝑥(𝑡) → 𝑦 ′ (𝑡) = 10 × 𝑘𝑥(𝑡) + 5
Comparing two output we can say that both are
not same and equal. Thus system does not follow
the law of homogeneity.
= 10 {𝑥1 (𝑡) + 𝑥2 (𝑡)} + 10
Again,
𝑦(𝑡) = 10 𝑥(𝑡) + 5
= 10 {𝑥1 (𝑡) + 𝑥2 (𝑡)} + 5
≠ 𝑦1 (𝑡) + 𝑦2 (𝑡)
Comparing two output we can say that both are not same and
equal. Thus system does not follow the law of additivity.
As the given system does not follow the law of additivity and homogeneity, the system is non-linear.
Example 07: Determine whether the following system is Linear or Non-linear
𝒚(𝒏) = 𝒏𝟐 𝒙(𝒏)
Solution:
Law of additivity:
Consider that 𝑥(𝑛) = 𝑥1 (𝑛) + 𝑥2 (𝑛)
So, 𝑦1 (𝑛) = 𝑛2 𝑥1 (𝑛)
𝑦2 (𝑛) = 𝑛2 𝑥2 (𝑛)
𝑦1 (𝑛) + 𝑦2 (𝑛) = 𝑛2 𝑥1 (𝑛) + 𝑛2 𝑥2 (𝑛)
= 𝑛2 {𝑥2 (𝑛) + 𝑥2 (𝑛)}
Law of homogeneity:
𝑘𝑦(𝑛) = 𝑘 𝒏𝟐 𝒙(𝒏)
𝑘𝑥(𝑛) → 𝑦 ′ (𝑛) = 𝑘 𝒏𝟐 𝒙(𝒏)
Comparing two output we can say that both are
same and equal. Thus system follows the law of
homogeneity.
Again,
y(n) = 𝒏𝟐 𝒙(𝒏)
= 𝒏𝟐 {𝑥1 (𝑛) + 𝑥2 (𝑛)}
= 𝑦1 (𝑛) + 𝑦2 (𝑛)
Comparing two output we can say that both are same and equal. Thus system follows the law of additivity.
As the given system follows the law of additivity and homogeneity, the system is linear.
Example 08: Determine the linearity of the following system
𝑡
𝑦(𝑡) = ∫ 𝑥(𝑟)𝑑𝑟
−∞
Solution:
Law of Additivity
Step-1
𝑡
𝑥1(𝑡)
𝑦1(𝑡) = ∫ 𝑥1(𝑟)𝑑𝑟
Sys
−∞
𝑡
𝑥2(𝑡)
𝑦2(𝑡) = ∫ 𝑥2(𝑟)𝑑𝑟
Sys
−∞
∴ 𝑦1(𝑡) + 𝑦2(𝑡)
𝑡
𝑡
= ∫ 𝑥1(𝑟)𝑑𝑟 + ∫ 𝑥2(𝑟)𝑑𝑟
−∞
−∞
𝑡
= ∫[𝑥1(𝑟) + 𝑥2(𝑟)]𝑑𝑟
−∞
Example 09: Determine whether the following system is Linear or Non-linear
𝒅𝟐 𝒚(𝒕)
𝒅𝒚(𝒕)
𝒅𝒙(𝒕)
−
𝟑
=
𝟓
𝒅𝒕𝟐
𝒅𝒕
𝒅𝒕
Solution:
Law of additivity:
Consider that 𝑥(𝑛) = 𝑥1 (𝑛) + 𝑥2 (𝑛)
So,
𝑑2 𝑦1 (𝑡)
𝑑𝑦1 (𝑡)
𝑑𝑥1 (𝑡)
−3
=5
… … … … … (1)
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑2 𝑦2 (𝑡)
𝑑𝑦2 (𝑡)
𝑑𝑥2 (𝑡)
−3
=5
… … … … … (2)
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
Adding Equation (1) and (2),
𝑑2 𝑦1 (𝑡)
𝑑𝑦1 (𝑡) 𝑑 2 𝑦2 (𝑡)
𝑑𝑦2 (𝑡)
𝑑𝑥1 (𝑡)
𝑑𝑥2 (𝑡)
−
3
+
−3
=5
+5
2
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
Again,
𝑑 2 𝑦(𝑡)
𝑑𝑦(𝑡)
𝑑
{𝑥 (𝑡) + 𝑥2 (𝑡)}
−
3
=
5
𝑑𝑡 2
𝑑𝑡
𝑑𝑡 1
= 5
𝑑𝑥1 (𝑡)
𝑑𝑥2 (𝑡)
+5
𝑑𝑡
𝑑𝑡
Comparing two output we can say that both are same and equal. Thus system follows the law of additivity.
Law of homogeneity:
𝑑 2 {𝑘 × 𝑦(𝑡)}
𝑑{𝑘 × 𝑦(𝑡)}
𝑑{𝑘 × 𝑥(𝑡)}
−
3
=
5
𝑑𝑡 2
𝑑𝑡
𝑑𝑡
𝑑2 {𝑦 ′ (𝑡)}
𝑑{𝑦 ′ (𝑡)}
𝑑{𝑘 × 𝑥(𝑡)}
−
3
=5
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
Comparing two output we can say that both are same and equal. Thus system follows the law of
homogeneity.
As the given system follows the law of additivity and homogeneity, the system is linear.
Home work related to linearity and Non-Linearity
Determine whether the following system is Linear or Non-linear
1.
y(𝑡) = 𝑥 (log 𝑡)
2.
y(𝑡) = 𝑒𝑡 𝑥 (𝑡)
3.
y(𝑡) = 2 + 𝑥 (𝑡)
4.
y(𝑡) = 3 log 𝑡 − sin 𝑡 − 𝑥 (𝑡2)
5.
y(𝑡) = 𝑥 (𝑡 − 1) + 𝑥 (𝑡 + 1)
6.
3
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
−2
𝑑𝑦(𝑡)
𝑑𝑡
=7
𝑑𝑥(𝑡)
𝑑𝑡
Time Variant and Time invariant System
Definition
1. A time invariant system is a system where any time delay provided to the input must be reflected to
the output.
2. A system is said to be time invariant if a time shift in the input signal causes an identical time shift
in the output.
Explanation
Sys
𝑥(𝑡)
𝑦(𝑡)
(𝑡 − 𝑡0)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
Output
Input
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
Sys
(𝑡 − 𝑡0)
𝑦/(𝑡)
Two Possibilities
1.
𝑦/(𝑡) = 𝑦(𝑡 − 𝑡0)
2.
𝑦/(𝑡) ≠ 𝑦(𝑡 − 𝑡0)
When it is equal to (𝑡 − 𝑡0) then we call the system as time invariant . Otherwise, it is time variant system.
Example-1
Determine whether the system is time variant or time invariant.
y(𝑡) = 𝑥(2𝑡)
Functionality of the system
System has the tendency to perform time scaling. Time scaling is performed on the variable 𝑡 not on the
constant 𝑡0. So time scaling factor 2 will be multiplied with 𝑡 and not with constant 𝑡0.
Step-1(Providing delay to output)
𝑦(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0) = [2(𝑡 − 𝑡0)]
Output
Note: Simple replace 𝑡 by 𝑡 − 𝑡0
Step-2(Providing delay to input)
Lt Col K M Mustafizur Rahman, EECE Dept, MIST
𝑦/(𝑡) = 𝑥(2𝑡 − 𝑡0)
(𝑡 − 𝑡0)
𝑥(𝑡)
Sys
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
Input
Comparing two output we see that they are not same. Thus the given system is time variant.
Example-2
Determine whether the system is time variant or time invariant.
y(𝑡) = 2 + 𝑥(𝑡)
Functionality of the system
System has the tendency to add dc value (2) with the input.
Step-1(Providing delay to output)
𝑦(𝑡)
(𝑡 − 𝑡0) = 2 + (𝑡 − 𝑡0)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
Output
Note: Simple replace 𝑡 by 𝑡 − 𝑡0
Step-2(Providing delay to input)
𝑥(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0)
Sys
𝑦/(𝑡) = 2 + 𝑥(𝑡 − 𝑡0)
Input
Comparing two output we see that they are same. Thus the given system is time invariant.
Important Conclusions
1.
Whenever system is performing time scaling then it is time variant.
2.
Whenever system is performing amplitude shifting the system is time invariant.
Example-3
Determine whether the system is time variant or time invariant.
y(𝑡) = 𝑥(cos 𝑡)
Functionality of the system
System replaces time 𝑡 in the input with cos 𝑡
Step-1(Providing delay to output)
𝑦(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0) = 𝑥 [cos(𝑡 − 𝑡0)]
Output
Note: Simple replace 𝑡 by 𝑡 − 𝑡0
Step-2(Providing delay to input)
𝑥(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0)
Sys
𝑦/(𝑡) = 𝑥(𝑐𝑜𝑠 𝑡 − 𝑡0)
Input
Comparing two output we see that they are not same. Thus the given system is time variant.
Example-4
Determine whether the system is time variant or time invariant.
𝑦(𝑡) = 𝑒−𝑡𝑥(𝑡)
Functionality of the system
System multiplies an exponential term with the input.
Step-1(Providing delay to output)
𝑦(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0) = 𝑒−(𝑡−𝑡0) 𝑥(𝑡 − 𝑡0)
Output
Note: Simple replace 𝑡 by 𝑡 − 𝑡0
Step-2(Providing delay to input)
𝑥(𝑡)
𝐷𝑒𝑙𝑎𝑦 𝑏𝑦 𝑡0
(𝑡 − 𝑡0)
Sys
𝑦/(𝑡) = 𝑒−𝑡𝑥( 𝑡 − 𝑡0)
Input
Comparing two output we see that they are not same. Thus the given system is time variant.
Home work related to Time variant and time invariant
Determine whether the following system is Time Invariant or Variant
1.
y(𝑡) = 𝑥 (tan 𝑡)
3.
y(𝑡) = tan(𝑡 + 1) . 𝑥 (𝑡)
4.
y(𝑡) = 𝑒−2𝑘 𝑥 (𝑡)
5.
y(𝑡) = cos 𝑡 . 𝑥 (𝑡)
6.
y(𝑡) = 2𝑡 + 𝑥 (𝑡)