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# Signals and Systems: Time Variant and Time-Invariant Systems

The property of a system which makes the behaviour of the system independent of time is known as time invariance. Time invariance means that the behaviour of the system does not depend on the time at which the input is applied to the system.

## Time-Invariant System

If the input and output characteristics of a system do not change with time, the system is called the **time-invariant system**.

### Continuous-time Case

The time-invariance property of a continuous time system can be tested as follows −

Let x(t) is the input and x(t-t_{0}) is the delayed input by t0 units. Then, the output of the system for the input x(t) is

𝑥(𝑡) → 𝑦(𝑡) = 𝑇[𝑥(𝑡)]

The output for the input x(t-t_{0}) is

𝑥(𝑡 − 𝑡_{0}) → 𝑦(𝑡, 𝑡_{0}) = 𝑇[𝑥(𝑡 − 𝑡_{0})] = 𝑦(𝑡)|_{𝑥(𝑡)=𝑥(𝑡−𝑡0)}

Also, the output delayed by t_{0} units is

𝑦(𝑡 − 𝑡_{0}) = 𝑦(𝑡)|_{𝑡=(𝑡−𝑡0)}

If

𝑦(𝑡, 𝑡_{0}) = 𝑦(𝑡 − 𝑡_{0})

That is, when the delayed output of the system is equal to the output due to delayed input for all possible values of t, then the given system is a timeinvariant system.

If the continuous-time system is described by a differential equation and if the coefficients of the differential equation are constants, then the system is called time-invariant system. For example,

$$\mathrm{5\frac{\mathrm{d} ^{2}y(t)}{\mathrm{d} t^{2}}+4\frac{\mathrm{dy}(t) }{\mathrm{d} t}+3y(t)=2x(t)}$$

The system expressed by the above differential equation is a time-invariant system because all its coefficients are constants.

### Discrete-time Case

In the discrete-time case, the time-invariance property is known as **shift invariance**.

A given system is time invariant or not can be tested as follows −

Consider x(n) is the input and x(n-k) is the delayed input to the given discrete time system. Then, the output of the system corresponding to the x(n) is given by

𝑥(𝑛) → 𝑦(𝑛) = 𝑇[𝑥(𝑛)]

And the output for the delayed input isf

𝑥(𝑛 − 𝑘) → 𝑦(𝑛, 𝑘) = 𝑇[𝑥(𝑛 − 𝑘)] = 𝑦(𝑛)|_{𝑥(𝑛)=𝑥(𝑛−𝑘)}

Also, the output of the system delayed by k units is

𝑦(𝑛 − 𝑘) = 𝑦(𝑛)|_{𝑛=(𝑛−𝑘)}

If

𝑦(𝑛, 𝑘) = 𝑦(𝑛 − 𝑘)

That is, when the delayed output of the system is equal to the output due to delayed input for all possible values of k, then the given system is a timeinvariant system.

If the discrete-time system is described by a difference equation and if the coefficients of the difference equation are constants, then the given system will be a time-invariant system. For example,

𝑦(𝑛) + 5𝑦(𝑛 − 2) + 4𝑦(𝑛 − 1) = 5𝑥(𝑛)

The system described by the above difference equation is a time-invariant system because all the coefficients are constants.

## Time-Variant System

A system whose input and output characteristics change with the time is known as **time-variant system**.

### Continuous-time Case

For a continuous-time, time-varying system, the delayed output of the system is not equal to the output due to delayed input, i.e.,

𝑦(𝑡, 𝑡_{0}) ≠ 𝑦(𝑡 − 𝑡_{0})

If a continuous time system is described by a differential equation and if its coefficients are the functions of time, then the system is a time-varying system. For example,

$$\mathrm{5t^{2}\frac{\mathrm{d} ^{2}y(t)}{\mathrm{d} t^{2}}+4t\frac{\mathrm{dy}(t) }{\mathrm{d} t}+3y(t)=2x(t)}$$

The system described by this differential equation is a time-varying system because all its coefficients are not constant or functions of time.

### Discrete-time Case

For a discrete-time, time varying system, the output of the system is not equal to the output due to delayed input, i.e.,

𝑦(𝑛, 𝑘) ≠ 𝑦(𝑛 − 𝑘)

If a difference equation is used to describe a discrete-time system, then the system will be a time-varying system if its coefficients are the functions of time.For example,

𝑦(𝑛) + 5𝑛𝑦(𝑛 − 2) + 4𝑛^{2}𝑦(𝑛 − 1) = 𝑥(𝑛) + 𝑥(𝑛 − 2)

This system is a time varying system because some of the coefficients are functions of time.

## Numerical Example

Determine whether the following systems are time-invariant or time-variant −

𝑦(𝑡) = 2𝑡

^{2}𝑥(𝑡)𝑦(𝑡) = 3𝑒

^{3𝑥(𝑡)}

### Solution (1)

The given system is,

𝑦(𝑡) = 2𝑡^{2} 𝑥(𝑡)

Here,

𝑦(𝑡) = 𝑇[𝑥(𝑡)] = 2𝑡^{2} 𝑥(𝑡)

The output of the system for the delayed input by 𝑡_{0} seconds is,

𝑦(𝑡, 𝑡_{0}) = 𝑇[𝑥(𝑡 − 𝑡_{0})] = 𝑦(𝑡)|_{𝑥(𝑡)=𝑥(𝑡−𝑡0)} = 2𝑡^{2} 𝑥(𝑡 − 𝑡_{0})

And the output of the system delayed by 𝑡_{0} seconds is,

𝑦(𝑡 − 𝑡_{0}) = 𝑦(𝑡)|_{𝑡=(𝑡−𝑡0) }= 2(𝑡 − 𝑡_{0})^{2} 𝑥(𝑡 − 𝑡_{0})

Therefore,

𝑦(𝑡, 𝑡_{0}) ≠ 𝑦(𝑡 − 𝑡_{0})

Hence, the given system is a time-varying system.

### Solution (2)

The given system is,

𝑦(𝑡) = 3𝑒^{3𝑥(𝑡)}

Here,

𝑦(𝑡) = 𝑇[𝑥(𝑡)] = 3𝑒^{3𝑥(𝑡)}

The output of the system for the input delayed by 𝑡_{0} seconds is,

𝑦(𝑡, 𝑡_{0}) = 𝑇[𝑥(𝑡 − 𝑡_{0})] = 𝑦(𝑡)|_{𝑥(𝑡)=𝑥(𝑡−𝑡0) }= 3𝑒^{3𝑥(𝑡−𝑡0)}

The output of the system delayed by 𝑡_{0} seconds is,

𝑦(𝑡- 𝑡_{0}) = 𝑦(𝑡)|_{t=(𝑡−𝑡0)}= 3𝑒^{3𝑥(𝑡−𝑡0)}

Hence for the given system,

𝑦(𝑡, 𝑡_{0}) = 𝑦(𝑡 − 𝑡_{0})

Therefore, the system is a time-invariant system.