When the system is linear as well as time-invariant, then it is called a linear time-invariant (LTI) system. 

Consider an LTI system with impulse response $h(t)$ whose input and output are $x(t)$ and $y(t)$ respectively, shown in Figure 1.

Figure 1: LTI system

When the input to LTI system is unit impulse $\delta (t)$ then the output of LTI  system is known as impulse response $h(t)$. Mathematically,

\[y(t) = x(t) * h(t)\]

Similarly, for the input to LTI system is unit step signal $ u(t)$ then the output of LTI system is known as step response $s(t)$. Mathematically,

\[s(t) = h(t) * u(t)\]

Using convolution property of thr Laplace transform,

\[S(s) = H(s) \cdot \frac{1}{s}\]

Also,

\[H(s) = s \cdot S(s)\]

Applying inverse laplace transform,

\[h(t) = \frac{d}{{dt}}s(t)\]

And

\[s(t) = \int\limits_{ – \infty }^t {h(\tau )d\tau } \]

Example : The unit impulse response of a system is $h(t) = – 4{e^{ – t}} + 6{e^{ – 2t}}$. What will be the step response of the same system for $t$ ≥ $0$.

Solution: Impulse response $h(t) = – 4{e^{ – t}} + 6{e^{ – 2t}}$

Step response is given by

\[s(t) = \int\limits_{ – \infty }^t {h(t)dt } \]

\[s(t) = \int\limits_{ – \infty }^t {( – 4{e^{ – t}} + 6{e^{ – 2t}})dt} \]

\[s(t) = {4{e^{ – t}} – 3{e^{ – 2t}}} {\text{ + constant}} \]

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When the system is linear as well as time-invariant, then it is called a linear time-invariant (LTI) system. When the input to LTI system is unit impulse $\delta (t)$ then the output of LTI system is known as impulse response $h(t)$. Consider an LTI system with impulse response $h(t)$ whose input and output are $x(t)$ and $y(t)$ respectively, shown in Figure 1.

Figure 1: LTI system

The ratio of the Laplace transform of output $y(t)$ to the Laplace transform of input $x(t)$ with zero initial conditions, is known as transfer function $H(s)$. So we write

\[H(s) = \frac{{Y(s)}}{{X(s)}}\]

Also

\[Y(s) = X(s)H(s)….(1)\]

Applying inverse laplace transform on Equation 1,

\[y(t) = x(t) * h(t)\]

Note:- * denotes convolution and LTI system is defined by impulse response in time domain and transfer function in frequency domain.

Example : The impulse response of a system is $h(t) = u(t)$ . What will be the output $y(t)$ for an input $\delta (t – 2)$, .

Solution: As we already discussed, the LTI system is defined by impulse response in the time domain and transfer function in the frequency domain.

\[Y(s) = X(s)H(s)\]

For the input $x(t) = \delta (t – 2)$ so applying laplace transfom gives

\[X(s) = {e^{ – 2s}}\]

Also the impulse response of a system is $h(t) = u(t)$ so applying laplace transfom gives

\[H(s) = \frac{1}{s}\]

Thus

\[Y(s) = {e^{ – 2s}} \cdot \frac{1}{s}\]

Applying Inverse laplace transfom gives

\[h(t) = u(t – 2)\]

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

Invertible System

A system is called invertible if there should be one to one mapping between input and output at a particular instant and when an invertible system cascaded with its inverse system then gain of will be equal to one. In other words, if the input can be recovered from the system output, the system is said to be invertible, it is shown in Figure 2. Mathematically the $y(t)$ signal write as

\[y(t) = {\mathbf{T}}x(t)\]

And the $z(t)$ signal write as

\[z(t) = {{\mathbf{T}}^{ – 1}}y(t)\]

\[ = {{\mathbf{T}}^{ – 1}}{\mathbf{T}}x(t)\]

\[ = x(t)\]

Thus

\[z(t)= x(t)\]

Figure 2: Invertible System.

Non Invertible System

A system is called non-invertible if there should be many to one mapping between input and output at a particular instant.

Example : Determine whether or not each of the following systems are invertible with input $x(t)$ and output $y(t)$.

(i) \[y(t) = 10 + x(t)\]

(ii) \[y(t) = {x^2}(t)\]\]

Solution : (i)  \[y(t) = 10 + x(t)\]

Let, $x(t) = 2$ which is the dc signal input so,

\[y(t) = 10 + 2 = 12\]

Let, $x(t) = 21$ which is the dc signal input so,

\[y(t) = 10 + 21 = 31\]

Since different inputs leads to different output hence system is invertible.

Solution : (ii)  \[y(t) = {x^2}(t)\]

Let, $x(t) = 5$ which is the dc signal input so,

\[y(t) = 5^2 = 25\]

Let, $x(t) = -5$ which is the dc signal input so,

\[y(t) = (-5)^2 = 25\]

Since different inputs leads to same output hence system is non-invertible.

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

Memoryless System

A system is called static if output of system is dependent on present value of input. It is also known as static system. Example of memoryless systems are

\[y(t) = x(t)\]

\[y(t) = tx(t) + 2x(t)\]

Memory System

A system is called dynamic if output of system dependents on past or future values of input at any instant of time. It is also known as dynamic. Example of dynamic systems are

\[y(t) = x(t + 1)\]

\[y(t) = tx(t) + x(t – 1)\]

Example : Determine whether or not each of the following systems are memoryless  with input $x(t)$ and output $y(t)$.

(i) \[y(t) = x(3t)\]

(ii) \[y(t) = x(-t)\]

(iii) \[x(\cos t)\]

Solution : (i)  \[y(t) = x(3t)\]

put $ t = 1$

\[ y(1) = x(3) \]

hence the system with memory as output $y(1)$ depends on future input $x(3)$.

Solution : (ii)   \[y(t) = 5x(t)\]

put $t$ = 1

\[ y(1) = 5x(1) \]

hence the system is memoryless as output $y(1)$ depends on present input $x(1)$.

Solution : (iii) \[x(\cos t)\]

put $t$ = 0

\[ y(0) = x(1) \]

hence the system with memory as output $y(0)$ depends on future input $x(1)$.

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

Static System

A system is called static if output of system is dependent on present value of input. It is also known as memory less system. Example of static systems are

\[y(t) = x(t)\]

\[y(t) = tx(t) + 2x(t)\]

Dynamic System

A system is called dynamic if output of system dependents on past or future values of input at any instant of time. It is also known as system with memory. Example of dynamic systems are

\[y(t) = x(t + 1)\]

\[y(t) = tx(t) + x(t – 1)\]

Example : Determine whether or not each of the following systems are static  with input $x(t)$ and output $y(t)$.

(i) \[y(t) = x(3t)\]

(ii) \[y(t) = x(-t)\]

(iii) \[x(\cos t)\]

Solution : (i)  \[y(t) = x(3t)\]

put $ t = 1$

\[ y(1) = x(3) \]

hence the system is Dynamic as output $y(1)$ depends on future input $x(3)$.

Solution : (ii)   \[y(t) = 5x(t)\]

put $t$ = 1

\[ y(1) = 5x(1) \]

hence the system is static as output $y(1)$ depends on present input $x(1)$.

Solution : (iii) \[x(\cos t)\]

put $t$ = 0

\[ y(0) = x(1) \]

hence the system is dynamic as output $y(0)$ depends on future input $x(1)$.

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Linear System

A system is called linear if it satisfies two properties

(i) Additivity

(ii) Homogeneity

Additivity

If an input $x_1(t)$ produces output $y_1(t)$ and another input $x_2(t)$ also acting along produces output $y_2(t)$, then, when both inputs acting on the system simultaneously, produces output $y_1(t) + y_2(t)$. Mathematically,

If

\[{x_1}(t)\xrightarrow{{system}}{y_1}(t)\]

and

\[{x_2}(t)\xrightarrow{{system}}{y_2}(t)\]

then

\[{x_1}(t) + {x_1}(t)\xrightarrow{{system}}{y_1}(t) + {y_2}(t)\]

Homogeneity

It states that if input is scaled by any scalar $k$, then output also scaled by the same amount.

If

\[x(t)\xrightarrow{{system}}y(t)\]

then

\[k \cdot x(t)\xrightarrow{{system}}k \cdot y(t)\]

Nonlinear System

Any system is called nonlinear that does not satisfy two properties

(i) Additivity

(ii) Homogeneity

Example : Determine whether or not each of the following systems are linear  with input $x(t)$ and output $y(t)$.

(i) \[y(t) = ax(t) + b\]

(ii) \[y(t) = xsin(t)\]

Solution : (i)  \[y(t) = ax(t) + b\]

Additivity

\[{y_1}(t) + {y_2}(t) = a{x_1}(t) + b + a{x_2}(t) + b\]

\[y(t) = a{x_1}(t) + a{x_2}(t) + b\]

\[y(t) \ne {y_1}(t) + {y_2}(t)\]

hence the system is Nonlinear.

Solution : (ii)  \[y(t) = xsin(t)\]

Additibity

\[{y_1}(t) + {y_2}(t) = {x_1}sin(t) + {x_2}sin(t)\]

\[y(t) = {x_1}sin(t) + {x_2}sin(t)\]

\[y(t) = {y_1}(t) + {y_2}(t)\]

Homogeneity

\[k \cdot y(t) = k \cdot xsin(t)\]

\[k \cdot x(t) = k \cdot xsin(t)\]

so

\[k \cdot y(t) = k \cdot x(t)\]

hence the system is Linear.

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

Causal System

A system is called causal if its output is independent of future values of input. Example of causal systems are

\[y(t) = x(t)\]

\[y(t) = x(t – 1)\]

\[y(t) = x(t) + x(t – 1)\]

Non-Causal System

A system is called noncausal if its output at the present time depends on future values of the input. Example of noncausal systems are

\[y(t) = x(t + 1)\]

\[y(t) = x(t) + x(t + 1)\]

Example : Determine whether or not each of the following systems are causal  with input $x(t)$ and output $y(t)$.

(i) \[y(t) = x(3t)\]

(ii) \[y(t) = x(-t)\]

(iii) \[y(t) = {e^{x(t)}}\]

Solution : (i)  \[y(t) = x(3t)\]

put $ t = 1$

\[ y(1) = x(3) \]

hence the system is Non-causal as output $y(1)$ depends on future input $x(3)$.

Solution : (ii)   \[y(t) = x(-t)\]

put $t$ = – 1

\[ y(-1) = x(1) \]

hence the system is Non-causal as output $y(-1)$ depends on future input $x(1)$.

Solution : (iii)   \[y(t) = {e^{x(t)}}\]

The system is causal as output $y(t)$ depends on present input $x(t)$ only.

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

A system is called time-invariant if a time shift in the input signal $x(t – {t_0})$ causes the same time shift in the output signal $y(t – {t_0})$, it is shown in Figure 2.

Figure 2: Time-Invariant System.

Example : Determine whether or not each of the following systems are time variant with input $x(t)$ and output $y(t)$.

(i) \[y(t) = tx(t)\]

(ii) \[y(t) = \cos t \cdot x(t)\]

(iii) \[y(t) = 2 + x(t)\]

Solution : (i)  \[y(t) = tx(t)\]

delay input by $t_0$, let the output be ${y_1}(t)$

\[{y_1}(t) = tx(t – {t_0})\]

delay output by $t_0$, let the output be ${y_2}(t)$

\[{y_2}(t) = y(t – {t_0}) = (t – {t_0})x(t – {t_0})\]

\[{y_1}(t) \ne {y_2}(t)\]

hence the system is time variant.

Solution : (ii)  \[y(t) = \cos t \cdot x(t)\]

delay input by $t_0$, let the output be ${y_1}(t)$

\[{y_1}(t) = \cos t \cdot x(t – {t_0})\]

delay output by $t_0$, let the output be ${y_2}(t)$

\[{y_2}(t) = y(t – {t_0}) = \cos (t – {t_0})x(t – {t_0})\]

\[{y_1}(t) \ne {y_2}(t)\]

hence the system is time variant.

Solution : (iii)  \[y(t) = 2 + x(t)\]

delay input by $t_0$, let the output be ${y_1}(t)$

\[{y_1}(t) = 2 + x(t – {t_0})\]

delay output by $t_0$, let the output be ${y_2}(t)$

\[{y_2}(t) = y(t – {t_0}) = 2 + x(t – {t_0})\]

\[{y_1}(t) = {y_2}(t)\]

hence the system is time-invariant.

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Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation

$y(t) = {\mathbf{T}}x(t)$

where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.

Figure 1: System with a single input and output signal.

A system is bounded-input/bounded-output (BIBO) stable if for any bounded input $x(t)$ results in the bounded output $y(t)$.

Mathematically,

if

$|x(t)| \leqslant {m_x} < \infty $

then

$|y(t)| \leqslant {m_y} < \infty $

So system is said to be stable. And where $m_x$ , and $m_y$, are finite real constants. Bounded means amplitude is finite and some examples of bounded inputs as sine function, cosine function, dc signal, etc.

Example : Determine whether or not each of the following systems are stable with input $x(t)$ and output $y(t)$.

(i) \[y(t) = tx(t)\]

(ii) \[y(t) = \frac{{dx(t)}}{{dt}}\]

(iii) \[y(t) = u[x(t)]\]

(iv) \[y(t) = \cos [x(t)]\]

Solution : (i)  \[y(t) = tx(t)\]

Let

\[x(t) = 2\]

Here 2 is the dc signal which is bounded input so,

\[y(t) = 2.t\]

The output $y(t)$ is unbounded and the bounded input produces unbounded output hence system is unstable.

(ii)  \[y(t) = \frac{{dx(t)}}{{dt}}\]

Let

\[x(t) = 2\]

Here 2 is the dc signal which is bounded input so,

\[y(t) = \frac{{d2}}{{dt}} = 0\]

The output $y(t)$ is bounded. Let

\[x(t) = u(t)\]

Here unit step function is bounded input so,

\[y(t) = \frac{{du(t)}}{{dt}} = \delta (t)\]

The unit impulse function $\delta (t)$ is unbounded output and the bounded input produces unbounded output hence the system is unstable.

(iii) \[y(t) = u[x(t)]\]

Let

\[x(t) = 2\]

Here 2 is the dc signal which is bounded input so,

\[y(t) = u[2] = 1\]

The output $y(t)$ is bounded and the bounded input produces bounded output hence system is stable.

(iv)\[y(t) = \cos [x(t)]\]

Let

\[x(t) = 2\]

Here 2 is the dc signal which is bounded input so,

\[y(t) = \cos (2)\]

The output $y(t)$ is bounded and the bounded input produces bounded output hence system is stable.

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Bounded Signal

A continuous-time signal $x(t)$ having finite value at any instant of time is said to be bounded signal i.e. if $x(t) < M$ ; where $M$ is the finite value for all time $t$. The bounded signal example with $M=1$ shown in Figure 1.

Figure 1: Bounded signal.

Unbounded Signal

A continuous-time signal $x(t)$ having infinite value at any instant of time is said to be the unbounded signal. The unbounded signal example is shown in Figure 2.

Figure 2: Unbounded signal.

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