Classification of systems

Continuous time LTI (Linear time invariant) systems

Discrete linear time invariant(LTI) system

Following are some of the properties of LTI systems

- The behavior of LTI system is completely characterized by its
**impulse response**. LTI systems can be defined uniquely in terms of transfer function or**impulse response**of the system.**Impulse response**of a system is the system response to input impulse function. In case of continuous time systems this impulse function is Dirac delta function, in case of discrete time systems it is Kronecker delta function.

The Dirac delta function is defined as δ(t-t_{o}) = 0 for t ≠ t_{o}

≠ 0 when t =t_{o}

and i.e the Dirac impulse is non zero only at the point where the argument inside δ() is zero.

Similarly Kronecker delta is defined as δ(n-n_{o}) = 0 for n ≠ n_{o}

≠ 1 when n =n_{o}

**Proof**: Any arbitrary continuous time function x(t) can be written as summation of products of samples of signal amplitude at a particular instant n*t_{o} and sampling functions shifted to right by n*t_{o }and sampling time t_{o} itself as follows

The sampling function can be defined as δ(t-nt_{o}) = 0 for t >(n+1)* t_{o},t<n*t_{o}

= 1/to for t <(n+1)* t_{o},t>n*t_{o}

For crosschecking the above equation put t =2*t_{0} then the equation reduces to

all the terms δ(2*t_{o}-nt_{o}) reduces to zero except for δ(2*t_{o}-2t_{o}) and by definition of sampling function δ(2*t_{o}-2t_{o})*t_{0} =1. Hence the equation reduces to

x(2*t_{o}) = x(2*t_{o})

As sampling time t_{o }tends to zero, sampling function approaches Dirac delta function.

Let us consider a system characterized by transformation represented by T{}. If an arbitrary input x(t) is applied to input the output is just the transformation of input. Hence

Y(t) = T{x(t)}

as t_{o} tends to zero then let n*t_{o} = τ and t_{o }=_{ }dτ(very small incremental variation in time), then the summation can be represented by integral and the equation for output can be written as

Hence the output is convolution of input and impulse response. If impulse response of the system is known then the output can be found for any input applied to it by using the convolution integral.

2. A LTI system is stable if and only if its impulse response is absolutely integrable i.e

**Proof: **A system is said to be stable in bounded input bounded output sense if for a bounded input output is also bounded.The output of a LTI system for a bounded input x(t) such that x(t) < ∞ for all t (i.e. x(t) is finite at all times) is Applying modulus on both sides

using Schwartz’s identity

By definition x(t) is bounded then x(t-τ)< ∞ for all t-τ. Then assume that he maximum value that input posses be m i.e. x(t-τ)≤m for all t-τ. Then the output is rewritten as From the above equation it is obvious that output will be finite only if the integral is finite i.e. the impulse response should be square integrable for the output to be finite so that the system is stable.

3. If we apply an exponential signal e^{st} as an input to LTI system, the output also will be exponential with a factor which depends on factor s. The factor s can be defined as

S = σ + j*ω

Where σ accounts for attenuation of exponential and is called as attenuation constant, ω is angular frequency in radians/second.

**Proof**: For an LTI system the output can be given as convolution of input x(t) and transfer function h(t). Hence

Y(t) = integral of X(t)

Where is the response of the system for exponential input in s domain. H(s) is known as transfer function of system and is Laplace transform of h(t). If s=j*ω the H(s) reduces to Fourier transform.

For example if you want to know the response of the system whose transfer function is 1/(s+2) for exponential input of e^{2t}, the straight forward method to find output is given as convolution of input and transfer function. For that we find out the inverse transform of transfer function in time domain. The inverse of 1/(s+2) is e^{-2t}*u(t) where u(t) is unit step function given as u(t) = 1 for t>0

= 0 for t<0

The output

The first integral goes to zero as u(τ) = 0 for τ<0, hence the output

Y(t) = e^{-2t}*(-1/4)*{(e^{-4τ})/τ=∞-(e^{-4τ})/τ=0} as e^{-∞} tends to zero and e^{0} =1

The output Y(t) for exponential input e^{2t} is (1/4)*e^{2t}.

It is obviously time taking procedure to perform. There is an elegant way of doing it remember that in the above derivation we found the response of the system for exponential input e^{st} here in the example input is e^{2t }comparing we get s=2. Substituting s=2 in the equation for H(s) we get H(s)_{/s=2}=1/4 the output is

Y(t) = H(s)*e^{2t}

Y(t) = e^{2t}/4

which matches with the output derived before based on convolution of input and output.

similarly for complex exponential input e^{j*ω*t }(s=j*ω)^{ }the output is y(t) = H(j*ω) *e^{j*ω*t }. For sinusoidal signals sin(ω*t) = (e^{j*ω*t} -e^{-j*ω*t}) /(2*j) the output is

y(t) = (H(j*ω) *e^{j*ω*t} -H(-j*ω) *e^{-j*ω*t} )/(2*j)

For cosine input signals cos(ω*t)= (e^{j*ω*t} +e^{-j*ω*t}) /2 the output is

y(t) = (H(j*ω) *e^{j*ω*t} +H(-j*ω) *e^{-j*ω*t })/2

From the theory of Fourier series and Fourier transform we know that any signal can be represented as sum of complex exponentials. From the above discussion it is clear that if an input e^{j*ω*t} is applied to a linear time invariant system the output is H(j*ω)*e^{j*ω*t} . Hence in a time invariant system it is guaranteed that the input frequencies are regenerated in the output.

4. For a LTI system to be causal the impulse response should be zero for t<0 i.e. h(t) = 0 for t<0.

**Proof**:The output of a LTI system is given as if x(t) = 0 for t<0 then in the integral the term x(t-τ) will be zero for t-τ <0 i.e. t< τ hence the integral can be written as

This integral is summation of x(t-τ)*h(τ)*dτ for incremental variations in τ from -∞ to t. It is obvious from the equation that the equation for output will be zero for t<0 only if h(τ) =0 for all t<0.