# Signals & Systems : Linear Time Invariant Systems ( LTI Systems)

Consider the response of a linear System to an arbitrary input x[n] . Suppose hk[n] denotes the response of the linear system to the shifted unit impulse δ[n-k] , then by using superposition property for a linear system , the response y[n] of the linear system to the input x[n] is simply weighted linear combination of their basic responses . That is with the input x[n] to the linear system , the output is expressed as ,

$\displaystyle y[n] = \Sigma_{k=-\infty}^{\infty} x[k] h_k[n]$   …(i)

with input $x[n] = \Sigma_{k=-\infty}^{\infty} \delta [n-k]$

Since δ[n-k] is a time shifted version of δ[n] , the response hk[n] is a time shifted version of h0[n]

$h_k[n] = h_0[n – k]$  …(ii)

Hence , $\displaystyle y[n] = \Sigma_{k=-\infty}^{\infty} x[k] h[n- k]$

### Continuous time LTI systems : The Convolution Integral

Any input x(t) to a Continuous time LTI systems can be represented in terms of impulses as ,

$\displaystyle x(t) = \int_{-\infty}^{\infty} x(\tau) . \delta (t – \tau) d\tau$

If h(t) is the response to δ(t) , then output y(t) can be expressed as :

$\displaystyle y(t) = \int_{-\infty}^{\infty} x(\tau) h (t – \tau) d\tau$

### Properties of Linear Time Invariant Systems

The Commutative Property :

In discrete time ,

$\displaystyle x[n] \ast h[n] = h[n] \ast x[n]$

$\displaystyle = \Sigma_{-\infty}^{\infty} h[k] x[n-k]$

In continuous time ,

$\displaystyle x(t) \ast h(t) = h(t) \ast x(t)$

$\displaystyle = \int_{-\infty}^{\infty} h(\tau) x(t – \tau) d\tau$

The Distributive Property :

In discrete time ,

$\displaystyle x[n] \ast ( h_1[n] + h_2[n]) = x[n] \ast h_1[n] + x[n] \ast h_2[n]$

In continuous time ,

$\displaystyle x(t) \ast [ h_1(t) + h_2(t) ] = x(t) \ast h_1(t) + x(t) \ast h_2(t) ]$

The Associative Property :

In discrete time ,

$\displaystyle x[n] \ast ( h_1[n] + h_2[n]) = ( x[n] \ast h_1[n] ) \ast h_2[n]$

In continuous time ,

$\displaystyle x(t) \ast [ h_1(t) + h_2(t) ] = [x(t) \ast h_1(t) ] \ast h_2(t) ]$

Example : Consider a casual LTI system whose input x[n] and output y[n] are related by the difference equation

$y[n] = \frac{1}{4} y[n-1] + x[n]$

Determine y[n] if x[n] = δ[n-1]

Solution :  y[n] = 0 for n < 1

Now , $y = \frac{1}{4} y + x$

= 0 + 1 = 1

$y = \frac{1}{4} y + x$

= 1/4 + 0 = 1/4

$y = \frac{1}{4} y + x$

= 1/16 + 0 = 1/16

….
….

$y[m] = ( \frac{1}{4} )^{m-1}$

Hence , $y[n] = ( \frac{1}{4} )^{n-1} u[n-1]$

### LTI System with & Without memory

A system is memoryless if its output at anytime depends only on the value of the input at that same time . The only way that this can be true for a discrete time LTI system is if h[n] = 0 for  n ≠ 0

In this case Impulse response has the form ,  h[n] = k δ[n]

When , k = h is a constant , and the convolution sum reduces to the relation , y[n] = k x[n]

If the discrete time LTI system has an impulse responses h[n] that is not identically zero for n ≠ 0 , then system has memory .

In particular , a continuous time LTI system is memoryless , if h(t) = 0 for t ≠ 0 and such a memoryless LTI system has the form ,

y(t) = k x(t)

for some constant k and has the impulse response , h(t) = k δ(t)

If k= 1 in equation (i) and (ii) then those systems becomes identity systems with output equal to the input and with unit impulse response equal to unit impulse .

In this case convolution Sum and Integral formula implies ,

$\displaystyle x[n] = \Sigma_{k=-\infty}^{\infty} x[k] \delta[n – k]$

$\displaystyle x(t) = \int_{k=-\infty}^{\infty}x[\tau] \delta(t – \tau)$

### Invertibility of LTI Systems:

The System is invertible only if an inverse system exists and when connected in series with the original system produces an output equal to the input to the first system . Again , if an LTI system is invertible then it has an LTI inverse .

### Casuality for LTI system :

The output of a casual system depends only on the present and past values of the input to the system and it requires that the impulse response of a casual discrete time LTI system satisfy the conditions

h[n] = 0 for n < 0

For casual discrete time LTI system , the condition implies that the convolution property

$\displaystyle y[n] = \Sigma_{k=-\infty}^{\infty} x[k] h[n-k]$

and , $\displaystyle y[n] = \Sigma_{k= 0}^{\infty} h[k] x[n-k]$

Similarly , a continuous time LTI system is casual ,

If h(t) = 0 for t < 0

and in this case the convolution integral is given by

$\displaystyle y(t) = \int_{-\infty}^{t} x(\tau) h(t – \tau) d\tau$

$\displaystyle y(t) = \int_{0}^{t} h(\tau) x(t – \tau) d\tau$

### Stability for LTI system :

A system is stable if every bounded input produces a bounded output .

If the impulse response is absolutely summable that is ,

$\displaystyle \Sigma_{k=-\infty}^{\infty}| h[k] | < \infty$

then y[n] is bounded in magnitude and hence the system is stable . Therefore a sufficient condition to guarantee the stability of a discrete time LTI system .

In continuous time , the system is stable , If impulse response is absolutely integrable . i.e.

$\displaystyle \int_{-\infty}^{\infty} | h(\tau) | d\tau < \infty$

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