math lessons - LTI system theory

In electrical engineering, specifically in signal processing and control theory, LTI system theory investigates the effects of a linear, time-invariant system on an arbitrary input.

Contents

1 Introduction

2 Impulse response

3 Complex exponentials as eigenfunctions

4 See also

Introduction

For example, suppose the input signal is $x (t)$ where its index set is the real line, i.e., $t \in \mathbb{R}$ . The linear operator $\mathbb{H}$ on this index set is a two-dimensional function

$h(t_1, t_2) \mbox{ where } t_1, t_2 \in \mathbb{R}$

The linear transformation of $x (t)$ is the superposition integral

$y(t_1) = \int_{-\infty}^{\infty} h(t_1, t_2) \, x(t_2) \, d t_2$

If the linear operator $\mathbb{H}$ is also time-invariant, the following property holds

$h(t_1, t_2) = h(t_1 + \tau, t_2 + \tau) = h(t_1 - t_2, 0) \,\, \forall \, \tau \in \mathbb{R}$

We usually drop the zero second argument to $h (t 1, t 2)$ for brevity of notation so that the superposition integral now becomes the familiar convolution integral used in filtering

$y(t_1) = \int_{-\infty}^{\infty} h(t_1 - t_2) \, x(t_2) \, d t_2 = (h * x) (t_1)$

Thus, the convolution integral represents the effect of a linear, time-invariant system on any input function. For a finite-dimensional analog, see the article on a circulant matrix.

Impulse response

If we input a Dirac delta function to this system, the result of the LTI transformation is known as the impulse response since the delta function is an ideal impulse. We illustrate this idea as follows:

$(h * \delta) (t) = \int_{-\infty}^{\infty} h(t - \tau) \, \delta (\tau) \, d \tau = h(t)$ (by definition of the delta function)

Note that

$h(t) = h(t_1 - t_2, 0) \,\!\mbox{ where } t = t_1 - t_2$

so that h(t) is the impulse response of the system. We can also think of the delta function as the identity operator for LTI operators.

Complex exponentials as eigenfunctions

The complex exponential functions $e^{j \omega t} \mbox{ where } \omega \in \mathbb{R}$ are eigenfunctions of a linear, time-invariant operator. A simple proof illustrates this concept.

Suppose the input is $x(t) = \,\!e^{j \omega t}$ . The transformation of this function is then

$\int_{-\infty}^{\infty} h(t - \tau) \, e^{j \omega \tau} \, d \tau$

which is equivalent to the following by the commutative property of convolution

$\int_{-\infty}^{\infty} h(\tau) \, e^{j \omega (t - \tau)} \, d \tau = e^{j \omega t} \int_{-\infty}^{\infty} h(\tau) \, e^{-j \omega \tau} \, d \tau = e^{j \omega t} \mathcal{F} \{ h(t)\} = e^{j \omega t} H(j \omega)$

where $\mathcal{F}$ is the Fourier transform.

So, $\,\!e^{j \omega t}$ is an eigenfunction of an LTI system because the system response is itself scaled by an amount $H (ω)$ . Therefore, the eigenvalue spectrum is the Fourier transform of the operator $\mathbb{H}$ .

Mathematics Encyclopedia and Lessons

Lessons

Popular
Subjects

References

LTI system theory

Introduction

Impulse response

Complex exponentials as eigenfunctions

See also

Mathematics Encyclopedia and Lessons

Lessons

Popular Subjects

References

LTI system theory

Introduction

Impulse response

Complex exponentials as eigenfunctions

See also

Popular
Subjects