Why using the impulse respond of an LTI to characterize its output?

Question

We all know that every LTI, is defined by its transfer function, so we can determine the output of any system, (using common transforms). It is also known that we can determine the output of any LTI for any given input signal just by having its impulse response (which gives us an output equal to its transfer function). I don't know if I'm following the point here, is it possible to determine the impulse respond of a system without having its transfer function? If so, how is it obtained? Can you give me a hand with this please?

Answer 1

For the two special classes of systems: $$ \textbf{Linear time and shift invariant systems } $$ An LTI and LSI system is one that obeys superposition and time and shift invariance. The LTI systems are described by linear ordinary differential equations and are your analog systems. The LSI systems are described by your linear constant coefficient difference equations. There are many analogues between them, as is to be expected as the discrete systems can be taken to be the discrete versions of the analog ones.

The dirac pulse and unit impulse sequence completely characterise the systems in both the time and frequency domain.

Let us first define what a system is, it is a relation that takes in an input signal $ x(t) $ or $ x[n] $ and produces an output $y(t)$ or $y[n]$. As is the case for analogue and discrete ones. The system is thus a relation, or an operator on the input to produce the output.

$$ y(t) = \textbf{T} \big \{x(t) \big \} $$

$$ y[n] = \textbf{T} \big \{x[n] \big \} $$ We define the response to a dirac pulse and a unit impulse respectively for system classes 1 and 2 as: $$ h(t) = \textbf{T} \big \{ \delta(t) \big \} $$

And the unit impulse response for discrete linear shift invariant systems as: $$ h[n] = \textbf{T} \big \{ \delta[n] \big \} $$

The impulse response for these two classes of systems is hence, the output of such systems when the input is a dirac pulse or a unit impulse sequence. Note the following properties of the dirac pulse and unit impulse sequence:

1. Every function can be expressed as a superposition or sum of scaled and shifted dirac pulses or unit impulses. $$ y(t) = \displaystyle \int_{- \infty}^{\infty} y(\tau) \cdot \delta(\tau - t) \,\,\,\, \text{d}t $$

$$ y[n]= \displaystyle \sum_{ k \to - \infty}^{k \to \infty} x[k] \cdot \delta[k - n] $$

2. The continous time fourier transform and the discrete fourier transform of the dirac pulse and unit impulse are both 1, hence, in the frequency domain, the pulse subjects your system to all possible frequencies, giving a fingerprint of the system.

$$ \textbf{The transfer function of a system (frequency domain or s domain) } $$

It is the fourier transform of the impulse response, or the laplace transform of the impulse response (s domain).

$$ H(j \omega) = \displaystyle \int_{- \infty}^{\infty} h(t) \cdot e^{-j(\omega t) } \,\,\,\, \text{d}t $$

And

$$ H(s) = \displaystyle \int_{- \infty}^{\infty} h(t) \cdot e^{-s (t) } \,\,\,\, \text{d}t $$

Now we have to talk about convolution and multiplication time duality. Multiplying two functions in the frequency domain is equal to convolving them in the frequency domain. Convolving in the time domain is the same as multiplying in the frequency domain.

Now to put it all togther, since your input function can be expressed as a sum of scaled and shifted dirac pulse or unit impulses, your output, due to linearity and time invariance, would be equally scaled and shifted impulse responses: $$ a h(t - t_{0} ) = \textbf{T} \big \{ a \delta(t - t_{0} ) \big \} $$ This is where convolution comes in, you express the input function as a sum of scaled and shifted dirac pulses or unit impulses, you then input these into your system, and to get the output you sum up or integrate all the scaled and shifted impulse responses, to get your output.

$$ y(t) = \displaystyle \int_{-\infty}^{\infty} f( \tau) \cdot h(t - \tau) \,\,\,\, \text{d} \tau $$

$$ y[n] = \displaystyle \sum_{k \to - \infty}^{ k \to \infty} x[k] \cdot h[n - k] $$ And in the frequency domain due to the theorems this is equal to $$ Y(s) = H(s) \cdot X(s) $$

Answer 2

The transfer function of an LTI system is the ratio of the transforms of the output and input signals. In continuous time, we use the Laplace transform, and in discrete time, we use the $\mathcal{Z}$-transform:

$$H(s)=\frac{Y(s)}{X(s)}\qquad\text{(continuous time)}$$ $$H(z)=\frac{Y(z)}{X(z)}\qquad\text{(discrete time)}$$

where $H(\cdot)$ denotes the transfer function, and $X(\cdot)$ and $Y(\cdot)$ are the transforms of the input and output signals, respectively.

Note that you can use the transfer function to describe a system but you don't need to. You can equivalently describe the system using its impulse response, which is just the inverse (Laplace / $\mathcal{Z}$) transform of the transfer function. The input-output relation is then given by a convolution.

Yet another way to describe a system is by a differential equation (in continuous time) or a difference equation (in discrete time). The latter is shown in Fat32's answer.

Answer 3

There are different ways to define systems and compute their impulse responses if they are LTI.

For example you can define a discrete-time system via its input-output relationship using a LCCDE (Linear Constant Coefficient Difference Equation):

$$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k] $$

and assuming initial rest and causality enables one to compute its impulse response as a solution of LCDDE assuming the input is $x[n] = \delta[n]$. For example if $N=0$ which signifies an FIR system then the impulse response $h[n]$ would directly be found from LCCDE by replacing $x[n]$ with $\delta[n]$ as:

$$h[n] = \frac{1}{a_0} \sum_{k=0}^{M} b_k \delta[n-k] $$

If $N \neq 0$ , then it (generally) signifies an IIR system and the solution procedure is outlined similarly in many posts here on DSP.SE.