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?

  • 1
    $\begingroup$ Not sure I understand the question; of course you can determine the impulse response without having the transfer function. Just apply an impulse to the input of the system and look at its response, that's the impulse response. $\endgroup$ – Matt L. Dec 1 '18 at 11:04
  • $\begingroup$ Thank you, that's what I was looking for, but mathematically speaking, i think it is required to have an input /output equation so you can apply an impulse to the input? $\endgroup$ – Atmane Lee Dec 1 '18 at 11:08
  • $\begingroup$ Of course you have to define the system before you can determine its impulse response. In continuous time this could be a differential equation, in discrete time it's a difference equation. $\endgroup$ – Matt L. Dec 1 '18 at 11:13
  • $\begingroup$ That definition, isn't it just its transfer function since we determine the output in function of the input? $\endgroup$ – Atmane Lee Dec 1 '18 at 11:19
  • $\begingroup$ No, the transfer function is the ratio of the transforms of the output and input signals. You can just describe the system in the time (or any other) domain by a difference/differential equation like e.g. $y[n]=ax[n]+bx[n-1]$. This defines the system without a transfer function. $\endgroup$ – Matt L. Dec 1 '18 at 11:21

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.


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.

  • $\begingroup$ @AtmaneLee hey hii ? you have two answers. Please select one as accepted and also upvote useful ones. $\endgroup$ – Fat32 Dec 3 '18 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.