Some confusion about impulse response

Question

I'm a new learner in signal processing. I've got a problem in understanding impulse response.

For a LTI system, the output of the combination of impulses equals the combination of every individual output of each impulse. My question is that can I say for every specified LTI system, the impulse response function is fixed due to the nature of this system?

My understanding about this problem is that for input impulse delta[n-k] there is a corresponding h[n-k], which means when delta[n] is shifted by k, and the h[n] will also be shifted by k, and the h(n) function curve are the same but shifted, right?

Answer 1

The answer is in your question: we're speaking of Linear Time-Invariant systems. Their properties are linearity and time invariance.

Linearity

$$\text{if } x_1(t) \text{ produces } y_1(t) \text{ and } x_2(t) \text{ produces } y_2(t)\\ \textbf{then}\\ \alpha\cdot x_1(t) + \beta\cdot x_2(t) \text{ produces } \alpha\cdot y_1(t) + \beta\cdot y_2(t)\\ \forall\ \alpha ,\ \beta\in\mathbb{R}$$

Time invariance

$$\text{if } x(t) \text{ produces } y(t)\\ \textbf{then}\\ x(t + \tau)\text{ produces } y(t + \tau)\\ \forall\ \tau\in\mathbb{R}$$

So if you know the impulse response, and call it \$h(t)\rightleftharpoons H(f)\$, then it is clear that for an arbitrary input, if you can express it as a sum of delayed impulses, the output will be a sum of delayed impulses responses.

Answer 2

My question is that can I say for every specified LTI system, the impulse response function is fixed due to the nature of this system?

This is bit subtle but a non-LTI system does not have an impulse response \$h(t)\$ (or an \$h[n]\$ for discrete time systems).

If the system is linear but not time invariant and the input is a delayed impulse \$\delta(t - \tau)\$, the output \$h(t,\tau)\$ is a function of both \$t\$ and \$\tau\$.

For example, consider the system defined by

$$y(t) = u(t)x(t)$$

For \$t \lt 0\$, the system output is zero regardless of the input. For \$t \gt 0\$, the system output is identical to the input.

Clearly, this system is not time invariant. If the input is \$x(t - \tau)\$, the output is

$$y(t, \tau) = u(t)x(t - \tau)$$

but

$$y(t - \tau) = u(t - \tau)x(t - \tau) \ne y(t,\tau)$$

so the output due to a delayed input is not equal to the delayed version of the output from a non-delayed input.

We can't meaningfully speak of an impulse response for this system since, for an impulse before \$t = 0\$, the output is zero whilst for an impulse after \$t = 0\$, the output is an impulse.

Likewise, we can't meaningfully speak of a frequency response for this time variant system.

The Fourier transform of the system is

$$Y(j \omega) = \frac{1}{j\omega} \ast X(j \omega) + \pi X(j\omega)$$

which is clearly not of the form

$$Y(j \omega) = H(j \omega) X(j \omega)$$

as required for there to be a transfer function (or frequency response) for the system.

All of the above is simply to show that an LTI system has an impulse response and linear time-variant systems do not.

Answer 3

You are right that an LTI system is completely characterized by its impulse response. Let me show this to you for the discrete-time case which you are referring to (the same holds of course also for the continuous-time case):

Any input sequence can be written as a sum of shifted and weighted impulses: $$x(n)=\sum_kx(k)\delta(n-k)\tag{1}$$

If \$h(n)\$ is the response to the impulse \$x(n)=\delta(n)\$, then - due to time-invariance - \$h(n-k)\$ is the response to a shifted impulse \$\delta(n-k)\$. Furthermore, due to linearity, the response to the weighted sum (1) is given by the weighted sum of the individual responses to the impulses \$\delta(n-k)\$. Consequently, the response to any input sequence \$x(n)\$ is

$$y(n)=\sum_kx(k)h(n-k)\tag{2}$$

where (2) is the convolution of the input signal with the impulse response. From (2) it is clear that the impulse response \$h(n)\$ completely characterizes the system, because the response to any input \$x(n)\$ can be computed by the convolution sum (2).