why only impulse signal for convolution?

Question

Why do we only use impulse response for convolution for getting output of LTI systems. why we don't use unit step or any other signal for this purpose?

Answer 1

Because the impulse response completely characterizes an LTI system. The reason is an arbitrary input $x(t)$ can be written as an infinite sum of time shifted and scaled Delta functions:

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

or alternatively, using the convolution operator, $$x(t) = x(t)*\delta(t)$$

Properties of linearity and time invariance lead to the following sequence of input-output pairs:

• $\delta(t) \longrightarrow h(t)$
• Time invariance: $\delta(t-\tau) \longrightarrow h(t-\tau)$
• Linearity: $x(\tau)\delta(t-\tau) \longrightarrow x(\tau)h(t-\tau)$
• Linearity: $\int_{-\infty}^{+\infty}x(\tau)\delta(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau$

The last input-output pair can be written as $$x(t) \longrightarrow y(t) = x(t)*h(t)$$

This way, any output can be computed using the input $x(t)$, the impulse response $h(t)$, and the convolution integral (or sum, in discrete time).

@a concerned citizen gave you some other hints regarding your question.

On the contrary, if you use the unit step function, you get

$$\int_{-\infty}^{+\infty}x(\tau)u(t-\tau)d\tau = \int_{-\infty}^t x(\tau)d\tau$$

or alternatively, using the convolution operator, $$\int_{-\infty}^t x(t) = x(t)*u(t)$$

Once again, just for comparison, you get the following sequence of input-output pairs:

• $u(t) = \int_{-\infty}^t \delta(\tau)d\tau \longrightarrow s(t) = \int_{-\infty}^t h(\tau)d\tau$
• Time invariance: $u(t-\tau) \longrightarrow s(t-\tau)$
• Linearity: $x(\tau)u(t-\tau) \longrightarrow x(\tau)s(t-\tau)$
• Linearity: $\int_{-\infty}^{+\infty}x(\tau)u(t-\tau)d\tau \longrightarrow \int_{-\infty}^{+\infty}x(\tau)s(t-\tau)d\tau$

where $s(t)$ is the step response, the system output when the unit step is presented at the input. The last input-output pair can be written as $$\int_{-\infty}^t x(\tau)d\tau \longrightarrow x(t)*s(t) = x(t)*u(t)*h(t)$$

The latter can be written as $$(x(t)*h(t))*u(t) = y(t)*u(t) = \int_{-\infty}^t y(\tau)d\tau$$

So after all, what you get at the output is the integral of the output you would get using the delta decomposition of an input signal. That is, your output must be differentiated to obtain the actual system output for a given $x(t)$.

Answer 2

If you try to pass a specific frequency through a system, the system will either stop that signal or attenuate/amplify it or it may even delay the signal(phase shift). The output determines how the system responds to that particular signal. 

What about another input signal with some arbitrary amplitude and frequency? The answer lies in the frequency response of the system.  Then the next question is what is frequency response? Frequency response is nothing but how the system responds to each individual frequency, provided the input amplitude is unity and zero phases (all signals aligned to the zero of the timeline),

To obtain the frequency response of a system what we need to do is to pass through the system is, a signal with all frequency components from $-\infty$ to $+\infty$ with unity amplitude and zero phases (which input signal is having such a property? this is your first answer.)

what the system will actually do is, it will act as a filter and pass only those frequency components which it can process, which is in effect gives you the frequency response. 

when you try to push any signal with an arbitrary spectrum,  the output will look like the multiplication of the input spectrum of the signal with the frequency response of the system, think of it as a product of point to point value multiplication for each frequency value. 

As you already know multiplication in the frequency domain is convolution in the time domain. ( Hope your second doubt got cleared.)

Answer 3

Unit impulse is a nice pedagogic construct in order to understand how we can measure and execute a LTI system.

The stimuli of choice when characterizing real-world near-LTI systems is often not a unit impulse. It may be a MLS sequence, a sine sweep or some other signal containing a range of frequencies and more energy than an impulse.

Answer 4

Because the impulse function is the neutral element for convolution? So it is a consequence of how convolution is defined. So why doesn't one define it differently? The Fourier transform, by and large, is an invertible transform. It transforms a constant 1 into a dirac pulse and vice versa. Dirac pulses are sort-of ugly since they are distributions rather than proper functions, but this makes for symmetric and reasonably powerful domains with regard to Fourier transforms, allowing to represent undampened perpetual continuous sine/cosine functions as an idealisation.

Once you move to other definitions that integrate or differentiate in one direction of the transform, either the transform domain or the time domain become vastly less useful.