Interpretation of the system response to a unit impulse

[lahanadar]

Hi,

I want to understand the meaning of unit impulse response when time flows. An example is given below:

In the example the left signal depicts a δ[n] function value of which is 1 only at n=0. The LHS signal is response of the system to the unit impulse.

I want to understand how to interpret this when time flows, i.e. at n=-4, n=-3, what is response of the system? OR at n=0, n=5, etc. what is response of the system?

Thnx

 

[Charles Link]

The linear response is of the form $V_{out}(t)=\int\limits_{-\infty}^{t} h(t-t')V_{in}(t') \, dt'$. When $V_{in}(t)=\delta(t)$ , (so that $V_{in}(t')=\delta(t')$), $V_{out}(t)=h(t)$. The function $h(t)$ is the response function of the linear system, and it connects $V_{out}(t)$ to $V_{in}(t)$ in a linear manner. The form of the function $h$ in the integral is such that $h$ often represents a delayed response to the input $V_{in}$. This form is also very useful for Fourier transforms of each of the component functions, but it would be presenting too much all at once to show you the details of that. $\\$ Additional note: In the graph they supplied you with, $h(t) , (h(n))$, is clearly a delayed response. It is also a delayed response that is spread out over an interval, as opposed to a delayed response that occurs all at one single time. $\\$ A good example of this found in nature is thunder that can occur as a delayed and extended noise after a flash of lightning. You could also ask the question, what would happen if we had multiple flashes of lightening, so the lightning was of the form $V_{in}(t)$ as opposed to being a single flash $\delta(t)$ function ? How would we compute the resulting thunder=$V_{out}(t)$? The above integral furnishes the answer.

 

[lahanadar]

Thank you for the replay. From the delay point of view, can we interpret it like: at time n=0 an event occurs at the input and no event occurs at the output. At n=1 through 4 again no output event occurs. Then at time n=5 an event occurs due to a "past" event in the input. Namely due to an event that occurred 5 seconds ago. And again remaining series of events occur at the output for n=6 to 11 due to a past event at the input?

 

[Charles Link]

Thank you for the replay. From the delay point of view, can we interpret it like: at time n=0 an event occurs at the input and no event occurs at the output. At n=1 through 4 again no output event occurs. Then at time n=5 an event occurs due to a "past" event in the input. Namely due to an event that occurred 5 seconds ago. And again remaining series of events occur at the output for n=6 to 11 due to a past event at the input?

In this case the response of the system was only sampled at integer times. The input came at $t=0$. Normally $h(t)$ is continuous, i.e. it occurs at all times, not just integer times, and the response can begin at any time $t \geq 0$. For the case they give you, the response is negligible until $n=5$, but it's difficult to assess exactly what it is for $0<t<5$. For the purposes at hand, they are trying to give you an introduction to the function $h(t)$, and the example they give is a good illustration to start to learn the concepts. $\\$ Editing: In fact, I like your interpretation better=in this case they are dealing with events that only happen at discrete (integer) times=perhaps not completely realistic, but that's what they are giving you. :)

 

[donpacino]

In this case the response of the system was only sampled at integer times. The input came at t=0 t=0 . Normally h(t) h(t) is continuous, i.e. it occurs at all times, not just integer times, and the response can begin at any time t≥0 t \geq 0 . For the case they give you, the response is negligible until n=5 n=5 , but it's difficult to assess exactly what it is for 0<t<5 0h(t) h(t) , and the example they give is a good illustration to start to learn the concepts. \\ Editing: In fact, I like your interpretation better=in this case they are dealing with events that only happen at discrete (integer) times=perhaps not completely realistic, but that's what they are giving you. :)

I would say this is not accurate. The input function specifies that there is 0 input until n=-4. This is important... It tells you that the system has an inherent latency.
for the output from 0<=n<=4, h = 0, its not nothing.

Also analysis in the discrete domain, and not the time domain is very important in certain fields of engineering. Saying it is just an illustration is not accurate.

Hi,

I want to understand the meaning of unit impulse response when time flows. An example is given below:

In the example the left signal depicts a δ[n] function value of which is 1 only at n=0. The LHS signal is response of the system to the unit impulse.

I want to understand how to interpret this when time flows, i.e. at n=-4, n=-3, what is response of the system? OR at n=0, n=5, etc. what is response of the system?

Thnx

like i said above, at n=-4 and n=-5 the system will be zero, assuming no other inputs. The input graph shows up to n=-4 to show that there is no input before the impulse.

If you want to know how to interpret this.

The system will respond to an impulse by waiting 5 timesteps after an input, then 4x gain over 4 timesteps, then gradually goes back to zero.

 

[Charles Link]

@lahanadar I do like the explanation that @donpacino provides. If you are looking to know the subject in any detail, his quantitative explanation of this discrete case is quite accurate, and is perhaps even a better explanation than I provided.

 

[lahanadar]

@donpacino and @Charles Link thanks for the replies. BTW there is a typo in the question, the LHS would be RHS, sorry if it caused any misunderstanding.

@donpacino you said there is a zero input until n=-4, I thought the first and only non-zero input is at n=0 for the input signal (LHS) and due to that unique input at that time instant, a series of outputs is only able to starts at n=5 and ends at n=11. Do I miss something?

 

[Charles Link]

@lahanadar I didn't even notice the LHS vs. RHS. These are very standard graphs to show the input and output functions. $\\$ For your question about n=-4, (and I believe @donpacino will concur)=I think the 2nd sentence in his response was a "typo". All-in-all, his response made perfect sense in how to interpret these graphs. :)

 

[donpacino]

@donpacino you said there is a zero input until n=-4, I thought the first and only non-zero input is at n=0 for the input signal (LHS) and due to that unique input at that time instant, a series of outputs is only able to starts at n=5 and ends at n=11. Do I miss something?

You are correct

 

[lahanadar]

Thank you for all help.