Trying to intuit the unit impulse response

  • #1

Homework Statement


Hi there, I've been trying to gain some intuition on how the convolution sum works, but as I dig deeper I am realizing that there is an issue with my intuition of signals and systems, in particular the unit impulse response.

My issue is trying to understand how a unit impulse signal can have a system response that is not a constant.

Homework Equations


For a signal x[n] the output from an lti system is h[n] when the input is the unit impulse function δ[n] .

The Attempt at a Solution


I am following a textbook (Oppenheim ) and I see in it, and in nearly every online source, that when a unit impulse is the signal then the output is some function h[n] but what I do not understand, is how this output h is not simply one value at one time (like the unit impulse passed into the system)?
Surely if the output of a system whose input was a unit impulse, the system cannot be a time invariant system if the output is a time dependent function!

Since I cannot find an explanation anywhere, and that every source I see seems to never mention this, I am guessing that there is some fundamental misunderstanding that I possess regarding systems and signals. Would appreciate any guidance!
 

Answers and Replies

  • #2
jambaugh
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Consider a simple real world example. The hammer striking a bell imparts an impulsive force on the bell. But a microphone attached to the bell will not output a single pulse but rather a ringing signal over time. To make this more of a signal processing filter consider a bell made of magnetic material and the "hammer" is just an electromagnet held very close to the bell. A pulse through the electromagnet will "strike" the bell (though less efficiently) and it will ring outputting a decaying ring.

Now by knowing the output at all times when the normalized (unit) impulse is fed into this filter you can (assuming it is acting as a linear filter) superpose the continuous linear combination of unit impulses and rightly assume you'll get the corresponding superposition of output signals... read this as a convolution.

Since as a property of the delta function:
[tex] f(x) = \int_{-\infty}^\infty f(y)\cdot \delta(y-x)dy [/tex]
You can view this as adding up a continuum of impulses at y: [itex]\delta_y(x)= \delta(y-x)[/itex] times the weighting function [itex]f(y)[/itex] which just so happens to be the original function.
If your filter output is [itex]G_y(x)=G(y-x)[/itex] for an input [itex]\delta_y(x)[/itex] then for an input of [itex]f(x)[/tex]
the output will be the integral:
[tex]\int_{-\infty}^\infty f(y)G(y-x)dy[/tex]
Since that is the corresponding integral of output for the integral of inputs
[tex] f(x) = \int_{-\infty}^\infty f(y)\cdot \delta(y-x)dy [/tex]

The convolution of two functions can basically be seen this way. Treating one of the functions as the response to a delta function impulse, and treating the other function as an integral of impulses, the response to the other function is the convolution, the integral of responses weighted by the values of the input function.

I don't know how clear that is for you but that's my intuit of the math.
 
  • #3
Consider a simple real world example. The hammer striking a bell imparts an impulsive force on the bell. But a microphone attached to the bell will not output a single pulse but rather a ringing signal over time. To make this more of a signal processing filter consider a bell made of magnetic material and the "hammer" is just an electromagnet held very close to the bell. A pulse through the electromagnet will "strike" the bell (though less efficiently) and it will ring outputting a decaying ring.

Now by knowing the output at all times when the normalized (unit) impulse is fed into this filter you can (assuming it is acting as a linear filter) superpose the continuous linear combination of unit impulses and rightly assume you'll get the corresponding superposition of output signals... read this as a convolution.

Since as a property of the delta function:
[tex] f(x) = \int_{-\infty}^\infty f(y)\cdot \delta(y-x)dy [/tex]
You can view this as adding up a continuum of impulses at y: [itex]\delta_y(x)= \delta(y-x)[/itex] times the weighting function [itex]f(y)[/itex] which just so happens to be the original function.
If your filter output is [itex]G_y(x)=G(y-x)[/itex] for an input [itex]\delta_y(x)[/itex] then for an input of [itex]f(x)[/tex]
the output will be the integral:
[tex]\int_{-\infty}^\infty f(y)G(y-x)dy[/tex]
Since that is the corresponding integral of output for the integral of inputs
[tex] f(x) = \int_{-\infty}^\infty f(y)\cdot \delta(y-x)dy [/tex]

The convolution of two functions can basically be seen this way. Treating one of the functions as the response to a delta function impulse, and treating the other function as an integral of impulses, the response to the other function is the convolution, the integral of responses weighted by the values of the input function.

I don't know how clear that is for you but that's my intuit of the math.
Hi jambaugh,

Thank you for your response. I never though about intuiting convolution this way, so this helps me visualize the impulse response.

I do have a more general question though--mathematically, what kind of a system could from an input of an impulse function, which is essentially an instantaneous value, produce a decaying function? Perhaps a decaying exponential function such as f(x) = (0.5)^t ?

And here I think is where my main misunderstanding comes.

Would it be incorrect for me to say that a system output: y[n] = (0.5)^n * x[n] would model such a behaviour? I can understand if the impulse RESPONSE produced some decaying exponential function, but is it incorrect for me to define an output y[n] = (0.5)^n * x[n] as the general system response to an input x[n] ?

I am guessing it is incorrect as y[n] = (0.5)^n * x[n] ,is time-variant. So what kind of a mathematical function could modify an impulse to produce a decaying function, yet still be time invariant?
 
  • #4
jambaugh
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Hi jambaugh,

Thank you for your response. I never though about intuiting convolution this way, so this helps me visualize the impulse response.

I do have a more general question though--mathematically, what kind of a system could from an input of an impulse function, which is essentially an instantaneous value, produce a decaying function? Perhaps a decaying exponential function such as f(x) = (0.5)^t ? [...]
I'm coming at this from a physics b.g. but again building the filter from a dampened harmonic oscillator with input and output coupling, when the oscillator is over dampened you'll get an exponential decaying response, when it's critically dampened you'll have a linear times exponential function. When it's under dampened you'll have the exponentially decaying harmonic oscillation at the resonant frequency as your response.

[...]
And here I think is where my main misunderstanding comes.

Would it be incorrect for me to say that a system output: y[n] = (0.5)^n * x[n] would model such a behaviour? I can understand if the impulse RESPONSE produced some decaying exponential function, but is it incorrect for me to define an output y[n] = (0.5)^n * x[n] as the general system response to an input x[n] ?

I am guessing it is incorrect as y[n] = (0.5)^n * x[n] ,is time-variant. So what kind of a mathematical function could modify an impulse to produce a decaying function, yet still be time invariant?
Pardon me for translating this to a continuous case but the examples are analogous.
If your general response y to input x was [itex]y(t) =0.5^t x(t)[/itex] then
that would indicate a response function, call it G such that:
[tex] 0.5^t x(t) = \int_\mathbb{R} G(t-s)x(s)ds[/tex]
for any x(t) and in particular when the input is an impulse:
[tex]0.5^t \delta(t)= \int_\mathbb{R}G(t-s)\delta(s)ds = G(t)[/tex]
What you got yourself there then is a time dependent amplifier/attenuator. Imagine a radio when you turn it off and the big hefty power capacitor exponentially discharges so the radio's output signal fades out exponentially.
 
  • #5
jambaugh
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Let me add, that if you're considering an exponential decay response, be sure you cut it off at the input time, you don't want a response that's exponentially growing backward in time before the impulse occurs. That's perfectly legitimate mathematically but not causal. The math relates the two signals via a linear transformation. The math doesn't know from causality i.e. which is input and which is output. You could just as easily be dealing with a two way filter with components traveling both directions. To keep it causal your responses should be multiplied by a step function.
 
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  • #6
Ray Vickson
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Homework Statement


Hi there, I've been trying to gain some intuition on how the convolution sum works, but as I dig deeper I am realizing that there is an issue with my intuition of signals and systems, in particular the unit impulse response.

My issue is trying to understand how a unit impulse signal can have a system response that is not a constant.

Homework Equations


For a signal x[n] the output from an lti system is h[n] when the input is the unit impulse function δ[n] .

The Attempt at a Solution


I am following a textbook (Oppenheim ) and I see in it, and in nearly every online source, that when a unit impulse is the signal then the output is some function h[n] but what I do not understand, is how this output h is not simply one value at one time (like the unit impulse passed into the system)?
Surely if the output of a system whose input was a unit impulse, the system cannot be a time invariant system if the output is a time dependent function!

Since I cannot find an explanation anywhere, and that every source I see seems to never mention this, I am guessing that there is some fundamental misunderstanding that I possess regarding systems and signals. Would appreciate any guidance!
The impulse gives the system a "kick start"; after that the system's dynamics takes over and governs its future behavior. It is like striking a stationary hanging pendulum bob with a hammer; that gets the pendulum started, and its motion continuous even after you put away the hammer.
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement


Hi there, I've been trying to gain some intuition on how the convolution sum works, but as I dig deeper I am realizing that there is an issue with my intuition of signals and systems, in particular the unit impulse response.

My issue is trying to understand how a unit impulse signal can have a system response that is not a constant.

Homework Equations


For a signal x[n] the output from an lti system is h[n] when the input is the unit impulse function δ[n] .

The Attempt at a Solution


I am following a textbook (Oppenheim ) and I see in it, and in nearly every online source, that when a unit impulse is the signal then the output is some function h[n] but what I do not understand, is how this output h is not simply one value at one time (like the unit impulse passed into the system)?
Surely if the output of a system whose input was a unit impulse, the system cannot be a time invariant system if the output is a time dependent function!

Since I cannot find an explanation anywhere, and that every source I see seems to never mention this, I am guessing that there is some fundamental misunderstanding that I possess regarding systems and signals. Would appreciate any guidance!
You are mis-representing the concept of time-invariance. Time-invariance means that the dynamics of the system are independent of time, but of course the "orbit" under such dynamics will be time-dependent.

Basically, if you hit an inert system with a unit impulse at time ##0## you get some future state values ##\{ x_0(t) \}## for ##t >0.## If, instead you hit the same inert system with a unit impulse at time ##\tau > 0## you get some future state values ##\{ x_{\tau}(t) \}## for ##t > \tau.## Time-invariance means that ##x_{\tau}(\tau + w) = x_0(w)## for all ##w > 0##; that is, the behavior at time ##w## units later than the impulse is exactly the same in both systems. It does not matter if you apply the impulse at noon on Monday or at 4 pm on Wednesday; 5 hours later the two systems would be in exactly the same states.
 

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