Can someone explain why the coverage(x, y) signal for triangles is not band-limited?

lucida

@mchoquet

I believe this is because the coverage(x,y) function is a square wave (step function) as opposed to a sinusoidal waveform. The change in the function values from its minimum of 0 to its maximum of 1 is instantaneous.

We can represent a square wave as an infinite sum of sinusoidal waves but it would require infinite bandwidth, which is why coverage(x,y) is not bandlimited.

kayvonf

@lucida. Nice answer!

ak-47

What is the meaning of "bandlimited?"

lucida

@ak-47 If we are talking about representing an image using frequencies as our basis vectors, then it makes sense talk about bandwidth. As mentioned above, since coverage is a step function (square wave), you need an infinite number of sinusoidal waves (of different frequencies) summed up to achieve the square wave form. Thus we need an unlimited bandwidth to fully represent the coverage signal.

Can someone explain why the coverage(x, y) signal for triangles is not band-limited?

@mchoquet

I believe this is because the coverage(x,y) function is a square wave (step function) as opposed to a sinusoidal waveform. The change in the function values from its minimum of 0 to its maximum of 1 is instantaneous.

We can represent a square wave as an infinite sum of sinusoidal waves but it would require infinite bandwidth, which is why coverage(x,y) is not bandlimited.

@lucida. Nice answer!

What is the meaning of "bandlimited?"

@ak-47 If we are talking about representing an image using frequencies as our basis vectors, then it makes sense talk about bandwidth. As mentioned above, since coverage is a step function (square wave), you need an infinite number of sinusoidal waves (of different frequencies) summed up to achieve the square wave form. Thus we need an unlimited bandwidth to fully represent the coverage signal.

Hope this helps.