Using the FT of the step function we have $H(\delta)=\pi\delta(\omega)+\frac{1}{j\omega}$, and it's magnitude is $\infty$ at $\omega=0$ and approaches $0$ as $\omega$ goes to both positive and negative infinity. Based on this, is the FT a low pass filter or a bandpass filter?


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The "unit step as a filter" is a filter whose impulse response is the unit step. In the time domain this would be

$$H(s) = 1/s$$

Which is an time domain integration. (Consider if we presented an impulse to the input of a "filter" constructed as a time domain integrator, it would immediately jump to 1 and then stay there for the rest of time-- which is a unit step as the "impulse response"). A time domain integration has a frequency response given as $1/(j\omega)$ so is indeed a low pass filter with a magnitude going down at -20 dB/decade, and a phase at -90° for all frequencies (and infinite gain at DC).

We can approximate this as a digital filter with different mapping techniques from the Laplace domain to the z domain, the most common is a simple accumulator given as:

$$H(z)= \frac{1}{z-1}$$

Converting this to the time domain in samples we get:

$$y[n] = x[n] + y[n-1]$$

With the continuous time integrator, if we passed in a unit step as the input, we would get a ramp out. Similarly with the accumulator, if we passed in a series of unit samples, the output would accumulate linearly (a ramp).

As a word of caution when dealing with systems in Laplace or z domains- do not confuse the transform of the waveform at the input or output with the transform of the impulse response of the system when considering the derived frequency response and behavior as a filter. Specifically when we concern ourselves with the question of the frequency response of a 2 port system- we are specifically interested in the impulse response for that system (what the output would be in time if only an impulse was presented at the input). We then take the Laplace Transform (or z transform if discrete time) of that impulse response to get the transfer function, and for that replace $s$ with $j\omega$ to get the frequency response. For further intuitive insight into the reason the impulse response is so interesting, see this post.


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