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What is difference between both implementations of low pass filter? Discrete/digital implementation and continuous implementation?

As far i am able to understand

In case of discrete implementation, passband spectrum exists between "-pi "and "pi"

While in case of continous implementation, passband spectrum exists between 0 and "f" . Where "f" is upper cut off frequency

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Difference in Implementation : Discrete time low pass filtering is done in DSP microprocessors using MAC(multiply and accumulate) operations for discrete convolution. Continuous time Low Pass filtering is done before ADC in analog domain using analog components.

Discrete domain signals can have frequencies only from $-\pi$ to $\pi$, these are digital frequency with units in radians/ sample. Continuous time signals have real frequencies in Hz. So, LPF in discrete domain must have a cutoff frequency less than $\pi$. But Continuous time LPF can have any cutoff frequency, it depends on application.

Designing a digital domain LPF is much easier than designing analog LPF.

When we sample a signal at $f_s$ sampling frequency, the maximum frequency that can be represented is $f_s/2$ and hence this maps to $\pi$ in digital domain. $\pi$ is the maximum frequency in digital domain.

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A discrete spectrum has a unique span extending from $-f_s/2$ to $+f_s/2$ where $f_s$ is the sampling rate (this is referred to as the first Nyquist Zone in the analog domain), thus in terms of normalized frequency when we divide the analog frequency by the sampling rate, it extends from $-0.5$ to $+0.5$ or $\pi$ to $\pi$ depending on which units we use as described further below.

In the case of a discrete implementation, the unique spectrum extends from $-\pi$ to $+\pi$ radians/sample. This is a common unit of frequency for sampled systems while cycles/sample is also used, which extends from $-.5$ to $+0.5$. The discrete frequency spectrum is periodic beyond these boundaries, so there is no reason to include them as no further information will be provided.

These units are the units of "normalized frequency" as they are arrived at by dividing the analog frequency units (either cycles/second or radians/second) by the sampling rate $f_s$. Cycles/Sec is equivalent to Hertz. The sampling rate is in units of samples/second therefore we get:

Normalized frequency: Cycles/Sec / Samples/Sec = Cycles/ Sample.

Normalized Angular Frequency: Radians/Sec/ Samples/Sec = Radians/ Sample.

That said, the passband of a discrete low pass filter must extend to an amount less the $\pi$ (in radians/sample units) since $\pi$ represents the highest frequency of the discrete spectrum.

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