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Excerpt from Discrete Time Signal Processing by Alan Oppenheim: enter image description here

The text says this can be considered as an approximation of a Low Pass Filter. A low pass filter in continuous signal system is something that would block frequencies above a cutoff. But here in the discrete system, how is this to be understood? In a discrete time system, "frequency" is just angle/sample. Does the picture above mean "the system allows samples taken at the rate (0 to $2\pi/5$) / sample and blocks the rest" ?

EDIT:

Is it like this: say there is signal with a frequency range 20Hz to 50Hz. The sampling frequency in 100Hz. The lower frequency components get smaller $\omega$ and for higher frequency (50Hz) it gets close to $\pi$. If I have a low pass filter, it would block out those samples with large $\omega$. And this functionality now resembles a continuous time low pass filter. Is this a correct interpretation?

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In discrete time, a frequency of zero is just a constant sequence, in complete analogy with continuous time. The main difference between discrete time and continuous time is that in discrete time you have a maximum frequency, whereas in continuous time you (theoretically) don't have one. That maximum frequency in discrete time is achieved for a sequence with alternating values, i.e., the sequence $(-1)^n$ has a maximum frequency of $\omega=\pi$. There can be no larger frequency.

If the sequence was obtained from sampling a continuous-time signal with a sampling frequency $f_s$, then $\omega=\pi$ translates to an actual frequency in Hertz via $f=\omega/(2\pi f_s)$.

Returning to the example of the discrete-time lowpass filter, from the frequency response you see that higher frequencies (up to the maximum frequency $\omega=\pi$) are attenuated. That means that a discrete-time sinusoid's amplitude is changed according to the magnitude of the filter's frequency response at the sinusoid's frequency.

So if the input signal equals

$$x[n]=A\sin(n\omega_0)\tag{1}$$

the amplitude of the corresponding output signal is given by $A|H(e^{j\omega_0})|$. If $\omega_0$ happens to be a multiple of $2\pi/5$ then the input is completely suppressed because these frequencies are the zeros of the filter's frequency response.

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  • $\begingroup$ Thank you. I have made an edit to the question. Could you please have a look? $\endgroup$ Dec 31 '20 at 15:15
  • $\begingroup$ @SatheeshPaul: Your interpretation sounds ok, except for the fact that you don't block out "samples with large $\omega$", but signal components with large frequencies. The notion of samples with a large frequency doesn't make sense. Just like in the continuous-time case, a lowpass filter attenuates signal components with higher frequencies. $\endgroup$
    – Matt L.
    Dec 31 '20 at 15:54
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You are right in your concerns here; as is obvious from the magnitude of the frequency response, a moving-average system does not block any portion of the spectrum except at those $M-1$ null frequencies, where $M$ is the length of the moving-average impulse response.

However, it's also quite evident from the shape of the spectrum that high frequencies are attenuated while low frequencies are passed better through. From this it's justfied to call the system as a low pass filter, which attenuates high frequencies while letting low frequencies.

Nevertheless, it's not a best kind of lowpass filter. The main advantage this filter is that it requires a very simple summation to implement. It doesn't need multiplications, whereas other better filter characteristics require multiplications (as the weights) to implement.

And also historically, for some similar reasons, the MA filter finds extensive usage in reducing noise variations on time-series data.

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