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$$h(t) = \frac{1}{1 + t^2}$$ and is it IIR or FIR filter. I tried finding the Laplace transform of this filter to get the data flow diagram with 5 taps and T=2s, however, I am unable to solve this. Probably the solution is simpler than I tried, however, being novice in DSP, I am unable to solve the same.

Thank You.

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  • $\begingroup$ You didn't explain how you obtain the 5 taps. In my answer below I assumed that you sample $h(t)$ in a way that preserves its symmetry. $\endgroup$ – Matt L. Dec 14 '15 at 8:27
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If

$$h(t)=\frac{1}{1+t^2}\tag{1}$$

denotes the filter's impulse response, then the filter is an infinite impulse response filter, simply because $h(t)$ has infinite support, i.e. it doesn't vanish anywhere (other than for $t\rightarrow \infty$). Note that if the definition $(1)$ of $h(t)$ is valid for all $t$, then the filter is not causal because $h(t)=0$ for $t<0$ doesn't hold.

Since the impulse response is real-valued and it satisfies $h(t)=h(-t)$, its Fourier transform must be real-valued (and symmetrical). Since the Fourier transform is real-valued, the filter's phase is zero.

If you sample that impulse response using $5$ taps at times $t_{-2}=-2T$, $t_{-1}=-T$, $t_0=0$, $t_1=T$, $t_2=2T$, you get an FIR filter, but the zero-phase property remains unchanged due to the symmetry of the filter coefficients. For implementation, the filter impulse response needs to be shifted by two samples to make it causal. This causes a delay of two samples, which changes the filter's phase $\phi(\omega)$ from zero to a linear phase $\phi(\omega)=-2\omega$.

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We can't answer your question definitively because we don't know the nature of your $t$ variable. In the world of signal processing the symbol $t$ typically represents continuous time (measured in seconds). So if your $t$ represents continuous time, and $h(t)$ is a filter's impulse response, then your filter is a continuous (analog) filter. In this case the question of "IIR or FIR?" has no meaning. (You mentioned the Laplace transform and that also implies your filter may be a continuous filter.) Did the person who gave you this problem describe the nature of your filter (analog or digital) and variable $t$? If so, please share that information with us.

Now if your $t$ variable is a sequence of numbers representing instants in time then that changes the whole scenario here. In that case we're now talking about a digital filter. If $t$ comprises discrete values and you can tell us the range of all possible values of $t$ then we can answer you question. You have what looks like variables $T$ and $s$ but, sadly, you expect us to read your mind to figure out what they represent. If you help us by defining your variables then we can help you.

The full story on how and why digital FIR filters can have linear phase can be found at: http://www.dsprelated.com/showarticle/808.php

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  • $\begingroup$ Taken literally, IIR and FIR both have meaning in continuous time as well as in discrete time. Conventional analog (well, continuous-time) filters are of course IIR, but you could come up with an idealized continuous-time impulse response which is FIR. If the latter makes any sense is a different matter. $\endgroup$ – Matt L. Dec 14 '15 at 21:17
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    $\begingroup$ Well, ...we should do our best to make sense here when we try to help DSP beginners. There's no need to encourage user 3602066 to think that the phrases IIR and FIR are meaningful in the field of continuous (analog) filters. Doing so would just cause him heartache in the future. It seems to me that, taken literally, the impulse responses of all real-world filters (both analog and digital) are finite in duration. It's hard to imagine an impulse response that would be non-zero until the end of time. $\endgroup$ – Richard Lyons Dec 15 '15 at 2:19
  • $\begingroup$ Matt L.: I should reword the last sentence in my recent comment to: "With the exception of resonators (oscillators) it's hard to imagine a real-world filter whose impulse response is non-zero until the end of time." $\endgroup$ – Richard Lyons Dec 16 '15 at 21:56
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    $\begingroup$ Matt L.: I'm glad that you're glad. A Goertzel filter is difficult to categorize. Its IIR in nature but we always halt its operation after a finite number of input samples. I'll let you decide if Goertzel filters are FIR or IIR. Standard digital filters are classified as being either FIR or IIR. We do so because FIR filters can have linear phase over their passband and transition bands. And as far as I know, no digital IIR or analog filter can have that characteristic. If you've ever encountered an analog FIR filter I'd sure be interested to hear about it. $\endgroup$ – Richard Lyons Dec 18 '15 at 10:53
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    $\begingroup$ Matt l.: Your urge to argue with everything I post is becoming an obsession. It's not healthy. If I write, "Batman is a hero." you reply with, "No. Batman is a wealthy socialite who beats up mentally ill people."go back and study Goertzel filters you'll see that the input to a Goertzel filter is a finite-length time-domain sequence. Goertzel filtering is not like traditional causal IIR filtering. $\endgroup$ – Richard Lyons Dec 19 '15 at 19:20

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