Impulse response is defined as 
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Its plot looks like this: 
Now I am basically supposed to say what does filter do with incomming signals based on this formula or plot. How can I approach this problem?
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1\$\begingroup\$ Hint 1: an input signal convolved with the impulse response of the filter gives its output response. Hint 2: the Fourier transform of the impulse response gives the frequency response. \$\endgroup\$– SynchrondyneCommented Oct 16, 2017 at 22:51
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\$\begingroup\$ @PeterK Thanks for your hints! In this particular case though, I'm looking for more intuitive way of estimating how does the filter behave. I'm trying to avoid integrating and time consuming calculations. Any ideas on that? \$\endgroup\$– John SmithCommented Oct 16, 2017 at 23:15
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2 Answers
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How can I approach this problem?
Here's a useful identity:
\$sin(x)cos(y)=\frac{1}{2}(sin(x+y)+sin(x−y))\$
\$ \begin{align} \end{align} \$
\$ \begin{align} h(n) &= sin(\pi\frac{1}{100}n)cos(2\pi\frac{12}{100}n) \\\\ & =\frac{1}{2}(sin(\pi\frac{1}{100}n+2\pi\frac{12}{100}n)+sin(\pi\frac{1}{100}n-2\pi\frac{12}{100}n))\\\\ & =\frac{1}{2}(sin(\pi\frac{25}{100}n)+sin(\pi\frac{-23}{100}n)) \end{align} \$
This together with the hints from Peter K should get you sorted.
I'm copy pasting Peter K's comment so it's gathered in one spot.
Hint 1: an input signal convolved with the impulse response of the filter gives its output response. Hint 2: the Fourier transform of the impulse response gives the frequency response. – Peter K
answered Oct 17, 2017 at 0:28
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As you're looking for an intuitive way of doing it ...
How 'boingy' does the response look?
If the response to an impulse was another impulse, the filter would simply be reproducing the input, with a flat frequency response.
But like a bell, this filter 'rings' when hit with an impulse. If it rang for a very long time, the ring frequency would be well-defined, and it would be a very narrow bandpass filter, centred at the ring frequency.
If it rang for only a cycle or two, then it's more of a 'clunk' than a 'doooiiiing', and the frequency is poorly defined, so it's a wide bandpass filter.
This response rings with significant energy for a handful of cycles, so the bandwidth is intermediate between the above two extremes.
The response width in the frequency domain is inversely proportional to the response width in the time domain, for some consistent definition of width, usually -3dB. To put actual numbers on that, to find the constant of proportionality, you will have to do the Fourrier thang.