# Interpolation and decimation

I have a sinusoid in continuous time, with a frequency of 18kHz, it is sampled ideally with a continuous to discrete convertor, with a frequency of 27kHz. After that, we change the sampling speed in discrete time, using interpolator system and decimate system,so that result frequency equivalent in discrete time is 13.5KHz.

The interpolation factors and decimation factors are:

a) I=3 and D=2.

b) I = 1 and D=2

c) I= 3 and D=4

d) I = 4 and D = 3

I see this example on the internet and I don't know what is the correct one. Can you please help me to know what is the true answer and how it is calculated?

• Can you edit your question and add a link to your example on the internet ? – Peter K. Mar 14 '18 at 13:26
• Is an exam of my university and the language is spanish, I translate it to english – victor26567 Mar 14 '18 at 13:47
• @victor26567 than that's not a "example on the internet" but an exam of your uni... – Marcus Müller Mar 14 '18 at 14:25
• Does it matter? – victor26567 Mar 14 '18 at 14:31
• @victor26567 It matters because you lied about the origin. – Peter K. Mar 19 '18 at 14:22

It seems to me that the only way in which one of those solutions is possible is without aliasing. Probably there is a typo error, and the actual sampling rate is greater than 36 kHz. In such a case, for example, if the sampling rate was actually 37 kHz, we can interpolate with a factor I = 4, and then decimate with a factor D = 3 (in that order): $$f' = f / I = 18.0~kHz / 4 = 4.5~kHz; \\ f'' = f' \times D = 4.5~kHz \times 3 = 13.5~kHz$$ Where $f'$ is the frequency component after interpolation, and $f''$ is the frequency component after decimation. Therefore, my final solution is (d) if, and ONLY if, there is a typo and the sampling rate ($f_s$) is actually greater than the Nyquist frequency: $$f_s > f_{Nyquist} = 2\times f = 2\times 18.0~kHz = 36.0~kHz$$.
Assuming we are working with an ideal cosine that has a frequency response that is a perfect delta, then if the original signal is at $18 kHz$ and it is sampled at $27kHz$ then there will be aliasing (any input above $fs/2 = 13.5 kHz$ in this case will be aliased).
Working in digital domain $13.5 kHz$ would be $\pi$ and the $18 kHz$ tone should be at $4*\pi/3$ and $-4*\pi/3$ while its aliased version (coming because they repeat every $2*\pi$) is at $-2*\pi/3$ and $2*\pi/3$ respectively. Analyzing only positive frequencies $2*\pi/3$ is $9000 Hz$ and $4*\pi/3$ is $18000 Hz$. In the end the exercise wants some tone to be at $13.5 kHz$. So interpolate by 4 (the $fs$ is now $27*4$ and $\pi$ is $13.5*4 kHz$ so your aliased moved from $9000$ to $9000/4 Hz$ and $18000$ to $18000/4$) and decimate by 3 (the $fs$ is now $27*4/3$ kHz and $\pi$ is $13.5*4/3 kHz$ so your aliased moved from $9000/4 Hz$ to $9000/4*3 Hz$ and $18000/4$ to $18000/4*3$). So this is $6750 Hz$ and $13500 Hz$ which would be what the designer wanted plus an aliased component due to original undersampling