Problem: The signal $cos(2\pi14100t)$ is sampled at $F_s = 400 Hz$. It is then upsampled with a factor 3 and then reconstructed ideally with a new frequency $F = 500 Hz$.

I now want to find the new signal that is created.

Incomplete solution: With the frequency before sampling $\pm 141000 Hz$. After sampling the normalized frequencies:

$f = \pm {14100 \above 1 pt 400} \pm k = \pm (35 + {1 \above 1 pt 4})\pm k= \pm {1 \above 1 pt 4} \pm k $ per sample

Interpolation with factor 3 yields the normalized frequencies:

$f = {\pm {1 \above 1 pt 4} \pm k \above 1 pt 3} = $ 3 frequencies $\pm k?$

I know how to solve it past that to get the final $F= \pm f * F_s$. I.e. $y(t) = cos(2\pi f_1t) + cos(2\pi f_2t) + cos(2\pi f_3t)$.

So my question is how do I calculate the 3 frequencies? Would greatly appreciate a simple solution and not just an answer so that I actually understand this.

  • 1
    $\begingroup$ there are so many things wrong with this question that there is little that can be done with it. $\endgroup$ May 15, 2014 at 14:59
  • $\begingroup$ Took most of it from a text book. With a given problem and partial solution given, so a bit surprised by it being that unsalvageable. But guess I probably translated or presented it incorrectly. Either way I think the original problem description is small and straight-forward enough that any issues with it can be pointed out or even just provided a complete solution like the answer I was given. $\endgroup$
    – Koslun
    Oct 8, 2014 at 14:01
  • $\begingroup$ so you don't see a problem sampling a 14100 Hz sinusoid at a sample rate of 400 Hz? $\endgroup$ Oct 8, 2014 at 14:03
  • $\begingroup$ For a real-life application I would imagine this could be a problem. This is however just an educational problem created for students to understand upsampling, downsampling and reconstruction of signals. $\endgroup$
    – Koslun
    Oct 8, 2014 at 14:06
  • $\begingroup$ this is not about "upsampling" or "downsampling". it is about under-sampling. it's about reconstructing an undersampled signal. why doesn't the Nyquist/Shannon sampling theorem apply here? are you assuming that you know it's a single sinusoid in a particular frequency range? without assumptions like that, you have no hope of reconstructing the signal. $\endgroup$ Oct 8, 2014 at 14:09

1 Answer 1


There are some mistakes in the question and your description.

When we sample a signal of frequency $f_m$ with a sampling rate $f_s$, the sampled signal contains the frequencies $f=f_m \pm nf_s$, where $n \in \mathbb{Z} $. Here the frequencies available in the sampled signal are calculated as,

$f_m=14100 Hz$ and $f_s=400 Hz$. we know that,

$ f=f_m \pm nf_s $, where $n = 0, \pm 1, \pm 2, \pm 3, ... $

ie., $ f= ..., -1500, -1100, -700, -300, 100, 500, 900, 1300, 1700, ... $

When we up sample the signal by 3, the frequencies will become $f/3$.

So here the new frequencies are, $ f_u = ..., -633.33, -500, -366.67, -233.33, -100, 33.33, 166.67, 300, 433.33, 566.67, ... $

After upsampling the signal is reconstructed by using an LPF of cutoff frequency 500. When we apply LPF to the above signal only the frequencies whose absolute value is below 500 remains and other frequencies will loss.

there for after applying LPF the frequency components available in the signal are, $ -500, -366.67, -233.33, -100, 33.33, 166.67, 300, 433.33 $.

I hope this solves your problem.

  • $\begingroup$ Yes thank you. Solved it through other means and ended up forgetting this, sorry for the late confirmation. $\endgroup$
    – Koslun
    Oct 8, 2014 at 13:55

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