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I need to transform this function:

$$ x(t) = 4\sin(20\pi t ) - 5\cos(24\pi t ) + 3\sin(120\pi t ) $$ into a sequence $x(n)$ given that the sampling frequency should be 50 Hz. So that means the sampling interval $T = \frac{1}{50} = 0.02\ s\ $. So would the sequence be:

$$ x[n] = 4\sin(20\pi0.02n ) - 5\cos(24\pi 0.02n ) + 3\sin(120\pi 0.02n) $$

I dont know how to compare the CTFT of $x(t)$ and the DTFT of $x(n)$. Any idea ?

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  • $\begingroup$ Your expression for the sampled sequence looks right. Note, however, that the frequency of the last $\sin$ term is 60 Hz. If you sample the continuous-time signal at 50 Hz, then you will experience aliasing. $\endgroup$ – Jason R Apr 18 '13 at 13:30
  • $\begingroup$ So i can't actually sample x(t) at 50hz witouth losing information ? $\endgroup$ – nerdy Apr 18 '13 at 13:38
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    $\begingroup$ See this answer for the relationship between the CTFT $X(f)$ of $x(t)$ and the DTFT of $x[n]$. The short answer is that the DTFT is the Fourier series representation of the frequency-domain periodic function $$G(f) = \sum_{k=-\infty}^{\infty} X\left(f-\frac{k}{T}\right)$$ which is the sum of replicas of $X(f)$ spaced $\frac{1}{T}$ Hz apart in the frequency domain. $\endgroup$ – Dilip Sarwate Apr 18 '13 at 13:43
  • $\begingroup$ Thanks Dilip. Can you please explain me why for different x(t) we got the same x(n) ?postimg.org/image/geld2veap $\endgroup$ – nerdy Apr 18 '13 at 14:06
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    $\begingroup$ Your signals have frequencies that rates are symmetrically spaced about the half-sampling frequency $\frac{1}{2}f_s$. This is the aliasing referred to in Matt's answer: signals between $\frac{1}{2}f_s$ Hz and $f_s$ Hz in frequency give the same sample values as the mirror-image signals between $0$ Hz and $\frac{1}{2}f_s$ Hz, that is, a signal at $\frac{1}{2}f_s+\Delta$ Hz has the same sample values as a signal at $\frac{1}{2}f_s-\Delta$ Hz $\endgroup$ – Dilip Sarwate Apr 18 '13 at 16:48
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Your sequence simply is a sampled version of the continuous signal $x(t)$. So

$$x_d(n)=x(nT),\quad T=\frac{1}{50}$$

The relation between the CTFT of $x(t)$ and the DTFT of $x_d(n)$ is that the DTFT is the sum of shifted versions of the CTFT. The spectra are shifted by the sampling frequency. That's the reason for aliasing.

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  • $\begingroup$ Can you please explain me why for different x(t) we got the same x(n) ?postimg.org/image/geld2veap $\endgroup$ – nerdy Apr 18 '13 at 14:05
  • $\begingroup$ That's a result of aliasing. As I said, the spectrum of the sampled signal is the sum of shifted spectra of the original signal. The spectra are shifted by the sampling frequency. So if your signal has a frequency that is greater than half of the sampling frequency, then you'll get aliasing. This means that your spectrum will be folded back into the frequency interval $[0,f_s/2]$. $\endgroup$ – Matt L. Apr 18 '13 at 15:15

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