The signal $$x(t)\;\;\;\;0\leq t\leq 0.2s $$

We know that the CTFT of $x(t)=0$ when $|w|>2*\pi*10^4$

We sample $x(t)$ in sample space of $$T=25\mu s$$ or $$F_s=1/T=40000Hz$$and we get a series with length of 8000.

$$x[n]=x(nT)\;\;\;\;0\leq n \leq 7999 $$

We calculate the DFT of $x[n]$ in size 8000 also.

What is the relation between the CTFT to the DFT

$$X^F(\omega) \rightarrow X_N^d[k]$$

The answer is:

$$ \begin{equation} X_N^d[k]=\begin{cases} \dfrac{N}{T}X^F(10\pi k), & 0\leq k\leq 3999\\ \\ \dfrac{N}{T}X^F(10\pi (k-8000)), & 4000\leq k\leq 7999 \end{cases} \end{equation} $$

  • $\begingroup$ Do you also submitted the answer? I am bit confused about your objective. $\endgroup$
    – jomegaA
    Feb 3, 2020 at 18:23
  • $\begingroup$ I have the answer but I don’t know how to solve the problem. $\endgroup$
    – ori
    Feb 3, 2020 at 18:42
  • $\begingroup$ The relationship between the spectrum of $X(f)$ of discrete time signal and the spectrum $X_a(F)$ of the analog signal is $$X(f)=F_s\sum_{k=-\infty}^{\infty}X_a\left[(f-k)F_s\right]$$ On right-hand side is the periodic repetition of the scaled spectrum, scaled with period $F_s$ $\endgroup$
    – jomegaA
    Feb 3, 2020 at 19:00
  • $\begingroup$ Proof is in Digital Signal Processing - Proakis $\endgroup$
    – jomegaA
    Feb 3, 2020 at 19:07
  • $\begingroup$ @jomegaA I don't understand how to get to the answer from what you wrote here. $\endgroup$
    – ori
    Feb 4, 2020 at 6:25

1 Answer 1


Let's give it a shot.

We know that during sampling, we obtained a discretized version, $x(nT_s)$ of a continuous time signal $x(t)$ sampled every $T_s$ seconds. I will denote their corresponding Fourier Transforms as $X_{DT}(e^{j\omega})$ and $X_{CT}(j\Omega)$, respectively, according to the majority of the bibliography. Indices denote discrete time (DT) and continuous-time (CT) Fourier Transforms.

The Discrete time Fourier Transform (DTFT) of the sampled signal is given by $$X_{DT}(e^{j\omega}) = \frac{1}{T_s}\sum_{l=-\infty}^{+\infty}X_{CT}\Big(j\Big(\frac{\omega}{T_s} - l\frac{2\pi l}{T_s}\Big)\Big)$$

Assuming that there is no aliasing (like in your case), over one period this turns into $$X_{DT}(e^{j\omega}) = \frac{1}{T_s}X_{CT}\Big(j\frac{\omega}{T_s}\Big)$$

Due to the $2\pi$-periodic nature of the DTFT and the fact that we're dealing with real-valued signals, we can rewrite this as $$X_{DT}(e^{j\omega}) = \left\{\begin{array}{ll} \frac{1}{T_s}X_{CT}\Big(j\frac{\omega}{T_s}\Big), & 0 \leq \omega < \pi \\ \frac{1}{T_s}X_{CT}\Big(j\frac{\omega-2\pi}{T_s}\Big), & \pi \leq \omega < 2\pi \\ \end{array}\right.$$

The Discrete Fourier Transform (DFT) is an $N-$point uniformly sampled version of the DTFT, so $$X_{DFT}[k] = X_{DT}(e^{j\omega})\Big|_{\omega = 2\pi k/N}$$ for $0 \leq k\leq N-1$.

Combining the last two equations we have $$X_{DFT}[k] = \left\{\begin{array}{ll} \frac{1}{T_s}X_{CT}\Big(j\frac{2\pi k}{NT_s}\Big), & 0 \leq k < N/2 \\ \frac{1}{T_s}X_{CT}\Big(j\frac{2\pi (k-N)}{NT_s}\Big), & N/2 \leq k \leq N-1 \\ \end{array}\right.$$

Except for the constant $N$ in your results - that I have failed to extract - I think what I wrote will help you.

  • $\begingroup$ The constant N is for approximately result. thanks! $\endgroup$
    – ori
    Feb 5, 2020 at 13:46

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