In a course i'm currently taking, the lecturer computed DTFT for the following signal: $$r[n] = \begin{cases} 1& 0 \le n \le N\\ 0& \mbox{otherwise} \end{cases} $$ For $N = 32$ i pictured $\frac{1}{N}\frac{\sin{\omega*N/2}}{\sin{\omega/2}}$: enter image description here

The lecturer told us that you can compute DTFT of any signal on finite support by using this formula:

\begin{align*} \bar{X}(e^{j\omega}) = \sum_{k=0}^{N-1}X[k]\bar{R}(e^{\omega - \frac{2\pi}{N} k})\\ \end{align*}

So, I tried to render DTFT of sawtooth (red) as the sum:

enter image description here

That's clearly wrong. What did I do wrong?


1 Answer 1


The discrete Fourier transform (DFT) coefficients of a sequence $x[n]$ of length $N$ is defined by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{1}$$

and the inverse transform is

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{2}$$

The discrete-time Fourier transform (DTFT) of a sequence $x[n]$ of length $N$ is given by


Plugging $(2)$ into $(3)$ gives

$$\begin{align}X(e^{j\omega})&=\sum_{n=0}^{N-1}\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{-jn(\omega-2\pi k/N)}\\&=\sum_{k=0}^{N-1}X[k]\frac{1}{N}\sum_{n=0}^{N-1}e^{-jn(\omega-2\pi k/N)}\tag{4}\end{align}$$

With the DTFT of a rectangular sequence $r[n]=1/N$ for $n\in [0,N-1]$ and zero otherwise


Eq. $(4)$ can be written as

$$\sum_{k=0}^{N-1}X[k]R\left(e^{j(\omega-2\pi k/N)}\right)\tag{6}$$

So $(6)$ is the correct version of the formula, and the DFT coefficients $X[k]$ are defined by $(1)$. If you use those formulas correctly it should work.

  • $\begingroup$ Thanks for the answer. I believe I used exactly these formulas to create figures. You can see stems of sawtooth DFT in the second picture and for this $R$ thing in the first. But they don't give me proper DTFT of sawtooth. $\endgroup$
    – Yola
    Jan 19, 2019 at 15:12
  • $\begingroup$ I wonder if I missed something while plotting $R$ and this is the reason that DTFT of sawtooth is wrong. $\endgroup$
    – Yola
    Jan 19, 2019 at 15:15
  • $\begingroup$ @Yola: Note that there's a linear phase term in $R(e^{j\omega})$ (Eq. 5) that I don't see in your formula. $\endgroup$
    – Matt L.
    Jan 19, 2019 at 15:40
  • $\begingroup$ I removed it because its absolute value is one anyway. Thus, my first picture is not $\mbox{abs}(R)$, but $\mbox{sign}(R)\mbox{abs}(R)$. Maybe that's wrong to do so, but otherwise, I get even worser picture for sawtooth's DTFT. $\endgroup$
    – Yola
    Jan 19, 2019 at 15:57

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