I have tried everything. If you actually know how to solve this could you provide a hint?

$$ e^{-2j\Omega}\frac{ \sin\left( \frac{7\Omega}{2}\right)}{ \sin\left( \frac{\Omega}{2} \right)}\star \frac{\sin\left( \frac{10 \Omega}{2} \right)}{\sin\left( \frac{\Omega}{2} \right) }$$

Ideally I would like to find the Fourier of every "fraction" in separate and then use properties: $x(n - n_o) \rightarrow e^{-jn_0\Omega}X(\omega)$ so I don't mind for $$ e^{-2j\Omega}$$ but I have 2 problems:

  1. I cannot use $\displaystyle \frac{\sin\left(\left(n+\frac 12\right)\Omega\right)}{\sin\left(\frac \Omega 2\right)}$ for $(n+1/2) = 10/2$ because $n \in Z$
  2. In DTFT , in my book there is no property like in continous time to transform convolution in $\Omega$ domain to multiplication in time domain so I don't know what to here as well.

After some comments and help from people who answered : I am going to try to do it as juch $\frac{sin(10\Omega/2)}{sin(\Omega/2)}= \frac{sin(10\Omega/2)}{sin(\Omega/2)}e^{-j\Omega(10-1)/2}e^{j\Omega(10-1)/2}=\Big[\frac{sin(10\Omega/2)}{sin(\Omega/2)}e^{-j\Omega(10-1)/2}\Big]e^{j9\Omega/2}$

I am to take advantage of the property : $\Big[\frac{sin(10\Omega/2)}{sin(\Omega/2)}e^{-j\Omega(10-1)/2}\Big]e^{j9\Omega/2} \rightarrow 2\pi F^{-1}{\Big[\frac{sin(10\Omega/2)}{sin(\Omega/2)}e^{-j\Omega(10-1)/2}\Big]} * F^{-1}[e^{j9\Omega/2}]$
The result is :
$F^{-1}[e^{j9\Omega/2}] =$ $\frac{1}{2\pi}int_{\pi}^{\pi}e^{j9\Omega/2}e^{j\Omega n}d\Omega = \frac{1}{2\pi}\frac{e^{j\Omega(9/2 +n)}}{j(9/2+n)}\Big|_{-\pi}^{\pi}=\frac{4(-1)^n}{2\pi(n+9)}$ ( i think)

and $F^{-1}[e^{j9\Omega/2}]=1$ for $n \in [0,9]$ and 0 anywhere else.
Now we need to compute the convolution of those 2:
the result should be non - zero only when $n \in [0,9]$ so:
$F^{-1}\Big[\Big[\frac{sin(10\Omega/2)}{sin(\Omega/2)}\Big]e^{-j\Omega(10-1)/2}\Big] = \begin{cases} \frac{4(-1)^n}{(n+9)} & n \in [0,9] \\ 0 & else \end{cases}$

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    $\begingroup$ I cannot view your formula. And what does $*$ means here? $\endgroup$ Dec 5, 2020 at 22:53
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    $\begingroup$ it is convolution $\endgroup$ Dec 6, 2020 at 8:11

2 Answers 2


It seems to me that this exercise is meant to combines the basic properties of the DTFT (gathered here: Table of DTFT Properties). Matt described the product/convolution property. You also get the time-shift/complex modulation. I suspect (thought I did not do the calculations) that the issue with the factor $10$ could be tackled with a change in variable: $10\Omega = 5\times (2\Omega)$, and the use of the time-scaling (expansion) property (see Time scaling of discrete-time sequences and the DTFT):

$$ S(c\Omega) \Longleftrightarrow \left\{ \begin{array}{ll} x[n/c] \textrm{ if } n/c \textrm{ is an integer } \\ 0 \textrm{ otherwise.} \end{array} \right.$$

Other hints hinge around Dirichlet kernels:

$$D_N(x) =\frac{\sin\left(\left(N +1/2\right) x \right)}{\sin(x/2)}$$

They are also called asinc or psinc (aliased or periodic cardinal sine or sinc), and related to finite-support discrete windows. If $*$ is the Convolution sign, the resolution may use the product/convolution properties of Fourier.

And more generally (at the bottom of the page on Dirichlet kernels), you have the identity:

$$\sum_{n=0}^{N-1} e^{jn\Omega} = e^{j(N-1)\Omega/2}\frac{\sin(N \, \Omega/2)}{\sin(\Omega/2)}\,,$$

which relates the DTFT of a discrete time window $w_{[0,N-1]}$ (from index $n=0$ to $n=N-1$) to ratios of sines with a phase correcting term. You can check details at Discrete-time Fourier transform of a window function.

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    $\begingroup$ the problem with this is that $n$ has to $\in Z$ but here $5=n+1/2 \rightarrow n = 9/2 $ which is not an integer to use this property $\endgroup$ Dec 6, 2020 at 8:12
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    $\begingroup$ Thank you though I haven't still found a proper way to use this $\endgroup$ Dec 6, 2020 at 19:01
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    $\begingroup$ you may take a look if you like , I tried a little bit harder $\endgroup$ Dec 6, 2020 at 22:25
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    $\begingroup$ I truly appreciate your efforts in solving the exercise (as honestly I did not redo it yet). I have updated the answer with a possible attack to the $10\Omega $ factor. Some colleagues here are possibly better than I am at verifying the calculations without making mistakes $\endgroup$ Dec 7, 2020 at 8:24

Also in discrete time we have the correspondence between multiplication in one domain and convolution in the other domain:

$$x[n]y[n]\Longleftrightarrow \frac{1}{2\pi}X(e^{j\omega})\star Y(e^{j\omega})\tag{1}$$

where convolution in the frequency domain is defined by

$$X(e^{j\omega})\star Y(e^{j\omega})=\int_{-\pi}^{\pi}X\big(e^{j\theta}\big)Y\big(e^{j(\omega-\theta)}\big)d\theta\tag{2}$$


$$H_N(e^{j\omega})=\frac{\sin\left(\frac{N\omega} {2}\right)}{\sin\left(\frac{\omega}{2}\right)},\qquad N\textrm{ odd}\tag{3}$$

corresponds to a very simple time domain sequence. I'm sure you can take it from here.

  • $\begingroup$ i still struggle.. I added some thought if you want to provide me feedback $\endgroup$ Dec 6, 2020 at 19:07
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    $\begingroup$ @brucebanner: If you denote the IDFT of Eq. $(3)$ in my answer by $h_N[n]$, then according to the convolution property, the result must be $2\pi h_7[n-2]\cdot h_{10}[n]$, right? $\endgroup$
    – Matt L.
    Dec 6, 2020 at 19:09
  • $\begingroup$ okay , if I get it right you say that : $X(\Omega) = e^{-2j\Omega}H_7[\Omega]H_10[\Omega] \rightarrow 2\pi h_7[n-2]h_{10}[n]$ but how do I find $h_n$? $\endgroup$ Dec 6, 2020 at 19:14
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    $\begingroup$ @brucebanner: I think you understood. The IDFT of $(3)$ is a very basic sequence you should have encountered already. It's also mentioned in Laurent's answer. $\endgroup$
    – Matt L.
    Dec 6, 2020 at 19:20
  • $\begingroup$ I can't figure out it , if you don't mind check the update in my post.. I can't see how can I break into in order to have a time varying value that is integer.. $\endgroup$ Dec 6, 2020 at 22:05

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