I was truing to solve an example of DTFT which is following multiplication property. The problem is $$ a^n \sin(\omega_0 n) u[n]$$ we know that the definition of DTFT is $$ X(j \omega) = \sum _ {n=-\infty} ^ {{+\infty}} x[n]e^{-j \omega n}$$ Multiplication in Time domain will be convolution in DTFT.

If we take the DTFT of $a^n u[n]$ we have $$\frac {1}{1-ae^{-j \omega}}$$ and DTFT of $\sin(\omega_0 n)\,u[n]$ will be $$\frac{\pi}{j}\sum _ {l=-\infty} ^ {{+\infty}} \delta(\omega + \omega_0 - 2\pi l) - \delta(\omega - \omega_0 - 2\pi l)$$

I have confusion how can I write it in the form of multiplication property.

  • $\begingroup$ What exactly is your problem? You know that multiplication in the time domain becomes convolution of the DTFTs, you know the DTFTs of both sequences, so why don't you just convolve them and see what you get? $\endgroup$
    – Matt L.
    Commented Dec 6, 2015 at 13:48

2 Answers 2


Your sequence is


First of all, note that the Discrete-Time Fourier Transform (DTFT) of $(1)$ only exists if $|a|<1$. (The case $|a|=1$ can be handled by using Delta impulses). Anyway, for $|a|>1$ the DTFT of $(1)$ does not exist. Second, you can write $x[n]$ as the multiplication of $x_1[n]=a^nu[n]$ and $x_2[n]=\sin(n\omega_0)$. The DTFTs of these two sequences are given by

$$\begin{align}X_1(e^{j\omega})&=\frac{1}{1-ae^{-j\omega}},\quad |a|<1\\ X_2(e^{j\omega})&=\frac{\pi}{j}\left[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)\right],\quad -\pi <\omega <\pi\end{align}\tag{2}$$

Of course, $X_2(e^{j\omega})$ is also $2\pi$-periodic, but in $(2)$ only the interval $-\pi<\omega<\pi$ is considered (assuming $0<\omega_0<\pi$). Note that in your expression for $X_2(e^{j\omega})$ you have a sign error. Furthermore, note that $X_2(e^{j\omega})$ is the DTFT of $x_2[n]=\sin(n\omega_0)$, and not of $x_2[n]=\sin(n\omega_0)u[n]$, as claimed in your question.

Now we have

$$X(e^{j\omega})=\frac{1}{2\pi}X_1(e^{j\omega})\star X_2(e^{j\omega})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_1(e^{j\theta})X_2(e^{j(\omega-\theta)})d\theta\tag{3}$$

The only thing you need to know to compute $(3)$ is the property

$$F(\omega)\star \delta(\omega-\omega_0)=F(\omega-\omega_0)\tag{4}$$

for any function $F(\omega)$. With $(4)$, evaluating $(3)$ with the DTFTs in $(2)$ results in


The two terms in $(5)$ can be combined resulting in


  • $\begingroup$ I have question that how you find the right most part in equation 3 $\endgroup$ Commented Dec 7, 2015 at 1:12
  • $\begingroup$ @AadnanFarooqA: This is just the definition of convolution for periodic functions. $\endgroup$
    – Matt L.
    Commented Dec 7, 2015 at 8:04
  • $\begingroup$ I am confused about this equation: 1) can we write $\frac{1}{2\pi}X_1(e^{j \omega}) * X_2(e^{j \omega})$ as $\frac{1}{2\pi}\int_{-\pi}^{\pi} X_1(e^{j \theta}) * X_2(e^{j (\omega - \theta})$?? 2) The definition of discrete time convolution is $\sum _{-\infty}^{\infty} x[k]h[n-k]$, then why we have integral, isn't should be summation? $\endgroup$ Commented Dec 7, 2015 at 8:59
  • 1
    $\begingroup$ @AadnanFarooqA: That's the definition. It is convolution of continuous functions, not of discrete sequences. Note that the DTFT $X(e^{j\omega})$ is a continuous function of $\omega$. $\endgroup$
    – Matt L.
    Commented Dec 7, 2015 at 9:05
  • $\begingroup$ @AadnanFarooqA: If you studied "Signals and Systems" by Oppenheim you should have come across it, called "Modulation Property" in the chapter "Fourier Analysis for Discrete-Time Signals and Systems". $\endgroup$
    – Matt L.
    Commented Dec 7, 2015 at 9:11

Yes, I have tried to convolve and i am getting this. Please let me know if there is any mistake Lets suppose $$ X_1(\omega)= \frac {1}{1-ae^{-j \omega}} $$ $$ X_2(\omega)=\frac {\pi}{j}[ \delta(\omega + \omega_0) - \delta(\omega - \omega_0)]$$ Then, $$ \frac {1}{2\pi}X_1(\omega) * X_2(\omega)$$ $$ \frac {1}{2\pi} [X_2(-\omega_0)X_1(\omega + \omega_0) + X_2(\omega_0)X_1(\omega - \omega_0)]$$ $$ \frac {1}{2\pi} [\frac {\pi}{j}X_1 (\omega + \omega_0) - \frac {\pi}{j} X_1(\omega - \omega_0)]$$

$$ \frac {1}{2\pi} [\frac {\pi}{j}\frac {1}{1-ae^{-j (\omega + \omega_0)}} - \frac {\pi}{j} \frac {1}{1-ae^{-j (\omega - \omega_0)}}] $$ $$ \frac {1}{2j}[\frac {1}{1-ae^{-j (\omega + \omega_0)}} - \frac {1}{1-ae^{-j (\omega - \omega_0)}}] $$

  • $\begingroup$ I suppose there is a "=" sign at the end of each line (?) Why $X_2(-\omega)$ and $X_2(\omega)$ in the second line, and how do they disappear in the third line? 3rd and 4th line are identical. The end result is correct apart from a sign error, which comes from a sign error in $X_2(\omega)$. $\endgroup$
    – Matt L.
    Commented Dec 6, 2015 at 15:16
  • $\begingroup$ i have tried to do with way of Discrete convolution and there is correction it should be $X_2= (-\omega_0)$ and $X_2= (\omega_0)$ in the second line. In the third line i used value at $X_2= (-\omega_0)$ and $X_2= (\omega_0)$ $\endgroup$ Commented Dec 7, 2015 at 1:08
  • $\begingroup$ I think that you don't understand how convolution of functions works (as opposed to discrete sequences). The corresponding formula is given in my answer. $\endgroup$
    – Matt L.
    Commented Dec 7, 2015 at 8:05

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