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So I know that the DTFT of $\{1,1\}$ is equivalent to $1+e^{j\omega }$.

But what is the DTFT of $\{1,-1\}$ equivalent to? Is it equivalent to $1-e^{j\omega }$?

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  • $\begingroup$ What do you mean by equivalent to? Are you calculating the DTFT of these sequences where x[0] = 1 and x[1] = 1 (or x[1] = -1 in the second case)? $\endgroup$ – Atul Ingle Dec 1 '16 at 1:22
  • $\begingroup$ Yes, the DTFT. I will revise my question. $\endgroup$ – Gary Dec 1 '16 at 1:23
  • $\begingroup$ Hi Gary- I assume you are describing the DTFT due to your tag. If so then yes, you are correct with your assumption. The first is a low pass frequency response while the second is a high pass frequency response. $\endgroup$ – Dan Boschen Dec 1 '16 at 1:24
  • $\begingroup$ how do you "know that the DTFT of $\{1,1\}$ is equivalent to $1+e^{j\omega }$"? $\endgroup$ – robert bristow-johnson Dec 1 '16 at 1:39
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You should consider the DTFT pair:

$$\delta[n] \xrightarrow{\text{DTFT}} 1$$

and the time-shifting property of the Fourier transform

$$x[n-n_0] \xrightarrow{\text{DTFT}} X(e^{j\omega})e^{-j\omega n_0}$$

plus the linearity of the FT. That is,

$$ax_1[n]+bx_2[n] \xrightarrow{\text{DTFT}} aX_1(e^{j\omega})+bX_2(e^{j\omega})$$


Now consider the discrete signal $\{1,1\}$ which can be represented by $x[n], n\in\mathbb{Z}$ as follows $$x[0]=1\\x[1]=1\\x[n]=0, \forall n\in\mathbb{Z} \ \backslash \ \{0,1\}$$

and can be denoted by $x[n]=\delta[n]+\delta[n-1]=x_1[n]+x_2[n]$. The DTFT is $$1+(1)e^{-j\omega(1)}=1+e^{-j\omega}$$ Similarly, the signal $\{1,-1\}$ can be represented by $$x[0]=1\\x[1]=-1\\x[n]=0, \forall n\in\mathbb{Z} \ \backslash \ \{0,1\}$$

which can be denoted by $x[n]=\delta[n]-\delta[n-1]=x_1[n]-x_2[n]$, and the DTFT is $$1-(1)e^{-j\omega(1)}=1-e^{-j\omega}$$

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  • $\begingroup$ I have an issue with your notation. Writing $\forall\notin S$ means you essentially allow any kind of object that is not in $S$. So you would have to define somewhere that you take $n$ from $\mathbb{Z}$. But in that case, you can simply write $\forall n \in \mathbb{Z} \setminus S$ or just $\forall n \in \bar{S}$. Also, you wouldn't really use the all-quantor like that. While it's clear what you mean, it would syntactically typically go in front of the statement. As a mathematician I would instead write $x[n]=0, n\in\bar{S}$ or $n\in\bar{S}\implies x[n]=0$. $\endgroup$ – Jazzmaniac Dec 1 '16 at 9:58
  • $\begingroup$ This is something you can safely ignore. But I think spreading a sense of good mathematical style is at least worth the attempt. No nitpicking intended! $\endgroup$ – Jazzmaniac Dec 1 '16 at 10:03
  • $\begingroup$ Thanks @Jazzmaniac. You are right that I didn't mention $n\in \mathbb{Z}$. However, it is something quite known from the context. Nonetheless, even with an incredible level of scrutiny $\forall n\notin\{0,1\}$ is perfectly sufficient for what I need here. It only defines the function in an over-constrained way. Although I don't need objects in $n\notin\mathbb{Z}$ and I set them to zero, eventually what I am looking for is a subset of that set. $\endgroup$ – msm Dec 1 '16 at 10:29
  • $\begingroup$ Yes, it is not wrong, but it is not good style either in my opinion. But like I said, feel free to ignore it. $\endgroup$ – Jazzmaniac Dec 1 '16 at 10:33
  • $\begingroup$ Sure, I will add $n\in \mathbb{Z}$ to the answer. Thanks again for the comment! $\endgroup$ – msm Dec 1 '16 at 10:36

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