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$$x\left[n\right]:=\text{discrete time signal}\tag{1}$$

The following plot is DTFT of$~x\left[n\right]~$

enter image description here

What I know so far are as below.

$$x\left[n\right]=\frac{1}{2\pi}\int_{0}^{2\pi}X\left(\exp\left(j\omega\right)\right)\exp\left(j\omega n\right)\,d\omega_{}\tag{2}$$

$$X\left(\exp\left(j\omega\right)\right)=\sum_{n=-\infty}^{\infty}x\left[n\right]\exp\left(-j\omega n\right)\tag{3}$$

I know the following property of DTFT.

$$\underbrace{x\left[n-k\right]}_\text{time domain}~~\leftrightarrow~~\underbrace{X\left(\exp\left(j\omega\right)\right)\exp\left(-j\omega k\right)}_\text{frequency domain}\tag{4}$$

The below plot is of DTFT of signal$~y\left[n\right]=x\left[n-2\right]~$

enter image description here

I can't get why right side takes$~\theta_{}\left(\omega_{}\right)=-2\omega_{}~$

Moreover, I even can't get why left side diagram can be held, since the following is held.

$$x\left[n-2\right]~~\leftrightarrow~~X\left(\exp\left(j\omega\right)\right)\exp\left(-j\omega 2\right)\tag{5}$$

What should I study first?

electrical apprentice
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1 Answers1

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First, you seem to be lacking some basics and should read a standard text book on the topic, e.g. the ever mentioned Oppenheim et al. "Discrete-time signal processing".

Just to help you on the way:

  • It is useful, to always keep in mind the basic propertys of the DTFT when dealing with it. Refer to the Wikipedia article on the topic. This also holds for all kinds of transforms (FT, DFT, Laplace, z-transform, etc.)
  • You should know the basic time-domain/frequency-domain transform pairs by heart (dirac, sinus/cosinus, rectangle, triangle and so on)
  • Armed with that, the math should clear out itself.

In this case:

  • Your nomenclature of the transform is kind of confusing. Just call it $X(\omega)$.
  • You correctly state in $(3)$, that a shift in time domain is equivalent to a modulation in frequency domain. A modulation means, mathematically, a multiplication by a complex pointer, which purely affects the phase of the signal. An alteration of a signal's phase cannot be seen in its magnitude plot but only in its phase plot which is the diagram on the right side.
  • The phase can be directly read off from the exponent, if the signal's transform appears in magnitude-phase-form.
  • The magnitude plot shows triangles. Referring to the table of basic transform pairs, the signal is of the $\text{sinc}^2$ category, of which $\theta$ is no function of $\omega$. So the pase of $x[n-2]$ can be read directly from the exponent.
Max
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