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Recall the definition of the DTFT and the inverse DTFT: $X(\omega) ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n}, ~~~~~~ -\pi \leq \omega \lt \pi$ $x[n] ~=~ \int_{-\pi}^\pi X(\omega)\, e^{j\omega n} \, \frac{d\omega}{2\pi} , ~~~~~~ \infty\lt n\lt\infty$ We refer to a signal and its corresponding DTFT as a DTFT signal pair:$ x[n]~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ X(\omega) $
Dtft periodicity
Although the DTFT of a signal is typically defined only on a $2\pi$ interval, it is nevertheless straightforward to show that the DTFT is actually $2\pi$ periodic: $X(\omega+2\pi k) ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j (\omega+2\pi k) n} ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n} \, e^{-j 2\pi k n} ~=~ X(\omega)~~\checkmark$
Dtft frequencies
Because $X(\omega)$ is essentially the inner product of a signal $x[n]$ with the signal $e^{j\omega n}$, we can say that $X(\omega)$ tells us how strongly the signal $e^{j\omega n}$ appears in $x[n]$. $X(\omega)$, then, is a measure of the "frequency content" of the signal $x[n]$. Consider the plot below of the DTFT of some signal $x[n]$:As the DTFT is $2\pi$ periodic, all the information in a DTFT is contained in any $2\pi$ length section. However, there are only two intervals that are typically used. One is to refer to the DTFT from the interval of $0$ to $2\pi$. The low frequencies are at the extremities of the plot, with the highest frequency in the center:
The dtft and time shifts
If a signal is shifted in time, what effect might this have on its DTFT? Supposing $x[n]$ and $X(\omega)$ are a DTFT pair, we have that$ x[n-m]~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ e^{-j\omega m} X(\omega) $So shifting a signal in time corresponds to a modulation (multiplication by a complex sinusoid) in frequency. We can use the DTFT formula to prove this relationship, by way of a change of variables $r=n-m$: $\begin{align*}\sum_{n=-\infty}^{\infty} x[n-m]\, e^{-j \omega n}&= \sum_{r=-\infty}^{\infty} x[r]\, e^{-j \omega (r+m)}\\&= \sum_{r=-\infty}^{\infty} x[r]\, e^{-j \omega r} \, e^{-j \omega m} \\[2mm]&= e^{-j \omega m} \sum_{r=-\infty}^{\infty} x[r]\, e^{-j \omega r}\\ amp;=e^{-j \omega m} \, X(\omega) ~~\checkmark \end{align*}$
The dtft and time modulation
We saw above how a shift in time corresponds to modulation in frequency. What do you suppose happens when a signal is modulated in time? If you guessed that it is shifted in frequency, you're right! If a signal $x[n]$ has a DTFT of $X(\omega)$, then we have this DTFT pair:$ e^{j\omega_0 n} \, x[n]~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ X(\omega-\omega_0) $Below is the proof: $\sum_{n=-\infty}^{\infty} e^{j\omega_0 n} \, x[n]\, e^{-j \omega n}~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j (\omega-\omega_0) n} ~=~X(\omega-\omega_0)~~\checkmark $
The dtft and convolution
Suppose that the impulse response of an LTI system is $h[n]$, the input to the system is $x[n]$, and the output is $y[n]$. Because the system is LTI, these three signals have a special relationship:$ y[n]~=~ x[n] * h[n]~=~ \sum_{m=-\infty}^{\infty} \: h[n-m] \, x[m]$ The output $y[n]$ is the convolution of $x[n]$ with $h[n]$. Just as with the other DTFT properties, it turns out there is also a relationship in the frequency domain. Consider the DTFT of each of those signals; call them $H(\omega)$, $X(\omega)$, and $Y(\omega)$. The convolution of the signals $x$ and $h$ in time corresponds to the multiplication of their DTFTs in frequency: $y[n]=x[n]*h[n]\stackrel{\mathrm{DTFT}}{\longleftrightarrow} Y(\omega)=X(\omega)H(\omega)$ This relationship is very important. It gives insight, showing us how LTI systems modify the frequencies of input signals. It is also useful, because it gives us an alternative way of finding the output of a system. We could take the DTFTs of the input and impulse response, multiply them together, and then take the inverse DTFT of the result to find the output. There are some cases where this process might be easier than finding the convolution sum.
The dtft and linearity
Since the DTFT is an infinite sum, it should come as no surprise that it is a linear operator. Nevertheless, it is a helpful property to know. Suppose we have the following DTFT pairs:$x_1[n] ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ X_1(\omega) ~~~~~~~~x_2[n] ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ X_2(\omega)$Then by the linearity of the DTFT we have that, for any constants $\alpha_1$ and $\alpha_2$: $\alpha_1 x_1[n] + \alpha_2 x[2]~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ \alpha_1 X_1(\omega)+ \alpha_2 X_2(\omega) $
Dtft symmetry
Looking at the plots of the DTFTs above, or perhaps of those you have computed on your own, you may notice that they are often symmetrical. It so happens that the DTFTs of real-valued signals always have a kind of symmetry, but there are other symmetry relationships as well. For example, signals that are purely imaginary and odd have DTFTs that are purely real and odd. These types of symmetry are a result of a property of the complex exponentials which build up the DTFTs. Any signal of the form $e^{j\omega n}$ is conjugate symmetric, meaning that its real part is even and its imaginary part is odd. Additionally, for conjugate symmetric signals, their magnitude is even and their phase is odd.As a result of the symmetry of the signals which compose a DTFT, the DTFTs of certain signals also have symmetries, according to the table below: ***DTFT SYMMETRY TABLE HERE***
Read also:
OpenStax, Discrete-time signals and systems. OpenStax CNX. Oct 07, 2015 Download for free at https://legacy.cnx.org/content/col11868/1.2
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