The DTFT Properties

Most of the DTFT properties are very similar to what we saw in the [continuous-time] Fourier Transform.

For each property, we consider two discrete-time signals $x(t)$ and $y(t)$ such that $$ x[n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) \quad \textrm{and} \quad y[n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} Y(\Omega) \; .$$

Linearity

$$ A x[n] + B y[n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} A X(\Omega) + B Y(\Omega) $$

Time Shifting

$$ x[n-n_0] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) e^{-j \Omega n_0} $$

Frequency shifting

$$ x[n] e^{j \Omega_0 n} \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega-\Omega_0) $$

Conjugation

$$ x^{*}[n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X^*(-\Omega) $$

Time Reversal

$$ x[-n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(-\Omega) $$

Convolution

$$ x[n] * y[n] = \int_{-\infty}^{\infty} x(\tau) y(t - \tau) \, d\tau \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) Y(\Omega) $$

Multiplication (circular convolution in frequency)

$$ x[n] y[n] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} \frac{1}{2 \pi} X(\Omega) \circledast Y(\Omega) = \frac{1}{2 \pi} \int_{2\pi} X(\theta) Y(\Omega - \theta) d \theta $$

Time Difference

$$ x[n] - x[n-1] \stackrel{\mathcal{DTFT}}{\longleftrightarrow} \left( 1 - e^{-j \Omega} \right) X(\Omega) $$

Conjugate Symmetry for Real Signals

$$ x[n] \textrm{ is real } \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) = X^*(-\Omega) $$

Real and even signals

$$ x[n] \textrm{ is real and even } \stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) \textrm{ is real and even}$$

Real and odd signals

$$ x[n] \textrm{ is real and odd }\stackrel{\mathcal{DTFT}}{\longleftrightarrow} X(\Omega) \textrm{ is purely imaginary and odd }$$

Parseval's Relation for Aperiodic signals

$$ \sum_{-\infty}^{\infty} |x[n]|^2 \, dt = \frac{1}{2 \pi} \int_{2 \pi} |X(\Omega)|^2 \, d \Omega $$

Circular Convolution

One notable difference with the continuous-time Fourier Transform is the multiplication property. Multiplication in time is circular convolution in frequency. Circular convolution occurs due to the periodic nature of the frequency domain.

Circular convolution causes signals to "wrap-around" in the frequency domain. This is easiest to see by illustration. Mathematically, we can demonstrate this below using a simple cosine signal.

No circular distortion

Consider the signal $$x[n] = \pi^{-1} \cos(\pi / 4 n) \; .$$ Also consider the system $$y[n] = x[n] \cos(\pi/2 n) = \pi^{-1} \cos(\pi / 4 n) \cos(\pi/2 n)$$ If we solve this using trignometric properties, we get $$\begin{eqnarray*} y[n] &=& (1/(2 \pi)) \left[ \cos((\pi / 4 + \pi /2 ) n) + \cos((\pi/2 - \pi/4) n) \right] \\ &=& (1/(2 \pi)) \left[ \cos((3 \pi / 4) n) + \cos(\pi/4 n) \right] \end{eqnarray*}$$ These two frequencies have a spacing of $\pi/2$, what we would expect from continuous-time modulation.

The frequency representation of $y[n] = \cos(\pi / 4 n)$.

Circular distortion

Now consider the signal $$x[n] = \pi^{-1} \cos(\pi / 4 n) \; .$$ Also consider the system $$y[n] = x[n] \cos(7 \pi/8 n) = \pi^{-1} \cos(\pi / 4 n) \cos(7 \pi/8 n)$$ If we solve this using trignometric properties, we get $$\begin{eqnarray*} y[n] &=& (1/(2 \pi)) \left[ \cos((\pi / 4 + 7 \pi /8 ) n) + \cos((7 \pi/8 - \pi/4) n) \right] \\ &=& (1/(2 \pi)) \left[ \cos((2 \pi / 8 + 7 \pi /8 ) n) + \cos((7 \pi/8 - 2 \pi/8) n) \right] \\ &=& (1/(2 \pi)) \left[ \cos((9 \pi / 8) n) + \cos(5 \pi / 8 n) \right] \end{eqnarray*}$$ Now note that since $n$ is an integer, $$\begin{eqnarray*} \cos((9 \pi / 8) n) &=& \cos((9 \pi / 8 - 2 \pi) n) \\ &=& \cos((- 7 \pi / 8) n) \\ &=& \cos((7 \pi / 8) n) \; . \end{eqnarray*}$$ Therefore, we get that $$\begin{eqnarray*} y[n] &=& (1/(2 \pi)) \left[ \cos((7 \pi / 8) n) + \cos(5 \pi / 8 n) \right] \end{eqnarray*}$$ These two frequencies have a spacing of $\pi/4$, different than we would expect from continuous-time modulation (we would expect spacing of $\pi / 2$).

Illustrating Circular Convolution

The figures to the side illustrate the circular convolution in frequency. Specifically, we illustrate the responses for the following system $$ \begin{eqnarray*} y[n] &=& \pi^{-1} \cos(\Omega_c n) \\ y[n] &=& \pi^{-1} \cos(0.18\pi n) \cos(\Omega_c n) \\ y[n] &=& x[n] \cos(\Omega_c n) \quad , \quad x[n] \textrm{ is a triangular pulse in frequency} \\ y[n] &=& x[n] \cos(\Omega_c n) \quad , \quad x[n] \textrm{ is a rectangular pulse in frequency} \\ \end{eqnarray*}$$ In these illustrations, we see how the frequency components wrap around $\Omega = \pi$ and add together. This can create some unexpected distorting effects, as shown in the above mathematical example.

The frequency representation of $y[n] = \cos(\Omega_c n)$ for various values of $\Omega_c$.

The frequency representation of $y[n] = x[n] \cos(\Omega_c n)$ for various values of $\Omega_c$ when $X(\Omega)$ is triangular.

Difference equations

Continuous-time systems can be characterized by linear, constant-coefficient differential equations. Similarly, discrete-time systems can be characterized by linear, constant-coefficient difference equations. That is a summation of shifted and weighted inputs $x[n]$ will equal a summation of shifted and weighted outputs $y[n]$ such that $$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{N} b_k x[n-k]$$ If we take the DTFT of both sides of this equation, we get that $$ \begin{eqnarray*} \sum_{k=0}^{N} a_k Y(\Omega) e^{-j \Omega k} &=& \sum_{k=0}^{N} b_k X(\Omega) e^{-j \Omega k} \\ Y(\Omega) \sum_{k=0}^{N} a_k e^{-j \Omega k} &=& X(\Omega) \sum_{k=0}^{N} b_k e^{-j \Omega k} \\ H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} &=& \frac{\sum_{k=0}^{N} b_k e^{-j \Omega k}}{\sum_{k=0}^{N} a_k e^{-j \Omega k}} \end{eqnarray*}$$ Therefore, we can represent a system by difference equations.

Examples

Difference equation

Question 1(a): Compute $H(\Omega)$ for the difference equation $$(1/2) y[n-2] + y[n-1] = x[n] + x[n-1]$$

Solution: If we apply the time-shifting property to both sides of the expression, we get $$\begin{eqnarray*} Y(\Omega) \left( (1/2) e^{-j 2 \Omega} + e^{-j \Omega} \right) &=& X(\Omega) \left( 1 + e^{-j \Omega} \right) \\ \frac{Y(\Omega)}{X(\Omega)} &=& H(\Omega) = \frac{1 + e^{-j \Omega}}{(1/2) e^{-j 2 \Omega} + e^{-j \Omega}} \end{eqnarray*}$$

Question 1(b): Compute $h[n]$ from the result of 1(a)

Solution: We can re-write $H(\Omega)$ as $$\begin{eqnarray*} H(\Omega) &=& \frac{1}{(1/2) e^{-j 2 \Omega} + e^{-j \Omega}} + \frac{e^{-j \Omega}}{(1/2) e^{-j 2 \Omega} + e^{-j \Omega}} \\ &=& \frac{1}{\left( (1/2) e^{-j \Omega} + 1 \right) e^{-j \Omega}} + \frac{e^{-j \Omega}}{\left( (1/2) e^{-j \Omega} + 1 \right) e^{-j \Omega}} \\ &=& \frac{e^{j \Omega}}{ 1 + (1/2) e^{-j \Omega} } + \frac{1}{ 1 + (1/2) e^{-j \Omega} } \\ &=& \frac{e^{j \Omega}}{ 1 - (-1/2) e^{-j \Omega} } + \frac{1}{ 1 - (-1/2) e^{-j \Omega} } \\ \end{eqnarray*}$$

From the table (with the time-shifting property), we can take the inverse Discrete-time Fourier transform of $$\begin{eqnarray*} h[n] &=& (-1/2)^{n+1} u[n+1] + (-1/2)^{n} u[n] \end{eqnarray*}$$

Class 23 (The Discrete-Time Fourier Transform Properties)