FREQUENCY-SHIFTING PROPERTY

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9.3 PROPERTIES OF THE DTFT

A close connection exists between the DTFT and the CTFT (continuous-time Fourier transform). For this reason, which Sec. 9.4 discusses, the properties of the DTFT are very similar to those of the CTFT, as the following discussion shows.

LINEARITY OF THE DTFT

If

x1[n] ⇐⇒ X1() and x2[n] ⇐⇒ X2()

then

$$a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1(\Omega)+a_2X_2(\Omega)$$

The proof is trivial. The result can be extended to any finite sums.

CONJUGATE SYMMETRY OF X()

In Eq. (9.20), we proved the conjugation property

$$x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.28}$$

^† To explain this point, consider the unit step function ^u[n] and its transforms. Both the ^z-transform and the DTFT synthesize ^x[n], using everlasting exponentials of the form ^zn. The value of ^z can be anywhere in the complex ^z-plane for the ^z-transform, but it must be restricted to the unit circle (^{z = e}j) in the case of the DTFT. The unit step function is readily synthesized in the z-transform by a relatively simple spectrum X[z] = z/(z − 1), by choosing z outside the unit circle (the ROC for u[n] is |z| > 1). In the DTFT, however, we are restricted to values of ^z only on the unit circle (^{z =} ej). The function ^u[n] can still be synthesized by values of z on the unit circle, but the spectrum is more complicated than when we are free to choose z anywhere, including the region outside the unit circle. In contrast, when x[n] is absolutely summable, the region of convergence for the z-transform includes the unit circle, and we can synthesize x[n] by using z along the unit circle in both the transforms. This leads to ^X[ej] = ^X().

868 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

We also showed that as a consequence of this, when x[n] is real, X() and X(−) are conjugates, that is,

$$X(-\Omega) = X^*(\Omega)$$

This is the conjugate symmetry property. Since X() is generally complex, we have both amplitude and angle (or phase) spectra

$$X(\Omega) = |X(\Omega)|e^{j\angle X(\Omega)}$$

Hence, for real x[n], it follows that

$$|X(\Omega)| = |X(-\Omega)| \quad \text{and} \quad \angle X(\Omega) = -\angle X(-\Omega)$$

Therefore, for real x[n], the amplitude spectrum |X()| is an even function of and the phase spectrum X() is an odd function of .

TIME AND FREQUENCY REVERSAL

Also called the reflection property, the time and frequency reversal property states that

$$x[-n] \Longleftrightarrow X(-\Omega) \tag{9.29}$$

Demonstration of this property is straightforward. From Eq. (9.19), the DTFT of x[−n] is

$$\text{DTFT}\{x[-n]\} = \sum_{n=-\infty}^{\infty} x[-n]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{j\Omega m} = X(-\Omega)$$

EXAMPLE 9.7 Using the Reflection Property

Use the time-frequency reversal property of Eq. (9.29) and pair 2 in Table 9.1 to derive pair 4 in Table 9.1.

Pair 2 states that

$$\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1$$

Hence, from Eq. (9.29),

$$\gamma^{-n}u[-n] = \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} \qquad |\gamma| < 1$$

Moreover, γ ^|n^| could be expressed as a sum of γ ⁿ u[n] and γ ⁻ⁿ u[−n], except that the impulse at n = 0 is counted twice (once from each of the two exponentials). Hence,

$$\gamma^{|n|} = \gamma^n u[n] + \gamma^{-n} u[-n] - \delta[n]$$

Combining these results and invoking the linearity property, we can write

$$\text{DTFT}\{\gamma^{|n|}\} = \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} + \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} - 1 = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2} \qquad |\gamma| < 1$$

which agrees with pair 4 in Table 9.1.

DR ILL 9.5 Using the Reflection Property

In Table 9.1, derive pair 13 from pair 15 by using the time-reversal property of Eq. (9.29).

MULTIPLICATION BY n: FREQUENCY DIFFERENTIATION

$$nx[n] \Longleftrightarrow j\frac{dX(\Omega)}{d\Omega} \tag{9.30}$$

The result follows immediately by differentiating both sides of Eq. (9.19) with respect to .

EXAMPLE 9.8 Using the Frequency-Differentiation Property

Use the frequency-differentiation property of Eq. (9.30) and pair 2 in Table 9.1 to derive pair 5 in Table 9.1.

Pair 2 states that

$$\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1$$

Hence, from Eq. (9.30),

$$n\gamma^{n}u[n] = j\frac{d}{d\Omega} \left\{ \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} \right\} = \frac{\gamma e^{j\Omega}}{(e^{j\Omega} - \gamma)^{2}} \qquad |\gamma| < 1$$

which agrees with pair 5 in Table 9.1.

TIME-SHIFTING PROPERTY If

x[n] ⇐⇒ X()

then

$$x[n-k] \Longleftrightarrow X(\Omega)e^{-jk\Omega} \qquad \text{for integer } k \tag{9.31}$$

This property can be proved by direct substitution in the equation defining the direct transform. From Eq. (9.19), we obtain

$$x[n-k] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n-k]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{-j\Omega[m+k]}$$ $$= e^{-j\Omega k} \sum_{n=-\infty}^{\infty} x[m]e^{-j\Omega m} = e^{-jk\Omega}X(\Omega)$$

This result shows that delaying a signal by k samples does not change its amplitude spectrum. The phase spectrum, however, is changed by −k. This added phase is a linear function of with slope −k.

PHYSICAL EXPLANATION OF LINEAR PHASE

Time delay in a signal causes a linear phase shift in its spectrum. The heuristic explanation of this result is exactly parallel to that for continuous-time signals given in Sec. 7.3 (see Fig. 7.22).

EXAMPLE 9.9 Demonstrating Linear Phase

To demonstrate the linear phase associated with a time shift, find the DTFT of x[n] = (1/4)sinc (π(n−2)/4), shown in Fig. 9.10a.

In Ex. 9.6, we found that

$$\frac{1}{4}\operatorname{sinc}\left(\frac{\pi n}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right)$$

Use of the time-shifting property [Eq. (9.31)] yields (for integer k)

$$\frac{1}{4}\operatorname{sinc}\left(\frac{\pi(n-2)}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega-2\pi m}{\pi/2}\right) e^{-j2\Omega}$$

The spectrum of the shifted signal is shown in Fig. 9.10b.

DR ILL 9.6 Using the Time-Shifting Property

Verify the result in Eq. (9.24) from pair 7 in Table 9.1 and the time-shifting property of the DTFT.

FREQUENCY-SHIFTING PROPERTY

If

$$x[n] \Longleftrightarrow X(\Omega)$$

then

$$x[n]e^{j\Omega_c n} \Longleftrightarrow X(\Omega - \Omega_c) \tag{9.32}$$

This property is the dual of the time-shifting property. To prove the frequency-shifting property, we use Eq. (9.19) as

$$x[n]e^{j\Omega_c n} \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n]e^{j\Omega_c n}e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega-\Omega_c)n} = X(\Omega-\Omega_c)$$

From this result, it follows that

$$x[n]e^{-j\Omega_c n} \Longleftrightarrow X(\Omega + \Omega_c)$$

872 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

Adding this pair to the pair in Eq. (9.32), we obtain

$$x[n]\cos(\Omega_c n) \Longleftrightarrow \frac{1}{2}\{X(\Omega - \Omega_c) + X(\Omega + \Omega_c)\}\tag{9.33}$$

This is the modulation property.

Multiplying both sides of pair (9.32) by ej^θ , we obtain

$$x[n]e^{j(\Omega_c n + \theta)} \Longleftrightarrow X(\Omega - \Omega_c)e^{j\theta}$$

Using this pair, we can generalize the modulation property as

$$x[n]\cos\left(\Omega_c n + \theta\right) \Longleftrightarrow \frac{1}{2}\left\{X(\Omega - \Omega_c)e^{i\theta} + X(\Omega + \Omega_c)e^{-i\theta}\right\}$$

EXAMPLE 9.10 Modulation Property

A signal x[n] = sinc (πn/4) modulates a carrier coscn. Find and sketch the spectrum of the modulated signal x[n] coscn for

• (a) ^c = π/2
• (b) ^c = 7π/8 = 0.875π

(a) For x[n] = sinc (πn/4), we find (Table 9.1, pair 8)

$$X(\Omega) = 4 \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right)$$

Figure 9.11a shows the DTFT X(). From the modulation property of Eq. (9.33), we obtain

$$x[n]\cos(0.5\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.5\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.5\pi - 2\pi m}{0.5\pi}\right)$$

Figure 9.11b shows half the X() shifted by π/2 and Fig. 9.11c shows half the X() shifted by −π/2. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.11d.

(b) Figure 9.12a shows X(), which is the same as that in part (a). For ^c = 7π/8 = 0.875π, the modulation property of Eq. (9.33) yields

$$x[n]\cos(0.875\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.875\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.875\pi - 2\pi m}{0.5\pi}\right)$$

Figure 9.11 Instance of modulation for Ex. 9.10a.

Figure 9.12b shows X() shifted by 7π/8 and Fig. 9.12c shows X() shifted by −7π/8. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.12d. In this case, the two shifted spectra overlap. Since the operation of modulation thus causes aliasing, it does not achieve the desired effect of spectral shifting. In this example, to realize spectral shifting without aliasing requires ^c ≤ 3π/4.

DR ILL 9.7 Using the Frequency-Shifting Property

In Table 9.1, derive pairs 12 and 13 from pair 11 and the frequency-shifting/modulation property.

TIME- AND FREQUENCY-CONVOLUTION PROPERTY If

x1[n] ⇐⇒ X1() and x2[n] ⇐⇒ X2()

then

$$x_1[n] * x_2[n] \Longleftrightarrow X_1(\Omega)X_2(\Omega) \tag{9.34}$$

and

$$x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi}X_1(\Omega)\circledast X_2(\Omega) \tag{9.35}$$

where

$$x_1[n] * x_2[n] = \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]$$

For two continuous, periodic signals, we define the periodic convolution, denoted by symbol -∗ as†

$$X_1(\Omega)\circledast X_2(\Omega) = \frac{1}{2\pi} \int_{2\pi} X_1(u)X_2(\Omega - u) du$$

The convolution here is not the linear convolution used so far. This is a periodic (or circular) convolution applicable to the convolution of two continuous, periodic functions with the same period. The limit of integration in the convolution extends only to one period.

Proof of the time-convolution property is identical to that given in Sec. 5.2 [Eq. (5.19)]. All we have to do is replace z with ej. To prove the frequency-convolution property of Eq. (9.35), we have

$$x_1[n]x_2[n] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x_1[n]x_2[n]e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x_2[n] \left[ \frac{1}{2\pi} \int_{2\pi} X_1(u)e^{-jnu} du \right] e^{-j\Omega n}$$

Interchanging the order of summation and integration, we obtain

$$x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi} \int_{2\pi} X_1(u) \left[ \sum_{n=-\infty}^{\infty} x_2[n] e^{-j(\Omega-u)n} \right] du = \frac{1}{2\pi} \int_{2\pi} X_1(u) X_2(\Omega-u) du$$

EXAMPLE 9.11 DTFT of an Accumulator System

If

$$x[n] \leftrightarrow X(\Omega)$$

, then show that $\sum_{k=-\infty}^{n} x[k] \leftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega)$ .

^† In Eq. (8.20), we defined periodic convolution for two discrete, periodic sequences in a different way. Although we are using the same symbol -∗ for both discrete and continuous cases, the meaning will be clear from the context.

876 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

To begin, we notice that

$$x[n] * u[n] = \sum_{k=-\infty}^{\infty} x[k]u[n-k] = \sum_{k=-\infty}^{n} x[k]$$

Applying the time-convolution property of Eq. (9.34) and pair 10 in Table 9.1, it follows that

$$\sum_{k=-\infty}^{n} x[k] = x[n] * u[n] \Longleftrightarrow X(\Omega) \left(\pi \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1}\right)$$

Because of 2π periodicity, X(0) = X(2πk). Moreover, X()δ( − 2πk) = X(2πk)δ( − 2πk) = X(0)δ(−2πk). Hence,

$$\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega)$$

DR ILL 9.8 Using the Frequency-Convolution Property

In Table 9.1, derive pair 9 from pair 8, assuming ^c ≤ π/2. Use the frequency-convolution property.

PARSEVAL’S THEOREM If

$$x[n] \Longleftrightarrow X(\Omega)$$

then Ex, the energy of x[n], is given by

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

\n(9.36)

To prove this property, we have from Eq. (9.28),

$$X^*(\Omega) = \sum_{n=-\infty}^{\infty} x^*[n]e^{i\Omega n}$$

Now,

$$\sum_{n=-\infty}^{\infty} |x[n]|^2 = \sum_{n=-\infty}^{\infty} x^*[n]x[n] = \sum_{n=-\infty}^{\infty} x^*[n] \left[ \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega \right]$$ $$= \frac{1}{2\pi} \int_{2\pi} X(\Omega) \left[ \sum_{n=-\infty}^{\infty} x^*[n] e^{i\Omega n} \right] d\Omega$$ $$= \frac{1}{2\pi} \int_{2\pi} X(\Omega) X^*(\Omega) d\Omega = \frac{1}{2\pi} \int_{2\pi} |X(\Omega)|^2 d\Omega$$

Table 9.2 summarizes Parseval’s theorem and the other important properties of the DTFT.

Operation	x[n]	X()
Linearity	a1x1[n] +a2x2[n]	a1X1()+a2X2()
Conjugation	x∗[n]	X∗(−)
Scalar multiplication	ax[n]	aX()
Multiplication by n	nx[n]	dX() j d
Time reversal	x[−n]	X(−)
Time shifting	x[n−k]	X()e−jk k integer
Frequency shifting	x[n] ejcn	X(−c)
Time convolution	x1[n] ∗ x2[n]	X1()X2()
Frequency convolution	x1[n]x2[n]	# 1 X1[u]X2[−u]du 2π 2π
Parseval’s theorem	= ”∞ x[n] 2 Ex n=−∞	# 1 2 d Ex X() = 2π 2π

TABLE 9.2 Properties of the DTFT

EXAMPLE 9.12 Using Parseval’s Theorem to Find Signal Energy

Find the energy of x[n] = sinc (cn), assuming ^c < π.

From pair 8, Table 9.1, the fundamental band spectrum of x[n] is

$$\operatorname{sinc}(\Omega_c n) \Longleftrightarrow \frac{\pi}{\Omega_c} \operatorname{rect}\left(\frac{\Omega}{2\Omega_c}\right) \qquad |\Omega| \le \pi$$

878 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

From Parseval’s theorem [Eq. (9.36)], we have

$$E_x = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{\pi^2}{\Omega_c^2} \left[ \text{rect}\left(\frac{\Omega}{2\Omega_c}\right) \right]^2 d\Omega$$

Because rect(/2c) = 1 over || ≤ ^c and is zero otherwise, the preceding integral yields

$$E_x = \frac{1}{2\pi} \left(\frac{\pi^2}{\Omega_c^2}\right) (2\Omega_c) = \frac{\pi}{\Omega_c}$$