9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL

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9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL

In Sec. 9.1 we succeeded in representing periodic signals as a sum of (everlasting) exponentials. In this section we extend this representation to aperiodic signals. The procedure is identical conceptually to that used in Ch. 7 for continuous-time signals.

Applying a limiting process, we now show that an aperiodic signal x[n] can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal x[n] such as the one illustrated in Fig. 9.4a by everlasting exponential signals, let us construct a new periodic signal xN⁰ [n] formed by repeating the signal x[n] every N⁰ units, as shown in Fig. 9.4b. The period N⁰ is made large enough to avoid overlap between the repeating cycles (N⁰ ≥ 2N+1). The periodic signal xN⁰ [n] can be represented by an exponential Fourier series. If we let N⁰ → ∞, the signal

Figure 9.4 Generation of a periodic signal by periodic extension of a signal x[n].

856 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

x[n] repeats after an infinite interval, and therefore,

$$\lim_{N_0 \to \infty} x_{N_0}[n] = x[n]$$

Thus, the Fourier series representing xN⁰ [n] will also represent x[n] in the limit N⁰ → ∞. The exponential Fourier series for xN⁰ [n] is given by

$$x_{N_0}[n] = \sum_{r = \langle N_0 \rangle} \mathcal{D}_r e^{jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0}$$

\n(9.11)

where

$$\mathcal{D}_r = \frac{1}{N_0} \sum_{n=-\infty}^{\infty} x[n] e^{-jr\Omega_0 n} \tag{9.12}$$

The limits for the sum on the right-hand side of Eq. (9.12) should be from −N to N. But because x[n] = 0 for |n| > N, it does not matter if the limits are taken from −∞ to ∞.

It is interesting to see how the nature of the spectrum changes as N⁰ increases. To understand this behavior, let us define X(), a continuous function of , as

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

\n(9.13)

From this definition and Eq. (9.12), we have

$$\mathcal{D}_r = \frac{1}{N_0} X(r\Omega_0) \tag{9.14}$$

This result shows that the Fourier coefficients D^r are 1/N⁰ times the samples of X() taken every ⁰ rad/s.† Therefore, (1/N0)X() is the envelope for the coefficients Dr. We now let N0→∞ by doubling N⁰ repeatedly. Doubling N⁰ halves the fundamental frequency 0, with the result that the spacing between successive spectral components (harmonics) is halved, and there are now twice as many components (samples) in the spectrum. At the same time, by doubling N0, the envelope of the coefficients D^r is halved, as seen from Eq. (9.14). If we continue this process of doubling N⁰ repeatedly, the number of components doubles in each step; the spectrum progressively becomes denser, while its magnitude D^r becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to X() in Eq. (9.13)]. In the limit, as N⁰ → ∞, the fundamental frequency ⁰ →0, and D^r →0. The separation between successive harmonics, which is 0, is approaching zero (infinitesimal), and the spectrum becomes so dense that it appears to be continuous. But as the number of harmonics increases indefinitely, the harmonic amplitudes D^r become vanishingly small (infinitesimal). We discussed an identical situation in Sec. 7.1.

We follow the procedure in Sec. 7.1 and let N⁰ → ∞. According to Eq. (9.13),

$$X(r\Omega_0) = \sum_{n=-\infty}^{\infty} x[n]e^{-jr\Omega_0 n}$$

^† For the sake of simplicity we assume ^Dr and therefore ^X() to be real. The argument, however, is also valid for complex D^r [or X()].

Using Eq. (9.14), we can express Eq. (9.11) as

$$x_{N_0}[n] = \frac{1}{N_0} \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} = \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} \left(\frac{\Omega_0}{2\pi}\right)$$

In the limit as N⁰ → ∞, ⁰ → 0 and xN⁰ [n] → x[n]. Therefore,

$$x[n] = \lim_{\Omega_0 \to 0} \sum_{r = \langle N_0 \rangle} \left[ \frac{X(r\Omega_0)\Omega_0}{2\pi} \right] e^{jr\Omega_0 n} \tag{9.15}$$

Because ⁰ is infinitesimal, it will be appropriate to replace ⁰ with an infinitesimal notation :

$$\Delta \Omega = \frac{2\pi}{N_0} \tag{9.16}$$

Equation (9.15) can be expressed as

$$x[n] = \lim_{\Delta\Omega \to 0} \frac{1}{2\pi} \sum_{r=\langle N_0 \rangle} X(r\Delta\Omega) e^{jr\Delta\Omega n} \Delta\Omega \tag{9.17}$$

The range r = #N0$ implies the interval of N⁰ number of harmonics, which is N0 = 2π according to Eq. (9.16). In the limit, the right-hand side of Eq. (9.17) becomes the integral

$$x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{in\Omega} d\Omega$$

\n(9.18)

where $ ^2π indicates integration over any continuous interval of 2π. The spectrum X() is given by [Eq. (9.13)]

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

\n(9.19)

The integral on the right-hand side of Eq. (9.18) is called the Fourier integral. We have now succeeded in representing an aperiodic signal x[n] by a Fourier integral (rather than a Fourier series). This integral is basically a Fourier series (in the limit) with fundamental frequency →0, as seen in Eq. (9.17). The amount of the exponential ejrⁿ is X(r)/2π. Thus, the function X() given by Eq. (9.19) acts as a spectral function, which indicates the relative amounts of various exponential components of x[n].

We call X() the (direct) discrete-time Fourier transform (DTFT) of x[n], and x[n] the inverse discrete-time Fourier transform (IDTFT) of X(). This nomenclature can be represented as

$$X(\Omega) = \text{DTFT}\{x[n]\}$$

and $x[n] = \text{IDTFT}\{X(\Omega)\}$

The same information is conveyed by the statement that x[n] and X() are a (discrete-time) Fourier transform pair. Symbolically, this is expressed as

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

The Fourier transform X() is the frequency-domain description of x[n].

9.2-1 Nature of Fourier Spectra

We now discuss several important features of the discrete-time Fourier transform and the spectra associated with it.

FOURIER SPECTRA ARE CONTINUOUS FUNCTIONS OF

Although x[n] is a discrete-time signal, X(), its DTFT is a continuous function of for the simple reason that is a continuous variable, which can take any value over a continuous interval from −∞ to ∞.

FOURIER SPECTRA ARE PERIODIC FUNCTIONS OF WITH PERIOD 2π

From Eq. (9.19), it follows that

$$X(\Omega + 2\pi) = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega + 2\pi)n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n}e^{-j2\pi n} = X(\Omega)$$

Clearly, the spectrum X() is a continuous, periodic function of with period 2π. We must remember, however, that to synthesize x[n], we need to use the spectrum over a frequency interval of only 2π, starting at any value of [see Eq. (9.18)]. As a matter of convenience, we shall choose this interval to be the fundamental frequency range (−π, π). It is, therefore, not necessary to show discrete-time-signal spectra beyond the fundamental range, although we often do so.

The reason for the periodic behavior of X() was discussed in Ch. 5, where we showed that, in a basic sense, the discrete-time frequency is bandlimited to || ≤ π. However, all discrete-time sinusoids with frequencies separated by an integer multiple of 2π are identical. This is why the spectrum is 2π periodic.

CONJUGATE SYMMETRY OF X()

From Eq. (9.19), we obtain the DTFT of x∗[n] as

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

In other words,

$$x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.20}$$

For real x[n], Eq. (9.20) reduces to x[n] ⇐⇒ X∗(−), which implies that for real x[n]

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

Therefore, for real x[n], X() and X(−) are conjugates. Since X() is generally complex, we have both amplitude and angle (or phase) spectra

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

Because of conjugate symmetry of X(), it follows that for real x[n],

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

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

PHYSICAL APPRECIATION OF THE DISCRETE-TIME FOURIER TRANSFORM

In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal x[n] as a sum of everlasting exponentials (or sinusoids). The Fourier spectrum of a signal indicates the relative amplitudes and phases of the exponentials (or sinusoids) required to synthesize x[n].

A detailed explanation of the nature of such sums over a continuum of frequencies is provided in Sec. 7.1-1.

EXISTENCE OF THE DTFT

Because |e−jn| = 1, from Eq. (9.19), it follows that the existence of X() is guaranteed if x[n] is absolutely summable; that is,

$$\sum_{n=-\infty}^{\infty} |x[n]| < \infty \tag{9.21}$$

This shows that the condition of absolute summability is a sufficient condition for the existence of the DTFT representation. This condition also guarantees its uniform convergence. The inequality

$$\left[\sum_{n=-\infty}^{\infty} |x[n]| \right]^2 \ge \sum_{n=-\infty}^{\infty} |x[n]|^2$$

shows that the energy of an absolutely summable sequence is finite. However, not all finite-energy signals are absolutely summable. Signal x[n] = sinc (n) is such an example. For such signals, the DTFT converges, not uniformly, but in the mean.†

To summarize, X() exists under a weaker condition

$$\sum_{n=-\infty}^{\infty} |x[n]|^2 < \infty \tag{9.22}$$

The DTFT under this condition is guaranteed to converge in the mean. Thus, the DTFT of the exponentially growing signal γ nu[n] does not exist when |γ | > 1 because the signal violates Eqs. (9.21) and (9.22). But the DTFT exists for the signal sinc(n), which violates Eq. (9.21) but does satisfy Eq. (9.22) (see later, Ex. 9.6). In addition, if the use of δ(), the continuous-time impulse function, is permitted, we can even find the DTFT of some signals that violate both Eq. (9.21) and Eq. (9.22). Such signals are not absolutely summable, nor do they have finite energy. For example, as seen from pairs 11 and 12 of Table 9.1, the DTFT of x[n] = 1 for all n and x[n] = ej0*ⁿ* exist, although they violate Eqs. (9.21) and (9.22).

$$\lim_{M \to \infty} \int_{-\pi}^{\pi} \left| X(\Omega) - \sum_{n=-M}^{M} x[n] e^{-j\Omega n} \right|^2 d\Omega = 0$$

^† This means

No.	x[n]	X()
1	δ[n−k]	e−jk	Integer k
2	γ nu[n]	ej ej −γ	γ < 1
3	−γ nu[−(n+1)]	ej ej −γ	γ > 1
4	γ n	1−γ 2 1−2γ cos+γ 2	γ < 1
5	nγ nu[n]	γ ej (ej −γ )2	γ < 1
6	γ n cos(0n+θ )u[n]	ej[ej cos θ −γ cos(0 −θ )] ej2 −(2γ cos0)ej +γ 2	γ < 1
7	u[n] −u[n− M]	sin(M/2) e−j(M−1)/2 sin(/2)
8	c π sinc (cn)	”∞ −2πk rect 2c k=−∞	c ≤ π
9	cn c 2π sinc2 2	”∞ −2πk 2c k=−∞	c ≤ π
10	u[n]	ej +π “∞ δ(−2πk) ej −1 k=−∞
11	1 for all n	2π “∞ δ(−2πk) k=−∞
12	ej0n	2π “∞ δ(−0 −2πk) k=−∞
13	cos0n	π ”∞ −2πk) +δ(+0 −2πk) δ(−0 k=−∞
14	sin0n	jπ “∞ δ(+0 −2πk)−δ(−0 −2πk) k=−∞
15	(cos0n)u[n]	ej2 −ej cos0 ”∞ π δ(−2πk−0)+δ(−2πk+0) +1 + ej2 −2ej cos0 2 k=−∞
16	(sin0n)u[n]	ej sin0 ”∞ π δ(−2πk−0)−δ(−2πk+0) +1 + ej2 −2ej cos0 2j k=−∞

TABLE 9.1 Select Discrete-Time Fourier Transform Pairs

EXAMPLE 9.3 DTFT of a Causal Exponential

Find the DTFT of x[n] = γ nu[n].

Using the definition, the DTFT is

$$X(\Omega) = \sum_{n=0}^{\infty} \gamma^n e^{-j\Omega n} = \sum_{n=0}^{\infty} (\gamma e^{-j\Omega})^n$$

This is an infinite geometric series with a common ratio γ e−j. Therefore (see Sec. B.8-3),

$$X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}}$$

provided |γ e−j| < 1. But because |e−j| = 1, this condition implies |γ | < 1. Therefore,

$$X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}} \qquad |\gamma| < 1$$

If |γ | > 1, X() does not converge. This result is in conformity with Eqs. (9.21) and (9.22). To determine magnitude and phase responses, we note that

$$X(\Omega) = \frac{1}{1 - \gamma \cos \Omega + j\gamma \sin \Omega} \tag{9.23}$$

so

$$|X(\Omega)| = \frac{1}{\sqrt{(1 - \gamma \cos \Omega)^2 + (\gamma \sin \Omega)^2}} = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}}$$

and

$$\angle X(\Omega) = -\tan^{-1}\left[\frac{\gamma \sin \Omega}{1 - \gamma \cos \Omega}\right]$$

Figure 9.5 shows x[n] = γ nu[n] and its spectra for γ = 0.8. Observe that the frequency spectra are continuous and periodic functions of with the period 2π. As explained earlier, we need to use the spectrum only over the frequency interval of 2π. We often select this interval to be the fundamental frequency range (−π,π).

The amplitude spectrum |X()| is an even function and the phase spectrum X() is an odd function of .

EXAMPLE 9.4 DTFT of an Anticausal Exponential

Figure 9.6 Exponential ^γ nu[−(n+1)].

Using the definition, the DTFT is

$$X(\Omega) = \sum_{n=-\infty}^{\infty} \gamma^n u [-(n+1)] e^{-j\Omega n} = \sum_{n=-1}^{-\infty} (\gamma e^{-j\Omega})^n = \sum_{n=-1}^{-\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^{-n}$$

Setting n = −m yields

$$x[n] = \sum_{m=1}^{\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^m = \frac{1}{\gamma} e^{j\Omega} + \left(\frac{1}{\gamma} e^{j\Omega}\right)^2 + \left(\frac{1}{\gamma} e^{j\Omega}\right)^3 + \cdots$$

This is a geometric series with a common ratio ej/γ . Therefore, from Sec. B.8-3,

$$X(\Omega) = \frac{1}{\gamma e^{-j\Omega} - 1} = \frac{1}{(\gamma \cos \Omega - 1) - j\gamma \sin \Omega}, \qquad |\gamma| > 1$$

Therefore,

$$|X(\Omega)| = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}} \quad \text{and} \quad \angle X(\Omega) = \tan^{-1} \left[ \frac{\gamma \sin \Omega}{\gamma \cos \Omega - 1} \right]$$

Except for the change of sign, this Fourier transform (and the corresponding frequency spectra) is identical to that of x[n] = γ nu[n]. Yet there is no ambiguity in determining the IDTFT of X() = 1/(γ e−j −1) because of the restrictions on the value of γ in each case. If |γ | < 1, then the inverse transform is x[n]=−γ nu[n]. If |γ | > 1, it is x[n] = γ ⁿ[−(n+1)].

EXAMPLE 9.5 DTFT of a Rectangular Pulse

Find the DTFT of the discrete-time rectangular pulse illustrated in Fig. 9.7a. This pulse is also known as the 9-point rectangular window function.

$$X(\Omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} = \sum_{n=-(M-1)/2}^{(M-1)/2} (e^{-j\Omega})^n \qquad M=9$$

This is a geometric progression with a common ratio e−j and (see Sec. B.8-3)

$$X(\Omega) = \frac{e^{-j[(M+1)/2]\Omega} - e^{j[(M-1)/2]\Omega}}{e^{-j\Omega} - 1}$$ $$= \frac{e^{-j\Omega/2} (e^{-j(M/2)\Omega} - e^{j(M/2)\Omega})}{e^{-j\Omega/2} (e^{-j\Omega/2} - e^{j\Omega/2})}$$ $$=\frac{\sin\left(\frac{M}{2}\Omega\right)}{\sin\left(0.5\Omega\right)}\tag{9.24}$$ $$=\frac{\sin(4.5\Omega)}{\sin(0.5\Omega)} \qquad \text{for } M=9$$

(9.25)

Figure 9.7b shows the spectrum X() for M = 9.

Figure 9.7 (a) Discrete-time gate pulse and (b) its Fourier spectrum.

DISCRETE-TIME FOURIER TRANSFORM USING MATLAB

Within a scale factor, the DTFS is identical to the DFT and, therefore, the FFT. That is, the DTFS is just the FFT scaled by ¹ N0 . Combined with Eq. (9.14), we see that the DFT Xr of finite-duration signal x[n] (repeated with period N⁰ large enough to avoid overlap) is just samples of the DTFT X() taken at = r0. That is, the length-N⁰ DFT of signal x[n] yields N⁰ samples of its DTFT X() as

$$X_r = X(r\Omega_0), \qquad \text{where } \Omega_0 = \frac{2\pi}{N_0} \tag{9.26}$$

This relationship provides a way to use MATLAB’s fft command to validate our DTFT calculations. By appropriately zero-padding x[n], we can obtain as many samples of X() as are desired. Let us demonstrate the process for the current example using N⁰ = 64. Notice that in taking the DFT, we modulo-N⁰ shift our rectangular pulse signal to occupy 0 ≤ n ≤ N⁰ −1.

>> Omega = linspace(0,2*pi,1000);

>> X = sin(4.5*Omega)./sin(0.5*Omega); X(mod(Omega,2*pi)==0) = 4.5/0.5;

>> N_0 = 64; M = 9; x = [ones(1,(M+1)/2) zeros(1,N_0-M) ones(1,(M-1)/2)];

>> Xr = fft(x); Omega_0 = 2*pi/N_0; r = 0:N_0-1;

>> plot(Omega,abs(X),'k-',Omega_0*r,abs(Xr),'k.'); axis([0 2*pi 0 9.5]);

xlabel(‘\Omega’); ylabel(‘|X(\Omega)|’);

As shown in Fig. 9.8, the FFT samples align exactly with our analytical DTFT result.

EXAMPLE 9.6 Inverse DTFT of a Rectangular Spectrum

Find the inverse DTFT of the rectangular pulse spectrum described over the fundamental band (|| ≤ π) by X() = rect(/2c) for ^c ≤ π. Because of the periodicity property, X() repeats at the intervals of 2π, as shown in Fig. 9.9a.

Figure 9.9 Periodic gate spectrum and its inverse discrete-time Fourier transform.

According to Eq. (9.18),

$$x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega) e^{jn\Omega} d\Omega = \frac{1}{2\pi} \int_{-\Omega_c}^{\Omega_c} e^{jn\Omega} d\Omega$$ $$= \frac{1}{j2\pi n} e^{jn\Omega} \Big|_{-\Omega_c}^{\Omega_c} = \frac{\sin(\Omega_c n)}{\pi n} = \frac{\Omega_c}{\pi} \text{sinc}(\Omega_c n)$$

The signal x[n] is depicted in Fig. 9.9b (for the case ^c = π/4).

DR ILL 9.4 Finding the DTFT

Find the DTFT and sketch the corresponding amplitude and phase spectra for

(a)

$$x[n] = \gamma^{|k|}$$

with $|\gamma| < 1$

(b)

$$y[n] = \delta[n+1] - \delta[n-1]$$

ANSWERS

(a)

$$X(\Omega) = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2}$$

\n(b) $|Y(\Omega)| = 2|\sin \Omega|$ and $\angle Y(\omega) = (\pi/2)[1 - \text{sgn}(\sin \Omega)]$

9.2-2 Connection Between the DTFT and the z**-Transform**

The connection between the (bilateral) z-transform and the DTFT is similar to that between the Laplace transform and the Fourier transform. The z-transform of x[n], according to Eq. (5.1), is

$$X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n}$$

\n(9.27)

Setting z = ej in this equation yields

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

The right-hand side sum defines X(), the DTFT of x[n]. Does this mean that the DTFT can be obtained from the corresponding z-transform by setting z = ej? In other words, is it true that X[ej] = X()? Yes, it is true in most cases. For example, when x[n] = anu[n], its z-transform is z/(z − a), and X[ej] = ej/(ej − a), which is equal to X() (assuming |a| < 1). However, for the unit step function u[n], the z-transform is z/(z − 1), and X[ej] = ej/(ej − 1). As seen from Table 9.1, pair 10, this is not equal to X() in this case.

We obtained X[ej] by setting z = ej in Eq. (9.27). This implies that the sum on the right-hand side of Eq. (9.27) converges for z = ej, which means the unit circle (characterized by z = ej) lies in the region of convergence for X[z]. Hence, the general rule is that setting z = ej in X[z] yields the DTFT X() only when the ROC for X[z] includes the unit circle. This applies for all x[n] that are absolutely summable. If the ROC of X[z] excludes the unit circle, X[ej] = X(). This applies to all exponentially growing x[n] and also x[n], which either is constant or oscillates with constant amplitude.

The reason for this peculiar behavior has something to do with the nature of convergence of the z-transform and the DTFT.†

This discussion shows that although the DTFT may be considered to be a special case of the z-transform, we need to circumscribe such a view. This cautionary note is supported by the fact that a periodic signal has the DTFT, but its z-transform does not exist.