[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER TRANSFORM

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PROBLEMS

• 6.1-1 For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric Fourier series and sketch the amplitude and phase spectra. If either the sine or cosine terms are absent in the Fourier series, explain why.

• 6.1-2 (a) Find the trigonometric Fourier series for y(t) shown in Fig. P6.1-2.

  • (b) The signal y(t) can be obtained by time reversal of x(t) shown in Fig. 6.2a. Use this fact to obtain the Fourier series for y(t) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a).
  • (c) Show that, in general, time reversal of a periodic signal does not affect the amplitude spectrum, and the phase spectrum is also unchanged except for the change of sign.

• 6.1-3 (a) Find the trigonometric Fourier series for the periodic signal y(t) depicted in Fig. P6.1-3.

  • (b) The signal y(t) can be obtained by time compression of x(t) shown in Fig. 6.2a by a factor 2. Use this fact to obtain the Fourier series for y(t) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a).
  • (c) Show that, in general, time compression of a periodic signal by a factor a expands the Fourier spectra along the ω axis by the same factor a. In other words C0,Cn, and θ*ⁿ* remain unchanged, but the fundamental frequency is increased by the factor a, thus expanding the spectrum. Similarly, time expansion of a periodic signal by a factor a compresses its Fourier spectra along the ω axis by the factor a.

• 6.1-4 (a) Find the trigonometric Fourier series for the periodic signal g(t) in Fig. P6.1-4. Take advantage of the symmetry.

• (b) Observe that g(t) is identical to x(t) in Fig. 6.4a left-shifted by 0.5 second. Use this fact to obtain the Fourier series for g(t) from the results in Ex. 6.2. Verify that the Fourier series thus obtained is identical to that found in part (a).

• (c) Show that, in general, a time shift of T seconds of a periodic signal does not affect the amplitude spectrum. However, the phase of the nth harmonic is increased or decreased nω0T depending on whether the signal is advanced or delayed by T seconds.

• 6.1-5 Determine the trigonometric Fourier series coefficients an and bn for the following signals. In each case, also determine the signals’ fundamental radian frequency ω0. No integration is required to solve this problem.

  • (a) xa(t) = cos(3πt)
  • (b) xb(t) = sin(7πt)
  • (c) xc(t) = 2+4 cos(3πt)−2jsin(7πt)
  • (d) xd(t) = (1+j)sin(3πt)+(2−j) cos(7πt)
  • (e) xe(t) = sin(3πt +1)+2 cos(7πt −2)
  • (f) xf(t) = sin(6πt)+2 cos(14πt)

• 6.1-6 If the two halves of one period of a periodic signal are identical in shape except that one is the negative of the other, the periodic signal is said to have a half-wave symmetry. If a periodic signal x(t) with a period T⁰ satisfies the half-wave symmetry condition, then

$$x\left(t - \frac{T_0}{2}\right) = -x(t)$$

In this case, show that all the even-numbered harmonics vanish and that the odd-numbered harmonic coefficients are given by

$$a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t \, dt$$

Figure P6.1-2

(b)

Figure P6.1-6

and

$$b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t \, dt$$

Using these results, find the Fourier series for the periodic signals in Fig. P6.1-6.

6.1-7 Over a finite interval, a signal can be represented by more than one trigonometric (or exponential) Fourier series. For instance, if we wish to represent x(t) = t over an interval 0 < t < 1 by a Fourier series with fundamental frequency ω⁰ = 2, we can draw a pulse x(t) = t over the interval 0 < t < 1 and repeat the pulse every π seconds so that T⁰ = π and ω⁰ = 2 (Fig. P6.1-7a). If we want the fundamental frequency ω⁰ to be 4, we repeat the pulse every π/2 seconds. If we want the series to contain only cosine terms with ω⁰ = 2, we construct a pulse x(t) = |t| over −1 < t < 1, and repeat it every π seconds (Fig. P6.1-7b). The resulting signal is an even function with period π. Hence, its Fourier series will have only cosine terms with ω⁰ = 2. The resulting Fourier series represents x(t) = t over 0 < t < 1, as desired. We do not care what it represents outside this interval.

Sketch the periodic signal x(t) such that x(t) = t for 0 < t < 1 and the Fourier series for x(t) satisfies the following conditions.

• (a) ω⁰ = π/2 and contains all harmonics, but cosine terms only
• (b) ω⁰ = 2 and contains all harmonics, but sine terms only
• (c) ω⁰ = π/2 and contains all harmonics, which are exclusively neither sine nor cosine
• (d) ω⁰ = 1 and contains only odd harmonics and cosine terms
• (e) ω⁰ = π/2 and contains only odd harmonics and sine terms

Figure P6.1-7

(f) ω⁰ = 1 and contains only odd harmonics, which are exclusively neither sine nor cosine.

[Hint: For parts (d), (e), and (f), you need to use half-wave symmetry discussed in Prob. 6.1-6. Cosine terms imply a possible dc component.] You are asked only to sketch the periodic signal x(t) satisfying the given conditions. Do not find the values of the Fourier coefficients.

• 6.1-8 State with reasons whether the following signals are periodic or aperiodic. For periodic signals, find the period and state which harmonics are present in the series.
  • (a) 3 sin t +2 sin 3t
  • (b) 2+5 sin 4t +4 cos 7t
  • (c) 2 sin 3t +7 cos πt
  • (d) 7 cos πt +5 sin 2πt
  • (e) 3 cos ^√2*^t* ⁺5 cos 2*^t*
  • (f) sin 5t ^{2 +}3 cos 6t ^{5 +}3 sin ^{t 7 +}30◦ t

(g)

$$\sin 3t + \cos \frac{15}{4}$$

• (h) (3 sin 2t +sin 5t)²
• (i) (5 sin 2t)³
• 6.3-1 For each of the periodic signals in Fig. P6.1-1, find exponential Fourier series and sketch the corresponding spectra.
• 6.3-2 A 2π-periodic signal x(t) is specified over one period as

$$x(t) = \begin{cases} \frac{1}{A}t & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \end{cases}$$

Sketch x(t) over two periods from t = 0 to 4π. Show that the exponential Fourier series coefficients Dn for this series are given by

$$D_n = \begin{cases} \frac{2\pi - A}{4\pi} & n = 0\\ \frac{1}{2\pi n} \left( \frac{e^{-jAn} - 1}{An} + je^{-jn\pi} \right) & n \neq 0 \end{cases}$$

6.3-3 A periodic signal x(t) is expressed by the following Fourier series:

$$x(t) = 3\cos t + \sin\left(t - \frac{\pi}{6}\right) - 2\cos\left(t - \frac{\pi}{3}\right)$$

• (a) Sketch the amplitude and phase spectra for the trigonometric series.
• (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra.
• (c) By inspection of spectra in part (b), write the exponential Fourier series for x(t).
• (d) Show that the series found in part (c) is equivalent to the trigonometric series for x(t).
• 6.3-4 The trigonometric Fourier series of a certain periodic signal is given by

$$x(t) = 3 + \sqrt{3}\cos 2t + \sin 2t$$ $$+ \sin 3t - \frac{1}{2}\cos\left(5t + \frac{\pi}{3}\right)$$

• (a) Sketch the trigonometric Fourier spectra.
• (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra.
• (c) By inspection of spectra in part (b), write the exponential Fourier series for x(t).
• (d) Show that the series found in part (c) is equivalent to the trigonometric series for x(t).
• 6.3-5 The exponential Fourier series of a certain function is given as

$$x(t) = (2+j2)e^{-j3t} + j2e^{-jt} + 3 - j2e^{jt} + (2-j2)e^{j3t}$$

• (a) Sketch the exponential Fourier spectra.

• (b) By inspection of the spectra in part (a), sketch the trigonometric Fourier spectra for x(t). Find the compact trigonometric Fourier series from these spectra.

• (c) Show that the trigonometric series found in part (b) is equivalent to the exponential series for x(t).

• (d) Find the signal bandwidth.

• 6.3-6 Figure P6.3-6 shows the trigonometric Fourier spectra of a periodic signal x(t).

  • (a) By inspection of Fig. P6.3-6, find the trigonometric Fourier series representing x(t).
  • (b) By inspection of Fig. P6.3-6, sketch the exponential Fourier spectra of x(t).
  • (c) By inspection of the exponential Fourier spectra obtained in part (b), find the exponential Fourier series for x(t).
  • (d) Show that the series found in parts (a) and (c) are equivalent.

• 6.3-7 Figure P6.3-7 shows the exponential Fourier spectra of a periodic signal x(t).

• (a) By inspection of Fig. P6.3-7, find the exponential Fourier series representing x(t).

• (b) By inspection of Fig. P6.3-7, sketch the trigonometric Fourier spectra for x(t).

• (c) By inspection of the trigonometric Fourier spectra found in part (b), find the trigonometric Fourier series for x(t).

• (d) Show that the series found in parts (a) and (c) are equivalent.

• 6.3-8 Let periodic signal x(t) have exponential Fourier series spectrum Dn. Prove the following properties.

  • (a) If x(t) has even symmetry, then Dn also has even symmetry.
  • (b) If x(t) has odd symmetry, then Dn also has odd symmetry.
  • (c) If x(t) is real, then Dn is conjugate symmetric (Dn = D^{∗ −}n).
  • (d) If x(t) is imaginary, then Dn is conjugate antisymmetric (Dn = −D^{∗ −}n).

• 6.3-9 (a) Find the exponential Fourier series for the signal in Fig. P6.3-9a.

  • (b) Using the results in part (a), find the Fourier series for the signal xˆ(t) in Fig. P6.3-9b,

Figure P6.3-7

Figure P6.3-9

which is a time-shifted version of the signal x(t).

• (c) Using the results in part (a), find the Fourier series for the signal x˜(t) in Fig. P6.3-9c, which is a time-scaled version of the signal x(t).
• 6.3-10 A periodic signal x(t) is expressed as an exponential Fourier series

$$x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t}$$

(a) Show that the exponential Fourier series for xˆ(t) = x(t −T) is given by

$$\hat{x}(t) = \sum_{n = -\infty}^{\infty} \hat{D}_n e^{jn\omega_0 t}$$

in which

$$|\tilde{D}_n| = |D_n| \quad \text{and} \quad \angle \tilde{D}_n = \angle D_n - n\omega_0 T$$

This result shows that time shifting of a periodic signal by T seconds merely changes the phase spectrum by nω0T. The amplitude spectrum is unchanged.

(b) Show that the exponential Fourier series for x˜(t) = x(at) is given by

$$\tilde{x}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn(a\omega_0)t}$$

This result shows that time compression of a periodic signal by a factor a expands its Fourier spectra along the ω axis by the same factor a. Similarly, time expansion of a periodic signal by a factor a compresses its Fourier spectra along the ω axis by the factor a. Intuitively explain this result.

6.3-11 (a) The Fourier series for the periodic signal in Fig. 6.7a is given in Drill 6.1. Verify Parseval’s theorem for this series, given that

$$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$$

• (b) If x(t) is approximated by the first N terms in this series, find N so that the power of the error signal is less than 1% of Px.
• 6.3-12 (a) The Fourier series for the periodic signal in Fig. 6.7b is given in Drill 6.1. Verify

Problems 675

Parseval’s theorem for this series, given that

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

• (b) If x(t) is approximated by the first N terms in this series, find N so that the power of the error signal is less than 10% of Px.
• 6.3-13 The signal x(t) in Fig. 6.17 is approximated by the first 2N + 1 terms (from n = −N to N) in its exponential Fourier series given in Drill 6.5. Determine the value of N if this (2N + 1)-term Fourier series power is to be no less than 99.75% of the power of x(t).
• 6.3-14 (a) A 2 rad/s periodic signal x1(t) has Fourier series spectrum D1[n]. Determine the Fourier series spectrum *^X*2[n] of *^x*2(t) ^{= 1 3} x1(−t −5) in terms of X1[n].
  • (b) A 2 rad/s periodic signal x1(t) has Fourier series spectrum D1[n]. Determine the Fourier series spectrum X2[n] of x2(t) = cos(10t)x1(t) in terms of X1[n].
  • (c) A 3 rad/s periodic signal x1(t) has Fourier series spectrum D1[n]. Determine the Fourier series spectrum X2[n] of x2(t) = x1(−t)−3x1(t +2) in terms of X1[n].
• 6.3-15 A 2-periodic signal x(t) is defined as

$$x(t) = \begin{cases} -t^2 - t + 0.25 & -1 \le t < 0\\ t^2 - t + 0.25 & 0 \le t < 1\\ x(t+2) & \forall t \end{cases}$$

• (a) Plot x(t) over −2 ≤ t ≤ 2.
• (b) Determine D0, the dc content of x(t).
• (c) Similar to Ex. 6.11, use properties and not integration to determine Dn for n = 0.
• (d) Plot the magnitude spectrum |Dn| over a suitable range of n.
• (e) How does the magnitude spectrum |Dn| compare to that of signal y(t) = cos(πt)? Note similarities as well as major differences.
• 6.3-16 A 3-periodic signal x(t) is defined as

$$x(t) = \begin{cases} \n|t| & -1 \le t \le 1\\ \n0 & 1 < |t| \le 1.5\\ \nx(t+3) & \forall t\n\end{cases}$$

(a) Plot

$$x(t)$$

over $-3 \le t \le 3$ .

• (b) Determine D0, the dc content of x(t).
• (c) Similar to Ex. 6.11, use properties and not integration to determine Dn for n = 0.
• (d) Plot the magnitude spectrum |Dn| over a suitable range of n. What is the most dominant frequency component of this signal?
• 6.4-1 Find the response of an LTIC system with transfer function

$$H(s) = \frac{s}{s^2 + 2s + 3}$$

to the periodic input shown in Fig. 6.2a.

• 6.4-2 A periodic signal x(t) = 1 + 2 cos(5πt) + 3 sin(14πt) is applied to an LTIC system to produce output y(t).
  • (a) Determine ω0, the fundamental radian frequency of x(t).
  • (b) Determine Dn, the exponential Fourier series spectrum of x(t).
  • (c) If the system is an ideal lowpass filter with cutoff frequency f^c = 2 Hz, what is the output y(t)?
  • (d) If the system is an ideal highpass filter with cutoff frequency f^c = 2 Hz, what is the output y(t)?
  • (e) If the system is an ideal bandpass filter with a 4 Hz passband centered at 4 Hz, what is the output y(t)?
  • (f) If the system is an ideal bandstop filter with a 5 Hz stopband centered at 10 Hz, what is the output y(t)?
  • (g) Describe the frequency response of a filter that, in response to x(t), would produce the output y(t) = 4 cos(5πt) −9 sin(14πt).
• 6.4-3 Consider a T⁰ = 1 periodic signal x(t) defined as

$$x(t) = \begin{cases} 1 - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}$$

• (a) Sketch x(t) for −2 ≤ t ≤ 2.
• (b) Determine Dn, the exponential Fourier series spectrum of x(t).
• (c) If x(t) is applied to an ideal bandpass filter with a 1 Hz passband centered at 3 Hz, determine the output y(t).
• 6.4-4 (a) Find the exponential Fourier series for a signal x(t) = cos 5t sin 3t. You can do this without evaluating any integrals.
  • (b) Sketch the Fourier spectra.

(c) The signal x(t) is applied at the input of an LTIC system with frequency response, as shown in Fig. P6.4-4. Find the output y(t).

Figure P6.4-4

• 6.4-5 (a) Find the exponential Fourier series for a periodic signal x(t) shown in Fig. P6.4-5a.
  • (b) The signal x(t) is applied at the input of an LTIC system shown in Fig. P6.4-5b. Find the expression for the output y(t).
• 6.4-6 A T-periodic τ/T duty-cycle square wave p(t) is defined as

$$p(t) = \begin{cases} 1 & |t| < \frac{\tau}{2} \\ 0 & \frac{\tau}{2} < |t| < \frac{T}{2} \\ p(t+T) & \forall t \end{cases}$$

where 0 <τ< T. Also consider the frequency response H(ω) of a lowpass communications channel with 10 rad/s bandwidth (e.g., |H(ω)| ≈ 0 for ω > 10). If T and τ are properly chosen, we can estimate H(ω) at points ω = nω⁰ as ^Hˆ (nω0) ^{= 1} P0 Yn, where Yn is the exponential FS spectrum of the channel output y(t) in response to input p(t) and P⁰ is the dc component of p(t).

• (a) Using direct integration, determine the exponential Fourier series coefficients Pn of signal p(t).
• (b) Determine a suitable value T so that p(t) applied to system H(ω) has 21 component frequencies over the system bandwidth 0 ≤ ω ≤ 10.
• (c) Assuming T is properly chosen, determine a suitable duty cycle τ/T so that H(nω0) ≈ Yn. Carefully justify your result.
• (d) From the perspective of using p(t) to help measure the system frequency response H(ω), what happens if T is properly chosen but τ/T is chosen too small?
• (e) From the perspective of using p(t) to help measure the system frequency response H(ω), what happens if T is properly chosen but τ/T is chosen too large?
• 6.5-1 Derive Eq. (6.32) in an alternate way by observing that e = (x−cy) and |e| ² = (x−cy)·(x − cy) = |x| ² +c2|y| ² −2cx · y.
• 6.5-2 A signal x(t) is approximated in terms of a signal y(t) over an interval (t1, t2):

$$x(t) \simeq cy(t) \qquad t_1 < t < t_2$$

where c is chosen to minimize the error energy.

Figure P6.4-5

• (a) Show that y(t) and the error e(t) = x(t) − cy(t) are orthogonal over the interval (t1, t2).
• (b) If possible, explain the result in terms of a signal-vector analogy.
• (c) Verify this result for the square signal x(t) in Fig. 6.23 and its approximation in terms of signal sin t.
• 6.5-3 If x(t) and y(t) are orthogonal, then show that the energy of the signal x(t) + y(t) is identical to the energy of the signal x(t) − y(t) and is given by Ex + Ey. Explain this result by using the vector analogy. In general, show that for orthogonal signals x(t) and y(t) and for any pair of arbitrary real constants c¹ and c2, the energies of c1x(t) + c2y(t) and c1x(t) − c2y(t) are both given by c^{2 1}Ex ⁺c^{2 2}Ey.
• 6.5-4 (a) For the signals x(t) and y(t) depicted in Fig. P6.5-4, find the component of the form y(t) contained in x(t). In other words, find the optimum value of c in the approximation x(t) ≈ cy(t) so that the error signal energy is minimum.
  • (b) Find the error signal e(t) and its energy Ee. Show that the error signal is orthogonal to y(t), and that Ex = c²Ey + Ee. Explain this result in terms of vectors.

Figure P6.5-4

• 6.5-5 For the signals x(t) and y(t) shown in Fig. P6.5-4, find the component of the form x(t) contained in y(t). In other words, find the optimum value of c in the approximation y(t) ≈ cx(t) so that the error signal energy is minimum. What is the error signal energy?

• 6.5-6 Represent the signal x(t) shown in Fig. P6.5-4a over the interval from 0 to 1 by a trigonometric Fourier series of fundamental frequency ω⁰ = 2π. Compute the error energy in the representation of x(t) by only the first N terms of this series for N = 1, 2, 3, and 4.

• 6.5-7 Represent x(t) = t over the interval (0, 1) by a trigonometric Fourier series that has

  • (a) ω⁰ = 2π and only sine terms
  • (b) ω⁰ = π and only sine terms
  • (c) ω⁰ = π and only cosine terms

You may use a dc term in these series if necessary.

• 6.5-8 In Ex. 6.15, we represented the function in Fig. 6.27 by Legendre polynomials.
  • (a) Use the results in Ex. 6.15 to represent the signal g(t) in Fig. P6.5-8 by Legendre polynomials.
  • (b) Compute the error energy for the approximations having one and two (nonzero) terms.

Figure P6.5-8

• 6.5-9 Walsh functions, which can take on only two amplitude values, form a complete set of orthonormal functions and are of great practical importance in digital applications because they can be easily generated by logic circuitry and because multiplication with these functions can be implemented by simply using a polarity-reversing switch. Figure P6.5-9 shows the first eight functions in this set. Represent x(t) in Fig. P6.5-4a over the interval (0, 1) by using a Walsh Fourier series with these eight basis functions. Compute the energy of e(t), the error in the approximation, using the first N nonzero terms in the series for N = 1, 2, 3, and 4. In Prob. 6.5-6 we found the trigonometric Fourier series for x(t). How does the Walsh series compare with the trigonometric series in Prob. 6.5-6 from the viewpoint of the error energy for a given N?
• 6.5-10 For the four-dimensional real space *^R*4, the so-called Walsh basis is given by: φ¹ = [1, 1, 1, 1], φ² = [1, 1,−1,−1], φ³ = [1,−1,−1, 1], and φ⁴ = [1,−1, 1,−1]. Denoting elements x = [x1, x2, x3, x4] and y = [y1, y2, y3, y4] (x, ^{y ∈} *^R*4), we can define orthogonality as

Figure P6.5-9

%^{4 k}=¹ xky^{∗ k} = 0. In linear algebra terminology, orthogonality here means that the inner product of vectors x and y is zero. Lastly, define a vector z = [−4, 0, 1,−7].

• (a) Show that the Walsh basis functions are mutually orthogonal. This requires a total of six calculations.
• (b) Are the Walsh basis functions normal? That is, does the inner product of each Walsh basis function with itself evaluate to 1?
• (c) Determine the coefficients [c1, c2, c3, c4] to represent z using Walsh basis functions as ^{zˆ =} %^{4 k}=¹ ckφk.
• (d) Determine the best three-dimensional approximation zˆ3*^D* to z in terms of the Walsh basis functions. That is, your estimate can only be a linear combination of three functions from [φ1,φ2,φ3,φ4]. Evaluate the three-term sum to determine the four elements of vector zˆ3D.
• 6.5-11 A function can be expanded in terms of many different types of basis functions, not just the complex exponentials of Fourier analysis. For example, Walsh functions are explored in Prob. 6.5-9. Laguerre polynomials Lk(t), which have support on the interval [0,∞), are another possible set of basis functions. The Laguerre expansion using any number of terms we choose. For the Laguerre expansion, we define orthogonality a little differently, as

$$\int_0^\infty e^{-t}x(t)y(t)dt = 0.$$

Notice the presence of the e−^t term in the integral. Using this definition, Laguerre polynomials are orthonormal.

(a) Show that L0(t) is normal. That is, show that

$$\int_0^\infty e^{-t} L_0(t) L_0^*(t) dt = 1$$

(b) Show that L1(t) is normal. That is, show that

$$\int_0^{\infty} e^{-t} L_1(t) L_1^*(t) dt = 1$$

• (c) Show that L0(t) is orthogonal to L1(t).
• (d) Compute the coefficient c⁰ that produces the best approximation xˆ0(t) = c0L0(t) of the function ^x(t) ^{= e}−^t u(t).
• (e) For the best estimate xˆ1(t) = c0L0(t) + ^c1L1(t) of the function ^x(t) ^{= e}−^t u(t), the coefficient c⁰ remains unchanged from part ()(d) and ^{c1 = 1 4} . Confirm that ^{c1 = 1} 4 and explain why the coefficient c⁰ does not change.
• 6.7-1 A periodic signal has ^{ω0 = 2 3}π and exponential Fourier series spectrum Dn = j cos (πn/10)(u[n+10] −u[n−11]). [Hint: Refer to Prob. 6.3-8 for some useful properties.]
  • (a) Determine the period T⁰ of the corresponding signal x(t).
  • (b) Is the time-domain signal real, imaginary, or neither? Justify your answer.
  • (c) Is the time-domain signal even, odd, or neither? Justify your answer.
  • (d) Use MATLAB to synthesize the timedomain signal x(t) and plot it over the interval [−T0,T0].
• 6.7-2 Repeat Prob. 6.7-1 for a periodic signal with ^{ω0 = 3 2}π and Dn = 2 sin(πn/10)(u[n + 10] − u[n−11]).
• 6.7-3 Consider the (T⁰ = 1)-periodic signal x(t):

$$x(t) = \begin{cases} 2t - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}.$$

• (a) Sketch x(t) for −2 ≤ t ≤ 2.

• (b) Using properties and minimal integration, determine Dn, the exponential Fourier spectrum of x(t).

• (c) Verify the correctness of Dn by using MAT-LAB to synthesize x(t) with a suitable truncation of Eq. (6.19).

• (d) Suppose x(t) is applied to an ideal bandpass filter with passband between 2.5 and 3.5 Hz. Determine the filter output y(t). Simplify your answer.

[Hint: Refer to Ex. 6.11.]

6.7-4 Section 6.7 discusses the construction of a phase-optimized multitone test signal with linearly spaced frequency components. This problem investigates a similar signal with logarithmically spaced frequency components.

A multitone test signal m(t) is constructed by using a superposition of N real sinusoids

$$m(t) = \sum_{n=1}^{N} \cos(\omega_n t + \theta_n)$$

where θ*ⁿ* establishes the relative phase of each sinusoidal component.

• (a) Determine a suitable set of N = 10 frequencies ω*ⁿ* that logarithmically spans [(2π ) ≤ ω ≤ 100(2π )] yet still results in a periodic test signal m(t). Determine the period T⁰ of your signal. Using θ*ⁿ* = 0, plot the resulting (T0)-periodic signal over −T0/2 ≤ t ≤ T0/2.
• (b) Determine a suitable set of phases θ*ⁿ* that minimize the maximum magnitude of m(t). Plot the resulting signal and identify the maximum magnitude that results.
• (c) Many systems suffer from what is called one-over-f noise. The power of this undesirable noise is proportional to 1/f . Thus, low-frequency noise is stronger than high-frequency noise. What modifications to m(t) are appropriate for use in environments with 1/f noise? Justify your answer.

CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM

We can analyze linear systems in many different ways by taking advantage of the property of linearity, whereby the input is expressed as a sum of simpler components. The system response to any complex input can be found by summing the system’s response to these simpler components of the input. In time-domain analysis, we separated the input into impulse components. In the frequency-domain analysis in Ch. 4, we separated the input into exponentials of the form est (the Laplace transform), where the complex frequency s = σ + jω. The Laplace transform, although very valuable for system analysis, proves somewhat awkward for signal analysis, where we prefer to represent signals in terms of exponentials ejω*^t* instead of est. This is accomplished by the Fourier transform. In a sense, the Fourier transform may be considered to be a special case of the Laplace transform with s = jω. Although this view is true most of the time, it does not always hold because of the nature of convergence of the Laplace and Fourier integrals.

In Ch. 6, we succeeded in representing periodic signals as a sum of (everlasting) sinusoids or exponentials of the form ejω*^t* . The Fourier integral developed in this chapter extends this spectral representation to aperiodic signals.

7.1 APERIODIC SIGNAL REPRESENTATION BY THE FOURIER INTEGRAL

Applying a limiting process, we now show that an aperiodic signal can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal x(t) such as the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal xT⁰ (t) formed by repeating the signal x(t) at intervals of T⁰ seconds, as illustrated in Fig. 7.1b. The period T⁰ is made long enough to avoid overlap between the repeating pulses. The periodic signal xT⁰ (t) can be represented by an exponential Fourier series. If we let T⁰ → ∞, the pulses in the periodic signal repeat after an infinite interval and, therefore,

$$\lim_{T_0 \to \infty} x_{T_0}(t) = x(t)$$

CHAPTER

7

Figure 7.1 Construction of a periodic signal: (a) signal x(t) and (b) periodic extension of x(t).

Thus, the Fourier series representing xT⁰ (t) will also represent x(t) in the limit T⁰ → ∞. The exponential Fourier series for xT⁰ (t) is given by

$$x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t}$$

\n(7.1)

where ^{ω0 = 2π T0} and

$$D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x_{T_0}(t) e^{-jn\omega_0 t} dt$$

\n(7.2)

Observe that integrating xT⁰ (t) over (−T0/2,T0/2) is the same as integrating x(t) over (−∞,∞). Therefore, Eq. (7.2) can be expressed as

$$D_n = \frac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-jn\omega_0 t} dt$$

\n(7.3)

It is interesting to see how the nature of the spectrum changes as T⁰ increases. To understand this fascinating behavior, let us define X(ω), a continuous function of ω, as

$$X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$$

\n(7.4)

A glance at Eqs. (7.3) and (7.4) shows that

$$D_n = \frac{1}{T_0} X(n\omega_0) \tag{7.5}$$

Figure 7.2 Change in the Fourier spectrum when the period T⁰ in Fig. 7.1 is doubled.

This means that the Fourier coefficients Dn are 1/T⁰ times the samples of X(ω) uniformly spaced at intervals of ω0, as depicted in Fig. 7.2a.† Therefore, (1/T0)X(ω) is the envelope for the coefficients Dn. We now let T⁰ → ∞ by doubling T⁰ repeatedly. Doubling T⁰ halves the fundamental frequency ω⁰ so that there are now twice as many components (samples) in the spectrum. However, by doubling T0, the envelope (1/T0)X(ω) is halved, as shown in Fig. 7.2b. If we continue this process of doubling T⁰ repeatedly, the spectrum progressively becomes denser while its magnitude becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to X(ω) in Eq. (7.4)]. In the limit as T⁰ → ∞, ω⁰ → 0 and Dn → 0. This result makes for a spectrum so dense that the spectral components are spaced at zero (infinitesimal) intervals. At the same time, the amplitude of each component is zero (infinitesimal). We have nothing of everything, yet we have something! This paradox sounds like Alice in Wonderland, but as we shall see, these are the classic characteristics of a very familiar phenomenon.‡

Substitution of Eq. (7.5) in Eq. (7.1) yields

$$x_{T_0}(t) = \sum_{n = -\infty}^{\infty} \frac{X(n\omega_0)}{T_0} e^{jn\omega_0 t}$$

\n(7.6)

As T⁰ →∞, ω⁰ becomes infinitesimal (ω⁰ →0). Hence, we shall replace ω⁰ by a more appropriate notation, ^ω. In terms of this new notation, ^{ω0 = 2π T0} becomes

$$\Delta \omega = \frac{2\pi}{T_0}$$

^† For the sake of simplicity, we assume Dn, and therefore X(ω), in Fig. 7.2, to be real. The argument, however, is also valid for complex Dn [or X(ω)].

^‡ If nothing else, the reader now has irrefutable proof of the proposition that 0% ownership of everything is better than 100% ownership of nothing.

Figure 7.3 The Fourier series becomes the Fourier integral in the limit as T⁰ → ∞.

and Eq. (7.6) becomes

$$x_{T_0}(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{X(n\Delta\omega)\Delta\omega}{2\pi} \right] e^{(jn\Delta\omega)t}$$

This equation shows that xT⁰ (t) can be expressed as a sum of everlasting exponentials of frequencies 0,±ω,±2ω,±3ω,… (the Fourier series). The amount of the component of frequency nω is [X(nω)ω]/2π. In the limit as T⁰ → ∞, ω → 0 and xT⁰ (t) → x(t). Therefore,

$$x(t) = \lim_{T_0 \to \infty} x_{T_0}(t) = \lim_{\Delta \omega \to 0} \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} X(n\Delta \omega) e^{(jn\Delta \omega)t} \Delta \omega$$

\n(7.7)

The sum on the right-hand side of Eq. (7.7) can be viewed as the area under the function X(ω)ejω*^t* , as illustrated in Fig. 7.3. Therefore,

$$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega$$

(7.8)

The integral on the right-hand side is called the Fourier integral. We have now succeeded in representing an aperiodic signal x(t) by a Fourier integral (rather than a Fourier series).† This integral is basically a Fourier series (in the limit) with fundamental frequency ω → 0, as seen from Eq. (7.7). The amount of the exponential ejnω*^t* is X(nω)ω/2π. Thus, the function X(ω) given by Eq. (7.4) acts as a spectral function.

We call X(ω) the direct Fourier transform of x(t), and x(t) the inverse Fourier transform of X(ω). The same information is conveyed by the statement that x(t) and X(ω) are a Fourier transform pair. Symbolically, this statement is expressed as

$$X(\omega) = \mathcal{F}[x(t)]$$

and $x(t) = \mathcal{F}^{-1}[X(\omega)]$

^† This derivation should not be considered to be a rigorous proof of Eq. (7.8). The situation is not as simple as we have made it appear [1].

or

$$x(t) \Longleftrightarrow X(\omega)$$

To recapitulate,

$$X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$$

\n(7.9)

and

$$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega$$

\n(7.10)

It is helpful to keep in mind that the Fourier integral in Eq. (7.10) is of the nature of a Fourier series with fundamental frequency ω approaching zero [Eq. (7.7)]. Therefore, most of the discussion and properties of Fourier series apply to the Fourier transform as well. The transform X(ω) is the frequency-domain specification of x(t).

We can plot the spectrum X(ω) as a function of ω. Since X(ω) is complex, we have both amplitude and angle (or phase) spectra

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

in which |X(ω)| is the amplitude and X(ω) is the angle (or phase) of X(ω). According to Eq. (7.9),

$$X(-\omega) = \int_{-\infty}^{\infty} x(t)e^{j\omega t}dt$$

Taking the conjugates of both sides yields

$$x^*(t) \Longleftrightarrow X^*(-\omega) \tag{7.11}$$

This property is known as the conjugation property. Now, if x(t) is a real function of t, then x(t) = x∗(t), and from the conjugation property, we find that

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

This is the conjugate symmetry property of the Fourier transform, applicable to real x(t). Therefore, for real x(t),

$$|X(-\omega)| = |X(\omega)| \quad \text{and} \quad \angle X(-\omega) = -\angle X(\omega) \tag{7.12}$$

Thus, for real x(t), the amplitude spectrum |X(ω)| is an even function, and the phase spectrum X(ω) is an odd function of ω. These results were derived earlier for the Fourier spectrum of a periodic signal [Eq. (6.22)] and should come as no surprise.

EXAMPLE 7.1 Fourier Transform of a Causal Exponential

Find the Fourier transform of e−atu(t).

By definition [Eq. (7.9)],

$$X(\omega) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt = \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty}$$

But |e−jω*^t* | = 1. Therefore, as t → ∞, e−(a+jω)^t = e−ate−jω*^t* = ∞ if a < 0, but it is equal to 0 if a > 0. Therefore,

$$X(\omega) = \frac{1}{a + j\omega} \qquad a > 0$$

Expressing ^a+j^ω in the polar form as ^{√ a2 +}ω² ejtan−1(ω/a) , we obtain

$$X(\omega) = \frac{1}{\sqrt{a^2 + \omega^2}} e^{-j \tan^{-1}(\omega/a)}$$

Therefore,

$$|X(\omega)| = \frac{1}{\sqrt{a^2 + \omega^2}} \quad \text{and} \quad \angle X(\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right)$$

The amplitude spectrum |X(ω)| and the phase spectrum X(ω) are depicted in Fig. 7.4b. Observe that |X(ω)| is an even function of ω, and X(ω) is an odd function of ω, as expected.

EXISTENCE OF THE FOURIER TRANSFORM

In Ex. 7.1 we observed that when a < 0, the Fourier integral for e−atu(t) does not converge. Hence, the Fourier transform for e−atu(t) does not exist if a < 0 (growing exponential). Clearly, not all signals are Fourier transformable.

Because the Fourier transform is derived here as a limiting case of the Fourier series, it follows that the basic qualifications of the Fourier series, such as equality in the mean and convergence conditions in suitably modified form, apply to the Fourier transform as well. It can be shown that if x(t) has a finite energy, that is, if

$$\int_{-\infty}^{\infty} |x(t)|^2 dt < \infty$$

then the Fourier transform X(ω) is finite and converges to x(t) in the mean. This means, if we let

$$\hat{x}(t) = \lim_{W \to \infty} \frac{1}{2\pi} \int_{-W}^{W} X(\omega) e^{j\omega t} d\omega$$

then Eq. (7.10) implies

$$\int_{-\infty}^{\infty} \left| x(t) - \hat{x}(t) \right|^2 dt = 0 \tag{7.13}$$

In other words, x(t) and its Fourier integral [the right-hand side of Eq. (7.10)] can differ at some values of t without contradicting Eq. (7.13). We shall now discuss an alternate set of criteria due to Dirichlet for convergence of the Fourier transform.

As with the Fourier series, if x(t) satisfies certain conditions (Dirichlet conditions), its Fourier transform is guaranteed to converge pointwise at all points where x(t) is continuous. Moreover, at the points of discontinuity, x(t) converges to the value midway between the two values of x(t) on either side of the discontinuity. The Dirichlet conditions are as follows:

1. x(t) should be absolutely integrable, that is,

$$\int_{-\infty}^{\infty} |x(t)| \, dt < \infty \tag{7.14}$$

If this condition is satisfied, we see that the integral on the right-hand side of Eq. (7.9) is guaranteed to have a finite value.

• 2. x(t) must have only a finite number of finite discontinuities within any finite interval.
• 3. x(t) must contain only a finite number of maxima and minima within any finite interval.

We stress here that although the Dirichlet conditions are sufficient for the existence and pointwise convergence of the Fourier transform, they are not necessary. For example, we saw in Ex. 7.1 that a growing exponential, which violates Dirichlet’s first condition in Eq. (7.14), does not have a Fourier transform. But the signal of the form (sinat)/t, which does violate this condition, does have a Fourier transform.

Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore has a Fourier transform. Thus, the physical existence of a signal is a sufficient condition for the existence of its transform.

LINEARITY OF THE FOURIER TRANSFORM

The Fourier transform is linear; that is, if

$$x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega)$$

then

$$a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(\omega) + a_2X_2(\omega) \tag{7.15}$$

The proof is trivial and follows directly from Eq. (7.9). This result can be extended to any finite number of terms. It can be extended to an infinite number of terms only if the conditions required for interchangeability of the operations of summation and integration are satisfied.

7.1-1 Physical Appreciation of the Fourier Transform

In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal in terms of everlasting sinusoids (or exponentials). The Fourier spectrum of a signal indicates the relative amplitudes and phases of sinusoids that are required to synthesize that signal. A periodic signal Fourier spectrum has finite amplitudes and exists at discrete frequencies (ω⁰ and its multiples). Such a spectrum is easy to visualize, but the spectrum of an aperiodic signal is not easy to visualize because it has a continuous spectrum. The continuous spectrum concept can be appreciated by considering an analogous, more tangible phenomenon. One familiar example of a continuous distribution is the loading of a beam. Consider a beam loaded with weights D1,D2,D3,…,Dn units at the uniformly spaced points y1, y2,…, yn, as shown in Fig. 7.5a.

The total load WT on the beam is given by the sum of these loads at each of the n points:

$$W_T = \sum_{i=1}^n D_i$$

Consider now the case of a continuously loaded beam, as depicted in Fig. 7.5b. In this case, although there appears to be a load at every point, the load at any one point is zero. This does not mean that there is no load on the beam. A meaningful measure of load in this situation is not the load at a point, but rather the loading density per unit length at that point. Let X(y) be the loading density per unit length of beam. It then follows that the load over a beam length y(y → 0), at some point y, is X(y)y. To find the total load on the beam, we divide the beam into segments of interval y(y → 0). The load over the nth such segment of length y is X(ny)y. The total load WT is given by

$$W_T = \lim_{\Delta y \to 0} \sum_{y_1}^{y_n} X(n\Delta y) \, \Delta y = \int_{y_1}^{y_n} X(y) \, dy$$

The load now exists at every point, and y is now a continuous variable. In the case of discrete loading (Fig. 7.5a), the load exists only at n discrete points. At other points, there is no load. On the other hand, in the continuously loaded case, the load exists at every point, but at any specific

Figure 7.5 Weight-loading analogy for the Fourier transform.

point y, the load is zero. The load over a small interval y, however, is [X(ny)]y (Fig. 7.5b). Thus, even though the load at a point y is zero, the relative load at that point is X(y).

An exactly analogous situation exists in the case of a signal spectrum. When x(t) is periodic, the spectrum is discrete, and x(t) can be expressed as a sum of discrete exponentials with finite amplitudes:

$$x(t) = \sum_{n} D_n e^{jn\omega_0 t}$$

For an aperiodic signal, the spectrum becomes continuous; that is, the spectrum exists for every value of ω, but the amplitude of each component in the spectrum is zero. The meaningful measure here is not the amplitude of a component of some frequency but the spectral density per unit bandwidth. From Eq. (7.7), it is clear that x(t) is synthesized by adding exponentials of the form ejnω*^t* , in which the contribution by any one exponential component is zero. But the contribution by exponentials in an infinitesimal band ω located at ω = nω is (1/2π )X(nω)ω, and the addition of all these components yields x(t) in the integral form:

$$x(t) = \lim_{\Delta\omega \to 0} \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} X(n\Delta\omega)e^{(jn\Delta\omega)t} \Delta\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{i\omega t} d\omega$$

Thus, nω approaches a continuous variable ω. The spectrum now exists at every ω. The contribution by components within a band dω is (1/2π )X(ω)dω = X(ω)df , where df is the bandwidth in hertz. Clearly, X(ω) is the spectral density per unit bandwidth (in hertz).† It also follows that even if the amplitude of any one component is infinitesimal, the relative amount of a component of frequency ω is X(ω). Although X(ω) is a spectral density, in practice, it is customarily called the spectrum of x(t) rather than the spectral density of x(t). Deferring to this convention, we shall call X(ω) the Fourier spectrum (or Fourier transform) of x(t).

A MARVELOUS BALANCING ACT

An important point to remember here is that x(t) is represented (or synthesized) by exponentials or sinusoids that are everlasting (not causal). Such conceptualization leads to a rather fascinating picture when we try to visualize the synthesis of a timelimited pulse signal x(t) [Fig. 7.6] by the sinusoidal components in its Fourier spectrum. The signal x(t) exists only over an interval (a,b) and is zero outside this interval. The spectrum of x(t) contains an infinite number of exponentials (or sinusoids), which start at t = −∞ and continue forever. The amplitudes and phases of these components add up exactly to x(t) over the finite interval (a,b) and to zero everywhere outside this interval. Juggling the amplitudes and phases of an infinite number of components to achieve

^† To stress that the signal spectrum is a density function, we shall shade the plot of ^|X(ω)^| (as in Fig. 7.4b). The representation of X(ω), however, will be a line plot, primarily to avoid visual confusion.

such a perfect and delicate balance boggles the human imagination. Yet the Fourier transform accomplishes it routinely, without much thinking on our part. Indeed, we become so involved in mathematical manipulations that we fail to notice this marvel.

7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS

For convenience, we now introduce a compact notation for the useful gate, triangle, and interpolation functions.

UNIT GATE FUNCTION

We define a unit gate function rect(x) as a gate pulse of unit height and unit width, centered at the origin, as illustrated in Fig. 7.7a† :

rect

$$(x)$$

=

$$\begin{cases} 0 & |x| > \frac{1}{2} \\ \frac{1}{2} & |x| = \frac{1}{2} \\ 1 & |x| < \frac{1}{2} \end{cases}$$

(7.16)

The gate pulse in Fig. 7.7b is the unit gate pulse rect(x) expanded by a factor τ along the horizontal axis and therefore can be expressed as rect(x/τ ) (see Sec. 1.2-2). Observe that τ , the denominator of the argument of rect (x/τ ), indicates the width of the pulse.

UNIT TRIANGLE FUNCTION

We define a unit triangle function (x) as a triangular pulse of unit height and unit width, centered at the origin, as shown in Fig. 7.8a

$$\Delta(x) = \begin{cases} 0 & |x| \ge \frac{1}{2} \\ 1 - 2|x| & |x| < \frac{1}{2} \end{cases}$$

(7.17)

Figure 7.7 A gate pulse.

^† At ^|x| = 0.5, we require rect(x) ⁼ 0.5 because the inverse Fourier transform of a discontinuous signal converges to the mean of its two values at the discontinuity.

Figure 7.8 A triangle pulse.

The pulse in Fig. 7.8b is (x/τ ). Observe that here, as for the gate pulse, the denominator τ of the argument of (x/τ ) indicates the pulse width.

INTERPOLATION FUNCTION SINC (x)

The function sinx/x is the “sine over argument” function denoted by sinc (x). † This function plays an important role in signal processing. It is also known as the filtering or interpolating function. We define

$$\operatorname{sinc}(x) = \frac{\sin x}{x} \tag{7.18}$$

Inspection of Eq. (7.18) shows the following:

• 1. sinc (x) is an even function of x.
• 2. sinc (x) = 0 when sin x = 0 except at x = 0, where it appears to be indeterminate. This means that sincx = 0 for x = ±π,±2π,±3π,…
• 3. Using L’Hôpital’s rule, we find sinc (0) = 1.
• 4. sinc (x) is the product of an oscillating signal sinx (of period 2π) and a monotonically decreasing function 1/x. Therefore, sinc (x) exhibits damped oscillations of period 2π, with amplitude decreasing continuously as 1/x.

Figure 7.9a shows sinc(x). Observe that sinc (x) = 0 for values of x that are positive and negative integer multiples of π. Figure 7.9b shows sinc (3ω/7). The argument 3ω/7 = π when ω = 7π/3. Therefore, the first zero of this function occurs at ω = 7π/3.

DR ILL 7.1 Sketching Basic Functions

Sketch: (a) rect(x/8), (b) (ω/10), (c) sinc (3πω/2), and (d) sinc (t)rect(t/4π ).

$$\operatorname{sinc}(x) = \frac{\sin \pi x}{\pi x}$$

^† sinc(x) is also denoted by Sa (x) in the literature. Some authors define sinc (x) as

Figure 7.9 A sinc pulse.

EXAMPLE 7.2 Fourier Transform of a Rectangular Pulse

Find the Fourier transform of x(t) = rect(t/τ ) (Fig. 7.10a).

$$X(\omega) = \int_{-\infty}^{\infty} \text{rect}\left(\frac{t}{\tau}\right) e^{-j\omega t} dt$$

Since rect(t/τ ) = 1 for |t| < τ/2, and since it is zero for |t| > τ/2,

$$X(\omega) = \int_{-\tau/2}^{\tau/2} e^{-j\omega t} dt$$

= $-\frac{1}{j\omega} (e^{-j\omega \tau/2} - e^{j\omega \tau/2}) = \frac{2 \sin(\frac{\omega \tau}{2})}{\omega}$
= $\tau \frac{\sin(\frac{\omega \tau}{2})}{(\frac{\omega \tau}{2})} = \tau \operatorname{sinc}(\frac{\omega \tau}{2})$

Figure 7.10 (a) A gate pulse x(t), (b) its Fourier spectrum X(ω), (c) its amplitude spectrum |X(ω)|, and (d) its phase spectrum X(ω).

Therefore,

$$\text{rect}\left(\frac{t}{\tau}\right) \Longleftrightarrow \tau \text{ sinc}\left(\frac{\omega\tau}{2}\right) \tag{7.19}$$

Recall that sinc(x) = 0 when x = ±nπ. Hence, sinc(ωτ /2) = 0 when ωτ/2 = ±nπ; that is, when ω = ±2nπ/τ ,(n = 1, 2, 3,…), as depicted in Fig. 7.10b. The Fourier transform X(ω) shown in Fig. 7.10b exhibits positive and negative values. A negative amplitude can be considered to be a positive amplitude with a phase of −π or π. We use this observation to plot the amplitude spectrum |X(ω)|=|sinc (ωτ /2)| (Fig. 7.10c) and the phase spectrum X(ω) (Fig. 7.10d). The phase spectrum, which is required to be an odd function of ω, may be drawn in several other ways because a negative sign can be accounted for by a phase of ±nπ, where n is any odd integer. All such representations are equivalent.

BANDWIDTH OF RECT t τ

The spectrum X(ω) in Fig. 7.10 peaks at ω = 0 and decays at higher frequencies. Therefore, rect(t/τ ) is a lowpass signal with most of the signal energy in lower-frequency components. Strictly speaking, because the spectrum extends from 0 to ∞, the bandwidth is ∞. However, much of the spectrum is concentrated within the first lobe (from ω = 0 to ω = 2π/τ ). Therefore, a rough estimate of the bandwidth of a rectangular pulse of width τ seconds is 2π/τ rad/s, or 1/τ Hz.† Note the reciprocal relationship of the pulse width with its bandwidth. We shall observe later that this result is true, in general.

^† To compute bandwidth, we must consider the spectrum for positive values of ω only. See the discussion in Sec. 6.3.

Find the Fourier transform of the unit impulse δ(t).

Using the sampling property of the impulse [Eq. (1.11)], we obtain

$$\mathcal{F}[\delta(t)] = \int_{-\infty}^{\infty} \delta(t)e^{-j\omega t}dt = 1 \quad \text{and} \quad \delta(t) \Longleftrightarrow 1$$

Figure 7.11 shows δ(t) and its spectrum.

EXAMPLE 7.4 Inverse Fourier Transform of the Dirac Delta Function

Find the inverse Fourier transform of δ(ω).

On the basis of Eq. (7.10) and the sampling property of the impulse function,

$$\mathcal{F}^{-1}[\delta(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega) e^{j\omega t} d\omega = \frac{1}{2\pi}$$

Therefore,

$$\frac{1}{2\pi} \Longleftrightarrow \delta(\omega) \quad \text{and} \quad 1 \Longleftrightarrow 2\pi \delta(\omega) \tag{7.20}$$

This result shows that the spectrum of a constant signal x(t) = 1 is an impulse 2πδ(ω), as illustrated in Fig. 7.12.

The result [Eq. (7.20)] could have been anticipated on qualitative grounds. Recall that the Fourier transform of x(t) is a spectral representation of x(t) in terms of everlasting exponential components of the form ejω*^t* . Now, to represent a constant signal x(t) = 1, we need a single

Figure 7.12 (a) A constant (dc) signal and (b) its Fourier spectrum.

everlasting exponential ejω*^t* with ω = 0.† This results in a spectrum at a single frequency ω = 0. Another way of looking at the situation is that x(t) = 1 is a dc signal that has a single frequency ω = 0 (dc).

If an impulse at ω = 0 is a spectrum of a dc signal, what does an impulse at ω = ω⁰ represent? We shall answer this question in the next example.

EXAMPLE 7.5 Inverse Fourier Transform of a Shifted Dirac Delta Function

Find the inverse Fourier transform of δ(ω −ω0).

Using the sampling property of the impulse function, we obtain

$$\mathcal{F}^{-1}[\delta(\omega - \omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega = \frac{1}{2\pi} e^{j\omega_0 t}$$

Therefore,

$$\frac{1}{2\pi}e^{j\omega_0 t} \Longleftrightarrow \delta(\omega - \omega_0) \quad \text{and} \quad e^{j\omega_0 t} \Longleftrightarrow 2\pi\delta(\omega - \omega_0) \tag{7.21}$$

This result shows that the spectrum of an everlasting exponential ej^ω0*^t* is a single impulse at ω = ω0. We reach the same conclusion by qualitative reasoning. To represent the everlasting

^† The constant multiplier 2^π in the spectrum [X(ω) ^{= 2}πδ(ω)] may be a bit puzzling. Since 1 ^{= e}jω*^t* with ω = 0, it appears that the Fourier transform of x(t) = 1 should be an impulse of strength unity rather than 2π. Recall, however, that in the Fourier transform x(t) is synthesized by exponentials not of amplitude X(nω)ω but of amplitude 1/2π times X(nω)ω, as seen from Eq. (7.7). Had we used variable f (hertz) instead of ω, the spectrum would have been the unit impulse.

exponential ej^ω0*^t* , we need a single everlasting exponential ejω*^t* with ω = ω0. Therefore, the spectrum consists of a single component at frequency ω = ω0. From Eq. (7.21) it follows that

$$e^{-j\omega_0 t} \Longleftrightarrow 2\pi \delta(\omega + \omega_0)$$

EXAMPLE 7.6 Fourier Transform of a Sinusoid Find the Fourier transform of the everlasting sinusoid cos ω0t (Fig. 7.13a). v⁰ 0 v⁰ x(t) cos v0t X(v) p ^{t v 0} (a) (b) Figure 7.13 (a) A cosine signal and (b) its Fourier spectrum.

Recall Euler’s formula

$$\cos \omega_0 t = \frac{1}{2} (e^{j\omega_0 t} + e^{-j\omega_0 t})$$

Applying Eq. (7.21), we obtain

$$\cos \omega_0 t \Longleftrightarrow \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)]$$

The spectrum of cos ω0t consists of two impulses at ω⁰ and −ω0, as shown in Fig. 7.13b. The result also follows from qualitative reasoning. An everlasting sinusoid cos ω0t can be synthesized by two everlasting exponentials, ej^ω0*^t* and e−jω0*^t* . Therefore, the Fourier spectrum consists of only two components of frequencies ω⁰ and −ω0.

EXAMPLE 7.7 Fourier Transform of a Periodic Signal

Determine the Fourier transform of a periodic signal x(t) using its Fourier series representation.

We can use a Fourier series to express a periodic signal as a sum of exponentials of the form ejn^ω0*^t* , whose Fourier transform is found in Eq. (7.21). Hence, we can readily find the Fourier transform of a periodic signal by using the linearity property in Eq. (7.15).