[6.5 GENERALIZED](#page-12-0) FOURIER SERIES: SIGNALS AS VECTORS

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6.5 GENERALIZED FOURIER SERIES: SIGNALS AS VECTORS

We now consider a very general approach to signal representation with far-reaching consequences.† There is a perfect analogy between signals and vectors; the analogy is so strong that the term analogy understates the reality. Signals are not just like vectors. Signals are vectors! A vector can be represented as a sum of its components in a variety of ways, depending on the choice of coordinate system. A signal can also be represented as a sum of its components in a variety of ways. Let us begin with some basic vector concepts and then apply these concepts to signals.

^† This section closely follows the material from the author’s earlier book [10]. Omission of this section will not cause any discontinuity in understanding the rest of the book. Derivation of Fourier series through the signal-vector analogy provides an interesting insight into signal representation and other topics such as signal correlation, data truncation, and signal detection.

6.5-1 Component of a Vector

A vector is specified by its magnitude and its direction. We shall denote all vectors by boldface. For example, x is a certain vector with magnitude or length |x|. For the two vectors x and y shown in Fig. 6.21, we define their dot (inner or scalar) product as

$$\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}| |\mathbf{y}| \cos \theta$$

where θ is the angle between these vectors. Using this definition, we can express |x|, the length of a vector x, as

$$|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x}$$

Let the component of x along y be cy as depicted in Fig. 6.21. Geometrically, the component of x along y is the projection of x on y and is obtained by drawing a perpendicular from the tip of x on the vector y, as illustrated in Fig. 6.21. What is the mathematical significance of a component of a vector along another vector? As seen from Fig. 6.21, the vector x can be expressed in terms of vector y as

$$\mathbf{x} = c\mathbf{y} + \mathbf{e}$$

However, this is not the only way to express x in terms of y. From Fig. 6.22, which shows two of the infinite other possibilities, we have

$$\mathbf{x} = c_1 \mathbf{y} + \mathbf{e}_1 = c_2 \mathbf{y} + \mathbf{e}_2$$

In each of these three representations, x is represented in terms of y plus another vector called the error vector. If we approximate x by cy,

$$\mathbf{x} \simeq c\mathbf{y}$$

the error in the approximation is the vector e = x − cy. Similarly, the errors in approximations in these drawings are e¹ (Fig. 6.22a) and e² (Fig. 6.22b). What is unique about the approximation in Fig. 6.21 is that the error vector is the smallest. We can now define mathematically the component of a vector x along vector y to be cy where c is chosen to minimize the length of the error vector e = x − cy. Now, the length of the component of x along y is |x| cos θ. But it is also c|y|, as seen from Fig. 6.21. Therefore,

$$c|\mathbf{y}| = |\mathbf{x}| \cos \theta$$

Multiplying both sides by |y| yields

$$c|\mathbf{y}|^2 = |\mathbf{x}||\mathbf{y}|\cos\theta = \mathbf{x}\cdot\mathbf{y}$$

Figure 6.21 Component (projection) of a vector along another vector.

Figure 6.22 Approximation of a vector in terms of another vector.

Therefore,

$$c = \frac{\mathbf{x} \cdot \mathbf{y}}{\mathbf{y} \cdot \mathbf{y}} = \frac{1}{|\mathbf{y}|^2} \mathbf{x} \cdot \mathbf{y}$$

(6.32)

From Fig. 6.21, it is apparent that when x and y are perpendicular, or orthogonal, then x has a zero component along y; consequently, c = 0. Keeping an eye on Eq. (6.32), we therefore define x and y to be orthogonal if the inner (scalar or dot) product of the two vectors is zero, that is, if

x · y = 0

6.5-2 Signal Comparison and Component of a Signal

The concept of a vector component and orthogonality can be extended to signals. Consider the problem of approximating a real signal x(t) in terms of another real signal y(t) over an interval (t1, t2):

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

The error e(t) in this approximation is

$$e(t) = \begin{cases} x(t) - cy(t) & t_1 < t < t_2 \\ 0 & \text{otherwise} \end{cases}$$

We now select a criterion for the “best approximation.” We know that the signal energy is one possible measure of a signal size. For best approximation, we shall use the criterion that minimizes the size or energy of the error signal e(t) over the interval (t1,t2). This energy Ee is given by

$$E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt$$

Note that the right-hand side is a definite integral with t as the dummy variable. Hence, Ee is a function of the parameter c (not t) and Ee is minimum for some choice of c. To minimize Ee, a necessary condition is

$$\frac{dE_e}{dc} = 0$$ $$\frac{d}{dc} \left[ \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt \right] = 0$$

d

or

644 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

Expanding the squared term inside the integral, we obtain

$$\frac{d}{dc} \left[ \int_{t_1}^{t_2} x^2(t) dt \right] - \frac{d}{dc} \left[ 2c \int_{t_1}^{t_2} x(t) y(t) dt \right] + \frac{d}{dc} \left[ c^2 \int_{t_1}^{t_2} y^2(t) dt \right] = 0$$

from which we get

$$-2\int_{t_1}^{t_2} x(t)y(t) dt + 2c \int_{t_1}^{t_2} y^2(t) dt = 0$$ $$c = \frac{\int_{t_1}^{t_2} x(t)y(t) dt}{\int_{t_1}^{t_2} y^2(t) dt} = \frac{1}{E_y} \int_{t_1}^{t_2} x(t)y(t) dt$$

(6.33)

We observe a remarkable similarity between the behavior of vectors and signals, as indicated by Eqs. (6.32) and (6.33). It is evident from these two parallel expressions that the area under the product of two signals corresponds to the inner (scalar or dot) product of two vectors. In fact, the area under the product of x(t) and y(t) is called the inner product of x(t) and y(t), and is denoted by (x, y). The energy of a signal is the inner product of a signal with itself, and corresponds to the vector length square (which is the inner product of the vector with itself).

To summarize our discussion, if a signal x(t) is approximated by another signal y(t) as

x(t) cy(t)

then the optimum value of c that minimizes the energy of the error signal in this approximation is given by Eq. (6.33).

Taking our clue from vectors, we say that a signal x(t) contains a component cy(t), where c is given by Eq. (6.33). Note that in vector terminology, cy(t) is the projection of x(t) on y(t). Continuing with the analogy, we say that if the component of a signal x(t) of the form y(t) is zero (i.e., c = 0), the signals x(t) and y(t) are orthogonal over the interval (t1, t2). Therefore, we define the real signals x(t) and y(t) to be orthogonal over the interval (t1, t2) if†

$$\int_{t_1}^{t_2} x(t)y(t) dt = 0$$

\n(6.34)

EXAMPLE 6.13 Sine-Wave Approximation of a Square Wave

For the square signal x(t) shown in Fig. 6.23, find the component in x(t) of the form sint. In other words, approximate x(t) in terms of sin t

x(t) csint 0 < t < 2π

so that the energy of the error signal is minimum.

^† For complex signals, the definition is modified as in Eq. (6.37), in Sec. 6.5-3.

Figure 6.23 Approximation of a square wave in terms of a single sinusoid.

In this case,

$$y(t) = \sin t$$

and $E_y = \int_0^{2\pi} \sin^2(t) dt = \pi$

From Eq. (6.33), we find

$$c = \frac{1}{\pi} \int_0^{2\pi} x(t) \sin t \, dt = \frac{1}{\pi} \left[ \int_0^{\pi} \sin t \, dt + \int_{\pi}^{2\pi} -\sin t \, dt \right] = \frac{4}{\pi}$$

Thus,

$$x(t) \simeq \frac{4}{\pi} \sin t$$

represents the best approximation of x(t) by the function sint, which will minimize the error energy. This sinusoidal component of x(t) is shaded in Fig. 6.23. By analogy with vectors, we say that the square function x(t) depicted in Fig. 6.23 has a component of signal sint and that the magnitude of this component is 4/π.

DR ILL 6.7 Sine Wave Approximation of a Ramp Function

Show that over an interval (−π < t < π), the “best” approximation of the signal x(t) = t in terms of the function sint is 2 sint. Verify that the error signal e(t) = t − 2 sint is orthogonal to the signal sint over the interval −π < t < π. Sketch the signals t and 2 sint over the interval −π < t < π.

6.5-3 Extension to Complex Signals

So far we have restricted ourselves to real functions of t. To generalize the results to complex functions of t, consider again the problem of approximating a signal x(t) by a signal y(t) over an

646 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

interval (t¹ < t < t2):

$$x(t) \simeq cy(t)$$

where x(t) and y(t) now can be complex functions of t. Recall that the energy Ey of the complex signal y(t) over an interval (t1, t2) is

$$E_{y} = \int_{t_1}^{t_2} |y(t)|^2 dt$$

In this case, both the coefficient c and the error

$$e(t) = x(t) - cy(t)$$

are complex (in general). For the “best” approximation, we choose c to minimize the energy Ee of the error signal e(t). Now,

$$E_e = \int_{t_1}^{t_2} |x(t) - cy(t)|^2 dt$$

\n(6.35)

Recall also that

$$|u + v|^2 = (u + v)(u^* + v^*) = |u|^2 + |v|^2 + u^*v + uv^* \tag{6.36}$$

After some manipulation, we can use this result to rearrange Eq. (6.35) as

$$E_e = \int_{t_1}^{t_2} |x(t)|^2 dt - \left| \frac{1}{\sqrt{E_y}} \int_a^{t_2} x(t) y^*(t) dt \right|^2 + \left| c \sqrt{E_y} - \frac{1}{\sqrt{E_y}} \int_{t_1}^{t_2} x(t) y^*(t) dt \right|^2$$

Since the first two terms on the right-hand side are independent of c, it is clear that Ee is minimized by choosing c so that the third term on the right-hand side is zero. This yields

$$c = \frac{1}{E_y} \int_{t_1}^{t_2} x(t) y^*(t) dt$$

In light of this result, we need to redefine orthogonality for the complex case as follows: two complex functions x1(t) and x2(t) are orthogonal over an interval (t¹ < t < t2) if

$$\int_{t_1}^{t_2} x_1(t) x_2^*(t) dt = 0 \qquad \text{or} \qquad \int_{t_1}^{t_2} x_1^*(t) x_2(t) dt = 0 \tag{6.37}$$

Either equality suffices. This is a general definition of orthogonality, which reduces to Eq. (6.34) when the functions are real.

DR ILL 6.8 Complex Exponential Approximation of a Square Wave

Show that over an interval (0 < t < 2π ), the “best” approximation of the square signal x(t) in Fig. 6.23 in terms of the signal ejt is given by (2/jπ ) ejt. Verify that the error signal e(t) = x(t)−(2/jπ )ejt is orthogonal to the signal ejt.

ENERGY OF THE SUM OF ORTHOGONAL SIGNALS

We know that the square of the length of a sum of two orthogonal vectors is equal to the sum of the squares of the lengths of the two vectors. Thus, if vectors x and y are orthogonal, and if z = x+y, then

$$|\mathbf{z}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2$$

We have a similar result for signals. The energy of the sum of two orthogonal signals is equal to the sum of the energies of the two signals. Thus, if signals x(t) and y(t) are orthogonal over an interval (t1, t2), and if z(t) = x(t)+y(t), then

$$E_z = E_x + E_y$$

We now prove this result for complex signals, of which real signals are a special case. From Eq. (6.36), it follows that

$$\int_{t_1}^{t_2} |x(t) + y(t)|^2 dt = \int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt + \int_{t_1}^{t_2} x(t)y^*(t) dt + \int_{t_1}^{t_2} x^*(t)y(t) dt$$

=

$$\int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt$$

The last result follows from the fact that because of orthogonality, the two integrals of the products x(t)y∗(t) and x∗(t)y(t) are zero [see Eq. (6.37)]. This result can be extended to the sum of any number of mutually orthogonal signals.

6.5-4 Signal Representation by an Orthogonal Signal Set

In this section we show a way of representing a signal as a sum of orthogonal signals. Here again we can benefit from the insight gained from a similar problem in vectors. We know that a vector can be represented as a sum of orthogonal vectors, which form the coordinate system of a vector space. The problem in signals is analogous, and the results for signals are parallel to those for vectors. So, let us review the case of vector representation.

ORTHOGONAL VECTOR SPACE

Let us investigate a three-dimensional Cartesian vector space described by three mutually orthogonal vectors x1, x2, and x3, as illustrated in Fig. 6.24. First, we shall seek to approximate a three-dimensional vector x in terms of two mutually orthogonal vectors x¹ and x2:

$$\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2$$

The error e in this approximation is

$$\mathbf{e} = \mathbf{x} - (c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2)$$

or

$$\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + \mathbf{e}$$

Figure 6.24 Representation of a vector in three-dimensional space.

As in the earlier geometrical argument, we see from Fig. 6.24 that the length of e is minimum when e is perpendicular to the x1–x² plane, and c1x¹ and c2x² are the projections (components) of x on x¹ and x2, respectively. Therefore, the constants c¹ and c² are given by Eq. (6.32). Observe that the error vector is orthogonal to both the vectors x¹ and x2.

Now, let us determine the “best” approximation to x in terms of all three mutually orthogonal vectors x1, x2, and x3:

$$\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 \tag{6.38}$$

Figure 6.24 shows that a unique choice of c1, c2, and c³ exists, for which Eq. (6.38) is no longer an approximation but an equality

$$\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3$$

In this case, c1x1, c2x2, and c3x³ are the projections (components) of x on x1,x2, and x3, respectively; that is,

$$c_i = \frac{\mathbf{x} \cdot \mathbf{x}_i}{\mathbf{x}_i \cdot \mathbf{x}_i} = \frac{1}{|\mathbf{x}_i|^2} \mathbf{x} \cdot \mathbf{x}_i \qquad i = 1, 2, 3$$

(6.39)

Note that the error in the approximation is zero when x is approximated in terms of three mutually orthogonal vectors: x1, x2, and x3. The reason is that x is a three-dimensional vector, and the vectors x1, x2, and x³ represent a complete set of orthogonal vectors in three-dimensional space. Completeness here means that it is impossible to find another vector x⁴ in this space, which is orthogonal to all three vectors, x1,x2, and x3. Any vector in this space can then be represented (with zero error) in terms of these three vectors. Such vectors are known as basis vectors. If a set of vectors {xi} is not complete, the error in the approximation will generally not be zero. Thus, in the three-dimensional case discussed earlier, it is generally not possible to represent a vector x in terms of only two basis vectors without an error.

The choice of basis vectors is not unique. In fact, a set of basis vectors corresponds to a particular choice of coordinate system. Thus, a three-dimensional vector x may be represented in many different ways, depending on the coordinate system used.

ORTHOGONAL SIGNAL SPACE

We start with real signals and then extend the discussion to complex signals. We proceed with our signal approximation problem, using clues and insights developed for vector approximation. As before, we define orthogonality of a real signal set x1(t), x2(t), …, xN(t) over interval (t1,t2) as

$$\int_{t_1}^{t_2} x_m(t) x_n(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases}$$

(6.40)

If the energies En = 1 for all n, then the set is normalized and is called an orthonormal set. An orthogonal set can always be normalized by dividing xn(t) by ^√En for all ⁿ.

Now, consider approximating a signal x(t) over the interval (t1, t2) by a set of N real, mutually orthogonal signals x1(t), x2(t),…, xN(t) as

$$x(t) \simeq c_1 x_1(t) + c_2 x_2(t) + \dots + c_N x_N(t) \simeq \sum_{n=1}^N c_n x_n(t)$$

(6.41)

In the approximation of Eq. (6.41), the error e(t) is

$$e(t) = x(t) - \sum_{n=1}^{N} c_n x_n(t)$$

and Ee, the error signal energy, is

$$E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt$$

(6.42)

According to our criterion for best approximation, we select the values of ci that minimize Ee. Hence, the necessary condition is ∂Ee/dci = 0 for i = 1, 2,…,N, that is,

$$\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^N c_n x_n(t) \right]^2 dt = 0$$

When we expand the integrand, we find that all the cross-multiplication terms arising from the orthogonal signals are zero by virtue of orthogonality: that is, all terms of the form $ xm(t)xn(t)dt with m = n vanish. Similarly, the derivative with respect to ci of all terms that do not contain ci is zero. For each i, this leaves only two nonzero terms:

$$\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ -2c_i x(t) x_i(t) + c_i^2 x_i^2(t) \right] dt = 0$$

or

$$-2\int_{t_1}^{t_2} x(t)x_i(t) dt + 2c_i \int_{t_1}^{t_2} x_i^2(t) dt = 0 \qquad i = 1, 2, \dots, N$$

Therefore,

$$c_i = \frac{\int_{t_1}^{t_2} x(t) x_i(t) dt}{\int_{t_1}^{t_2} x_i^2(t) dt} = \frac{1}{E_i} \int_{t_1}^{t_2} x(t) x_i(t) dt \qquad i = 1, 2, ..., N$$

(6.43)

A comparison of Eq. (6.43) with Eq. (6.39) forcefully brings out the analogy of signals with vectors.

Finality Property. Equation (6.43) shows one interesting property of the coefficients of c1, c2, …, cN: the optimum value of any coefficient in Eq. (6.41) is independent of the number of terms used in the approximation. For example, if we used only one term (N = 1) or two terms (N = 2) or any number of terms, the optimum value of the coefficient c¹ would be the same [as given by Eq. (6.43)]. The advantage of this approximation of a signal x(t) by a set of mutually orthogonal signals is that we can continue to add terms to the approximation without disturbing the previous terms. This property of finality of the values of the coefficients is very important from a practical point of view.†

ENERGY OF THE ERROR SIGNAL

When the coefficients ci in Eq. (6.41) are chosen according to Eq. (6.43), the error signal energy is minimized. This minimum value of Ee is given by Eq. (6.42):

$$E_e = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt$$

=

$$\int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 \int_{t_1}^{t_2} x_n^2(t) dt - 2 \sum_{n=1}^{N} c_n \int_{t_1}^{t_2} x(t) x_n(t) dt$$ $$x(t_1) = a_0 + a_1t_1$$

and $x(t_2) = a_0 + a_1t_2$

Solution of these equations yields the desired values of a⁰ and a1. For a three-point approximation, we must choose the polynomial a⁰ +a1t +a2t ² with

$$x(t_i) = a_0 + a_1t_i + a_2t_i^2$$

$i = 1, 2, and 3$

The approximation improves with a larger number of points (higher-order polynomial), but the coefficients a0, a1, a2, … do not have the finality property. Every time we increase the number of terms in the polynomial, we need to recalculate the coefficients.

^† Contrast this situation with a polynomial approximation of x(t). Suppose we wish to find a two-point approximation of x(t) by a polynomial in t; that is, the polynomial is to be equal to x(t) at two points t¹ and t2. This can be done by choosing a first-order polynomial a⁰ +a1t with

Substitution of Eqs. (6.40) and (6.43) in this equation yields

$$E_e = \int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 E_n - 2 \sum_{n=1}^{N} c_n^2 E_n = \int_{t_1}^{t_2} x^2(t) dt - \sum_{n=1}^{N} c_n^2 E_n$$

(6.44)

Observe that because the term c^{2 k}Ek is nonnegative, the error energy Ee generally decreases as N, the number of terms, is increased. Hence, it is possible that the error energy →0 as N → ∞. When this happens, the orthogonal signal set is said to be complete. In this case, Eq. (6.41) is no more an approximation but an equality

$$x(t) = c_1 x_1(t) + c_2 x_2(t) + \dots + c_n x_n(t) + \dots = \sum_{n=1}^{\infty} c_n x_n(t) \qquad t_1 < t < t_2 \tag{6.45}$$

where the coefficients cn are given by Eq. (6.43). Because the error signal energy approaches zero, it follows that the energy of x(t) is now equal to the sum of the energies of its orthogonal components c1x1(t), c2x2(t), c3x3(t), …

The series on the right-hand side of Eq. (6.45) is called the generalized Fourier series of x(t) with respect to the set {xn(t)}. When the set {xn(t)} is such that the error energy Ee → 0 as N → ∞ for every member of some particular class, we say that the set {xn(t)} is complete on (t1, t2) for that class of x(t), and the set {xn(t)} is called a set of basis functions or basis signals. Unless otherwise mentioned, in the future we shall consider only the class of energy signals.

Thus, when the set {xn(t)} is complete, we have the equality of Eq. (6.45). One subtle point that must be understood clearly is the meaning of equality in Eq. (6.45). The equality here is not an equality in the ordinary sense, but in the sense that the error energy, that is, the energy of the difference between the two sides of Eq. (6.45), approaches zero. If the equality exists in the ordinary sense, the error energy is always zero, but the converse is not necessarily true. The error energy can approach zero even though e(t), the difference between the two sides, is nonzero at some isolated instants. The reason is that even if e(t) is nonzero at such instants, the area under e²(t) is still zero; thus the Fourier series on the right-hand side of Eq. (6.45) may differ from x(t) at a finite number of points.

In Eq. (6.45), the energy of the left-hand side is Ex, and the energy of the right-hand side is the sum of the energies of all the orthogonal components.† Thus,

$$\int_{t_1}^{t_2} x^2(t) dt = c_1^2 E_1 + c_2^2 E_2 + \dots = \sum_{n=1}^{\infty} c_n^2 E_n$$

(6.46)

This is Parseval’s theorem expressed for energy signals. In Eqs. (6.26) and (6.27), we have already encountered Parseval’s theorem for power signals. Recall that the signal energy (area under the squared value of a signal) is analogous to the square of the length of a vector in the vector-signal analogy. In vector space, we know that the square of the length of a vector is equal to the sum of

^† Note that the energy of a signal cx(t) is c²Ex.

652 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES

the squares of the lengths of its orthogonal components. Parseval’s theorem of Eq. (6.46) is the statement of this fact as it applies to signals.

GENERALIZATION TO COMPLEX SIGNALS

The foregoing results can be generalized to complex signals as follows: a set of functions x1(t), x2(t), …, xN(t) is mutually orthogonal over the interval (t1, t2) if

$$\int_{t_1}^{t_2} x_m(t) x_n^*(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases}$$

If this set is complete for a certain class of functions, then a function x(t) in this class can be expressed as

$$x(t) = c_1 x_1(t) + c_2 x_2(t) + \cdots + c_i x_i(t) + \cdots$$

where

$$c_n = \frac{1}{E_n} \int_{t_1}^{t_2} x(t) x_n^*(t) dt$$

\n(6.47)

EXAMPLE 6.14 Approximating a Square Wave with a Set of Harmonic Sine Waves

In Ex. 6.13, the square signal x(t) in Fig. 6.23 is approximated by a single sinusoid sin t. In this example, we approximate x(t) using the set of harmonic sine waves sin t, sin 2t, …, sin nt, …, and see how the approximation improves with the number of terms.

To begin, we note that the set of harmonic sine waves sin t, sin 2t,…, sin nt,… is orthogonal over any interval of duration 2π. † The reader can verify this fact by showing that for any real number a,

$$\int_{a}^{a+2\pi} \sin mt \sin nt dt = \begin{cases} 0 & m \neq n \\ \pi & m = n \end{cases}$$

(6.48)

Using this set, we approximate x(t) as

x(t) c¹ sin t +c² sin 2t +···+cn sin Nt

^† This sine set, along with the cosine set cos 0t, cos t, cos 2t,…, cosnt,…, forms a complete set. In this case, however, the coefficients ci corresponding to the cosine terms are zero. For this reason, we have omitted cosine terms in this example. This composite sine and cosine set is the basis set for the trigonometric Fourier series.

Figure 6.25 Approximation of a square wave by a sum of harmonic sinusoids.

Therefore,

$$x(t) \simeq \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \dots + \frac{1}{N} \sin Nt \right)$$

(6.49)

Note that coefficients of terms sinkt are zero for even values of k. Figure 6.25 shows how the approximation improves as we increase the number of terms in the series.

Let us investigate the error signal energy as N → ∞. From Eq. (6.44),

$$E_e = \int_0^{2\pi} x^2(t) dt - \sum_{n=1}^{\infty} c_n^2 E_n$$

Note that

$$\int_0^{2\pi} x^2(t) dt = \int_0^{\pi} 1^2 dt + \int_{\pi}^{2\pi} -1^2 dt = 2\pi$$ $$c_n^2 = \begin{cases} \frac{16}{n^2 \pi^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$$

and from Eq. (6.48),

$$E_n=\pi$$

Therefore,

$$E_e = 2\pi - \sum_{n=1,3,5,\dots}^{N} \frac{16}{n^2 \pi^2} \pi = 2\pi - \frac{16}{\pi} \sum_{n=1,3,5,\dots}^{N} \frac{1}{n^2}$$

For a single-term approximation (N = 1),

$$E_e = 2\pi - \frac{16}{\pi} = 1.1938$$

For a two-term approximation (N = 3),

$$E_e = 2\pi - \frac{16}{\pi} \left( 1 + \frac{1}{9} \right) = 0.6243$$

Continuing this process, we compute the error energy Ee for various values of N as

N	1	3	5	7	99	$\infty$
$E_e$	1.1938	0.6243	0.4206	0.3166	0.02545	0

Clearly, x(t) can be represented by the infinite series

$$x(t) = \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \cdots \right) = \frac{4}{\pi} \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n} \sin nt$$

The equality exists in the sense that the error signal energy → 0 as N → ∞. In this case, the error energy decreases rather slowly with N, indicating that the series converges slowly. This is to be expected because x(t) has jump discontinuities and consequently, according to discussion in Sec. 6.2-2, the series converges asymptotically as 1/n.

DR ILL 6.9 Approximating a Ramp Signal with a Set of Harmonic Sine Waves

Approximate the signal x(t) = t − π (Fig. 6.26) over the interval (0, 2π ) in terms of the set of sinusoids {sin nt}, n = 0, 1, 2,…, used in Ex. 6.14. Find Ee, the error energy. Show that Ee → 0 as N → ∞.

4π

ANSWERS

%^N

1

x(t) −2 n=1 ^N sin nt and Ee ^{= 2} 3π^{3 −}%^N n=1 n2 x(t) p p p 2p t Figure 6.26 Ramp signal for Drill 6.9.

SOME EXAMPLES OF GENERALIZED FOURIER SERIES

Signals are vectors in every sense. Like a vector, a signal can be represented as a sum of its components in a variety of ways. Just as vector coordinate systems are formed by mutually orthogonal vectors (rectangular, cylindrical, spherical), we also have signal coordinate systems (basis signals) formed by a variety of sets of mutually orthogonal signals. There exist a large number of orthogonal signal sets that can be used as basis signals for generalized Fourier series. Some well-known signal sets are trigonometric (sinusoid) functions, exponential functions, Walsh functions, Bessel functions, Legendre polynomials, Laguerre functions, Jacobi polynomials, Hermite polynomials, and Chebyshev polynomials. The functions that concern us most in this book are the trigonometric and the exponential sets discussed earlier in this chapter.

LEGENDRE FOURIER SERIES

A set of Legendre polynomials Pn(t) (n = 0, 1, 2, 3,…) forms a complete set of mutually orthogonal functions over an interval (−1 < t < 1). These polynomials can be defined by the Rodrigues formula:

$$P_n(t) = \frac{1}{2^n n!} \frac{d^n}{dt^n} (t^2 - 1)^n \qquad n = 0, 1, 2, \dots$$

It follows from this equation that

$$P_0(t) = 1, P_1(t) = t, P_2(t) = \left(\frac{3}{2}t^2 - \frac{1}{2}\right), P_3(t) = \left(\frac{5}{2}t^3 - \frac{3}{2}t\right)$$

, and so on

We may verify the orthogonality of these polynomials by showing that

$$\int_{-1}^{1} P_m(t) P_n(t) dt = \begin{cases} 0 & m \neq n \\ \frac{2}{2m+1} & m = n \end{cases}$$

We can express a function x(t) in terms of Legendre polynomials over an interval (−1 < t < 1) as

$$x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots$$

\n(6.50)

where

$$c_r = \frac{\int_{-1}^{1} x(t)P_r(t) dt}{\int_{-1}^{1} P_r^2(t) dt} = \frac{2r+1}{2} \int_{-1}^{1} x(t)P_r(t) dt$$

\n(6.51)

Note that although the series representation is valid over the interval (−1, 1), it can be extended to any interval by the appropriate time scaling (see Prob. 6.5-8).

EXAMPLE 6.15 Legendre Fourier Series

Determine the Legendre Fourier series of the square signal shown in Fig. 6.27.

Figure 6.27 Square signal for Ex. 6.15.

From Eq. (6.50), we know that the Legendre Fourier series takes the form

$$x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots$$

The coefficients c0, c1, c2,…, cr may be found from Eq. (6.51). We have

$$x(t) = \begin{cases} 1 & \cdots -1 < t < 0 \\ -1 & \cdots 0 < t < 1 \end{cases}$$

and

$$c_0 = \frac{1}{2} \int_{-1}^{1} x(t) dt = 0$$

\n

$$c_1 = \frac{3}{2} \int_{-1}^{1} tx(t) dt = \frac{3}{2} \left( \int_{-1}^{0} t dt - \int_{0}^{1} t dt \right) = -\frac{3}{2}$$

\n

$$c_2 = \frac{5}{2} \int_{-1}^{1} x(t) \left( \frac{3}{2} t^2 - \frac{1}{2} \right) dt = 0$$

This result follows immediately from the fact that the integrand is an odd function of t. In fact, this is true of all cr for even values of r, that is,

$$c_0 = c_2 = c_4 = c_6 = \cdots = 0$$

Also,

$$c_3 = \frac{7}{2} \int_{-1}^{1} x(t) \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt = \frac{7}{2} \left[ \int_{-1}^{0} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt - \int_{0}^{1} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt \right] = \frac{7}{8}$$

In a similar way, coefficients c5, c7,… can be evaluated. We now have

$$x(t) = -\frac{3}{2}t + \frac{7}{8}\left(\frac{5}{2}t^3 - \frac{3}{2}t\right) + \cdots$$

TRIGONOMETRIC FOURIER SERIES

We have already proved [see Eqs. (6.4), (6.5), and (6.6)] that the trigonometric signal set

{1, cosω0t, cos 2ω0t, …, cosnω0t, …; sinω0t, sin 2ω0t, …, sinnω0t, …}

is orthogonal over any interval of duration T0, where T⁰ = 1/f⁰ is the period of the sinusoid of frequency f0. This is a complete set for a class of signals with finite energies [11, 12]. Therefore, we can express a signal x(t) by a trigonometric Fourier series over any interval of duration T⁰ seconds as

$$x(t) = a_0 + a_1 \cos \omega_0 t + a_2 \cos 2\omega_0 t + \cdots$$ $$+ b_1 \sin \omega_0 t + b_2 \sin 2\omega_0 t + \cdots$$

or

$$x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \qquad t_1 < t < t_1 + T_0$$

where

$$\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0}$$

We can use Eq. (6.43) to determine the Fourier coefficients a0, an, and bn. Thus,

$$a_n = \frac{\int_{t_1}^{t_1 + T_0} x(t) \cos n\omega_0 t \, dt}{\int_{t_1}^{t_1 + T_0} \cos^2 n\omega_0 t \, dt}$$

\n(6.52)

The integral in the denominator of Eq. (6.52) has already been found to be T0/2 when n = 0 [Eq. (6.4) with m = n]. For n = 0, the denominator is T0. Hence,

$$a_0 = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) dt \quad \text{and} \quad a_n = \frac{2}{T_0} \int_{t_1}^{t_1+T_0} x(t) \cos n\omega_0 t dt \qquad n = 1, 2, 3, \dots \tag{6.53}$$

Similarly, we find that

$$b_n = \frac{2}{T_0} \int_{t_1}^{t_1 + T_0} x(t) \sin n\omega_0 t \, dt \qquad n = 1, 2, 3, \dots \tag{6.54}$$

Note that the Fourier series in Eq. (6.49) of Ex. 6.14 is indeed the trigonometric Fourier series with T⁰ = 2π and ω⁰ = 2π/T0. In this particular example, it is easy to verify from Eq. (6.53) that an = 0 for all n, including n = 0. Hence, the Fourier series in that example consisted only of sine terms.

EXPONENTIAL FOURIER SERIES

As shown in the footnote on page 622, the set of exponentials ejn^ω0*^t* (n = 0,±1,±2,…) is a set of functions orthogonal over any interval of duration T⁰ = 2π/ω0. An arbitrary signal x(t) can now be expressed over an interval (t1,t¹ +T0) as

$$x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad t_1 < t < t_1 + T_0$$

where [see Eq. (6.47)]

$$D_n = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) e^{-jn\omega_0 t} dt$$

WHY USE THE EXPONENTIAL SET?

If x(t) can be represented in terms of hundreds of different orthogonal sets, why do we exclusively use the exponential (or trigonometric) set for the representation of signals or LTI systems? It so

happens that the exponential signal is an eigenfunction of LTI systems. In other words, for an LTI system, only an exponential input est yields the response that is also an exponential of the same form, given by H(s)est. The same is true of the trigonometric set. This fact makes the use of exponential signals natural for LTI systems in the sense that the system analysis using exponentials as the basis signals is greatly simplified.