Signals Defined on ℝ

Abstract

In Chap. 2, continuous-time signals were introduced and developed following a classical approach. In this chapter they are reconsidered as signals defined on ℝ, following the UST. The target is twofold: (1) to review this fundamental class of signals and to get a deeper insight into it, and (2) to introduce concepts that are specific to this signal class. For example, differentiation and integration, which are not possible in a discrete domain, are developed. Also, for signals defined on ℝ, the Laplace transform, which plays a role similar to the Fourier transform, and the Hilbert transform are introduced.

Appendices

Appendix A: Proof of the Theorem 9.1 on Band-Duration Incompatibility

We recall the following statements from the theory of analytic functions [2]:

• (1) Let γ be a curve (open or closed) of finite length ℓ, and Ω a region of the complex plane ℂ. Let f(t,z) be a function of two complex variables defined for t∈γ, z∈Ω, continuous in γ×Ω, analytic with respect to z in the region Ω. Then, the integral

  $$F(z) = \int_\gamma f(t,z) \,\mathrm{d}t$$

  represents an analytic function in Ω, and

  $$F'(z) = \int_\gamma f_z'(t,z) \,\mathrm{d}t .$$

• (2) (restricted identity principle) If f(z) is analytic in Ω and zero in a subset E and if an accumulation point of E belongs to Ω, then f(z) is identically zero in Ω.

Applying (1) and (2), we sketch a proof of the Theorem 9.1. Specifically, we prove that if a signal s(t) is absolutely integrable, s∈L ₁(ℝ), and is strictly band-limited, and if the support complement $\bar{e}_{0}(s)$ contains an accumulation point of E, then $\bar{e}_{0}(s) = \mathbb{R}$, that is, the signal is identically zero. In fact, the absolute integrability assures the existence of the FT, so that we can write

$$s(t) = \int_{-B}^B S(f) \mathrm{e}^{\mathrm{i}2\pi ft} \, \mathrm {d}f\quad \stackrel{\mathcal{F}}{\longrightarrow}\quad S(f) = \int_{-\infty}^{+\infty} s(t) \mathrm{e}^{-\mathrm{i}2\pi ft} \, \mathrm{d}t . $$

(9.67)

Therefore, since s∈L ₁(ℝ), from the second of (9.67) it follows that S(f) is a continuous function. Then, application of statement (1) to the first of (9.67) yields that s(t) is an analytic function for every t∈ℝ. Moreover, from statement (2) it follows that if $\bar{e}_{0}(s)$ contains an accumulation point of E, then $\bar{e}_{0}(s) = \mathbb{R}$.

Appendix B: Proof of Theorems on Asymptotic Behavior

Proof of Lemma 9.1

If h→0, we have

$$\bigl| s(t+h)-s(t)\bigr|\leq\int_{-\infty}^{+\infty} \bigl| S(f) \bigr|\bigl| \mathrm{e}^{\mathrm{i}2\pi f(t+h)}-\mathrm{e}^{\mathrm {i}2\pi ft}\bigr| \, \mathrm{d}f ,$$

where the integrand converges to 0 as h→0. Moreover,

$$\bigl| S(f) \bigr| \bigl| \mathrm{e}^{\mathrm{i}2\pi f(t+h)}-\mathrm{e}^{\mathrm{i}2\pi ft} \bigr|\leq2 \bigl| S(f) \bigr| \in L_1(\mathbb{R}).$$

Now, the dominate convergence theorem assures that also the left-hand side →0 and s(t) is a continuous function. Moreover,

$$\bigl\| s(t) \bigr\|_{\infty}=\max \bigl\| s(t)\bigr\|=\max\biggl| \int_{-\infty}^{+\infty} S(f)\mathrm{e}^{\mathrm{i}2\pi ft} \mathrm{d}t \biggr| \leq\int_{-\infty }^{+\infty}\bigl| S(f)\bigr|\, \mathrm{d}f=\bigl\|S(f)\bigr\|_1.$$

Finally, from

$$ \begin{array}{l} - s\left( t \right) = {\rm e}^{ - i\pi } \int_{ - \infty }^{ + \infty } {S\left( f \right){\rm e}^{{\rm i}2\pi ft} {\rm d}t = \int_{ - \infty }^{ + \infty } {S\left( f \right){\rm e}^{i2\pi \left( {f - 1/\left( {2t} \right)} \right)} {\rm d}f} }\\ \quad \quad= \int_{ - \infty }^{ + \infty } {s\left( {f + 1/\left( {2t} \right)} \right){\rm e}^{{\rm i}2\pi ft} {\rm d}f}\\ \end{array} $$

it follows that

$$ \begin{array}{l} 2\left| {s\left( t \right)} \right| = \left| {\int_{ - \infty }^{ + \infty } {\left[ {S\left( f \right) - S\left( {f + 1/\left( {2t} \right)} \right)} \right]} {\rm e}^{{\rm i}2\pi ft} df} \right| \\ \quad \quad \quad=\le \int_{ - \infty }^{ + \infty } {\left| {S\left( f \right) - S} \right|\left( {f + 1/\left( {2t} \right)} \right)\left| {df = \left\| {S - S_{1/\left( {2t} \right)} } \right\|} \right.1.}\\ \end{array} $$

As t→±∞, the right-hand side goes to 0 by the continuity of the shift operator, and hence the conclusion. □

Proof of Lemma 9.2 and of Theorem 9.2

Since S(f) and S′(f) are absolutely integrable, their inverse FTs s(t) and s ₁(t) exist. Next, we integrate by parts:

$$ \begin{array}{*{20}l} {s(t)} & { = \int_{ - \infty }^{ + \infty } {S(f){\rm e}^{{\rm i}2\pi ft} } {\rm d}f = \frac{1}{{{\rm i}2\pi t}}S(f){\rm e}^{{\rm i}2\pi ft} \left| {_{ - \infty }^{ + \infty } } \right. - \frac{1}{{{\rm i}2\pi t}}\int_{ - \infty }^{ + \infty } {S'(f){\rm e}^{{\rm i}2\pi ft} } {\rm d}f} \\ {} & { = - \frac{1}{{{\rm i}2\pi t}}\int_{ - \infty }^{ + \infty } {S'(f){\rm e}^{{\rm i}2\pi ft} } {\rm d}f = \frac{1}{{{\rm i}2\pi t}}s_1 (t),} \\ \end{array} gg$$

where the first part of the integral vanishes because S(f) is absolutely integrable. Then lim  _f→±∞ S(f)=0. Finally, from Lemma 9.1 we have $\lim_{t \to \pm\infty} t s(t)=\lim_{t \to \pm\infty} -\frac {1}{\mathrm{i}2\pi} s_{1}(t)=0$.

The proof of Theorem 9.2 is based on a recursive application of Lemma 9.2. □

Proof of Theorem 9.3

The regularity degree of order n assures that the (n−1)th derivative has the following decomposition:

$$S^{(n-1)}(f) = S_c^{(n-1)}(f) + S_d^{(n-1)}(f) ,$$

where

$$S_d^{(n-1)}(f) = \sum_i \frac{1}{2}, d_i \operatorname{sgn}(f - f_i).$$

Then, the inverse FT of $S_{d}^{(n-1)}(f)$ is

$$s_{d,n-1}(t) = \sum_i \frac{1}{2}d_i\frac{\mathrm{e}^{\mathrm{i}2\pi f_it}}{-\mathrm{i}2\pi t} .$$

From the frequency differentiation rule it follows that s _d,n−1(t)=(−i2πt)ⁿ⁻¹ s _d (t), and therefore s _d (t)=O( t ⁻ⁿ). On the other hand, for the continuous part s _c,n (t), Theorem 9.2 assures that s _c (t)≤O(t ⁻ⁿ). Then, by combination s(t)=s _c (t)+s _d (t) is O(t ⁻ⁿ). □

Appendix C: Proof of Uncertainty Principle Inequality

We recall the Schwartz–Gabor inequality (see Sect. 4.5)

$$\biggl| \int_{-\infty}^{+\infty}\bigl[x(t) y^{\ast}(t)+x^{\ast}(t) y(t)\bigr] \,\mathrm{d}t \biggr|^2\leq4 \int_{-\infty}^{+\infty} \bigl|x(t)\bigr|^2\, \mathrm{d}t\int_{-\infty}^{+\infty}\bigl|y(t)\bigr|^2\, \mathrm{d}t , $$

(9.68)

where the equality holds if and only if x(t) and y(t) are proportional: y(t)=Kx(t) with K real. We apply (9.68) to the signals x(t)=s′(t), y(t)=ts(t), namely

$$ \begin{array}{*{20}l} {\left| {\int_{ - \infty }^{ + \infty } {[ts(t)s'(t)^* + ts'(t)s(t)^* ]} \,{\rm d}t\left| {^2 } \right.} \right.} & { \le 4\int_{ - \infty }^{ + \infty } {\left| {s'(t)} \right|^2 } {\rm d}t\int_{ - \infty }^{ + \infty } {t^2 \left| {s(t)} \right|^2 } {\rm d}t} \\ {} & { = 4\int_{ - \infty }^{ + \infty } {\left| {s'(t)} \right|^2 } {\rm d}t\,D_q^2 E_s ,} \\ \end{array} $$

(9.69)

where s′(t) has FT i2πfS(f), and therefore,

$$\int_{-\infty}^{+\infty} \bigl|s'(t)\bigr|^2 \, \mathrm{d}t=(2\pi )^2 \int_{-\infty}^{+\infty} f^2\bigl|S(f)\bigr|^2\, \mathrm{d}f = (2\pi)^2 B_q^2 E_S .$$

Considering that s(t)s′(t)^∗+s′(t)s(t)^∗=d|s(t)|²/dt and integrating by parts, the first term of (9.69) yields

$$\int_{-\infty}^{+\infty} t \frac{\mathrm{d} |s(t)|^2}{\mathrm {d}t} \,\mathrm{d}t = t \bigl|s(t)\bigr|^2 \big|_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \bigl|s(t)\bigr|^2 \, \mathrm{d}t =- E_s ,$$

where we have used assumption (9.32). Combination of the results yields $E_{s}^{2} \leq4(2\pi)^{2} D_{q}^{2} B_{q}^{2} E_{s}^{2}$, that is, (9.33). Considering (9.68), the equality holds when s′(t)=Kts(t) with K real. The unique solution of this differential equation is $s(t)=C \exp( \frac{1}{2} K t^{2} )$, but condition (9.32) implies that K<0.

In (9.37) ΔD _q ≤D _q and ΔB _q ≤B _q , and hence the first inequality. To prove the second, we apply (9.33) to the Fourier pair

$$s_0(t)=s(t+t_c) \mathrm{e}^{-\mathrm{i}2\pi f_c t} \quad \stackrel {\mathcal{F}}{\longrightarrow}\quad S_0(f)=S(f+f_c) \mathrm {e}^{\mathrm{i}2\pi(f+f_c)t_c}.$$

So we get B _q (S ₀)D _q (s ₀)≤1/(4π), where B _q (S ₀) and D _q (s ₀) refer to the signal s ₀(t). But $E_{s_{0}}=E_{s}$ and

$$ \begin{array}{l} E_{s0} D_q^2 \left( {s_0 } \right) = \int_{ - \infty }^{ + \infty } {t^2 } \left| {s_0 \left( t \right)} \right|^2 {\rm d}t = \int_{ - \infty }^{ + \infty } {t^2 } \left| {s\left( {t + t_c } \right)} \right|^2 {\rm d}t \\ \quad \quad \quad \quad= \int_{ - \infty }^{ + \infty } {\left( {t - t_c } \right)} ^2 \left| {s\left( t \right)} \right|^2 dt = E_s \Delta D_q^2 \left( s \right), \\ \end{array} $$

and analogously we find $E_{S_{0}} B_{q}^{2}(S_{0})=E_{S} \varDelta B_{q}^{2}(S)$. Hence, ΔB _q (s)ΔD _q (S)≤1/(4π).

Next, we prove (9.38), showing that

$$D_q^2=\frac{E_{S'}}{(2\pi)^2 E_S},\qquad B_q^2=\frac{E_{s'}}{(2\pi )^2 E_s} . $$

(9.70)

The energy of $S'(f)=\mathcal{F}[-\mathrm{i}2\pi t s(t) | f]$ is

$$E_{S'}=\int_{-\infty}^{+\infty}\bigl|S'(f)\bigr|^2 \,\mathrm{d}f =\int_{-\infty}^{+\infty}\bigl |-\mathrm{i}2\pi t s(t)\bigr|^2 \, \mathrm{d}t =(2\pi)^2 D_q^2 E_s ,$$

which proves the first (9.70). The proof of the second one is similar, considering the energy of s′(t). For the proof of (9.39), consider the auxiliary signal x(t)=|s(t)|². Then, applying the FT rules, we have

$$E_s t_c=\int_{-\infty}^{+\infty} t x(t) \,\mathrm{d}t=\frac{1}{-\mathrm{i}2\pi} \frac{\mathrm{d}X(f)}{\mathrm{d}f} \bigg|_{f=0} ,$$

but X(f) is given by the correlation C _S (f) of S(f), and its derivative at f=0 is given by the cross-energy (see Sect. 5.7). The second of (9.37) follows by symmetry.

Appendix D: Inverse Fourier Transform of Raised Cosine

We prove the Fourier pairs (9.26a), (9.26b). The proof, based on FT definition (9.7b) is cumbersome. An easier proof is based on the frequency integration rule.

Letting R ₀(f)=rcos (f,α), we find that the derivative $X(f)=R'_{0}(f)$ is different from zero only on the two roll-off parts (Fig. 9.27) and can be written in the form

$$X(f)=R'_0(f)=P\biggl(f +\frac{1}{2}\biggr)-P\biggl(f-\frac {1}{2}\biggr) , $$

(9.71)

where (see (9.22))

$$P(f)= \frac{\pi}{2 \alpha} \cos\biggl( \frac{\pi}{\alpha} f\biggr)\operatorname{rect}\biggl(\frac{f}{\alpha} \biggr). $$

(9.72)

Hence,

$$R_0(f)= \int_{-\infty}^{f} X(\lambda) \, \mathrm{d}\lambda.$$

But from the frequency integration rule we obtain

$$r_0(t)= \frac{1}{-\mathrm{i}2 \pi t} x(t) +\frac{1}{2} x(0) \delta (t)= \frac{1}{-\mathrm{i}2 \pi t}x(t) , $$

(9.73)

where x(0)=area (X)=0. Then, from (9.71) we find

$$x(t)=p(t) \mathrm{e}^{-\mathrm{i}\pi t} - p(t) \mathrm{e}^{\mathrm {i}\pi t} = -2\mathrm{i}\sin\pi t p(t)$$

and by combination r ₀(t)=p(t)sinc (t). Next, we evaluate the inverse FT P(f) starting from (9.72). Considering that sinc (αt)→(1/α)rect (f/α) and expressing the cosine by the Euler formula, one gets

$$p(t)= \frac{\pi}{2} \operatorname{sinc}\biggl( \alpha t -\frac {1}{2}\biggr) + \frac{\pi}{2} \operatorname{sinc}\biggl(\alpha t+\frac{1}{2}\biggr),$$

and (9.26a) follows.

Fig. 9.27

Derivative $R_{0}'(f)$ of raised cosine for α=0.6

For the second formula, we use the same technique. We let $R_{1}(f) \,\buildrel \varDelta \over= \,\operatorname{rrcos}(f, \alpha)$ and write the derivative in the form (Fig. 9.28)

$$ \begin{array}{*{20}l} {X_1 (f)\,\mathop = \limits^\Delta \,R'_1 (f)} & { = - Q\left( {f - \frac{1}{2}} \right) + Q\left( { - f - \frac{1}{2}} \right)} \\ {} & { = - Q\left( {f - \frac{1}{2}} \right) + Q^* \left( { - f - \frac{1}{2}} \right),} \\ \end{array} $$

(9.74)

where

$$Q(f)=\frac{\pi}{2 \alpha} \sin\biggl( \frac{\pi}{2 \alpha} f +\frac{\pi}{4}\biggr) \operatorname{rect}( f/\alpha). $$

(9.75)

Next, the integration rule again yields (9.73), where now

$$x(t)=-q(t) \mathrm{e}^{-\mathrm{i}\pi t}+q^*(t) \mathrm{e}^{-\mathrm {i}\pi t}= -2\mathrm{i}\Im\bigl[q(t) \mathrm{e}^{\mathrm{i}\pi t}\bigr].$$

Hence, $r_{1}(t)= \frac{1}{\pi t} \Im[{q(t) \mathrm{e}^{\mathrm{i}\pi t}}]$. The evaluation of q(t), which is not immediate, finally gives (9.26b). Note that

$$\lim_{t \to 0} r_1(t) = 1- \alpha\biggl(\frac{4}{\pi}-1 \biggr)$$

as we can check by calculating the area of R ₁(f).

Fig. 9.28

Derivative $R'_{1}(f)$ of root–raised cosine, for α=0.6