DISCRETE-SIGNAL ANALYSIS AND DESIGN- P27

DISCRETE-SIGNAL ANALYSIS AND DESIGN- P27:Electronic circuit analysis and design projects often involve time-domainand frequency-domain characteristics that are difÞcult to work with usingthe traditional and laborious mathematical pencil-and-paper methods offormer eras. This is especially true of certain nonlinear circuits and sys-tems that engineering students and experimenters may not yet be com-fortable with.
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DISCRETE-SIGNAL ANALYSIS AND DESIGN- P27116 DISCRETE-SIGNAL ANALYSIS AND DESIGNVL(k ), P L(k) = V L(k)2 [GB (k) + G(k)] watts. If G(k) + GB (k) = 0, theimaginary power (vars) P L(k) = ±j B(k) × V L(k)2 . The real part of the power PL(k ) is converted to radio or sound waves orheat dissipation of some kind, and the imaginary part is cycled back andforth between energy storage elements (lumped components) or stand-ing waves (transmission lines) of some kind. This energy cycling al-ways involves slightly lossy storage elements that dissipate a little of thereal power.Example 7-1: The Use of Eq. (7-2)Figure 7-2 is an example of the use of Eq. (7-2). The x (n) input signalvoltage waveform in part (a) is a complex time sequence of cosine andsine waves. This Þgure uses steps of 0.1 in the (n) values for better visualresolution, and this is the only place where x (n) is plotted. The two plotsin part (a) are I (n) (real) and Q(n) (imaginary) sequences that we havelooked at previously. Parts (b) and (c) are the DFT of part (a) that showthe two-sided phasor frequency X (k ) voltage values. The DFT uses (k )steps of 1.0 to avoid spectral leakage between (k ) integers (Chapter 3).If a dc voltage is present, it shows up at k = 0 (see Fig. 1-2). In thisexample there is no dc, but it will be considered later. The integer valuesare sufÞcient for a correct evaluation if there are enough of them to satisfythe requirements for adequate sampling. In part (d) the two-sided phasors are organized into two groups. Onegroup collects phasor pairs that have even symmetry about N /2 and areadded coherently (Chapter 1). These are the cosine (or j cosine) terms.The other pairs that have odd symmetry about N /2 are the sine (or thej sine) terms and are subtracted coherently. This procedure accounts forall phasor pairs in any signal, regardless of its even and odd components,and the results agree with Fig. 2-2. Plots (f) and (g) need only the positivefrequencies. Note also that the frequency plots are not functions of time,like x (n), so each observation at frequency (k ) is a steady-state measure-ment and we can take as much time at each (k ) as we like, after the x (n)time sequence is obtained. Part (e) calculates the load admittance Y (k) = G(k) ± j B(k) at each(k ) for the frequency dependence that we have speciÞed. The plot in part(f) shows the complex value of Y (k ) at each (k ). THE POWER SPECTRUM 117 N := 32 n := 0, 0.1 .. N − 1 k := 0, 1 .. N − 1 R := 1 x(n) := 3⋅cos 2π⋅ n ⋅1 + 4⋅j⋅sin 2⋅π⋅ n ⋅3 + 5⋅j⋅cos 2⋅π⋅ n ⋅5 − 6⋅j⋅sin 2⋅π⋅ n ⋅7 N N N N 10 Re(x(n)) Im(x(n)) 0 −10 0 5 10 15 20 25 30 n (a) N−1 X(k) := 1 . N ∑ n=0 x(n)⋅exp −j⋅2⋅π⋅ n ⋅k N (b) 5 Re(X(k)) 0 −5 0 5 10 15 20 25 30 k (c) XE(k) := X(k) + X(N − k) XO(k) := X(k) − X(N − k) (d) (Y(k)) := 1 + j· k 4.5 (e) 2 1.4 0.8 Re(Y(k)) Im(Y(k)) 0.2 −0.4 −1 0 1 2 3 4 5 6 7 8 k (f ) (XE(k))2 + (XO(k))2 PL(k) := ·Y(k) (1 + R·Y(k))2 15 10 Re(PL(k)) 5 Im(PL(k)) 0 −5 0 1 2 3 4 5 6 7 8 k (g)Figure 7-2 Power spectrum of a complex signal: (a) complex two-sidedtime domain, real and imaginary; (b) complex two-sided phasor voltagespectrum; (c) complex two-sided phasor voltage spectrum; (d) even andodd parts of phasor ...