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Unformatted text preview: lements YI, Y 2, . .. , Y n).
We shall demonstrate t he use of this rule with a n example. (B.28)
- 60°) touches _ 4e- 2t a t i ts negative peaks. Therefore, _ 4e- 2t
is a n envelope for negative amplitudes of 4 e- 2t cos (6t - 60°). T hus, t o sketch
4 e- 2t cos (6t - 60°), we first draw t he envelopes 4 e- 2t a nd _ 4e- 2t ( the mirror image
of 4 e- 2t a bout t he horizontal axis), a nd t hen sketch t he sinusoid cos (6t - 60°), w ith
these envelopes a cting as constraints on t he sinusoid's amplitude (see Fig. B .llc).
In general, K e-atcos(wot+lI) c an b e sketched in this manner, with K e- at a nd
_ Ke- at c onstraining t he a mplitude of cos (wot + 11).
4 e- 2 tcos(6t Yn We d enote t he m atrix o n t he l eft-hand side formed by t he elements a ij as A . T he
d eterminant of A is d enoted by IAI. I f t he d eterminant IAI is n ot zero, t he s et of
equations (B.29) has a unique solution given b y C ramer's formula 3 -1 =4 Cramer's Rule T his is a v ery convenient rule used t o solve simultaneous linear equations.
Consider a s et o f n linear simultaneous equations in n unknowns X l, X 2, . .. , X n: Since IAI = 4 # 0, a unique solution exists for
by Cramer's rule (B.31) as follows: X l, X 2, and 3
Xl = I !I 7 3 -1 =~ = -1 = 4: = 1 2 (B.29)
1 X2 These equations can be expressed in m atrix form as = IAI 7 4 X 3. This solution is provided .............. -----------------------------.;~-------------------Background 24 B.5 P artial F raction Expansion 25 Therefore, F (x) c an be expressed as
2 X3 = o 1 jAj' 3 37= ~S = -2 F (x) • x2 2x x -I +I + x 2 + 4x ~ p olynomial in x (B.34b) +3 ""--v-' A proper function can be further expanded into partial fractions. T he remaining
discussion in this section is concerned with various ways of doing this. B .5-1 A = [2 1 1 ;1 3 - 1;1 1 I J; b =[3 7 I J'; for k =I:3 Partial Fraction Expansion: Method o f Clearing Fractions T his m ethod consists of writing a rational function as a sum of appropriate partial fractions with unknown coefficients, which are determined by clearing fractions
a nd e quating t he coefficients of similar powers on t he two sides. This procedure is
d emonstrated by t he following example. A l=A;
x (k)=det(D} J det(A);
x =x' • E xample B .8
Expand the following rational function F (x) into partial fractions: x=2 8 .5 + 4x + 3 p roper f unction E xample C B.5
Using a Computer, solve Example B.7. 1
-2 = 2 x 3 + 9 x 2 + l lx + 2 = 3
x + 3x + 4x + 6
( ) - (x + l )(x + 2 )(x + 3)2 0 This function can be expressed as a sum of partial fractions with denominators (x
( x + 2), ( x + 3), and ( x + 3)2, as shown below. Partial Fraction Expansion I n t he a nalysis of linear time-invariant systems, we encounter functions t hat
are ratios of two polynomials in a certain variable, say x . Such functions are known
as r ational f unctions. A r ational function F (x) c an be expressed as + b m_lX m - l + .. , + b lx + bo
x n + a n_lX n - l + ... + a lx + aO b mxm Fx=:"...
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