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When we plug x = 2/3 into the final polynomial equation we found in solution to Prob 3 and then grind through the arithmetic, the process goes like this, step-by-step: 105x 3 + 211x 2 + 112x + 12 = 0 105 ( 2/3)3 + 211 ( 2/3)2 + 112 ( 2/3) + 12 = 0 105 ( 8/27) + 211 (4/9) + 112 ( 2/3) + 12 = 0 840/27 + 844/9 224/3 + 12 = 0 840/27 + 2,532/27 2,016/27 + 324/27 = 0 ( 840 + 2,532 2,016 + 324) / 27 = 0 840 + 2,532 2,016 + 324 = 0 0=0 To check the root x = 6/5, we go through this sequence of calculations: 105x 3 + 211x 2 + 112x + 12 = 0 105 ( 6/5)3 + 211 ( 6/5)2 + 112 ( 6/5) + 12 = 0 105 ( 216/125) + 211 (36/25) + 112 ( 6/5) + 12 = 0 22,680/125 + 7,596/25 672/5 + 12 = 0 22,680/125 + 37,980/125 16,800/125 + 1,500/125 = 0 ( 22,680 + 37,980 16,800 + 1,500) / 125 = 0 22,680 + 37,980 16,800 + 1,500 = 0 0=0 To check the root x = 1/7, we go through this arithmetic, step-by-step: 105x 3 + 211x 2 + 112x + 12 = 0 105 ( 1/7)3 + 211 ( 1/7)2 + 112 ( 1/7) + 12 = 0 105 ( 1/343) + 211 (1/49) + 112 ( 1/7) + 12 = 0 105/343 + 211/49 112/7 + 12 = 0 105/343 + 1,477/343 5,488/343 + 4,116/343 = 0 ( 105 + 1,477 5,488 + 4,116) / 343 = 0 105 + 1,477 5,488 + 4,116 = 0 0=0

t

R10 33k ohm, 14w / 5% resistor R11,R12 820 ohm, 14w / 5% resistor R13 10k ohm potentiometer (panel) R14,R17 10 ohm, 14w, / 5% resistor R15,R16 22k ohm, 14w, / 5% resistor

25

h(t )E w( ) d h(t ) w ( )d (441)

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where the transition from the second to the third line uses the following facts: expectation and time-integration are linear operators and the impulse response h is deterministic Eqn (441) shows that the mean of the output process is the convolution of the impulse response with the mean of the input process In the case where the system is causal and the mean of the input process is constant, the mean of the output reduces to y (t) = H(0) w

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5 Here s the general binomial-trinomial equation once again: (a1x + b1)(a2x 2 + b2x + c) = 0 Using the product of sums rule, we can rewrite this as a1a2x 3 + a1b2x 2 + a1cx + b1a2x 2 + b1b2x + b1c = 0 When we bring the terms for x 2 next to each other, and then do the same thing with the terms for x, we get a1a2x 3 + a1b2x 2 + b1a2x 2 + a1cx + b1b2x + b1c = 0 Let s use the commutative law for multiplication on the third term to get a before b in the interest of elegance! That gives us a1a2x 3 + a1b2x 2 + a2b1x 2 + a1cx + b1b2x + b1c = 0 Finally, we can use the distributive law for multiplication over addition to consolidate the coefficients for x 2 and x, giving us the equation in true polynomial standard form: a1a2x 3 + (a1b2 + a2b1)x 2+ (a1c + b1b2)x + b1c = 0 6 If the coefficients and constants are real numbers, and if a1 and a2 are both nonzero, then the cubic (a1x + b1)(a2x 2 + b2x + c) = 0 has at least one real root, which is x = b1/a1 That s the root that we get when we create a first-degree equation from the binomial term by setting it equal to 0 The cubic might have no more real roots (therefore one real root in total), one more (two in total), or two more (three in total) To find out which of these situations is the true case, we can look at the discriminant d for the quadratic a2x 2 + b2x + c = 0 In this notation, d = b22 4a2c If d > 0, then the quadratic has two real roots, so the original cubic has three If d = 0, then the quadratic has one real root with multiplicity 2, so the original cubic has two real roots, one of which has multiplicity 2 If d < 0, then the quadratic has no real roots, so the original cubic has only one 7 For reference, here s the equation again: (3x + 5)(16x 2 56x + 49) = 0.

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We found the real roots x = 5/3 or x = 7/4

The output autocorrelation function is slightly more complicated to derive Its derivation is addressed in two steps First, from eqn (429), we nd an expression for the cross correlation between y and w,

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