Signal Processing and Linear Systems Solution Ch.3

Chapter 3 3.1-1 Trivial. 3.1-2 (a) In this case E_{x} = integrate 1 dt from 0 to 1 = 1 and c= 1 E * integrate f(i) * x(t) dt from 0 to 1 = 1/1 * integrate t dt from 0 to 1 = 0.5 (b) Thus, f(t) \approx 0.5x(t) and the error e(t) = t - 0.5 over (0≤≤ 1), and zero outside this interval. Also Ey and E. (the error) are E_{f} = integrate f ^ 2 * (t) dt from 0 to 1 = integrate t ^ 2 dt from 0 to 1 = 1/3 The error (-0.5) is orthogonal to x(t) becaцве and E_{e} = integrate (t - 0.5) ^ 2 dt from 0 to 1 = 1/12 integrate (t - 0.5)(1) dt from 0 to 1 = 0 Note that E f = c ^ 2 * E_{x} +E * . To explain these results in terms of vector concepts we observe from Fig. 3.1 that the error vector to the component cx. Because of this orthogonality, the length-square of f energy of f(t) ] is equal to the sum of the square of the lengths of cx and e sum of the energies of cx(t) and e(t) * j 3.1-3 In this case E_{f} = integrate f ^ 2 * (t) dt from 0 to 1 = integrate t ^ 2 dt from 0 to 1 = 1/3 and c = 1/E_{f} * integrate z(t) * f(t) dt from 0 to 1 = 3 * integrate t dt from 0 to 1 = 1.5 Thus, x(t) \approx 1.5f(t) and the errore e(t) = x(t) - 1.5f(t) = 1 - 1.5t over (0 <= t <= 1) and zero outside this interval E_{e} energy of the error E_{e} = integrate (1 - 1.5t) ^ 2 dt from 0 to 1 = 1/4 3.1-4 (a) In this case E_{x} = integrate sin^2 2 * pi*t dt from 0 to 1 = 0.5 and c = 1/E_{x} * integrate f(t) * x(t) dt from 0 to 1 = 1/0.5 * integrate t * sin 2pi*t dt from 0 to 1 = - 1 / pi (b) Thus, f(t) \approx - (1 / x) * x(t) and the error sin 2rt over and zero outside this interval. Also E_{f} and E. (the energy of the E_{f} = integrate f ^ 2 * (t) dt from 0 to 1 = integrate t ^ 2 dt from 0 to 1 = 1/3 e(t) = t + (1 / s) and (0 <= t <= 1] E_{e} = integrate [t - (1 / pi) * sin 2pi*t] ^ 2 dt from 0 to 1 = 1/3 - 1/(2pi ^ 2) The error [t + (1 / pi) * sin 2pi*t] is orthogonal to x(t) because int 0 ^ 1 sin 2 pi t| t + (1 / pi) * sin 2pi*t )dt=0 Note that E_{f} = c ^ 2 * E_{x} + E_{e} To explain these results in terms of vector concepts we observe from Fig. 3.1 that the error vector e is orthogonal to the component cx. Because of this orthogonality, the length of f energy of f(t) \ is equal to the sum of the square of the lengths of cx and e sum of the energies of cx(t) and e\{t\} \ 3.1-5 (a) If x(t) and y(t) are orthogonal, then we 24 showed see Eq. (3.22)) the energy of x(t) + y(t) is Ex+ E. We now find the energy of z(t)y(t): int - infty ^ infty |x(t)-y(t)|^ 2 dt= integrate |x(t)| ^ 2 dt from - ∞ to ∞ + integrate |y(t)| ^ 2 dt from - ∞ to ∞ - int - infty ^ infty x(t)y^ * (t)dt- int - infty ^ infty x^ * (t)y(t)dt = integrate |x(t)| ^ 2 dt from - ∞ to ∞ + integrate |y(t)| ^ 2 dt from - ∞ to ∞ (3.22) The last result follows from the fact that because of orthogonality, the two integrals of the cross products (1)*(t) and z*(t)u(t) are zero see Eq. (3.20)]. Thus the energy of (t)+ y(t) is equal to that of (t)y(t) if (1) and y(t) are orthogonal. (b) Using similar argument, we can show that the energy of ex(t)+czy(t) is equal to that of er(t)cay(t) if z(1) and (1) are orthogonal. This energy is given by cEs+calEy (c) If (t) = x(t) (t), then int - infty ^ infty |x(t) pm y(t)|^ 2 dt= integrate |x(t)| ^ 2 dt from - ∞ to ∞ + integrate |y(t)| ^ 2 dt from - ∞ to ∞ pm int - infty ^ infty x(t)y^ * (t)dt pm int - infty ^ infty x^ * (t)y(t)dt E+E+ (Es + Eye) 3.2-1 We shall compute ca using Eq. (3.25) for each of the 4 cases. Let us first compute the energies of all the signals. E_{z} = integrate sin^2 2 * pi*t dt from 0 to 1 = 0.5 In the same way we find EE БЕ, 0.5. Using Eq. (3.25), the correlation coefficients for four cases are found as (1) 1/(sqrt((0.3)(0.5))) * integrate sin 2pi*t * sin 4nt dt from 0 to 2 = 0 (2) 1/(sqrt((0.5)(0.5))) * integrate (sin 2pi*t)(- sin 2pi*t) dt from 0 to 1 = - 1 (3) 1/(sqrt((0.5)(0.5))) * integrate 0.707sin 2pi*t dt from 0 to 1 = 0 (4) 1 sqrt (0.5)(0.5) [ integrate 0.707sin 2pi*t dt from 0 to 0.5 - int 0.2 ^ 1 0,707 sin 2 pi t dt]=1.414/ pi maximum 3.3-1 f(2,-1), f(-1,2), fs(0,-2), fa(1,2), fs (2, 1), and fo(3.0). From the figure, we ace that pairs (fs.fa). (fr.) and (fa, fs) are orthogonal. We can verify this also analytically.