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Let x(t) = 3 .7 + 2.6 cos(48πt + 2.3) + 5.9 cos(80πt − 1 .2) + 1.5 cos(112πt + 0.9) (i) (3 pts.) Determine the fundamental period T 0 of x(t). (ii) (4 pts.) Define (i.e., give the numerical values of) a scalar c and a vector b having the smallest possible length, such that c =b = [ ; ].’ ; z = c*ifft(b,256) ;x = real(z) ;
generates a vector x consisting of 256 samples of x(t) uniformly spaced in [0, T 0 ). (iii) (4 pts.) If b and c are as specified in (ii) above, the code b_new = [0 ; 4 ; abs(b)exp(-j1.3)] ; % ABS(.) : MODULUS z = c*ifft(b_new,256) ; s = real(z) ; produces 256 samples, uniformly spaced in [0, 2 T 0 ), of a periodic signal s(t) with fundamental period 2 T 0. Assuming s(t) has no harmonics higher than the 128th, write an equation for s(t). (iv) (4 pts.) Suppose that x(t) (not s(t)) is the input to an ideal lowpass filter with unit gain, zero delay and cutoff frequency 30 Hz. The filter output y(t) is processed by a squarer to produce v(t) = y^2 (t) Ifthat series (sum)? v(t) is expressed as a Fourier series using real sinusoids, which frequencies (in Hz) will be present in
(i) x(t) is a sum of sinusoids with frequencies 0, 24, 40 and 56 Hz. ThereforeNontrivial f_0 harmonics = GCD (24,40,56) at k = 0, =3, 8 5 Hz, and and 7. T_0 = 1/8 sec.
(ii) b = [3.7 0 0 2.6exp(j2.3) 0 5.9exp(-j1.2) 0 1.5exp(j0.9) c = 256 (iii) Fundamental frequency in this case is f_0/2 = 4 Hz => W_0 = 8pi rad/sec Nontrivial harmonics at k = 1, 2, 5, 7 and 9. s(t) = 4cos(8pit) + 3.7cos(16pit-1.3) + 2.6cos(40pit - 1.3) ...+ 5.9cos(56pit - 1.3) + 1.5cos(72pit - 1.3)
(iv) y(t) = 3.7 + 2.6cos(48pit + 2.3) v(t) = (y(t))^2 contains frequencies 0, 24 and 224 = 48 Hz. This follows from 2(cos(theta))^2 = cos(2theta) + 1 ; and can also be obtained by expanding (a + bexp(jtheta) + bexp(-j*theta))^
The signal r(t) shown below (top graph) is periodic with period T 0 = 8 and complex Fourier series expansion r(t) =
∑^ ∞
k=−∞^ Rke
jkΩ 0 t (^) ,
where Ω 0 is the fundamental angular frequency. r ( t ) ... 1 ... − 8 1 8 t x ( t )
... ...
− 1
y ( t )
1
... ...
−10 8 −6 −2 2 6 8 10 t
− 8 −4 4 8 t
− 1
2
(i) (2 pts.) Are all Rk’s purely real, purely imaginary, or neither? Explain. (ii) (3 pts.) Express x(t) (middle graph) in terms of r(t). (iii) (4 pts.)to the extent possible, and distinguish between cases if necessary. Express the Fourier series coefficients Xk of x(t) in terms of Rk’s. Simplify your answer
Let y(t) also be periodic with period T 0 = 8 , and such that for all k, Yk = j(Rk− 8 − Rk+8) (iv) (4 pts.) Express y(t) in terms of r(t). (v) (2 pts.) Sketch y(t) using the axes provided (bottom graph).
(i) r(-t) = -r(t), thus also R[-k] = -R[k]. Furthermore, r(t) is real, thus R[-k] = conj(R[k]). It follows that R[k] is purely imaginary. (ii) x(t) = 1 - r(t+1) - r(t-1) (iii) X[0] = 1 - 2R[0] = 1. For k other than 0,X[k] = -R[k](exp(jkW_0) + exp(-jkW_0)) = -2R[k]cos(kpi/4) (iv) y(t) = jr(t)(exp(j8W_0t) - exp(-j8W_0t))= -2r(t)sin(2pit)
The magnitude and phase response of a linear filter are as plotted below. The nonzero portion of the graph on the left (for |ω| ≤ 2 π/3) is one period of a sinusoid with a suitable DC offset. H ( e j ω) 2π
−2π
H ( e j ω)
−π −2π/3 0 2π/3 π −π −2π/3 2π/3 π
2
(i) (3 pts.) Write an equation for the nontrivial portion of |H(ejω)|, i.e., for |ω| ≤ 2 π/3. (ii) (8 pts.) Determine the output y[n] if the input is given by x[n] = 2 cos^ (^ πn 9 − π 4 ) + 6 cos^ (^ πn 2 + π 5 ) + 3 cos(2. 6 n + 1.7) Simplify your answer. (iii) (4 pts.) Are x[n] and y[n] periodic, and if so, what are their fundamental periods?
(i) For |w| =< 2pi/3, |H(exp(jw))| = 1 + cos(aw), where a4pi/3 = 2pi (i.e., one cycle). Therefore a = 3/2. (ii) Also, angle(H(exp(jw)) = -3w. Hence H(exp(jpi/9)) = (1 + cos(pi/6))exp(-jpi/3) H(exp(jpi/2) = (1 + cos(3pi/4))exp(-j3pi/2) H(exp(j2.6) = 0, and y[n] = (2+sqrt(3))cos(pin/9 - 7pi/12)...+(6-3sqrt(2))cos(pin/2 - 13*pi/10)
(iii) x[n] is not periodic (since w = 2.6 is not a rational multiple of pi). y[n] is periodic with period = LCM(18,4) = 36
Let h = [ 4 -1 1 -4 ].’ ; s = [ 1H = fft(h,9) 2 3 -1 -2 -3 ].’ ; ; S = fft(s,9)C = H.S ;; c = ifft(C) ; (i) (9 pts.) Determine the vector c. (ii) (6 pts.) Without performing another convolution, determine the vector y obtained by h = [ 4 -1x = [ 1 2 1 -4 ].’ ; 3 1 2 3 -2 -4 -6 ].’ ; G = fft(h,12) ;X = fft(x,12) ; Y = G.Xy = ifft(Y) ;;
k: 0 __________________________________________________________s[k]: 1 2 3 -1 -2 -3 c
h[-k]h[1-k] -4 (^) -4 1 -1 1 -1 (^4 4 ) h[2-k]h[3-k] -4 (^) -4 1 -1 1 -1 (^4 4) -9 11 h[4-k]h[5-k] -4 (^) -4 1 -1 1 -1 4 4 -12- h[6-k]h[7-k] -4 (^) -4 1 -1 1 -1 (^4 4 ) h[7-k] -4 1 -1 4 12 c = conv(h,s) = [ 4 7 11 -9 -12 -23 5 5 12 ].’ (ii) x = [s; zeros(3,1)] + 2[zeros(3,1) ; s] The sought vector y equals [conv(h,x) ; 0] = [c; zeros(3,1)] + 2[zeros(3,1) ; c] = [ 4 7 11 -9 -12 -23 5 5 12 0 0 0 ].’ +[ 0 0 0 8 14 22 -18 -24 -46 10 10 24].’ = [ 4 7 11 -1 2 -1 -13 -19 -34 10 10 24].’
