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With t in seconds, let x(t) = − 4 .1 + 2.3 cos(28πt − 1 .5) + 1.9 cos(42πt − 0 .7) + 3.7 cos(70πt + 1.4) (i) Determine the fundamental period T 0 (in sec) of x(t). (ii) 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,200) ;x = real(z) ;
generates a vector x consisting of 200 samples of x(t) uniformly spaced in [0, T 0 ). (iii) If b and c are as specified in (ii) above, the code b_new = [1.7 ; 0 ; conj(b)] ; % CONJ(.) : COMPLEX CONJUGATE z = c*ifft(b_new,100) ; % 100, not 200 s = real(z) ; produces 100 samples, uniformly spaced in [0, T 0 ), of a periodic signal s(t) with the same fundamental period. Assuming s(t) has no harmonics higher than the 50th, write an equation for s(t). (iv) Let y(t) = x(t) cos(6πt). What is the fundamental frequency (in Hz) of y(t)?
(i) x(t) is a sum of sinusoids with frequencies 0, 14, 21 and 35 Hz. Therefore f_0 = GCD(14,21,35) = 7 Hz, and T_0 = 1/7 sec.Nontrivial harmonics at k = 0, 2, 3 and 5.
(ii) b = [ -4.1 0 2.3exp(-j1.5) 1.9exp(-j0.7) 0 3.7exp(j1.4) ].’ c = 200 (iii) Nontrivial harmonics at k = 1, 2, 4, 5 and 7. Since c=200 is used (instead of 100 = number of samples generated in this case), s(t) = 3.4 - 8.2cos(28pit) + 4.6cos(56pit + 1.5) + 3.8cos(70pit + 0.7) ...+ 7.4cos(98pit - 1.4)
(iv) x(t) consists of complex sinusoids of frequencies 0 and (+/-) 14, 21 and 35 Hz. Each will be multiplied by a complex sinusoid at (+/-) 3 Hz, so y(t) will be a sumof real sinusoids of frequencies 3, 11, 17, 18, 24, 32 and 38 Hz. Thus fundamental frequency of y(t) = 1 Hz.
The signal r(t) shown below (top graph) is periodic with period T 0 = 12 and complex Fourier series expansion r(t) =
∑^ ∞
k=−∞^ Rke
jkΩ 0 t (^) ,
where Ω 0 is the fundamental angular frequency.
−12 −6 − 2 2 6 12 t
y ( t ) 1 ... ... t −
r ( t ) ... 1 ... −
−3 3
(i) Evaluate R 0. (ii) For what value of D in (0, 12) does s(t) = r(t − D) have purely real Fourier series coefficients Sk? (iii) Let x(t) be periodic with the same period as r(t), and such that X 0 = − 2 / 3 and Xk = 2 Rk cos(kπ/2) (k 6 = 0) Express x(t) in terms of r(t). (iv) Express the real-valued periodic signal y(t) (bottom graph) in terms of r(t). The curved sections of the graph are sinusoidal half-cycles. (v) Express the kth^ Fourier series coefficient Yk of y(t) in terms of of Rk’s.
(i) R[0] = (Net area under graph)/(T_0) = (4-2)/12 = 1/ (ii) r(t) is real and even-symmetric, so the same holds for R[k]. S[k] = R[k]exp(-jW0D) will be real only if W0D = pi, i.e.,D = T_0/2 = 6. (Note that the resulting graph would also be even-symmetric.)
(iii) X[0] = -2/3 = 2R[0] - 1. Therefore for all k, X[k] = R[0]( exp(jkW03) + exp(-jkW03) ) - delta[k] The -delta[k] term represents a DC shift of -1, thus x(t) = r(t+3) + r(t-3) - 1 (iv) y(t) = -r(t)sin(6W0t) = (j/2)r(t)(exp(j6W0t) - exp(-j6W0t)) (v) By modulation property, Y[k] = (j/2)( R[k-6] - R[k+6] )
PROBLEM 4 Grade: The magnitude and phase response of a linear filter are as plotted below. H ( e j ω) 4π/
−4π/
A
H ( e j ω)
−π −2π/3 0 2π/3 π −π −2π/3 2π/3 π
(i) Determine the output y[n] if the input is given by
x[n] = 5 + 2 cos^ (^ πn 4 − π 3 ) + 3 cos
( (^5) πn 7 −^
2 π 5
Simplify your answer. (ii) Are x[n] and y[n] periodic, and if so, what are their fundamental periods? (iii) Suppose y[n] is the input to a second filter, resulting in the output v[n] = y[n] + y[n − 4] Determine v[n], simplifying your answer.
(i) Magnitude and phase response values, from graphs: w = 0 : mag = A, phase = 0 w = pi/4 : mag = A(2/3 - 1/4)/(2/3) = 5A/8, phase = -pi/ w = 5pi/7 : mag = 0 (since 5pi/7 > 2pi/3), phase irrelevant Therefore y[n] = 5A + (5A/4)cos(pin/4 - 5pi/6) (ii) Period of x[n] = LCM(8,14) = 56 Period of y[n] = 8 (iii) The second filter has complex frequency response 1 + exp(-j4w), equal to 2 at w = 0 ; and 0 at w = pi/ Therefore v[n] = 10*A, all n
Consider the FIR filter with input-output relationship y[n] = 3 x[n] − 2 x[n − 1] + 2x[n − 2] − 3 x[n − 3] (i) The input x[0 : 3] = [ 1 2 4 8 ]T^ ; x[n] = 0 for all other n is applied to the filter. Determine the vector c = y[0 : 6] of corresponding output samples. What is the value of y[n] for n < 0 and n > 6? (ii) Suppose a different input, namely x′[0 : 6] = [ 1 2 4 6 − 4 − 8 − 16 ]T^ ; x′[n] = 0 for all other n is applied to the same filter. Express the nontrivial part of the resulting output sequencethe vector c defined in part (i) above, with zero-padding as appropriate. (You should not use a convolution y′[n] in terms of table for this part.) (iii) Finally, consider the input sequence x˜[n] =
{ (^2) n, n ≥ 0 0 , n < 0 Determine the output ˜y[n] at all times n, distinguishing between different cases. (Use the result of (i) where appropriate and simplify any expression involving the general parameter n.)
(i) Convolution table: _________________________________________________________________x[k]^1 2 4 8 c h[-k]h[1-k] -3 (^) -3 2 -2 2 -2 (^3 3 ) h[2-k]h[3-k] -3 (^) -3 2 -2 2 -2 (^3 3 ) h[4-k] -3 2 -2 3 - h[5-k]h[6-k] -3 (^) -3 2 -2 2 -2 (^3 3) -24 4 y[0:6] = c (column on right); y[n] = 0 for n<0 and n> (ii) x’[n] = x[n] - 2x[n-3] , therefore y’[n] = y[n] - 2y[n-3] The nontrivial portion of y’[.] is y’[0:9] = [c ; 0 ; 0 ; 0] - [0 ; 0 ; 0 ; 2c] (y’[n] = 0 for n<0 and n>0) (iii) y_tilde[n] = 0 for n< y_tilde[0:3] = y[0:3] = [3 4 10 17].’ Starting at n=3, y[n] = (3 - 2/2 + 2/4 -3/8)2^n = (17/8)2^n
