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# solved02 - S 2.1 (P 1.10) _ Since cos(t)+j*sin(t) =...

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S 2.1 (P 1.10) _____________ Since cos(t)+j*sin(t) = exp(j*t), we have z = exp(j*pi/N) * exp(j*2*pi/N) * . .. * exp(j*(N-1)*pi/N) which reduces to exp(j*t) for t = (1+2+. ..+N-1)*pi/N Since 1+2+. ..+N-1 = N(N-1)/2, we have z = exp(j*(N-1)*pi/2) To express z in Cartesian form, note that w = exp(j*pi/2) = cos(pi/2) + j*sin(pi/2) = 0 + j*1 = j and thus z = w^(N-1) = j^(N-1) Hence z = -j if N = 4*k; = 1 if N = 4*k+1; = j if N = 4*k+2; = -1 if N = 4*k+3; where k is an integer. S 2.2 _____ We know that exp(j*t) + exp(-j*t) = 2*cos(t) Pairing up terms with opposite (negative) angles, we have: f = 1 - 2*cos(t) + 2*cos(2*t) S 2.3 _____ Let z = exp(j*t) = cos(t) + j*sin(t) (i.e., z is on the unit circle). Then z^3 = exp(j*3*t) = cos(3*t) + j*sin(3*t) (eqn. 1) But also, by expanding the cube: z^3 = (cos(t) + j*sin(t))^3

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= (cos(t))^3 + 3*j*(cos(t))^2*sin(t) + 3*(-1)*cos(t)*(sin(t))^2 + (-j)*(sin(t))^3 (eqn. 2) Equating the real and imaginary parts of z^3 as given by equations 1 and 2: cos(3*t) = (cos(t))^3 - 3*cos(t)*(sin(t))^2
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## This note was uploaded on 10/18/2011 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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solved02 - S 2.1 (P 1.10) _ Since cos(t)+j*sin(t) =...

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