S 1.1 (P 1.9) _____________ Cartesian form: Numerator: a = (1 + j*sqrt(3))*(1 - j) = (1+sqrt(3)) + j*(-1+sqrt(3)) Denominator b = (sqrt(3) + j)*(1 + j) = (-1+sqrt(3)) + j(1+sqrt(3)) b' (complex conjugate) = (-1+sqrt(3)) - j(1+sqrt(3)) |b|^2 = (sqrt(3)-1)^2 + (sqrt(3)+1)^3 = 8 a*b' = 4 - j*4*sqrt(3) a/b = a*b'/|b|^2 = 1/2 - j*sqrt(3)/2 Polar form: 1+j*sqrt(3): modulus = 2, angle = pi/3 1-j: modulus = sqrt(2), angle = -pi/4 sqrt(3) + j: modulus = 2, angle = pi/6 1+j: modulus = sqrt(2), angle = pi/4 a: modulus = 2*sqrt(2), angle = pi/12 b: modulus = 2*sqrt(2), angle = 5*pi/12 a/b: modulus = 1, angle = pi/12 - 5*pi/12 = -pi/3 (Note that |a/b|=1, which follows directly from the identities |z1*z2| = |z1|*|z2| and |z1/z2| = |z1|/|z2| .) S 1.2 _____ z = (1+j*2)/(3-j*b). Multiplying numerator and denominator by 3+j*b, we obtain z = (1+j*2)(3+j*b)/(9+b^2) = (3-2*b)/(9+b^2) + j*(6+b)/(9+b^2) i) z is purely real if its imaginary part is zero, i.e., if b=-6. Then z = 1/3. ii) z is purely imaginary if its real part is zero, i.e., if b = 3/2. Then z = j*(3/2). iii) The modulus of z is the ratio of the moduli of 1+2*j and 3-j*b. Thus |z|^2 = 5/(9+b^2) If |z|^2 = 1/2, then 9+b^2 = 10, i.e., b = +1 or -1. S 1.3 _____ We first express z in polar form: |z| = r = sqrt( 9/25 + 81/100 ) = 1.081665 angle(z) = t = atan(9/6) = 0.9827934 radians Then |z^17| = r^17 = 3.798 angle(z^17) = 17*t = 16.707, which reduces to 4.141 by subtracting 4*pi from it. S 1.4 _____ i) |z-1+j|=1 represents a circle centered at z0 = 1-j = (1,-1) and having radius equal to 1. From elementary geometry, the point on the circle closest to the origin will be z = (c,-c) where c = 1-sqrt(2)/2 = 0.2929. The modulus of z equals sqrt(2)-1 = 0.4142. ii) |z-2| = |z-3*j| represents the perpendicular bisector of the linear segment joining (2,0) and (0,3). The distance of the perpendicular bisector from the origin can be found geometrically by considering two similar triangles. It equals 5/(2*sqrt(13)) = 0.6934 S 1.5 (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 1.6 _____ 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 1.7 _____ 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 = (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 = (cos(t))^3 - 3*cos(t)*(1-(cos(t))^2) = 4*(cos(t))^3 - 3*cos(t) and sin(3*t) = 3*(cos(t))^2*sin(t) - (sin(t))^3 = 3*(1-(sin(t))^2)*sin(t) - (sin(t))^3 = 3*sin(t) - 4*(sin(t))^3 S 1.8 _____ The numerator of the ratio is |z - w'| = |z'- w| Following the hint, |z|=1 implies that z'=1/z, therefore |z - w'| = |z'- w| = |(1/z) - w| Factoring out |w/z|, we have |(1/z) - w| = |w/z|*|(1/w) - z| Note that the second factor equals the denominator of the ratio, and thus |z - w'|/|z - (1/w)| = |w/z| = |w| (since |z|=1). Thus the ratio is constant for all values of z on the unit circle. S 1.9 (P 1.11) ______________ (i) The phase shift (phase at t=0) equals 0.25 < pi/2. The first zero will occur when 500*pi*t + 0.25 = pi/2 or t = (pi/2 - 0.25)/(500*pi) = 0.841 ms (ii) The period of x(t) is T = (2*pi)/(500*pi) = 4.0 ms Thus time interval [0, 0.01] corresponds to 2.5 periods of the sinusoid. The zeros will occur for values of t such that 500*pi*t + 0.25 = (2*k+1)*(pi/2) where k is integer (i.e., when the LHS equals an odd multiple of pi/2. Solving for t (as in problem 1), we have t = (2*k+1)*(0.001) - 0.0005/pi The first term on the RHS above is a rational number, the second one is irrational. Therefore each value of t given by the above equation is irrational. The values of t generated by MATLAB are multiples of 0.001, hence rational. Therefore, none of the 101 values in the vector x are zero. This can be verified using the MATLAB command min(abs(x)) S 1.10 (P 1.12) ______________ (i) Let the amplitude be A (>0). Then 2 < |x(t)| < A for 30% of each cycle (period). Starting a cycle at a positive peak, the signal therefore takes 7.5% of the cycle to drop from A to +2 35% of the cycle to drop from +2 to -2 15% of the cycle to go from -2 to -A to -2 (in the trough) 35% of the cycle to rise from -2 to +2 7.5% of the cycle to complete it (return to A) Thus 2/A = cos(0.075*2*pi), i.e., A = 2.245 (ii) From the cycle breakdown shown above, we have 300 ms = 0.35*T, i.e., T = 857 ms (iii) A phase shift of zero (i.e., x(t) = A*cos(W*t)) would result in the first downward crossing of the level -2.0 at t = (0.075 + 0.35)*T = 364ms or equivalently, an angle W*t = (0.075 + 0.35)*2*pi Since the first crossing actually occurs at t=40 ms, this means we have a phase advance equal to (364 - 40)/857 of a cycle, or (364 - 40)*2*pi/857 = 2.3771 radians. Thus x(t) = 2.245*cos(7.330*t + 2.3771) S 1.11 (P 1.13) ______________ v(t) = A*cos(W*t + q) (i) During the first period of y = cos(theta), the value of y is >1/2 for 0 <= theta < arccos(1/2) = pi/3 and then again for (5*pi)/3 < theta < 2*pi This represents exactly one third of the period of cos(theta). Thus in the long run, the diode stays on 33.3% of the time. (ii) Using the same facts as in (i) above, we have W*(1.5*10^(-3)) + q = pi/3 W*(9.5*10^(-3)) + q = 5*pi/3 Therefore W*(8*10^(-3)) = 4*pi/3 and W = (500*pi)/3 rad/sec (i.e., frequency = 83.3 Hz) q = pi/3 - 0.75*pi/3 = pi/12