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Electrical Engineering 20N - Spring 2000 - Midterm 2

# Electrical Engineering 20N - Spring 2000 - Midterm 2 - EECS...

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Unformatted text preview: EECS 20. Midterm No. 2 Solution April 5, 2000. 1. (a) The fundamental frequency is !0 = 1. The Fourier series coefﬁcients are A0 A2 = 1 and 1 = ,=2, with everything else having value 0. = A 1 = (b) The output will have Fourier series coefﬁcients Ak scaled by the frequency response H !, so A0 = 1 and A1 = 1=2, and all others are 0. The phases of the frequency response add to those of the input, so the output will have 1 = ,=2 + =2 = 0. I.e., k = 0 for all k . Thus, the output is yt = 1 + cost=2: (c) The frequency, magnitude and phase responses of the cascade composition are G! = H !; jG!j = jH !j ; 6 G! = 26 H !: 2 2 Here is a sketch: ∠G(ω) |G(ω)| 2π 1 0 1 ω 2 0 1 2 ω (d) The Fourier series coefﬁcients of the input will now be scaled by G!  instead of H ! , getting A0 = 1 A1 = 1=4, and 1 = ,=2 +  = =2. Thus, the output is yt = 1 + cost + =2=4 = 1 , sint=4: 2. (a) The sketches are shown below: 1 u(n) ... ... n ... n ... n ... n δ(n) ... y(n) ... (b) Note that n  = un , un , 1: Since the system is LTI, it must therefore be true that hn = yn , yn , 1 = n , n , 4: This is sketched below: h(n) ... 3. (a) Note that 2 cos  t=6 = = = eit= + e,it=  =4 i t= + 2 + e,i t= =4 e 1 + cos2t=6=2: 6  2 6 2 6 2 6 Moreover, sin t=6 = cost=6 , =2: Therefore (b) xt = 0:5 + cost=6 , =2 + 0:5 cos2t=6: Using the results of part (a), ! = =6; A = 0:5, A = 1, A k 2; and = ,=2, and k = 0 for k 1. 0 0 1 2 1 2 : = 0 5, and Ak = 0 for (c) Rewriting the result from part (a), : t=6 , =2 + 0:5 cos2t=6 i t= ,= + 0:5e,i t= ,= + 0:25ei t= + 0:25e,i t= : = 0:5 + 0:5e From this, we can read of the Fourier series coefﬁcients, X = 0:5, X = 0:5e,i= = ,j=2, X, = 0:5ei= = j=2, X = X, = 0:25, and Xk = 0 for k 2 or k ,2. xt = 0 5 + cos  1 4. 2 6 2  6 2 2 0 2 6 1 2 6 2 2 (a) Note from the difference equation that what we need to remember about the past is yn , 1. Thus, deﬁne the state to be sn = yn , 1: (You could equally well choose to deﬁne the state to be ,0:9y n , 1, among other possible choices.) Thus, this is a one-dimensional SISO system. The state update equation becomes sn + 1 = ,0:9sn + xn because sn + 1 = y n. Thus, A = ,0:9, a 1  1 matrix, and b = 1. The output equation is (b) yn = ,0:9sn + xn from which we recognize c = ,0:9 and d = 1. Let the input x = , the Kronecker delta function, and note that 8 0; n 0 1; n=0 yn = ,0:9 n = 1 :81 2 : 0,0:9n n = 2  n This can be written more compactly as y n = ,0:9n un, where un is the unit step function. 3 ...
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