EE 20N - Spring 2002 - Varaiya - Midterm 2 Solutions

EE 20N - Spring 2002 - Varaiya - Midterm 2 Solutions -...

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EECS20, Spring 2002 – Solutions to Midterm 2 1. 30 points (a) By inspection, we obtain that w 0 = 1 , A 0 = 3 , A 3 = 2 , A 4 = 3 , φ 3 = - π/ 2 . All other A k and φ k are equal to zero. (b) Using the relations X 0 = A 0 , X k = 0 . 5 A k e k , k = 1 , 2 , ··· X - k = X k * = 0 . 5 A k e - k , k = 1 , 2 , ··· we get X k = 3 , k = 0 - i, k = 3 i, k = - 3 1 . 5 , k = - 4 , 4 0 , otherwise (c) y ( t ) = X k = -∞ X k H ( w 0 k ) e ikw 0 t = X 0 H (0) + X 3 H (3) e i 3 t + X - 3 H ( - 3) e - i 3 t + X 4 H (4) e i 4 t + X - 4 H ( - 4) e - i 4 t = 4sin(3 t ) + 6cos(4 t ) . 2. 20 points (a) s ( n ) = x ( n - 1) x ( n - 2) y ( n - 1) , A = 0 0 0 1 0 0 0 2 1 . 1 , b = 1 0 0 , c = h 0 2 1 . 1 i , d = 0 . (b) Given that x ( n ) = δ ( n ) , we can compute y ( n ) directly from the relation y ( n ) = 2 x ( n - 2) + 1 . 1 y ( n - 1) . We get y ( n ) = ( 2(1 . 1) n - 2 , n 2 0 , otherwise (c) No, this system is not stable. From part (b) above, we see that for one
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Unformatted text preview: particular input ( x ( n ) = ( n ) ), the output y ( n ) as n . 1 3. 20 points (a) TI (b) TI (c) N (d) TI (e) L 4. 15 points The new input (call it x 2 ( t ) ) can be expressed in terms of the old input (call it x 1 ( t ) ) as x 2 ( t ) = x 1 ( t )-x 1 ( t-1) . Using the linearity and time-invariance properties, we obtain that y 2 ( t ) = y 1 ( t )-y 1 ( t-1) , which can be simplied to y 2 ( t ) = ( sin( t ) , t < 4 , otherwise 2...
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EE 20N - Spring 2002 - Varaiya - Midterm 2 Solutions -...

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