hw07-s06-sol-redacted - EECS 20N: Structure and...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY HW 7.1 (a) x(t) = e-at u(t), a > 0 Problem Set 7 SOLUTIONS X() = - e-at u(t)e-jt dt e-at e-jt dt = 0 e-(a+j)t = | -(a + j) 0 0-1 = -(a + j) 1 = a + j (b) x(t) = e-a|t| X() = - 0 x(t)e-jt dt eat e-jt dt + e(a-j)t dt + 0 = - 0 e-at e-jt dt e-(a+j)t dt = = e e |0 + | - a - j -(a + j) 0 1-0 0-1 = + a - j -(a + j) 1 1 = + a + j a - j 2a = 2 a + 2 - (a-j)t 0 -(a+j)t 1 (c) x(t) = sgn(t) dx(t) = 2(t) dt y(t) = 2(t) Y () = - 2(t)e-jt dt = 2 dx(t) dt differential property: CT F T = 2 dx(t) d 1 y(t) = X()ejt d = dt dt 2 - 1 = X()jejt d 2 - Y () = X()j For = 0 Y () X() = j For = 0 X(0) = X() = (d) 0 sgn(t)e-jt dt|=0 = =0 =0 sgn(t)dt = 0 - - 2 j x(t) = u(t) = Let y(t) = 1 Then, 2 1 1 = Y ()ejt dt 2 2 - Y () = () Using 1(c), 1 =0 j X() = () = 0 2 sgn(t) + 1 2 (e) x(t) = k=- (t - kT ) Xk eik0 t , k=- T 2 x(t) = Xk 1 = T 1 T 0 = 2 CTFS Expansion T 1 T (t)e-ik0 t dt = eik0 t -T 2 x(t) = X() = HW 7.2 (a) 2 T k=- ( - k k=- 2 ) T x(t) = x(at) ^ ^ X() = if Let t a>0t t a<0t t = = = = = - x(at)e-it dt at, dt = adt when t = - when t = - - when t = when t = - t x(t )e-i a dt a - t - x(t )e-i a dt a -i a t ^ X() = = a>0 a<0 1 x(t )e |a| - 1 ^ X() = X( ) |a| a If x(t) = x(-t), i.e.a = -1 ^ then dt ^ X() = 1 X( ) = X(-) | - 1| -1 Reversing time reverses frequency. 3 (b) x(t) = x(t) = ^ = = = 1 X()eit d 2 - d x(t) dt d 1 X()eit d dt 2 - 1 d X()( eit )d 2 - dt 1 (iX())eit d 2 - ^ X() = iX() (c) (i) - x(t)y (t)dt = = = = < x, y > = (ii) Let y = x 1 Y (-)eit d dt 2 - - Using conjugating property x(t) X(), x (t) X (-) 1 x(t)eit dt d Y (-) 2 - - - 1 x(t)e-i t dt (-d ) Y ( ) 2 - changing variable = - 1 Y ( )X( )d 2 - 1 < X, Y > 2 x(t) LHS =< x, x >= - x(t)x (t)dt = - |x(t)|2 dt - 1 1 RHS = < X, X >= 2 2 1 X()X ()d = 2 - 1 < X, X > 2 |X()|2 d < x, x >= 4 (d) x(t) = 2x(-t ) = - 1 2 - X()eit d X()e-it d, changing variable - t = t X(t )e-i t dt , changing variable = t , t = 2x(- ) = - This is analogous to the CTFT analysis equation: X() = - f x(t)e-it dt X(t) 2x(-) HW 7.3 (a) X() = n=- x(n)e-in d d i X() = i d d = i x(n)e-in x(n) d -in e d n=- n=- = i n=- x(n)(-in)e-in (nx(n))e-in = n=- x(n) = nx(n) (b) Upsampling 5 X() = n=- x(n)e-in x(n)e-in n=- X() = but x(n) is none-zero only at n = kN, k = . . . - 2, -1, 0, 1, 2, . . . X() = n=- x(kN )e-ikN x(k)e-i(N )k n=- = = X(N ) Echo system Remember that (n - n0 ) e-in0 , and that for LTI system Y () = H()X() From the difference equation y(n) = x(n) + y(n - N ) Take Fourier transform of both sides Y () = X() + e-iN Y () (1 - e-iN )Y () = X() H() = Y () X() 1 = 1 - e-iN F For single sample echo system y(n) = x(n) + y(n - 1) We have frequency response H() = 1-e1-iN by letting N = 1 in previous part. So, H() = H(N ) Find impulse response: h(n) is, by definition, the output when the input is an impulse, so let x(n) = (n) and find y(n) 6 = = = = . . . y(n) = h(n) = y(n) y(n) y(0) y(1) (n) + y(n - 1) 0, n < 0 1 n , n 0 n u(n) where u(n) is the unit step function. By upsampling property, h(n) = N 0 n if n 0 and n mod N = 0 otherwise HW 7.5 (a) (I) h(n) = 0 (II) (i) A() = H()ei2 a(n) = h(n + 2) (ii) a(n) = a(0) = (iii) a(n)is symmetric. so a(n) and h(n)are both length 5 . 1 2 1 2 or h(n) = a(n - 2) when n<0 A()ein d <2> A()d = <2> 1 (12) = 6 2 9 (iv) 2 A() = n=-2 2 a(n)e-in a(n) n=-2 A(0) = = = = A() = = n=-2 2 a(-2) + a(-1) + a(0) + a(1) + a(2) 2a(2) + 2a(1) + 6 8 (from the figure) A(-) 2 a(n)e-in a(n)ein n=-2 2 = = n=-2 a(n)(-1)n = a(-2) - a(-1) + a(0) - a(1) + a(2) = 2a(2) - 2a(1) + 6 = 12 We have 2 equations for 2 unlcnowns...solving them gives a(2) = a(-2) = 2, a(1) = a(-1) = -1 a(n) 6 h(n) 6 2 2 n 2 2 n 0 1 -1 2 3 -1 4 -2 -1 -1 0 1 -1 2 (b) y(n) = (x h)(n) = 2x(n) - x(n - 1) + 6x(n - 2) - x(n - 3) + 2x(n - 4) 10 from the sifting property of the delta funtion (c) x(n) 2 z-1 q1(n) z-1 q2(n) z-1 q3(n) z-1 q4(n) 2 -1 6 -1 + y(n) + + + (d) q1 (n + 1) 0 0 q2 (n + 1) 1 0 q3 (n + 1) = 0 1 q4 (n + 1) 0 0 A 0 0 0 1 0 q1 (n) 0 q2 (n) 0 q3 (n) q4 (n) 0 1 0 + x(n) 0 0 B y(n) = -1 6 -1 2 C q1 (n) q2 (n) q3 (n) + 2 x(n) D q4 (n) 11 (e) 4 H() = n=0 h(n)e-in 2 - e-i + 6e-2i - e-3i + 2e-4i e-2i (2e2i - ei + 6 - e-i + 2e-2i ) ei(-2) (6 - 2 cos() + 4 cos(2)) -2 (6 - 2 cos() + 4 cos(2)) = |A()| So the given figure can be used for (f) because |H(0)| = 8, H(0) = 0 = = = H() = |H()| = (f) (i) y(n) = 8, (ii) x(n) = ein y(n) = ei(-2) (6 - 2 cos() + 4 cos(2))ein = 12ein = 12(-1)n (iii) y(n) = (6 - 2 cos( ) + 4 cos( )) cos( n - ) 4 2 4 2 = (6 - 2) sin( n) 4 ...
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This note was uploaded on 04/02/2008 for the course EE 20N taught by Professor Ayazifar during the Fall '08 term at University of California, Berkeley.

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