Bài t p ch ng Cho chu i x(n) có bi n đ i Z X(z) Hãy xác đ nh bi n đ i Z c a tín hi u sau (theo X(z)) ⎧ n ⎪ x ( ) n chan a x1 ( n ) = ⎨ ⎪⎩ n le b x2 ( n ) = x (2n ) Xác đ nh tín hi u nhân qu x(n) n u bi t bi n đ i Z c a + 3z −1 a X ( z ) = + 3z −1 + z −2 b X ( z ) = −1 − z + 12 z −2 z −6 + z −7 − z −1 + z −2 d X ( z ) = + z −2 + z −1 + z −2 e X ( z ) = + z −1 + z −2 − 12 z −1 f X ( z ) = + 12 z −1 Xác đ nh t t c tín hi u x(n) có th n u bi n đ i Z c a 5z −1 a X ( z ) = (1 − z −1 )(3 − z −1 ) b X ( z ) = −1 − 1.5z + 0.5z −2 c X ( z ) = −1 − z + 14 z −2 Dùng bi n đ i Z đ xác đ nh tích ch p c a c p tín hi u sau ⎧ x ( n ) = ( 14 ) n u ( n − 1) a ⎨ 1 n ⎩ x2 ( n ) = [1 + ( ) ]u( n ) c X ( z) = ⎧ x (n ) = u(n ) b ⎨ 1 n ⎩ x2 ( n ) = δ ( n ) + ( ) u ( n ) ⎧ x ( n ) = ( 12 ) n u( n ) c ⎨ ⎩ x2 ( n ) = cos(π n )u ( n ) ⎧ x ( n ) = nu ( n ) d ⎨ n ⎩ x2 ( n ) = u ( n − 1) N u X(z) bi n đ i Z c a x(n), ch ng minh r ng a Z{x*(n)} = X*(z*) ⎧ x ( n ) neu n/k la so nguyen thi X k ( z ) = X ( z k ) b Neu xk ( n ) = ⎨ k truong hop khac ⎩ c Z {e jw0n x ( n )} = X ( ze − jw0 ) Xác đ nh x(n) có bi n đ i Z X ( z ) = n u bi t X(z) h i t − 103 z −1 + z −2 vòng tròn đ n v Dùng bi n đ i Z m t phía đ xác đ nh y(n) (n≥0) tr ⎧ y ( n ) + 12 y ( n − 1) − 14 y ( n − 2) = a ⎨ ⎩ y ( −1) = y ( −2) = ⎧ y ( n ) − 1.5 y ( n − 1) + 0.5 y ( n − 2) = ⎪ b ⎨ y ( −1) = ⎪ y ( −2) = ⎩ ⎧ y ( n ) = 12 y ( n − 1) + x ( n ) ⎪ c ⎨ x( n ) = ( 13 ) n u( n ) ⎪ y ( −1) = ⎩ ⎧ y ( n ) = 14 y ( n − 2) + x ( n ) ⎪ x(n ) = u(n ) ⎪ d ⎨ ⎪ y ( −1) = ⎪⎩ y ( −2) = Ch ng minh h th ng sau t ng đ ng y (n ) = 0.2 y ( n − 1) + x( n ) − 0.3x ( n − 1) + 0.02 x ( n − 2) y (n ) = x (n ) − 0.1x ( n − 1) Xác đ nh đáp ng tr ng thái không c a h th ng sau a h ( n ) = ( 13 )n u( n ); x ( n ) = ( 12 ) n (cos π3 n )u( n ) ng h p sau b h ( n ) = ( 12 ) n u( n ); x ( n ) = ( 13 ) n u( n ) + ( 12 ) − n u( − n − 1) ⎧ y ( n ) = −0.1 y ( n − 1) + 0.2 y ( n − 2) + x ( n ) + x ( n − 1) c ⎨ n ⎩ x(n ) = ( ) u(n ) ⎧ y ( n ) = 12 x ( n ) − 12 x ( n − 1) d ⎨ π ⎩ x ( n ) = 10(cos n )u ( n ) ⎧ y ( n ) = − y ( n − 2) + 10 x ( n ) e ⎨ π ⎩ x ( n ) = 10(cos n )u ( n ) f h ( n ) = ( 25 ) n u( n ); x ( n ) = u ( n ) − u( n − 7) g h ( n ) = ( 12 ) n u( n ); x ( n ) = ( −1) n ; −∞ < n < +∞ h h ( n ) = ( 12 ) n u (n ); x ( n ) = (n + 1)( 14 ) n u ( n ) − z −1 +2 z −2 − z −3 10 Xét h th ng có H ( z ) = (1 − z −1 )(1 − 0.5z −1 )(1 − 0.2 z −1 ) ROC : 0.5 ≤ z ≤ a Phác th o s đ m c c m không c a h th ng H th ng có n đ nh không? b Xác đ nh đáp ng xung đ n v c a h th ng