93下信號與系çµ&plusmn

93下信號與系統

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Signals and Systems, Midterm Exam Solutions Spring 2005, Edited by bypeng 1. (12) Consider a continuous-time linear system with the input-output pairs depicted below. Inputs Outputs 1 () x t t 2 0 1 3 1 2 x t t 2 01 3 1 3 x t t 2 3 1 1 yt t 2 0 1 3 1 2 t 2 3 1 3 t 2 3 1 1 Answer the following question and justify your answer. (a) (3) Is the system causal? (b) (4) Is the system memoryless? (c) (5) What would be the response of the system to the following signal x t ? Give a sketch of the response. x t t 2 3 1 2 Solution: The system is linear, therefore (a) The system is NOT CAUSAL since 2 y t is non-zero in the interval (0,1) but 2 x t is zero in the interval (, 1 ) −∞ . (b) A memoryless system implies a causal system, so the system is NOT MEMORYLESS by (a). (c) Observing that 123 () 2 () x tx t x t x t =+ , we have y ty t y t y t = +− . The sketch is given as the following figure. 2. (12) Consider a discrete-time LTI system with unit sample response [ ] ( 1) [ ], n hn n un α where 1 < . Determine the step response of the system by (a) (6) performing the convolution sum (b) (6) using the discrete-time Fourier transform and its properties. Solution: (a) 0 []*[] []*[] [][ ] ( 1 ) [ ] nk kk hn un un hn ukhn k n k un k ∞∞ =−∞ = == = + ∑∑ . if 0 n < , then [ ]* [ ] 0 = ;
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
if 0 n , then 1 00 [ ]* [ ] ( 1) [ ] ( ( ... 2 1 n nk n n kk hn un n k un k n n n αα α −− == = −+ − = + = + + ++ + ∑∑ 1 111 1 1 1 1 [1 ( ) ] [1 ( ) ] [1 ] (1 ) 1 ... ( 1 ... ) 1 1 nnnnn n n n ααα + + =+ + = + + + + −−− 1 11 1 1 1 2 ) ) ) 1 1 ) 1 )(1 ) 1 ( 1)(1 ) ) 1 n nn n n n n + + +++ ++ + =− = = Therefore, 2 1( 1 ) []*[] [] ) 1 n un ⎛⎞ ⎜⎟ ⎝⎠ . (b) 1 () ) j j He e ω −2 = , 1 ( 2) 1 j j k Ue k e πδω π =−∞ , N 22 2 1 (1 ) (1 ) 1 1when 2 (1 ) 1 2 1 ( ) 1 ) 1 (2 ) ( 1 ) ( 1 ) ) ) 1 1 jj j k j j k k k He Ue k ee e k e e k ωω ωπ πδ ω 2 2 = + + + + ) j e 1 2 1 ) { ( ) ( ) } ( 1 ) ( ( 1 ( 1 n n n F He Ue n −( + + = . 3. (12) Consider two sequences ,0 6 0, otherwise n n xn = and 1, 0 3 0, otherwise n yn = Calculate the convolution of the two signals. Solution: 3 0 [][ ] [ ] [] [ 1 ] [ 2 ] [ 3 ] xn yn yn xn ykxn k xn k xn xn = = = −= −= + −+ −+ − n 1 ≤− 0 1 2 3 to 6 7 8 9 10 x ny n 0 1 1 + 2 1 321 nnnn + 456 + + 56 + 6 0 4. (12) Let the input x t to an LTI system with impulse response ht be given in the following figure. Find the output yt . x t t 0 1 1 t 1 0 1 3 1 2 1 Solution: N 0 ()* () ( ) ( ) ( ) ( ) t t t y tx t h t x h t d h t d hd ττ τ τ τ ∞+ −∞ = = ∫∫ ± , so the output is as the following figure.
Background image of page 2
5. (12) Consider a first-order system described by 1 [] [ 1 ] [] 4 y ny n x n −− = .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/21/2012 for the course EE 101 taught by Professor 張捷力 during the Spring '07 term at National Taiwan University.

Page1 / 6

93下信號與系統

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online