96下信號與系çµ&plusmn

96下信號與系çµ&plusmn

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Signals and Systems, Midterm Exam Solutions Spring 2008, Edited by bypeng 1. [10] Consider a system H to be tested as being memoryless , causal , linear , time invariant , and invertible . Three signals 1 () x t , 2 x t , and 3 x t are sent to the system, and the corresponding output signals 1 y t , 2 y t , and 3 y t are obtained as shown in Figure 1. Based on the three input-output pairs, is it possible to determine each of the five properties for system H ? If yes, what is it? If no, why? Justify your answer. Figure 1 Solution: i. NEITHER memoryless NOR causal. Observe that 13 x tx t = for 2 t < , but y ty t for 02 t << . ii. NOT linear. Observe that 312 x t =+ , but y t + . iii. NOT time invariant. Observe that 21 ( 1 ) xt xt = , but ) yt yt . iv. NOT invertible. Observe that 12 x t , but y t = . 2. Consider a system as shown in Figure 2, where ht is the impulse response of the LTI sub-system in the block, and 2D is the operation of time delay for 2 units. H H H 1 2 3 4 5 1 2 3 0 4 x 1 ( t ) 1 2 3 4 5 1 2 3 0 4 x 2 ( t ) 1 2 3 4 5 1 2 3 0 4 x 3 ( t ) 1 2 3 4 5 1 2 3 0 4 y 1 ( t ) 1 2 3 4 5 1 2 3 0 4 y 2 ( t ) 1 2 3 4 5 1 2 3 0 4 y 3 ( t ) h ( t ) 2D + x ( t ) y ( t )
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Figure 2 (a) [4] Plot the impulse response of the overall system. (b) [7] Plot the output () y t of the system for input x t shown in Figure 2. (c) [5] Repeat x t in time with a period of 6, and let ) ( ~ t x be the corresponding periodic version of ( ) x t . Plot the output ) ( ~ t y of the system for input ) ( ~ t x . Solution: (a) The impulse response of the block 2D is ( 2) t δ , so the plot of the impulse response of the overall system is as the right graph. (b) For the upper part, [ ] ( )* ( ) ( ) ( 2) 2 ( 4) * ( ) ( )* ( ) ( 2)* ( ) 2 ( 4)* ( ) x th t u t u t h t h t h t =+ = + and 2 1 3 00 ()* () ( ) 0 3 33 t t ut ht h d t t t ττ −∞ < = =≤ < So t interval (2 ) * ( ) ut ht 2( 4 )*() x t 0 t < 0 0 0 0 02 t ≤< 2 1 3 t 0 0 2 1 3 t 23 t 2 1 3 t 22 11 4 4 3 3 ) tt t −=− + 0 2 24 4 3 −+ 34 t 3 2 14 4 3 0 2 13 3 45 t 3 2 4 3 16 32 3 3 (4 ) t =− + 2 19 1 4 57 t 3 3 2 16 32 2 3 + 2 16 21 4 3 −+− 7 t 3 3 6 0 Adding the lower part, the plot is as following left graph. (c) Since ) ( ~ t x is the periodic version of ( ) x t with the period of 6, ) ( ~ t y is the periodic version of y t with the same period 6. The plot is as the above right graph. 12345 1 2 0 h total ( t ) 1 2 3 4 5 1 2 0 x ( t ) 1 2 0 h ( t )
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3. [6] Let [ ] x n be a periodic discrete-time sequence with period 8 N = and Fourier series coefficients 4 = k k a a . Now generate a sequence ] 1 [ 2 ) 1 ( 1 ] [ + = n x n y n with period 8 N = based on [ ] x n . Denoting the Fourier series coefficients of [ ] y n as k b , find a function ] [ k f such that k k a k f b ] [ = . Solution: 2 4 8 1(1 ) 1 1 [] [ 1 ] ] (1 ) [ 1 ] 22 2 11 [1 ] ] ] ] n n jn yn xn e xn e π ⎛⎞ +− =− = + ⎜⎟ ⎝⎠ + + So (4 ) 88 44 4 4 1 1 () 2 2 jk j k jk j kk k k k k ba e a e e a a e e e a ππ −− =+ = + = That is, 4 fk e = 4. Consult tables of Fourier transform pairs and answer the following questions: (a)
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96&auml;&cedil;‹&auml;&iquest;&iexcl;&egrave;™Ÿ&egrave;ˆ‡&ccedil;&sup3;&raquo;&ccedil;&micro;&plusmn

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