# ps4sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 4 Solution Home Study Exercise - O&W 3.63 For an LTI system whose frequency response is: 1 , | | ← W H ( j ) = 0 , | | > W. and which has a continuous-time periodic input signal x ( t ) with the following Fourier series representation: x ( t ) = ± | k | e jk ( / 4) t , where ± is a real number between 0 and 1 k = How large must W be in order for the output of the system to have at least 90% of the average energy per period of x ( t )? Basically, H ( j ) is an ideal Low Pass Filter (LPF) and we need to ±nd how wide it needs to be, in order to pass at least 90% of its input’s average energy per period (i.e. average power). First, let’s rewrite the condition above relating the average powers of the input and out- put, with Fourier series coeﬃcients a k and b k , respectively: | b k | 2 P x = | a k | 2 , P y = k = k = The required condition, then,would be: | b k | 2 P y RP x R | a k | 2 , where R=0.9 ( ) k = k = Then, let’s calculate the Fourier series coeﬃcients of the output, b k : a k , | k 0 | ← W a k , | k | ← W/ 0 b k = a k H ( 0 ) b k = = 0 , | k 0 | > W 0 , | k | > W/ 0 1

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± ² ³ ´ And fnally, we plug the expressions oF a k and b k in the required condition and simpliFy. By matching the the expression oF x ( t ) with the synthesis equation, we can conclude that 0 = and a k = ± | k | 4 P x = | a k | 2 = | ± | k | | 2 = 2 ± 2 k 1 ( ± is real) k = k =0 k = 2 = 1 ( 0 < | ± | < 1) 1 ± 2 N P y = | b k | 2 = | a k | 2 , where N is the largest integer, such that N W/ 0 k = k = N N N = | ± | k | | 2 = 2 ± 2 k 1 ( ± is real) k = N k =0 1 ( ± 2 ) N +1 M ² 1 ² M +1 n = 2 1 = , For any complex ² = 0 1 ± 2 1 ² n =0 Plugging in ( ): 1 ( ± 2 ) N +1 ³ ´ 2 2 1 R 1 1 ± 2 1 ± 2 2 2 ± 2 N +2 2 R R + 1 1 ± 2 1 ± 2 2 2 ± 2 N +2 2 R + (1 R )(1 ± 2 ) 1 ± 2 1 ± 2 2 2 ± 2 N +2 R + 1 (1 R ) ± 2 ( 1 ± 2 > 0) 2 2 ± 2 N +2 ± 1 . 9 0 . 1 ± 2 (plugging in R=0.9) 2 N +2 0 . 05 + 0 . 05 ± 2 (simpliFying a bit) (2 N + 2) log( ± ) log(0 . 05 + 0 . 05 ± 2 ) log(0 . 05 + 0 . 05 ± 2 ) N + 1 (recall that ± < 1 log( ± ) < 0) 2 log( ± ) log(0 . 05 + 0 . 05 ± 2 ) N 1 2 log( ± ) AFter choosing an integer N that satisfes the inequality above, W can be chosen such that W N 0 . 2
± ± Problem 1 Consider the LTI system with impulse response given in O&W 3.34. Find the Fourier series representation of the output y ( t ) for the following input. x ( t ) 4 3 2 3 5 6 · · · 5 2 1 2 1 1 4 · · · t From O & W 3.34, the impulse response of the system is: t h ( t ) = e 4 | | . the ±gure above, we can see that x ( t ) has a period T = 3 0 = 2 3 . First, we calculate the frequency response: t e j±t dt H ( j ) = h ( t ) e dt = e 4 | | 0 e 4( t ) e dt + e 4( t ) e dt = 0 0 (4 ) t dt + ( 4 ) t dt = e e 0 ± ± 0 1 ± 1 ± (4 ) t ± + ( 4 ) t = e e 4 j ± 4 j 0 1 1 = (1 0) + (0 1) (remember that e + ja = 0 for any real a) 4 j 4 j 1 1 4 + j + 4 j = + = 4 j 4 + j 16 + 2 8 H ( j ) = .

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