midsem_solutions

midsem_solutions - Department of Electrical Engineering EE...

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Unformatted text preview: Department of Electrical Engineering EE 308: Communication Systems (Spring 2009) Solutions to Midsem 1. (a) Since p ( t ) is a periodic signal with period 1 f c , it can be represented as a fourier series of the form: p ( t ) = a 2 + ∞ summationdisplay k =1 [ a k cos(2 πkf c t ) + b k sin(2 πkf c t )] . Since x c ( t ) = x ( t ) cos(2 πf c t ), X ( f ) f c- f c X c ( f ) p ( t ) .x c ( t ) = [ a 2 + ∞ summationdisplay k =1 [ a k cos(2 πkf c t ) + b k sin(2 πkf c t )]] .x ( t ) cos(2 πf c t ) = a 2 .x ( t ) cos(2 πf c t ) + 1 2 ∞ summationdisplay k =1 [ a k cos(2 π ( k + 1) f c t ) + a k cos(2 π ( k − 1) f c t ) + b k sin(2 π ( k + 1) f c t ) + b k sin(2 π ( k − 1) f c t )]] = a 1 2 .x ( t ) + a 2 .x ( t ) cos(2 πf c t ) + g ( t ) . where g ( t ) has terms with frequencies higher than or equal to 2 f c . Therefore, a 1 2 .x ( t ) can be recovered by using a low pass filter of bandwidth W , where W is the bandwidth of x ( t ) as shown in the figure. − w/ 2 w/ 2 1 / 2 α 1 (b) Since p ( t ) is a periodic signal with period n f c , it can be represented as a fourier series of the form: p ( t ) = a 2 + ∞ summationdisplay k =1 [ a k cos(2 πk f c n t ) + b k sin(2 πk f c n t )] . p ( t ) .x c ( t ) = [ a 2 + ∞ summationdisplay k =1 [ a k cos(2 πk f c n t ) + b k sin(2 πk f c n t )]] .x ( t ) cos(2 πf c t ) = a 2 .x ( t ) cos(2 πf c t ) + 1 2 ∞ summationdisplay k =1 [ a k cos(2 π ( k f c n + f c ) t ) + a k cos(2 π ( k f c n − f c ) t ) + b k sin(2 π ( k f c n + f c ) t ) + b k sin(2 π ( k f c n − f c ) t )]] = a n 2 .x ( t ) + a 2 .x ( t ) cos(2 πf c t ) + g ( t ) , where g ( t ) has terms with frequencies higher than or equal to W 2 assuming f c n ≥ W . Therefore, a n 2 .x ( t ) can be recovered by using a low pass filter of bandwidth W , where W is the bandwidth of x ( t )....
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This note was uploaded on 02/05/2012 for the course EE EE308 taught by Professor B.k.dey during the Spring '09 term at IIT Bombay.

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midsem_solutions - Department of Electrical Engineering EE...

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