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# ps4soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal...

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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Solution of Problem Set 4 Assigned: Oct. 2, 2008 Due: Oct. 9, 2008 Problem 1: We have shown in class (see the Fourier handout): F { sin( ω c t ) } = ( δ ( ω ω c ) δ ( ω + ω c )) F { cos( ω c t ) } = π ( δ ( ω ω c ) + δ ( ω + ω c )) (a) When f audio 0, f AM ( t ) = sin( ω c t ), and from above � � 1 (b) There are several ways of doing this. For example (i) Expand f AM ( t ) into sinusoidal components using this trigonometric relationship: cos α sin β = (sin ( α + β ) + sin ( β α )) 2 f AM ( t ) = (1 + af audio ( t )) sin ( ω c t ) = (1 + 0 . 5 (0 . 5 cos(2 π 1000 t ) + 0 . 25 cos(2 π 2000 t )) sin( ω c i ) · · = sin( ω c t ) +0 . 25 cos(2 π 1000 t ) sin( ω c t ) · +0 . 125 cos(2 π 2000 t ) sin( ω c t ) · = sin( ω c t ) +0 . 125 (sin(( ω c + 2000 π ) t ) + sin(( ω c 2000 π ) t )) +0 . 0625 (sin(( ω c + 4000 π ) t ) + sin(( ω c 4000 π ) t ))
and take the Fourier transform of each of the five components: F AM ( ) = { ( δ ( ω ω c ) δ ( ω + ω c )) + 0 . 125 ( δ ( ω ( ω c 2000 π )) δ ( ω + ( ω c 2000 π ))) + 0 . 125 ( δ ( ω ( ω c + 2000 π )) δ ( ω + ( ω c + 2000 π ))) + 0 . 0625 ( δ ( ω ( ω c 4000 π )) δ ( ω + ( ω c 4000 π ))) + 0 . 0625 ( δ ( ω ( ω c + 4000 π )) δ ( ω + ( ω c + 4000 π ))) } giving a total of 10 impulse components in the spectrum. (ii) Alternatively you can recognize that the expansion to f AM ( t ) = sin( ω c t ) + 0 . 25 cos(2 π 1000 t ) sin( ω c t ) + 0 . 125 cos(2 π 2000 t ) sin( ω c t ) · · involves time-domain products and these will result in frequency-domain convo- lutions, so that 1 1 F AM ( ) = F c ( ) + F c ( ) F 1000 ( ) + F c ( ) F 2000 ( ) 2 π 2 π where F c ( ) = F { sin( ω c t ) } , F 1000 ( ) = F { 0 . 25 cos(2000 πt ) } , and F 2000 ( ) = F { 0 . 125 cos(4000 πt ) } . The same result as in (i) will follow. H 0 ω c ω −2000π ω −ω c −ω −2000π −ω +2000π +2000π c c c c −ω c −4000π −ω c +4000π ω c −4000π ω +4000π c (c) The following generalizes the results of part (b) |a| |a| H 0

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(d) From the above figure it can be seen that the required bandwidth is 2 ω u rad/s. Problem 2: This problem uses the time-reversal property of the Fourier transform, if F { f ( t ) } = F ( ) then F { f ( t ) } = F ( ), and if f ( t ) is real then F ( ) = F ( ), so that F { f ( t ) } = F ( ).
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