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Unformatted text preview: 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 . MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing  Continuous and Discrete Fall Term 2008 Solution of Problem Set 8: FIR Linear Filters Assigned: November 6, 2008 Due: November 18, 2008 Problem 1: The standard Hilbert transformer is defined as: j > 0 H ( j ) = + j < 0 MATLAB uses the same definition and this can be verified by finding the phase of + . The implemented filters phase has a linear component added to the / 2 value (for > 0). Our filter has / 2 phase shift and this corresponds to: A sin (( a ) t ) = A (sin( a t ) cos( t ) cos( a t ) sin( t )) = A sin( a t ) sin( t / 2) A cos( a t ) sin( t ) We have to be careful in implementing the scrambler. The scrambler diagram corresponds to the below figure. However, note that the audio play is not affected by sign of output (i.e. a signal is played the same as its negative version). a a (a) The delay in implementation does not affect this discussion. So we ignore it for this analysis. We consider a sinusoidal component of input as f ( t ) = sin( o t ). The output of filter would be  H ( j o )  sin( o t + negationslash H ( j o )) = (1+ ) sin( o t 2 ). We follow this through above diagram to compute g ( t ): g ( t ) = (1 + ) sin( o t / 2) sin( a t ) cos( a t ) sin( o t ) = +(1 + ) cos( o t ) sin( a t ) cos( a t ) sin( o t ) = sin(( a o ) t ) + sin( a t ) cos( o t ) = sin(( a o ) t ) + (sin(( a o ) t ) + sin(( a + o ) t )) 2 = (1 + ) sin(( a o ) t ) + sin(( a + o ) t ) 2 2 As a result due to practical implementation, some spurious components are introduced at a + o . The original signal has a (desired) audio frequency content of o = 300 3000 Hz. In the beginning, we can preprocess the audio file with a bandpass filter passing 300 3000 Hz contents and then follow it with the scrambling process. We use a Hilbert transformer with a passband equal to 300 to ( F s / 2 300) Hz, where F s is the sampling frequency . The extended passband of the filter, is because a highorder (lowripple) firpmbuilt Hilbert filter requires a symmetric passband (see MATLAB doc umentation). After passing the audio file through our transformer with a = 3500 Hz, the output will have a large (desired) spectrum content at 500 3200 Hz and a spurious content at 3800 6500 Hz (which could be folded to some lower frequency). If is not small enough, we might wish to postprocess the filter output with a passband filter passing 500 3200 Hz contents and then save it as scrambled signal. This post processing step of scrambling, can...
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 Fall '08
 DerekRowell

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