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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 Ω > 0 H ( j Ω) = + j Ω < 0 MATLAB uses the same definition and this can be verified by finding the phase of ω → + . The implemented filter’s phase has a linear component added to the − π/ 2 value (for ω > 0). Our filter has − π/ 2 phase shift and this corresponds to: A sin ((Ω a − Ω) t ) = A (sin(Ω a t ) cos(Ω t ) − cos(Ω a t ) sin(Ω t )) = − A sin(Ω a t ) sin(Ω t − π/ 2) − A cos(Ω a t ) sin(Ω 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). Ω a Ω 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(Ω o t ). The output of filter would be  H ( j Ω o )  sin(Ω o t + negationslash H ( j Ω o )) = (1+ δ ) sin(Ω o t − π 2 ). We follow this through above diagram to compute g ( t ): g ( t ) = − (1 + δ ) sin(Ω o t − π/ 2) sin(Ω a t ) − cos(Ω a t ) sin(Ω o t ) = +(1 + δ ) cos(Ω o t ) sin(Ω a t ) − cos(Ω a t ) sin(Ω o t ) = sin((Ω a − Ω o ) t ) + δ sin(Ω a t ) cos(Ω o t ) δ = sin((Ω a − Ω o ) t ) + (sin((Ω a − Ω o ) t ) + sin((Ω a + Ω o ) t )) 2 δ δ = (1 + ) sin((Ω a − Ω o ) t ) + sin((Ω a + Ω o ) t ) 2 2 As a result due to practical implementation, some “spurious” components are introduced at Ω a + Ω o . The original signal has a (desired) audio frequency content of Ω 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 Ω 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 δ 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
 Digital Signal Processing, Highpass filter, Bandpass filter, Lowpass filter, Passband

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