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Contents : Problem Set 2 Solutions ECE 403 Spring 2005 February 18 2005 Problem 4.1 a. Let P (r ) 1 r P1 ( )e jkr . Then P 1 jk + r r 2P r2 P (r ) 1 1 + jk + r2 r 2 P (r ) By plugging these into the spherical wave equation we con rm that this solution works. b. U (r ) 1 P j r Using the derivative from part (a) we con rm that Z (r ) 0 c 1 j kr c. I (r ) 1 P1 2 P (r ) 2 2z0 2z0 r2 d. RR ( ) z0 (kr)2 1 + (kr)2 XR ( ) z0 (kr) 1 + (kr)2 Problem 4.3 H ( ) Qs ( ) A (1 sec(kL)) Us ( ) H (0) 0. H ( ) at c/2L and 3 c/2L. In the range c/2L c/2L H ( ) 2. Problem 4.6 a. x 0.34cm Fs 1/T 100kHz b. The function can be written with N 10 samples (at 0.5 x 1.5 x . . .) or with N 11 samples (at 0 x . . .). Here is the version with N 11. 1 function PP PM tubemodel(PP0 PM0 PIN) N length(PP0) M length(PIN) PP zeros(M N) PM zeros(M N) PP(1 :) PP0 PM(1 :) PM0 PP(1 2:N) PP0(1:(N-1)) PM(1 1:(N-1)) PM0(2:N) PP(1 1) PM(1 1) PM(1 N) PIN(1)-PP(1 N) for m 2:M PP(m 2:N) PP(m-1 1:(N-1)) PM(m 1:(N-1)) PM(m-1 2:N) PP(m 1) PM(m 1) PM(m N) PIN(m)-PP(m N) end Impulse and Frequency Responses of Middle Ear to Mid Canal 1 Pressure (Pa) 0.5 0 0.5 1 0 0.5 1 1.5 2 2.5 Time (sec) 3 2.5 Freq (Hz) 3 3.5 4 4.5 5 3 x 10 100 Pressure (Pa) 50 0 50 100 150 0 0.5 1 1.5 2 3.5 4 4.5 5 4 x 10 c. The spectrum contains multiple harmonics of a fundamental frequency at about 2500Hz. d. Lowpass ltering at 5kHz removes all but the lowest harmonic so the resulting signal is a sinusoid. 2

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