Homework _8 Spring 11

Homework _8 Spring 11 - BME6360 Homework #8 Spring 2011...

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Unformatted text preview: BME6360 Homework #8 Spring 2011 Neural Engineering Prof. Wheeler This is a demonstration assignment. It's purpose is to review some basic operations with Fourier transforms. It's primary goal is to help you look at a Fourier transform, especially one done with the Fast Fourier Transform (e.g. discrete time, discrete frequencies), and to interpret it. Some of the functions used here will be used on Homework #9. The following program files are available under: ifftplot.m, fftplot.m, ftp.m, xcorcirc.m, hw8data.mat The following vectors are available in hw8data.mat and may be loaded by executing load hw8data F1a, F1b, F1c, F1d, F1e, F1f, F1g, F1h-- Frequency domain values for problem 1. F2a, F2b, F2c, F2d-- Frequency domain values for problem 2. s1, c1, s2, c32, c31, c33, mix, c6401, c3_5, s16_3, czeros-- time domain values for problem 3 y4a, pt1, pt2-- time domain values for problem 4 Rpt1, Rpt2-- autocorrelations for problem 4 t5, y5, y5n-- time domain values for problem 5 You do not need to execute the instructions below which create the vectors. They are repeated here to remind you as to their content. 1. In order to understand FFTs and power spectra, you should understand that each element of an FFT of a signal represents a complex exponential. This problem demonstrates that correspondence by using the inverse transform to reconstruct the time domain signal of single frequency domain elements. Using the function ifftplot demonstrate the correspondence between the following sequences in the frequency domain and their time domain: a. F1a=16*[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]' ; % should be DC (constant) b. F1b=16*[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]' ; % exp(jwt); one cycle % the result should have both real and imaginary components c. F1c=16*[0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0] '; % exp(j2wt); two cycles d. F1d=16*[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1] '; % exp(-jwt); one cycle; why? e. F1e=16*[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0] '; % exp(-j2wt); two cycles f. F1f=16*[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]'; % max frequency g. F1g=16*[0 -j 0 0 0 0 0 0 0 0 0 0 0 0 0 0].'; % (-j)*exp(jwt); one cycle; real part is sin(wt) h. F1h=16*[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 j].'; % (j)*exp(-jwt); one cycle; real part is sin(wt) NOTES: We assume that all signals are column vectors, not row vectors. For problems 1g and 1h notice that the transpose operator is .' not ' The operator .' does a simple transpose, changing a row vector to a column vector. The operator ' does a conjugate transpose, changing the row vector to a column vector and changing the sign of the imaginary components. All but parts g and h used real elements only, so that it did not matter which transpose operator you used....
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This note was uploaded on 01/23/2012 for the course EEL 6502 taught by Professor Principe during the Spring '08 term at University of Florida.

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Homework _8 Spring 11 - BME6360 Homework #8 Spring 2011...

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