BME343_fall2009_HW6_Solutions

# BME343_fall2009_HW6_Solutions - Homework 6 Solution 6.1-1...

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Unformatted text preview: Homework 6 Solution 6.1-1 " b) = 10 , # " = \$ #" =   2 I= 10 I= \$ #"  cos Even Symmetry, therefore all sine terms are zero { {= I" = (by inspection) ' \$ J 5  sin  ' (# = I cos J 5 15 J F sin 5J 5 + I sin | − = J 5 J 2 J J{ { 5 J = 0 (integrand is an odd function of t) Here I = 0 and we allow Cn to take negative values. Note that Cn = an for n = 0,1,2,3 … Cn vs. w 0.4 0.35 0.3 0.25 Cn 0.2 0.15 0.1 0.05 0 -0.05 -0.1 0 6.1-3 a) " 0.5  = \$, " 1 =4 1.5 2 w 2.5 3 3.5 4 I" = I= I= 2 4 4 { { = I" +  \$ "  \$ "  \$ " (# cos{4J { + I sin{4J { I = 0.504 cos {4J { = 0.504{ sin {4J { = 0.504{ Therefore, C0 = I" = 0.504, 2 { 1 + 16J\$ 8J { 1 + 16J\$ = I \$ +I \$ = 0.504{ \$ ## {, = − tan # 4J b) This Fourier Series is identical to Eqn. (6-15a) with t replaced with 2t c) If { { = " + cos{J{I " { + { (# Thus, time scaling by a factor a merely scales the fundamental frequency by the same factor a. Everything else remains unchanged. If we compress (or time expand) a periodic signal by a factor a, its fundamental frequency increases by the same factor a (or decreases by the same factor a). Comparison of the results in part (a) with those in Example 6.1 confirms this conclusion. This result applies equally well. 6.1-7 c) Not periodic therefore no e) Not periodic therefore no %  g) Periodic; " = &, " = % h) Periodic; " = 1, " = 2 6.3-1 c) { { = = ( # \$ " \$ so that | |= # \$ " or period " or period eD H , where by inspection D0 = 0.5 , and < = \$ = ,  \$ J>0  −\$ J < 0 |Dn| vs. w 0.5 0.45 0.4 0.35 |Dn| 0.3 0.25 0.2 0.15 0.1 0.05 0 -10 -8 -6 -4 -2 0 w 2 4 6 8 10 2 4 6 8 10 <Dn vs. w 2 1.5 1 <Dn 0.5 0 -0.5 -1 -1.5 -2 -10 -8 -6 -4 -2 0 w " = 6, "  = % , where by inspection D0 = 0.5 { { = 0.5 + 1 =[ 6 # \$ ( { + 2{ 3 J = \$ \$ {cos J 3 e  % D % H + # # 2J − cos F{ 3  % + \$ # {− + 2{  % |Dn| vs. w 0.5 0.45 0.4 0.35 0.3 |Dn| f) 0.25 0.2 0.15 0.1 0.05 0 -15 -10 -5 0 w 5 10 15 <Dn vs. w 3.5 3 2.5 <Dn 2 1.5 1 0.5 0 -15 -5 a) Rewrite { { = 3 + 2 cos 2 − trigonometric identities. 0 w  5 10  15 # + cos 3 − \$ + \$ cos {5 − \$ % {, using Cn vs. w 5 4.5 4 3.5 3 Cn 6.3-4 -10 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 w 3 3.5 4 4.5 5 Theta vs. w n 0 -0.5 Thetan -1 -1.5 -2 -2.5 0 0.5 1 1.5 2 2.5 w 3 3.5 4 4.5 5 b) By inspection of trigonometric spectra, we can sketch the exponential Fourier spectra: |Dn| vs. w 3 2.5 |Dn| 2 1.5 1 0.5 0 -5 -4 -3 -2 -1 0 w 1 2 3 4 5 <Dn vs. w 2.5 2 1.5 1 <Dn 0.5 0 -0.5 -1 -1.5 -2 -2.5 -5 6.4-1 " =, H{ -4 " {= Therefore, { {= =2 { ( -3 %{ -2 \$ -1 0 w 1 , and from Eqn. (6.30b) H{jnw" {eD KH = ( 2 = 3 4 5 ".'"& # & {# & {{ #." % \${ \$ ...
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## This note was uploaded on 12/14/2011 for the course BME 343 taught by Professor Emelianov during the Spring '09 term at University of Texas at Austin.

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