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Example 2.4 The one-sided Fourier series

Example 2.4 The one-sided Fourier series. 3. 2. 1. -10. -5. 0. 5. 10. Example 2.4.

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Example 2.4 The one-sided Fourier series

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  1. Example 2.4 The one-sided Fourier series

  2. 3 2 1 -10 -5 0 5 10 Example 2.4 Use the one-sided form of the Fourier series to represent the signal x(t) shown in Example 2.3 (reproduced below) as a constant plus a series of sinusoids, with only one term at each harmonic frequency. Draw the magnitude and phase spectra of the signal. volts x(t) · · · · · · seconds

  3. Solution to Example 2.4 In Example 2.3 we determined the an and bn coefficients of the trigonometric form of the Fourier series for x(t). We can thus easily calculate the coefficients of the one-sided form. (continued)

  4. dc term (X ) = 0.6 o f a b Harmonic (n) Magnitude X Phase (degrees) n n n n 1 0.489829 0.795775 0.934446041 -58.38617756 2 -0.05782 0.397887 0.402066024 -98.26769855 3 0.038544 0.265258 0.268044016 -81.73230145 4 -0.12246 0.198944 0.23361151 -121.6138224 5 0 0 0 * 6 0.081638 0.132629 0.155741007 -58.38617756 7 -0.01652 0.113682 0.114876007 -98.26769855 8 0.014454 0.099472 0.100516506 -81.73230145 9 -0.05443 0.088419 0.103827338 -121.6138224 10 0 0 0 * 11 0.04453 0.072343 0.08494964 -58.38617756 12 -0.00964 0.066315 0.067011004 -98.26769855 13 0.008895 0.061213 0.061856311 -81.73230145 14 -0.03499 0.056841 0.066746146 -121.6138224 15 0 0 0 * Solution to Example 2.4(continued) The Xn and fn coefficients are given here: * Technically, phase is irrelevant if magnitude = 0. These terms will, however, be plotted as having 0o phase on the phase spectrum.

  5. Phase in degrees 180 135 90 45 ... 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 0 -45 -90 -135 -180 Frequency in Hz Magnitude and Phase Spectra of x(t) Magnitude in volts Magnitude spectrum of x(t) ... Phase spectrum of x(t)

  6. -b n a n arctan ( )and arctan( ) Calculating the Arctangent Warning: Extra care is needed when calculating the arctangent function. In Example 2.4, for instance, many calculators will produce f2 = 81.73 rather than the correct value f2 = -98.27. Reason for potential errors:When asked to calculate arctan ( ), many calculators will evaluate the fraction first and then just obtain the arctangent of the result. Thus, concerning the f2 calculation in Example 2.4, many calculators will not differentiate between (continued)

  7. Calculating the Arctangent(continued) Advice: Remember that for any value of x, there are two solutions to arctan(x). To find out which value is appropriate if x was produced by a fraction, see which quadrant contains the actual numerator and denominator. For Example 2.4, f2 lies in the quadrant between -90 and -180. Thus f2 = -98.27, not 81.73. A2=-0.507 f2 b2=0.398

  8. volts volts 2.5 2.5 2 2 1.5 1.5 1 1 0.5 0.5 sec sec -2 -2 -1 -1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 -0.5 -0.5 Channel with 1Hz bandwidth passes first five harmonics Channel with 1Hz bandwidth

  9. volts volts 2.5 2.5 2 2 1.5 1.5 1 1 0.5 0.5 sec sec -2 -2 -1 -1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 -0.5 -0.5 Channel with 2Hz bandwidth Channel with 2Hz bandwidth passes first ten harmonics

  10. volts volts 2.5 2.5 2 2 1.5 1.5 1 1 0.5 0.5 sec sec -2 -2 -1 -1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 -0.5 -0.5 Channel with 3Hz bandwidth Channel with 3Hz bandwidth passes first fifteen harmonics

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