HW04s09_soln

HW04s09_soln - SOLUTIONS TO ECE2025 PROBLEM SET #4, JANUARY...

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S OLUTIONS TO ECE2025 P ROBLEM S ET #4, J ANUARY 30, 2009 PROBLEM 4.1*. From the picture: (a) Concepts — Periodic signals, fundamental frequency, harmonic frequencies, GCD Approach — A sum of sinuoids is periodic if the sinusoid frequencies are harmonically related; i.e., if they can all be expressed as an integer multiple of a common frequency called the fundamental frequency. When this is true, the fundamental frequency can be found using the greatest-common-divisor (GCD) as shown below. Solution — From the picture we see two nonzero frequencies: 3.2 π rad / s, and 8.4 π rad / s. The fundamental frequency can be computed as: ω 0 = 0.1 π GCD { 32 , 84 } = 0.1 π (4) = 0.4 π rad / s. Note how 0.1 π was factored out from both frequencies to ensure that the arguments to GCD were integers. (b) Concepts — Periodic signals, fundamental frequency, fundamental period Approach — The fundamental period T 0 is related to ω 0 by T 0 = 2 π / ω 0 Solution — Using the answer from part (a) yields: T 0 = 2 π / ω 0 = 2 π /0.4 π = 5 seconds. (c) Concepts — DC value, constant value, average value, zero frequency Approach — From the formula A cos( ω t + φ ) of a generic sinusoid, we see that a “sinusoid” with zero frequency ( ω = 0 ) reduces to a constant value that is independent of time. The DC value is precisely specified by the spectrum at zero frequency. Solution — From the picture we see that the spectrum shows an amplitude of 5 at zero frequency; therefore, the DC value is 5 . (d) Concepts — Periodic signals, Fourier series representation, Spectrum of Fourier Series Approach — The Fourier series coefficients of a periodic signal may be determined easily from the signal’s spectrum; specifically, the k -th Fourier series coefficient a k is the complex amplitude at frequency k ω 0 . Solution — Translating the spectrum (from left to right) into an equation for the signal waveform yields: x ( t ) = 2 e j π /3 e j  ω 0 t + 3 e j 3 π /4 e j ω 0 t + 5 + 3 e j 3 π /4 e j ω 0 t + 2 e j π /3 e j  ω 0 t . This has the desired form of a Fourier series, i.e. x ( t ) = a k e jk ω 0 t , where a  = 2 e j π /3 , a = 3 e j 3 π /4 , a 0 = 5 , a 8 = 3 e j 3 π /4 , a  = 2 e j π /3 .
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HW04s09_soln - SOLUTIONS TO ECE2025 PROBLEM SET #4, JANUARY...

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