5 a5115 b50 10 ohms problem 9 derive the trigonometric

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Unformatted text preview: nω o t nπ n 2 cos nω t o sin nπ n 2 cos nω t o E) None of the above 3 TF = 1/50, G(t) = 1 rms A0=1.5, A5=1.15, B5=0 10 ohms Problem 9 Derive the trigonometric form of the FSE of the waveform f (t) shown A Solution: The function is even, and a0 = C0 = 1 T A 4A × A × = ; an = T 4 4 T T /4 ⎛ ∫ 0 4 ⎞ ⎜ − t + 1⎟ cos nω0 tdt ⎝ T ⎠ T/2 T /4 -T/4 ⎤ 4A ⎡ 4 1 4 t 1 cos nω0 t − sin nω0 t − sin nω0 t ⎥ = ⎢− 2 2 T nω 0 nω 0 T ⎣ T n ω0 ⎦0 = 16 A 2 T n 2ω 0 2 f (t ) = nπ ⎤ nπ ⎞ 4A ⎛ ⎡ ⎢1 − cos 2 ⎥ = π 2 n 2 ⎜1 − cos 2 ⎟ . Hence, ⎝ ⎠ ⎦ ⎣ 1 1 A 4A ⎛ 1 ⎞ + 2 ⎜ cos ω0 t + cos 2ω0 t + cos 3ω0 t + cos 5ω0 t + ... ⎟ . 4 π ⎝ 2 9 25 ⎠ T/4 t T 1. in the circuit shown, each source is 15cos10t V. The 1Ω 1Ω power dissipated in R is 50 W. If the frequency of one of the sources is doubled, the power dissipated in R is: + A. 100 W R – B. 50W 1Ω + – C. 25 W D. 12.5 W E. None of the above. Solution: The current due to each source is 1 ⎛ 15 ⎞ ⎜ ⎟ cos 10t = 5 cos 10t A. The power is 2 ⎝ 1 .5 ⎠ 2 ⎛ 5 ⎞ ⎜ ⎟ = 12.5 W. the power dissipated due to both ⎜ ⎟ ⎝ 2⎠ f1(t) sources is 25 W. 3 2. f2(t) is the function f1(t) lowered by 1 unit, as t shown. If F1rms and F2rms are the rms values of T f1(t) and f2(t), respectively, then: f2(t) A. F1rms = F2rms 2 B. F1rms > F2rms T t C. F1rms < F2rms Solution: The AC components of f1(t) and f2(t) are the same. The DC component of f1(t) is larger than that of f2(t). Hence, F1rms > F2rms. 5. The Fourier coefficients ak and bk for the periodic function shown are: A. ak = 0 for all k; bk = 0 for k odd and is non-zero for k even v V 10π 2.5π B. bk = 0 for all k; ak = 0 for k even and is non-zero -2.5π for k odd C. bk = 0 for all k; ak = 0 for k = 0, ak = 0 for k odd 3T/4 T/4 T/2 -10π and is non-zero for k even D. ak = 0 for all k; bk = 0 for k even and is non-zero for k odd E. None of the above. Solution: The function is odd, half-wave symmetric. Its average is zero; it contains no cosine terms, only odd sine terms. 1/1 t T 11. The current through an inductor of 1 H is i, A given by the periodic triangula...
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This test prep was uploaded on 03/11/2014 for the course EECE 210 taught by Professor Riadchedid during the Spring '07 term at American University of Beirut.

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