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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 nonzero
for k even v V
10π
2.5π B. bk = 0 for all k; ak = 0 for k even and is nonzero 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 nonzero for k even
D. ak = 0 for all k; bk = 0 for k even and is nonzero for k odd
E. None of the above.
Solution: The function is odd, halfwave 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.
 Spring '07
 RiadChedid

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