-
x (t) is a real valued function of a real variable with period T. Its trigonometric Fourier Series expansion contains no terms of frequency
ω = 2π (2k)/T; k = 1, 2,.... Also, no sine terms are present. Then x (t) satisfies the equation
-
- x (t) =– x (t – T)
- x (t) = x (T – t) = – x (– t)
- x (t) = x (T – t) = – x (t – T/2)
- x (t) = x (t – T) = x (t – T/2)
Correct Option: D
Since, the fourier expansion of x(t) contains no sine terms, therefore,
x(t) = x(–t)
or, x(t) = x(T – t)
as x(t) is periodic with T
Now, as signal x(t) contains odd harmonics.
Then x(t) = -x | ![]() | t - | ![]() | |
2 |
Thus x(t) = x(T - t) = -x | ![]() | t - | ![]() | |
2 |