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If x1(t) = 2 sin πt + cos 4πt
x2(t) = sin 5πt + 3 sin 13πt, then—
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- x1(t) and x2(t) both are periodic
- x1(t) and x2(t) both are non-periodic
- x1 is periodic, but x2 is not periodic
- x1 is not periodic, but x2 is periodic
Correct Option: A
Given x1(t) = 2 sin πt + cos 4πt
Fundamental period of
| x1′(t)=T1′ = | = 2 | π |
Fundamental period of
| x1′′(t)=T1′′ = | = | |||
| 4π | 2 |
| Ratio = | = | = | = integer (or rational) | |||
| T2′′ | 1 | |||||
| 2 |
Therefore, x1(t) is periodic.
Now, for signal
x2(t) = sin 5πt + 3 sin 13πt
Fundamental period of
| x2′(t)=T2′ = | = | |||
| 5π | 5 |
Fundamental period of
| x2′′(t) T2′′ = | = | |||
| 13π | 13 |
| Ratio | = | = | = 2·6 | |||
| T1′ | ||||||
| T2′′ | 5 | |||||
| 13 |
i.e., rational. Therefore, x2(t) is positive.
Hence, alternative (A) is the correct answer.
