# Fundamental period of Cubed sinusoidal signal?

Joined Apr 21, 2015
2
Hi
The fundamental period of a composite signal can be found by LCM(individual time periods of the components of signal)...
If I have a cubed signal like sin^3(2t)
using this concept I use trigonometric identity (sin^3(X) = (3/4)sinX - (1/4)sin(3X) ) to convert the signal into summation and then apply the same rule of LCM .
(3/4) sin(2t)-(1/4)(sin(6t))
T1=2pi/2 = pi
T2= 2pi/6 = pi/3
LCM(pi,pi/3) = pi
T= pi sec
But the answer to this question is (1/pi) sec...
How can we find the fundamental period of this signal ??

(This question is from the book signals and systems by Simon Haykin(page#22, Pb 1.5(b)))

#### WBahn

Joined Mar 31, 2012
26,398
Hi
The fundamental period of a composite signal can be found by LCM(individual time periods of the components of signal)...
If I have a cubed signal like sin^3(2t)
using this concept I use trigonometric identity (sin^3(X) = (3/4)sinX - (1/4)sin(3X) ) to convert the signal into summation and then apply the same rule of LCM .
(3/4) sin(2t)-(1/4)(sin(6t))
T1=2pi/2 = pi
T2= 2pi/6 = pi/3
LCM(pi,pi/3) = pi
T= pi sec
But the answer to this question is (1/pi) sec...
How can we find the fundamental period of this signal ??

(This question is from the book signals and systems by Simon Haykin(page#22, Pb 1.5(b)))
Have you plotted the signals to see what the answer should be visually?

Have you thought about the problem to see what the answer should be intuitively?

It might also help if you bothered to track your units properly -- or, more accurately in this case, if the person that came up with the solution had bothered to track the units properly.

Joined Apr 21, 2015
2
In the book there is no solution to this question. Only the answer is given.
I plotted the signal and the time period is "pi" for sin^3(2t).
So I think that the answer given in the book may be wrong .

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#### WBahn

Joined Mar 31, 2012
26,398
It is. Had they tracked their units, they would probably have ended up with units of 1/seconds and caught their error.

#### MrAl

Joined Jun 17, 2014
8,157
Hi,

If we reduce sin(t)^2, then multiply that result by sin(t) and reduce again, we get:
(3*sin(t)-sin(3*t))/4

and with this it is immediately clear that the period is 2*pi because the period of sin(t) is 2*pi. Since you have sin(2*t) then the period must be 1/2 of that, so it must be pi.

I think the rule for powers of sin(w*t) is that even powers produce the second harmonic as lowest and odd powers just retain the fundamental as lowest. In fact, i think ALL the harmonics follow the odd or even numbers as well, so for sin(t)^8 we would have harmonics 2,4,6, and 8, and for sin(t)^9 we would have harmonics 1,3,5,7, and 9 for example. We could check this by finding the formula for the reduction of sin(t)^n.

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