# Period of signal

Printable View

• Oct 21st 2010, 04:45 AM
jordanrs
Period of signal
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid

given that $x(t) = cos(600\pi t)$ and $m(t) = 4 + 3cos(20\pi t)$ what is the period (if any) of the resulting signal

Also the signal $x(t) = 10cos(600\pi t + \frac{\pi}{3}) + 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal

Im not sure how to calculate the period when theres more than one sinusoid wave

Any help most appreciated
• Oct 21st 2010, 05:44 AM
CaptainBlack
Quote:

Originally Posted by jordanrs
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid

Also the signal $x(t) = 10cos(600\pi t + \frac{\pi}{3}) 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal

Im not sure how to calculate the period when theres more than one sinusoid wave

Any help most appreciated

First question snipped as it not complete.

The product of two sinusoids has sinusoidal components at the sum and diffrence frequencies. This for some amplitudes (non-zero) and phases is:

$x(t)=A\sin(300\pi t+\phi_1)+B\sin(1500\pi t+\phi_2)$

and since the larger period is an integer multiple of the smaller the final period is $1/150$ s

CB
• Oct 22nd 2010, 02:45 AM
jordanrs
Quote:

Originally Posted by jordanrs
Stuck on the following problem, aint sure how to work out the signal of a composed sinusoid

given that $x(t) = cos(600\pi t)$ and $m(t) = 4 + 3cos(20\pi t)$ what is the period (if any) of the resulting signal

Also the signal $x(t) = 10cos(600\pi t + \frac{\pi}{3}) 6cos(900\pi t - \frac{\pi}{4}) - 4cos(1500\pi t)$ whats the period of the signal

Im not sure how to calculate the period when theres more than one sinusoid wave

Any help most appreciated

cheers for your reply, the first part should have read

when $y(t) = x(t)m(t)$ and $x(t) = cos(600\pi t)$ and $m(t) = 4 + 3cos(20\pi t)$ what is the resulting period of $(y(t))$

also i still dont get how to calculate the periodm i dont see where you got 1/150 s from
• Oct 22nd 2010, 08:27 PM
CaptainBlack
Quote:

Originally Posted by jordanrs
cheers for your reply, the first part should have read

when $y(t) = x(t)m(t)$ and $x(t) = cos(600\pi t)$ and $m(t) = 4 + 3cos(20\pi t)$ what is the resulting period of $(y(t))$

also i still dont get how to calculate the periodm i dont see where you got 1/150 s from

You decompose the signal into a sum of sinusoids, then the period is the lcm of the individual periods of the components.

CB