Use Fourier analysis to construct an amplitude spectrum

[luckyduck]

Homework Statement

A sensor yields a signal y(t) = |sin(120\pit)|
a. Using Fourier analysis please construct an amplitude spectrum for this signal.

Homework Equations

A₀ = \frac{1}{T}\int ^{-T/2}_{T/2}y(t) dt
A_n =\frac{2}{T}\int^{-T/2}_{T/2}y(t)cos\frac{2n\pi t}{T}dt

The Attempt at a Solution

Because y(-t) = y(t), the function is even, and we can ignore B_n.

My question is: what is the period? In prior examples, we used T = 2\pi.
Therefore, will:
A₀ = \frac{1}{2\pi} \int^{-\pi}_{\pi} |sin(120\pit)| dt?

 

[LCKurtz]

Homework Statement

A sensor yields a signal y(t) = |sin(120\pit)|
a. Using Fourier analysis please construct an amplitude spectrum for this signal.

Homework Equations

A₀ = \frac{1}{T}\int ^{-T/2}_{T/2}y(t) dt
A_n =\frac{2}{T}\int^{-T/2}_{T/2}y(t)cos\frac{2n\pi t}{T}dt

The Attempt at a Solution

Because y(-t) = y(t), the function is even, and we can ignore B_n.

My question is: what is the period? In prior examples, we used T = 2\pi.
Therefore, will:
A₀ = \frac{1}{2\pi} \int^{-\pi}_{\pi} |sin(120\pit)| dt?

The period of $\sin(bt)$ is $2\pi /b$ so the period of your function without the absolute values would be$$
T=\frac{2\pi}{120\pi}=\frac 1 {60}$$I would use that in half range formula. Note that by using the half range formula you can drop the absolute value.

 

[luckyduck]

Thank you!

By half range, do you mean integrating between 0 and 30?

 

[luckyduck]

Also, would this mean that the function is now odd?

 

[LCKurtz]

Thank you!

By half range, do you mean integrating between 0 and 30?

No. Half the period would be 1/120.

Also, would this mean that the function is now odd?

No. You use the half range cosine formulas which gives the even extension of the function, which has the same effect as the absolute values. Look at the half range formulas in your text.

 

[luckyduck]

Duh! Brain fart. Sorry, I don't have a text for this class, so I've been relying on google. Thanks again for your help!