Amplitude and phase spectra from fundamental frequency?

[TheSodesa]

Homework Statement

Let
\begin{equation*}
f(t) = 2 + \cos\left( 3t - \frac{\pi}{6} \right) + \frac{1}{4}\cos\left( \frac{1}{2}t + \frac{\pi}{3} \right) + \sin^2(t)
\end{equation*}
Determine the period $T$ and fundamental frequency $\omega_0$ of $f$ and draw images of its amplitude and phase spectra.

Homework Equations

If
\begin{align}
\hat{f}
&= \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(n \omega t) + b_n \sin(n \omega t)\\
&= c_0 + \sum_{n = 1}^{\infty} \bigg( c_n e^{jn \omega t} + c_n^* e^{-j n \omega t}\bigg)=c_0 + \sum_{-\infty}^{\infty} c_n e^{jn \omega t}\\
&= c_0 + \sum_{n = 1}^{\infty} 2|c_n| \cos(n \omega t + \theta_n)
\end{align}
is the Fourier-series of the function $f$, then the amplitude spectrum is defined as the sequence
\begin{equation}
\ldots ,|c_3^*|, |c_2^*|, |c_1^*|,|c_1|,|c_2|,|c_3|,|c_4|,\ldots
\end{equation}
and the phase spectrum as the corresponding sequence
\begin{equation}
\ldots, -\theta_3,-\theta_2, -\theta_1, \theta_1,\theta_2,\theta_3,\theta_4,\ldots
\end{equation}

The Attempt at a Solution

I had no trouble answering the first two questions: By drawing the following picture I was able to quess that the smallest period $T$ was $4\pi$ and test for it.

The test came out positive. This gave me the fundamental frequency of $\omega_0 = \frac{2\pi}{T} = \frac{1}{2}$, and therefore the upper harmonics as its integer multiples.

Now, however, I'm at a standstill. I'm a bit reluctant to start deriving the Fourier series for this function, since my professor has been in the habit of giving excercises, where taking shorcuts is possible using fancy threoms mentioned in the course handout, where I've still gone and wasted time deriving the series because I didn't notice a certain footnote, or something was not adequately well explained for my needs. Not to mention, that the integrals are going to be rather nasty.

My question therefore is: "Is there a way to extract the amplitude and phase spectra from just knowing the fundamental frequency of the given function?"

 

[Charles Link]

$sin^2 t=1-cos^2 t=1-(cos(2 t)+1)/2=(1/2)-(1/2)cos(2t)$. The amplitudes can be seen from inspection. You got the fundamental $\omega_o=1/2$ correct. Besides a $c_o$ term, and a $c_1$ (corresponding to $\omega_o$), there is an $\omega=4 \omega_o$ term (from the $cos(2t)$ that came from the $sin^2 t$), and a $6 \omega_o$ term (from the $cos(3t)$ term). It is not necessary to do any integrals to see this result. $\\$ Note: the amplitude spectrum is the sequence $|c_o|,2|c_1|, 2|c_2|,2|c_3|,...$ , but all you need to do in this case is look at $f(t)$, once you rewrite $sin^2 t$, and read them off.

 

[Ray Vickson]

Homework Statement

Let
\begin{equation*}
f(t) = 2 + \cos\left( 3t - \frac{\pi}{6} \right) + \frac{1}{4}\cos\left( \frac{1}{2}t + \frac{\pi}{3} \right) + \sin^2(t)
\end{equation*}
Determine the period $T$ and fundamental frequency $\omega_0$ of $f$ and draw images of its amplitude and phase spectra.

If you write $t/2 = \tau$ your function becomes
$$f = \cos\left( 6\tau - \frac{\pi}{6} \right) + \frac{1}{4}\cos\left( \tau + \frac{\pi}{3} \right) + \sin^2(2 \tau)$$
You ought to be able to find the $\tau$-period (and then the $t$-period) easily from that.

 

[TheSodesa]

$sin^2 t=1-cos^2 t=1-(cos(2 t)+1)/2=(1/2)-(1/2)cos(2t)$. The amplitudes can be seen from inspection. You got the fundamental $\omega_o=1/2$ correct. Besides a $c_o$ term, and a $c_1$ (corresponding to $\omega_o$), there is an $\omega=4 \omega_o$ term (from the $cos(2t)$ that came from the $sin^2 t$), and a $6 \omega_o$ term (from the $cos(3t)$ term). It is not necessary to do any integrals to see this result. $\\$ Note: the amplitude spectrum is the sequence $|c_o|,2|c_1|, 2|c_2|,2|c_3|,...$ , but all you need to do in this case is look at $f(t)$, once you rewrite $sin^2 t$, and read them off.

So basically, if I'm asked to find out the spectra of a given function, I should separate its harmonic components and look at their coefficients and phase angles... I don't know why I made it so difficult for myself.

I was able to put together these images.

Thanks for the help.

 

[Charles Link]

It is worth mentioning that if you get two terms of the same frequency of the form $A cos(\omega_n t)+B sin(\omega_n t)$, you can write it as $\sqrt{A^2+B^2} cos(\omega_n t-\phi)$ where $\phi=tan^{-1}(\frac{B}{A})$. $\\$ This is because $cos(\theta-\phi)=cos(\theta)cos(\phi)+sin(\theta)sin(\phi)$. You can factor out $\sqrt{A^2+B^2}$ from the first expression to get $\sqrt{A^2+B^2} [\frac{A}{\sqrt{A^2+B^2}}cos(\omega_n t)+\frac{B}{\sqrt{A^2+B^2}}sin(\omega_n t)]$. Then let $cos(\phi)=\frac{A}{\sqrt{A^2+B^2}}$ and $sin(\phi)=\frac{B}{\sqrt{A^2+B^2}}$. $\\$ The term $\sqrt{A^2+B^2}$ becomes your amplitude at frequency $\omega_n$.

 

[Ray Vickson]

Homework Statement

Let
\begin{equation*}
f(t) = 2 + \cos\left( 3t - \frac{\pi}{6} \right) + \frac{1}{4}\cos\left( \frac{1}{2}t + \frac{\pi}{3} \right) + \sin^2(t)
\end{equation*}
Determine the period $T$ and fundamental frequency $\omega_0$ of $f$ and draw images of its amplitude and phase spectra.

Now, however, I'm at a standstill. I'm a bit reluctant to start deriving the Fourier series for this function, since my professor has been in the habit of giving excercises, where taking shorcuts is possible using fancy threoms mentioned in the course handout, where I've still gone and wasted time deriving the series because I didn't notice a certain footnote, or something was not adequately well explained for my needs. Not to mention, that the integrals are going to be rather nasty.

My question therefore is: "Is there a way to extract the amplitude and phase spectra from just knowing the fundamental frequency of the given function?"

If you use the double-angle formula, you can write $\sin^2(t)$ in terms of $\sin(2t)$ and/or $\cos(2t)$. When you do that, and expand $\cos(3t- \pi/6)$ and $\cos(t/2 + \pi/3)$, you will already have the Fourier series for the function. There is no need to do any integrals, etc.

 

[TheSodesa]

If you use the double-angle formula, you can write $\sin^2(t)$ in terms of $\sin(2t)$ and/or $\cos(2t)$. When you do that, and expand $\cos(3t- \pi/6)$ and $\cos(t/2 + \pi/3)$, you will already have the Fourier series for the function. There is no need to do any integrals, etc.

Because the Fourier series of the cosine/sine function is the function itself?

 

[TheSodesa]

It is worth mentioning that if you get two terms of the same frequency of the form $A cos(\omega_n t)+B sin(\omega_n t)$, you can write it as $\sqrt{A^2+B^2} cos(\omega_n t-\phi)$ where $\phi=tan^{-1}(\frac{B}{A})$. $\\$ This is because $cos(\theta-\phi)=cos(\theta)cos(\phi)+sin(\theta)sin(\phi)$. You can factor out $\sqrt{A^2+B^2}$ from the first expression to get $\sqrt{A^2+B^2} [\frac{A}{\sqrt{A^2+B^2}}cos(\omega_n t)+\frac{B}{\sqrt{A^2+B^2}}sin(\omega_n t)]$. Then let $cos(\phi)=\frac{A}{\sqrt{A^2+B^2}}$ and $sin(\phi)=\frac{B}{\sqrt{A^2+B^2}}$. $\\$ The term $\sqrt{A^2+B^2}$ becomes your amplitude at frequency $\omega_n$.

I might have just ran into this exact case in the next problem of this week's problem set...

 

[Ray Vickson]

Because the Fourier series of the cosine/sine function is the function itself?

Exactly.