# Amplitude and phase spectra from fundamental frequency?

• TheSodesa
In summary, the function has a period of 4pi, a fundamental frequency of 1/2, and a spectrum of cos(2t), cos(3t), cos(4t), ...
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

\ldots ,|c_3^*|, |c_2^*|, |c_1^*|,|c_1|,|c_2|,|c_3|,|c_4|,\ldots

and the phase spectrum as the corresponding sequence

\ldots, -\theta_3,-\theta_2, -\theta_1, \theta_1,\theta_2,\theta_3,\theta_4,\ldots

## 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?"

Last edited:
## 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.

Last edited:
TheSodesa
TheSodesa said:

## 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.

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 ##.

TheSodesa said:

## 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.

Ray Vickson said:
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?

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...

Last edited:
TheSodesa said:
Because the Fourier series of the cosine/sine function is the function itself?
Exactly.

TheSodesa

## 1. What is amplitude and phase spectra from fundamental frequency?

Amplitude and phase spectra from fundamental frequency refer to the representation of a signal or wave in terms of its amplitude and phase at different frequencies. This helps in understanding the frequency components present in the signal and their relative strengths and phases.

## 2. How is the amplitude spectrum calculated?

The amplitude spectrum is calculated by taking the absolute value of the Fourier transform of the signal. This transforms the signal from the time domain to the frequency domain, where the amplitude at each frequency component can be observed.

## 3. What do the peaks in the amplitude spectrum represent?

The peaks in the amplitude spectrum represent the dominant frequencies present in the signal. The height of each peak indicates the strength or amplitude of that frequency component.

## 4. What is the phase spectrum and how is it different from the amplitude spectrum?

The phase spectrum represents the phase shift of each frequency component in the signal. It is calculated by taking the complex argument of the Fourier transform. Unlike the amplitude spectrum, which only shows the strength of each frequency, the phase spectrum also provides information about the timing and alignment of each frequency component.

## 5. What is the significance of analyzing the amplitude and phase spectra from fundamental frequency?

Analyzing the amplitude and phase spectra from fundamental frequency can help in understanding the characteristics of a signal, such as its frequency content, dominant frequencies, and phase relationships between different frequency components. This can be useful in various fields, including signal processing, audio engineering, and vibration analysis.

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