0
$\begingroup$

I am reading book signals and systems Laboratory with MATLAB where I am studing chapter 5, Fourier series and i trying to understand magnitude spectrum and phase spectrum but i have certain confuisons regarding attached snapshot

  1. How we can interpret and understand magnitude spectrum (especially in this scenarios where both positve and negative frequencies are shown) Is it showing how much contribution is made by each frequency component and maximum magintude contribution is made by Dc(0 Hz) frequency component

  2. How we can interpret and understand phase spectrum(especially in this scenarios where both positve and negative frequencies are shown) Is it showing how much phase shift contribution is made by each frequency component and minimum phase shift contribution is made by Dc(0 Hz) frequency component

enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

We probably all know that the Fourier Series Expansion will decompose an arbitrary waveform that is analytic over a time interval $T$ into a sum of infinite frequency components, discrete in the frequency domain as they only occur at the frequencies given by $\frac{N}{T}$ where $N$ is any integer. What confuses many is what "negative frequencies" represent and the significance of the phase result.

My intention is that the following explanation will remove all confusion with negative frequency and the significance of discrete impulses in the frequency domain (in magnitude and phase). If any of this is still confusing, please comment below so that we can further clarify this!

First Point: when you see a positive and negative frequency axis, each discrete line shown is the magnitude (for the magnitude plot) and phase (for the phase plot) for coefficients of $e^{j\omega t}$. It is NOT the coefficients of sines or cosines.

Second Point: The general expression $Ke^{j\theta}$ with $K$ and $\theta$ as real numbers, represents a phasor on the complex plane with magnitude $K$ and angle $\theta$. We could also write this equivalently as $K \angle \theta$, or $K\cos(\theta)+jK\sin(\theta)$.

complex phasor

Third Point: If the phasor shown above does not change with time, this is "DC", and regardless of what $\theta$ is, if it's not moving, it is the DC term. If the phasor was rotating at a constant rate counter-clockwise with phase growing in the positive direction as $\omega$ radians/sec, this represents a positive frequency. If instead the phasor was rotating at a constant rate clock-wise, with phase growing in the negative direction at $-\omega$ radians/sec, this represents a negative frequency. This is the meaning of $Ke^{j\omega t}$ and $Ke^{-j\omega t}$. Any such phasor in the time domain; with a constant magnitude versus time and rotating at a constant rate, will be a single impulse in the frequency domain. The magnitude of that impulse will be the magnitude of the rotating phasor, and the phase of that impulse will be the starting phase of the rotating phasor at $t=0$.

Fourth Point: If the frequency domain is complex conjugate symmetric the resulting waveform in the time domain MUST be real. Complex conjugate symmetric means for every impulse at frequency $\omega_c$, there is another impulse at frequency $-\omega_c$ and they both have the same magnitude and opposite phase. The impulse at DC ($f=0$) is considered complex conjugate symmetric if it has a phase of $0$ radians specifically. If there is any deviation from this condition, then the signal will be complex, or possibly completely imaginary (under other conditions I am not listing). A great example of this is Euler's formula which relates real sinusoids to the complex exponential form:

$$2\cos(\omega t) = e^{j\omega t} + e^{-j\omega t} \tag{1}\label{1}$$

Note if we were to scale \ref{1} by $K$ and add a phase offset $\phi$ to the above relationship, we would get:

$$2K\cos(\omega t + \phi) = Ke^{j(\omega t + \phi)} + Ke^{-j(\omega t - \phi)}\tag{2}\label{2}$$

And walah! We see how the Fourier Transform of a cosine is complex conjugate symmetric with two impulses in frequency, one in the positive frequency axis and one in the negative frequency axis.

It is satisfying to review a plot of the sum of the two "spinning phasors" as given in equation $\ref{2}$, knowing that when we add such phasors geometrically, we place one on the end of the other and in doing this we see, in this case of complex conjugate symmetry, that the result will always stay on the real axis! The plot below is an example of this, showing the phasors given by $2K\cos(\omega t+\phi)$ with the phasors as shown where they would be at $t=0$.

phasors

Conclusion: From this we should now have a clear understanding of the depiction of impulses on a negative and positive frequency axis, and with observation of the OP's spectrums, we observe a strong DC term with a phase of 0 radians and pairs of impulses that are complex conjugate symmetric, therefore each pair represents a sinusoidal components with magnitudes and phase offsets as given by the relationship between each exponential term in \ref{2}. This is the Fourier Series Expansion which will always be discrete in frequency with impulses representing the magnitude and starting phase for each of the "spinning phasors" in the time domain, that when all added would recreate the original time domain waveform over interval $T$. These same concepts are extended further to the Fourier Transform when we allow $T$ to extend to infinity, which causes the frequency domain to be a continuous function rather than discrete as in this case.

$\endgroup$
2
  • $\begingroup$ Sir, please clarify,In your conclusion,you used the term"starting phase", is it same thing as phase shift? $\endgroup$
    – DSP_CS
    Commented Apr 27, 2022 at 6:55
  • $\begingroup$ @engr yes you can call it phase shift, but many then confuse it with a delay (phase and delay are not the same thing). Specifically and clearly each single tone you see in your plot represents a phasor in time with a starting phase and magnitude as given in your plot at t=0. As time moves forward, each phasor maintains the same magnitude but rotates (phase increases) according to the phase rate of change given on the frequency axis (in rad/sec). If you combine two phasors that are rotating at the same rate and opposite direction, these as a sum will be a sinusoid. $\endgroup$ Commented Apr 27, 2022 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.