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While solving Fourier series coefficients in example, i found couple of things which confuse me.

  1. How the minus sign changes to plus sign $a_1= 1-\frac{1}{2j} = 1+\frac{1}{2}j$?
  2. After plotting the magnitude and phase part of all the Fourier coefficients, what kind of information we can get from magnitude and phase plot?
  3. Also, how $4je^{j3(2\pi/8)t} - 4je^{-j3(2\pi/8)t} = 8\sin(6\pi/8)t$
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    $\begingroup$ Are you talking about point no 1 of your question. we know that (1/j) = -j. So (1-(1/(2j)))=1+(j/2) $\endgroup$ – Karan Talasila Oct 8 '15 at 10:17
  • $\begingroup$ @Talasila, can you please check the 3rd part i just added $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 11:04
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    $\begingroup$ Farooq Multiply and divide the left hand side of equation in 3rd part by (2j). then use exponential form of sine function and it can be simplified as 8j^2sin((6pi/8)t) which is equal to -8sin((6pi/8)t). $\endgroup$ – Karan Talasila Oct 8 '15 at 11:15
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    $\begingroup$ $e^{j\theta} = \cos\theta + j \sin\theta$ $\endgroup$ – geometrikal Oct 8 '15 at 15:45
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    $\begingroup$ @AadnanFarooqA $\frac{1}{2}e^{j\pi /4} = \frac{1}{2}\left(\cos(\pi /4) + j\sin(\pi /4) \right) = \frac{1}{2}\left(\frac{\sqrt{2}}{2} + j\frac{\sqrt{2}}{2}\right)$ which you can factorize further to that. $\endgroup$ – Gilles Oct 8 '15 at 17:06
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Frankly I don't deserve to get any credit for this answer. That's the reason I was commenting on the question. What you are asking as a question is too trivial.

Part 1

$$a_1=1-\frac{1}{2j}= 1-\frac{j}{2j^2}$$ We know that $j^2=-1$ from complex number arithmetic. So $a_1$ simplifies to $1+\frac{j}{2}$

Part 3

$$4je^{j3\frac{2\pi}{8}t} - 4je^{-j3\frac{2\pi}{8}t}$$

Multiply and divide the above expression by $2j$.

we get $$4j\left(e^{j3\frac{2\pi}{8}} - e^{-j3\frac{2\pi}{8})t}\right) \frac{2j}{2j}$$

It simplifies to $$8j^2\left(e^{j3\frac{2\pi}{8}t} - e^{-j3\frac{2\pi}{8}t}\right)\frac{1}{2j}$$

which is equal to $$8j^2 \sin\left(\frac{6\pi}{8}t\right) = -8 \sin \left(\frac{6\pi}{8}t\right)$$

because from exponential form of definition of $\sin(x)$ $$\sin(x) = \dfrac{e^{jx} + e^{-jx}}{2j}$$

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  • $\begingroup$ if we have complete equation than can we solve in parts like half part we will be multiple and divide by 2 to make $cos$ equation and half will be multiple and divide by 2j to make $sin$ equation?? $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 11:59
  • $\begingroup$ @Aadnan I didn't get you. can you be a bit more clear? $\endgroup$ – Karan Talasila Oct 8 '15 at 12:11
  • $\begingroup$ Like if we have $4je^{(j3(2pi/8)t)} - 4je^{(-j3(2pi/8)t}$ and $2e^{(j(pi/8)t)} - 2e^{(-j(pi/8)t}$, then the first term will be multiplied and divide by $2j$ to make equation for $sin$ and second term will be multiplied and divide by 2 to make $cos$.. right? So i want to know that if both these are in one equation like this $4je^{(j3(2pi/8)t)} - 4je^{(-j3(2pi/8)t} + 2e^{(j(pi/8)t)} - 2e^{(-j(pi/8)t}$ then can we multiple and $2j$ separatelyand $2$ separately in same equation?? $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 12:24
  • $\begingroup$ @AadnanFarooqA Ya you can. Note that multiplying and dividing by the same term makes no difference to it's value. Your second term is a mistake.Cos(x) =((e^(jx) +e^(-jx))/2). There is no negative sign between the exponential terms. So given the expression that you have written above, you should multiply and divide by (2j) for the second term also and express it in sine form. $\endgroup$ – Karan Talasila Oct 8 '15 at 13:03
  • $\begingroup$ actually i have to express both term in $cos$ so i was asking.. Also can you tell me that how they converted $\frac{1}{2}e^{j\pi/4} = \frac{\sqrt{2}}{4}(1+j)$ $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 13:08
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You can write a phase-shifted sinusoid as $A\cos(\omega t + \phi)$ as

$$ \frac{A}{2} e^{i\phi}e^{i\omega t} + \frac{A}{2} e^{-i\phi}e^{-i\omega t}$$

The first part is the 'positive frequency' part, the second is the negative frequency part.

  1. For a real valued signal, the coefficients come in a conjugate pair:

$$ \frac{A}{2} e^{i\phi} \text{ and } \frac{A}{2} e^{-i\phi}$$

This is why the minus sign changes to a plus sign; the imaginary parts have to cancel out.

It is just a mathematical convenience; when expressed this way we can do linear operations like addition and multiplication. We could instead have the transform spit out magnitude and phase values, but then when applying filters we would have to convert these to their components, add / multiply, then convert back.

  1. The magnitude tells you which frequencies are strongest. The phase contains most of the structural information about the signal, but the phase values by themselves aren't really useful to analyse.
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  • $\begingroup$ I have edited the first part of the question, can you please check now $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 10:10
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1) What you've written is complex conjugate. In the Fourier case it could be interesting while

  • we are working with real input signal so it's Fourier domain image is Hermitian function, see wiki for example
  • In matrix form discrete Fourier transform could be written as a matrix $W$ so for inverse transform it will be $W^H$, where $()^H$ denotes Hermitian transpose.

2) Magnitude and phase of Fourier coefficients determine signal's spectrum in Fourier domain. The physical meaning of Fourier domain depends on which dimension are used for signal representation. The simplest examples are time and space. If we have signal represented as a time serie, after Fourier we will obtain frequency spectrum, while if we have signal represented as plane wave in space, Fourier domain will reflect spatial (or angular) spectrum.

Usually the most interest is in magnitude part of Fourier because it is used for estimating power spectrum. But sometimes phase is also important - filter analysis, frequency domain equalization, etc.

Hope it helps.

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  • $\begingroup$ I have edited the first part of the question, can you please check now $\endgroup$ – Aadnan Farooq A Oct 8 '15 at 10:10

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