• sahil_time
In summary: However, for finite N, the Fourier series S_N(t) can be calculated as 1/T times the summation from n = -N to N of e^(i * n * 2π * t/T), which can be simplified to (sin((2π/T)(N+1/2)t))/(T*sin((π/T)t)). As N gets larger, the oscillations in the series get more closely spaced, so when multiplied by a continuous, smooth function and integrated, there is no contribution to the integral in the neighborhood of t=0.5 as N goes to infinity.

sahil_time

http://en.wikipedia.org/wiki/Dirac_comb

Please have a look at the Fourier Series section, and its last equation.

Let T = 1.

After expanding the Equation

x(t) = 1 + 2cos(2∏t) + 2cos(4∏t) + 2cos(6∏t) ...

Now this does not give the original Dirac Comb.
Eg: at t = 1/2

x(1/2) = 0

But RHS

= 1 + 2cos(2∏*1/2) + 2cos(4∏*1/2) + 2cos(6∏*1/2) ...
= 1 + 2cos(∏) + 2cos(2∏) + 2cos(3∏) ...
= 1 -2 + 2 -2 + 2...
≠ LHSWhat is the problem?

The mathematics is tricky here. You need to set it up as a function of t in closed form and take limits for t = multiples of 1/2.

mathman said:
The mathematics is tricky here. You need to set it up as a function of t in closed form and take limits for t = multiples of 1/2.

Agreed Mathman, after doing that you will get the LHS ≠ RHS. Where LHS is the dirac comb graph, and RHS is the computation via Fourier series. Can you please elaborate?

sahil_time said:
Agreed Mathman, after doing that you will get the LHS ≠ RHS. Where LHS is the dirac comb graph, and RHS is the computation via Fourier series. Can you please elaborate?

What I was trying to say is that you need to see what the Fourier series sums to first.

I'll give a simple example of the procees I was talking about.

1 + x + x2 + ... = 1/(1-x) for |x| < 1. for x = -1, the l.h.s is 1 - 1 + 1 ..., while the r.h.s = 1/2. This makes sense only if you get the sum first where it converges and then extend it.

What I am suggesting is the closed form for values where it converges and then extend it.

mathman said:
What I was trying to say is that you need to see what the Fourier series sums to first.

I'll give a simple example of the procees I was talking about.

1 + x + x2 + ... = 1/(1-x) for |x| < 1. for x = -1, the l.h.s is 1 - 1 + 1 ..., while the r.h.s = 1/2. This makes sense only if you get the sum first where it converges and then extend it.

What I am suggesting is the closed form for values where it converges and then extend it.

Mathman if we just look at the second equation on the wikipedia page, which has absolutely no convergence issues because we can choose T as per our convenience, and if we let T=1, and expand it, it will not give LHS = RHS for values other than t = nT. Where n is an integer.

mathman said:
What I was trying to say is that you need to see what the Fourier series sums to first.

I'll give a simple example of the procees I was talking about.

1 + x + x2 + ... = 1/(1-x) for |x| < 1. for x = -1, the l.h.s is 1 - 1 + 1 ..., while the r.h.s = 1/2. This makes sense only if you get the sum first where it converges and then extend it.

What I am suggesting is the closed form for values where it converges and then extend it.

I had taken the limits for T=1 between t=1/2 and t=-1/2.

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mathman said:
The mathematics is tricky here. You need to set it up as a function of t in closed form and take limits for t = multiples of 1/2.
I think i found the answer , it is unusual.

Where 1-1+1-1+1... = 1/2

http://en.wikipedia.org/wiki/Grandi's_series#Heuristics

So the original RHS = 1 -2 + 2 - 2 + 2...

= 1 -2( 1 - 1 + 1 - 1...)

=1 -2( 1/2 )...Using Grandi Series

=1 -1

=0

=LHS.

I am am an engineer not a mathematician, so my answer will not pretend to be rigorous.

Anyway, when you are dealing with animals like delta functions it does not make sense to try to evaluate what they "equal to" point-wise. They only make any sense when multiplied by a "nice" function (continuous, differentiable, falls of fast as t-> infinity) and integrated. Thus, we would expect that the Fourier series doesn't make sense evaluated at individual points either. Also, since the Dirac comb has nasty discontinuities (in some sense), we expect Gibbs phenomenon to be present in the Fourier series.

Going back to your problem, for finite $N$, if I did my math correctly (just using sum of geometric series here and basic trig identities, nothing fancy),
$$S_N(t) = \frac{1}{T} \sum_{n=-N}^{N} \exp\left( i n \frac{2 \pi}{T} t \right) = \frac{\sin\left( \frac{2 \pi}{T} (N + \frac{1}{2}) t \right)}{T \sin\left( \frac{\pi}{T}t\right)}$$
With T=1 and t=0.5 for example, this is either 1 or -1, depending upon the value of N and no matter how big N gets. So pointwise the sequence $S_N (t)$ does not converge at all as N goes to infinity. However, if you look at the oscillations in the series as a function of t, as N gets bigger and bigger, the oscillations get more closely spaced in time, so when you multiply by a continuous, smooth function and integrate there is no contribution to the integral in the neighborhood of t=0.5 as N goes to infinity. Likewise for every point in between the locations of the delta functions.

It can be instructive to actually plot the Fourier series $S_N(t)$ over some time interval for a few values of N (say 10, 30, 100) just to see how it behaves.

jason

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1. What is a Fourier Series of Dirac Comb?

A Fourier Series of Dirac Comb is a mathematical representation of a periodic function that is composed of an infinite sum of equally spaced Dirac delta functions.

2. How is a Fourier Series of Dirac Comb used in science?

In science, a Fourier Series of Dirac Comb is often used to analyze and model periodic phenomena, such as signals and waves, in fields such as physics, engineering, and mathematics.

3. What is the difference between a Fourier Series and a Fourier Series of Dirac Comb?

A Fourier Series represents a periodic function as a sum of sines and cosines, while a Fourier Series of Dirac Comb represents a periodic function as a sum of Dirac delta functions, which are impulses. This allows for a more precise representation of a periodic function with sharp peaks or discontinuities.

4. How is a Fourier Series of Dirac Comb calculated?

A Fourier Series of Dirac Comb can be calculated using the formula: f(x) = 1/T * ∑n=-∞∞ f(nT)δ(x-nT), where T is the period of the function and δ(x) is the Dirac delta function.

5. What are some real life applications of Fourier Series of Dirac Comb?

Fourier Series of Dirac Comb is commonly used in signal processing, image processing, and audio analysis to compress and analyze data in a more efficient way. It is also used in telecommunications to transmit data over long distances. Additionally, it has applications in quantum mechanics, where it is used to describe the wave nature of particles.