Fourier expansion of periodic signal proof
$begingroup$
Definitions
Period: The period is the smallest value of T satisfying $$g(t + T) = g(t)label{0}tag{0}$$ for all t. The period is defined so because if g(t + T) = g(t) for all t, it can be verified that g(t + T') = g(t) for all t where T' = 2T, 3T, 4T, ... In essence, it's the smallest amount of time it takes for the function to repeat itself. If the period of a function is finite, the function is called "periodic". Functions that never repeat themselves have an infinite period, and are known as "aperiodic functions". The period of a periodic waveform will be denoted with a capital T. The period is measured in seconds.
Kronecker delta function:
$$
delta_{ij}=
begin{cases}
0&text{if}, ineq j\
1&text{if}, i= j
end{cases}
label{1}tag{1}
$$Discrete-time signal x: $x(n)=sum_{k=-infty}^{infty}x(k)delta(n-k)label{2}tag{2}$
- Continuous-time signal x: $$x(t)=int_{-infty}^{infty}x(tau)delta(t-tau)delta{tau}label{3}tag{3}$$
Euler formula
derivations:
$$e^{jtheta}=cos(theta)+jsin(theta)label{4}tag{4}$$
$$e^{-jtheta}=cos(theta)-jsin(theta)label{5}tag{5}$$
$$cos(theta) = frac{1}{2}(e^{-jtheta}+e^{jtheta})label{6}tag{6}$$
$$sin(theta) = frac{1}{2j}(e^{-jtheta}-e^{jtheta})label{7}tag{7}$$
Problem
The fourier expansion of a periodic signal $x_T(t)=x_T(t+T)$ is
$$x_T(t)=F^{-1}[X[k]]=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{8}tag{8}$$
where $X[k]$ is the fourier coefficient
$$X[k]=F[x_T(t)]=frac{1}{T}int_Tx_T(t)e^{-jkw_0t}dt label{9}tag{9}$$
where $k=0, pm1,pm2,...$
Question
I'd like to prove this:
$$x_T(t)=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{10}tag{10}$$
I believe by considering the above definitions it should be possible to get a nice simple proof but I don't know how to proceed here.
proof-writing fourier-analysis fourier-series
$endgroup$
add a comment |
$begingroup$
Definitions
Period: The period is the smallest value of T satisfying $$g(t + T) = g(t)label{0}tag{0}$$ for all t. The period is defined so because if g(t + T) = g(t) for all t, it can be verified that g(t + T') = g(t) for all t where T' = 2T, 3T, 4T, ... In essence, it's the smallest amount of time it takes for the function to repeat itself. If the period of a function is finite, the function is called "periodic". Functions that never repeat themselves have an infinite period, and are known as "aperiodic functions". The period of a periodic waveform will be denoted with a capital T. The period is measured in seconds.
Kronecker delta function:
$$
delta_{ij}=
begin{cases}
0&text{if}, ineq j\
1&text{if}, i= j
end{cases}
label{1}tag{1}
$$Discrete-time signal x: $x(n)=sum_{k=-infty}^{infty}x(k)delta(n-k)label{2}tag{2}$
- Continuous-time signal x: $$x(t)=int_{-infty}^{infty}x(tau)delta(t-tau)delta{tau}label{3}tag{3}$$
Euler formula
derivations:
$$e^{jtheta}=cos(theta)+jsin(theta)label{4}tag{4}$$
$$e^{-jtheta}=cos(theta)-jsin(theta)label{5}tag{5}$$
$$cos(theta) = frac{1}{2}(e^{-jtheta}+e^{jtheta})label{6}tag{6}$$
$$sin(theta) = frac{1}{2j}(e^{-jtheta}-e^{jtheta})label{7}tag{7}$$
Problem
The fourier expansion of a periodic signal $x_T(t)=x_T(t+T)$ is
$$x_T(t)=F^{-1}[X[k]]=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{8}tag{8}$$
where $X[k]$ is the fourier coefficient
$$X[k]=F[x_T(t)]=frac{1}{T}int_Tx_T(t)e^{-jkw_0t}dt label{9}tag{9}$$
where $k=0, pm1,pm2,...$
Question
I'd like to prove this:
$$x_T(t)=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{10}tag{10}$$
I believe by considering the above definitions it should be possible to get a nice simple proof but I don't know how to proceed here.
proof-writing fourier-analysis fourier-series
$endgroup$
$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32
add a comment |
$begingroup$
Definitions
Period: The period is the smallest value of T satisfying $$g(t + T) = g(t)label{0}tag{0}$$ for all t. The period is defined so because if g(t + T) = g(t) for all t, it can be verified that g(t + T') = g(t) for all t where T' = 2T, 3T, 4T, ... In essence, it's the smallest amount of time it takes for the function to repeat itself. If the period of a function is finite, the function is called "periodic". Functions that never repeat themselves have an infinite period, and are known as "aperiodic functions". The period of a periodic waveform will be denoted with a capital T. The period is measured in seconds.
Kronecker delta function:
$$
delta_{ij}=
begin{cases}
0&text{if}, ineq j\
1&text{if}, i= j
end{cases}
label{1}tag{1}
$$Discrete-time signal x: $x(n)=sum_{k=-infty}^{infty}x(k)delta(n-k)label{2}tag{2}$
- Continuous-time signal x: $$x(t)=int_{-infty}^{infty}x(tau)delta(t-tau)delta{tau}label{3}tag{3}$$
Euler formula
derivations:
$$e^{jtheta}=cos(theta)+jsin(theta)label{4}tag{4}$$
$$e^{-jtheta}=cos(theta)-jsin(theta)label{5}tag{5}$$
$$cos(theta) = frac{1}{2}(e^{-jtheta}+e^{jtheta})label{6}tag{6}$$
$$sin(theta) = frac{1}{2j}(e^{-jtheta}-e^{jtheta})label{7}tag{7}$$
Problem
The fourier expansion of a periodic signal $x_T(t)=x_T(t+T)$ is
$$x_T(t)=F^{-1}[X[k]]=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{8}tag{8}$$
where $X[k]$ is the fourier coefficient
$$X[k]=F[x_T(t)]=frac{1}{T}int_Tx_T(t)e^{-jkw_0t}dt label{9}tag{9}$$
where $k=0, pm1,pm2,...$
Question
I'd like to prove this:
$$x_T(t)=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{10}tag{10}$$
I believe by considering the above definitions it should be possible to get a nice simple proof but I don't know how to proceed here.
proof-writing fourier-analysis fourier-series
$endgroup$
Definitions
Period: The period is the smallest value of T satisfying $$g(t + T) = g(t)label{0}tag{0}$$ for all t. The period is defined so because if g(t + T) = g(t) for all t, it can be verified that g(t + T') = g(t) for all t where T' = 2T, 3T, 4T, ... In essence, it's the smallest amount of time it takes for the function to repeat itself. If the period of a function is finite, the function is called "periodic". Functions that never repeat themselves have an infinite period, and are known as "aperiodic functions". The period of a periodic waveform will be denoted with a capital T. The period is measured in seconds.
Kronecker delta function:
$$
delta_{ij}=
begin{cases}
0&text{if}, ineq j\
1&text{if}, i= j
end{cases}
label{1}tag{1}
$$Discrete-time signal x: $x(n)=sum_{k=-infty}^{infty}x(k)delta(n-k)label{2}tag{2}$
- Continuous-time signal x: $$x(t)=int_{-infty}^{infty}x(tau)delta(t-tau)delta{tau}label{3}tag{3}$$
Euler formula
derivations:
$$e^{jtheta}=cos(theta)+jsin(theta)label{4}tag{4}$$
$$e^{-jtheta}=cos(theta)-jsin(theta)label{5}tag{5}$$
$$cos(theta) = frac{1}{2}(e^{-jtheta}+e^{jtheta})label{6}tag{6}$$
$$sin(theta) = frac{1}{2j}(e^{-jtheta}-e^{jtheta})label{7}tag{7}$$
Problem
The fourier expansion of a periodic signal $x_T(t)=x_T(t+T)$ is
$$x_T(t)=F^{-1}[X[k]]=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{8}tag{8}$$
where $X[k]$ is the fourier coefficient
$$X[k]=F[x_T(t)]=frac{1}{T}int_Tx_T(t)e^{-jkw_0t}dt label{9}tag{9}$$
where $k=0, pm1,pm2,...$
Question
I'd like to prove this:
$$x_T(t)=sum_{k=-infty}^{+infty}X[k]e^{jkw_0t}label{10}tag{10}$$
I believe by considering the above definitions it should be possible to get a nice simple proof but I don't know how to proceed here.
proof-writing fourier-analysis fourier-series
proof-writing fourier-analysis fourier-series
edited Feb 2 at 19:58
BPL
asked Jan 24 at 16:30
BPLBPL
6817
6817
$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32
add a comment |
$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32
$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.
In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.
Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.
Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.
$endgroup$
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
add a comment |
$begingroup$
Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as
$$
delta(x)=frac1{2pi}sum_{n=-infty}^infty e^{inx}.
$$
(cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have
begin{equation}begin{aligned}
frac1Tsum_{k=-infty}^inftyint_0^Tdt'x(t')e^{2pi ikfrac{t-t'}T}&=frac1Tint_0^Tdt'x(t')sum_{k=-infty}^infty e^{2pi ikfrac{t-t'}T}\
&=x(t).
end{aligned}end{equation}
Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.
$endgroup$
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
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2 Answers
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2 Answers
2
active
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$begingroup$
I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.
In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.
Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.
Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.
$endgroup$
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
add a comment |
$begingroup$
I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.
In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.
Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.
Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.
$endgroup$
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
add a comment |
$begingroup$
I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.
In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.
Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.
Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.
$endgroup$
I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.
In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.
Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.
Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.
edited Feb 3 at 19:18
answered Feb 3 at 17:40
Alex RavskyAlex Ravsky
42.5k32383
42.5k32383
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
add a comment |
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
1
1
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
$begingroup$
Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established
$endgroup$
– BPL
Feb 3 at 18:16
add a comment |
$begingroup$
Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as
$$
delta(x)=frac1{2pi}sum_{n=-infty}^infty e^{inx}.
$$
(cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have
begin{equation}begin{aligned}
frac1Tsum_{k=-infty}^inftyint_0^Tdt'x(t')e^{2pi ikfrac{t-t'}T}&=frac1Tint_0^Tdt'x(t')sum_{k=-infty}^infty e^{2pi ikfrac{t-t'}T}\
&=x(t).
end{aligned}end{equation}
Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.
$endgroup$
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
|
show 2 more comments
$begingroup$
Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as
$$
delta(x)=frac1{2pi}sum_{n=-infty}^infty e^{inx}.
$$
(cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have
begin{equation}begin{aligned}
frac1Tsum_{k=-infty}^inftyint_0^Tdt'x(t')e^{2pi ikfrac{t-t'}T}&=frac1Tint_0^Tdt'x(t')sum_{k=-infty}^infty e^{2pi ikfrac{t-t'}T}\
&=x(t).
end{aligned}end{equation}
Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.
$endgroup$
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
|
show 2 more comments
$begingroup$
Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as
$$
delta(x)=frac1{2pi}sum_{n=-infty}^infty e^{inx}.
$$
(cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have
begin{equation}begin{aligned}
frac1Tsum_{k=-infty}^inftyint_0^Tdt'x(t')e^{2pi ikfrac{t-t'}T}&=frac1Tint_0^Tdt'x(t')sum_{k=-infty}^infty e^{2pi ikfrac{t-t'}T}\
&=x(t).
end{aligned}end{equation}
Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.
$endgroup$
Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as
$$
delta(x)=frac1{2pi}sum_{n=-infty}^infty e^{inx}.
$$
(cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have
begin{equation}begin{aligned}
frac1Tsum_{k=-infty}^inftyint_0^Tdt'x(t')e^{2pi ikfrac{t-t'}T}&=frac1Tint_0^Tdt'x(t')sum_{k=-infty}^infty e^{2pi ikfrac{t-t'}T}\
&=x(t).
end{aligned}end{equation}
Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.
edited Feb 2 at 9:32
answered Feb 1 at 13:20
Alec B-GAlec B-G
49519
49519
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
|
show 2 more comments
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :)
$endgroup$
– BPL
Feb 1 at 18:39
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
I've edited the question... in case you're still interested
$endgroup$
– BPL
Feb 2 at 18:35
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $sum_k X[k]e^{jkomega_0 t}$. I've written $i$ for $j$ and used $omega_0=frac{2pi}{T}$.
$endgroup$
– Alec B-G
Feb 2 at 18:43
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out.
$endgroup$
– Alec B-G
Feb 2 at 18:44
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
$begingroup$
If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $ref{0}$ & $ref{2}$ (discrete) or $ref{0}$ & $ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones.
$endgroup$
– BPL
Feb 2 at 19:14
|
show 2 more comments
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$begingroup$
This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter.
$endgroup$
– jmerry
Jan 28 at 13:42
$begingroup$
When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain.
$endgroup$
– BPL
Jan 28 at 17:32