Fourier Transform of Dirac Comb/Impulse Train

[Terocamo]

With Dirac Comb is defined as follow:
$$III(t)=\sum_{n=-\infty}^\infty\delta(t-nT)$$
Fourier Transform from t domain to frequency domain can be obtained by:
$$F(f)=\int_{-\infty}^{\infty}f(t)\cdot e^{-i2\pi ft}dt$$
I wonder why directly apply the above equation does not work for the Dirac Comb:
$$F(III(t))=\sum_{n=-\infty}^\infty\int_{-\infty}^{\infty}\delta(t-nT)\cdot e^{i2\pi ft}dt$$
$$III(f)=\sum_{n=-\infty}^\infty e^{-i2\pi fnT}dt\cdots\cdots [1]$$
Where the correct way to obtain the FT of Dirac Comb is to first find the Fourier series, and then do the Fourier Transform for each term in the summation.

Writing III(t) as Fourier Series:
$$III(t)=\frac{1}{T}\sum_{n=-\infty}^\infty e^{i2\pi nt/T}$$
Doing Fourier Transform:
$$III(f)=\frac{1}{T}\sum_{n=-\infty}^\infty \int_{-\infty}^{\infty} e^{i2\pi nt/T}\cdot e^{-i2\pi ft} dt\cdots\cdots [2]$$
$$III(f)=\frac{1}{T}\sum_{n=-\infty}^\infty\delta(f-\frac{n}{T})\cdots\cdots [3]$$

I copied the upper section from a open lecture slide, and I don't even understand how it goes from [2] to [3], not to mention [1] and [3] are totally not the same thing. Any hints guys?

 

[blue_leaf77]

$$III(f)=\sum_{n=-\infty}^\infty e^{-i2\pi fnT}dt\cdots\cdots [1]$$

The $dt$ shouldn't be there.

[1] and [3] are totally not the same thing.

They are equivalent. To see this, consider a function $f(x) = \sum_n \delta (x-na)$. This is a periodic function with periodicity $a$, which means it can be expanded in Fourier series. So
$$
f(x) = \sum_{n=-\infty}^{\infty} \delta (x-na) = \sum_{k=-\infty}^{\infty} c_k \hspace{0.5mm}e^{-i2\pi x/a}
$$
Find $c_k$ and you will see that [1] and [3] are equivalent.

I don't even understand how it goes from [2] to [3]

That's the Fourier definition of delta function
$$
\delta(x-a) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-a)} \hspace{1mm} dk
$$

 

[Terocamo]

They are equivalent. To see this, consider a function $f(x) = \sum_n \delta (x-na)$. This is a periodic function with periodicity $a$, which means it can be expanded in Fourier series. So
$$
f(x) = \sum_{n=-\infty}^{\infty} \delta (x-na) = \sum_{k=-\infty}^{\infty} c_k \hspace{0.5mm}e^{-i2\pi x/a}
$$
Find $c_k$ and you will see that [1] and [3] are equivalent.

$$

Oh I get it, with III(f) having a period of 1/T, the Fourier series should write:
$$\frac{1}{T}\sum_{n=-\infty}^\infty\delta(f-\frac{n}{T})=\sum_{k=-\infty}^\infty c_k\cdot e^{i2\pi kfT}\cdots\cdots[4]$$
$$c_k= T\cdot\int_{0}^{1/T} \sum_{n=-\infty}^\infty\delta(f-\frac{n}{T}) \cdot e^{-i2\pi kfT}=1$$

Putting c_k=1 back to [4] results in:
$$\sum_{k=-\infty}^\infty e^{i2\pi kfT} = \text{(1)}$$

Just to clear my concept, s this equivalent to (1) since the exponent missed out a -ve sign, but k ranged from -inf to inf so its just the same?

 

[blue_leaf77]

$$\frac{1}{T}\sum_{n=-\infty}^\infty\delta(f-\frac{n}{T})=\sum_{k=-\infty}^\infty c_k\cdot e^{i2\pi kfT}\cdots\cdots[4]$$

The used definition for Fourier series in complex form is to use negative exponent.