DTFT

DTFT

$$\begin{aligned} X(f) &= \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt\\ x^*(t) &= \int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df\\ \end{aligned}$$

When $\hat{x}(t) = x(t) p(t) = x(t) \cdot \sum \delta(t-nT)= \sum x(nT) \delta(t-nT)$, then we have $\lambda < 1$ that

$$\begin{aligned} \hat{X}(f) &= \int_{-\infty}^{\infty} \hat{x}(t) e^{-j2\pi ft} dt = \sum_n x(nT) \int_{-\infty}^{\infty} \delta(t-nT)e^{-j2\pi ft} dt\\ &=\sum_n x(nT) e^{-j2 \pi n (f/f_s)}\\ \hat{x}^*(t) &= \int_{-\infty}^{\infty} \hat{X}(f) e^{j2\pi ft} df\\ x(nT) &=\int_{-\infty}^{\infty} \hat{x}^*(t) [\text{u}(t-(n-\lambda)T)-\text{u}(t-(n+\lambda)T)] dt\\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \hat{X}(f) e^{j2\pi ft} df [\text{u}(t-(n-\lambda)T)-\text{u}(t-(n+\lambda)T)] dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \int_{-\infty}^{\infty} e^{j2\pi ft} [\text{u}(t-(n-\lambda)T)-\text{u}(t-(n+\lambda)T)] dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \int_{(n-\lambda)T}^{(n+\lambda)T} e^{j2\pi ft} dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \int_{(n-\lambda)}^{(n+\lambda)} e^{j2\pi (f/f_s)t'} dt'/f_s \quad[t' = f_s t]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) \frac{e^{j2\pi (f/f_s)(n+\lambda)}-e^{j2\pi (f/f_s)(n-\lambda)}}{j2\pi (f/f_s)}\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} \frac{e^{j2\pi (f/f_s)\lambda}-e^{-j2\pi (f/f_s)\lambda}}{j2\pi (f/f_s)}\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} \frac{2\lambda j\sin(2\pi(f/f_s)\lambda)}{j2\pi (f/f_s)\lambda}\\ &= 2\lambda \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} \frac{\sin(2\pi(f/f_s)\lambda)}{2\pi (f/f_s)\lambda}\\ \end{aligned}$$

If we replace kernel function: $[\text{u}(t-(n-\lambda)T)-\text{u}(t-(n+\lambda)T)]$ with function

$$\frac{\sin(\pi f_s(t-nT))}{\pi f_s(t-nT)} = \frac{e^{j2\pi (f_s/2)(t-nT)}-e^{-j2\pi (f_s/2)(t-nT)}}{j2\pi f_s(t-nT)}$$

We still have:

$$\int_{-\infty}^{\infty} \hat{x}^*(t) \frac{\sin(\pi f_s(t-nT))}{\pi f_s(t-nT)} dt\\ = \int_{-\infty}^{\infty} \sum_{n'} x(n'T) \delta(t-n'T) \frac{\sin(\pi f_s(t-nT))}{\pi f_s(t-nT)} dt\\ = \sum_{n'} x(n'T) \int_{-\infty}^{\infty} \delta(t-n'T) \frac{\sin(\pi f_s(t-nT))}{\pi f_s(t-nT)} dt\\ = \sum_{n'} x(n'T) \frac{\sin(\pi (f_s/f_s)(n'-n))}{\pi (f_s/f_s)(n'-n)} \int_{-\infty}^{\infty} \delta(t-n'T) dt\\ = \sum_{n'} x(n'T) \delta(n'-n) = x(nT)$$

So, we have:

$$\begin{aligned} x(nT)&=\int_{-\infty}^{\infty} \hat{x}^*(t) \frac{\sin(\pi(t-nT))}{\pi(t-nT)} dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \int_{-\infty}^{\infty} e^{j2\pi ft}\frac{\sin(2\pi (f_s/2)(t-nT))}{2\pi (f_s/2)(t-nT)}dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \int_{-\infty}^{\infty} e^{j2\pi ft} \frac{e^{j2\pi (f_s/2)(t-nT)}-e^{-j2\pi (f_s/2)(t-nT)}}{2j2\pi (f_s/2)(t-nT)}dt\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \cdot [e^{-j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi ft} \frac{e^{j2\pi (f_s/2) t}}{j2\pi f_s(t-nT)}dt - e^{ j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi ft} \frac{e^{-j2\pi (f_s/2) t}}{j2\pi f_s(t-nT)}dt]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) df \cdot [e^{-j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s)t'} \frac{e^{j\pi t'}}{j2\pi (t'-n)}dt'/f_s - e^{ j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s)t'} \frac{e^{-j\pi t'}}{j2\pi (t'-n)}dt'/f_s]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) \cdot [e^{-j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s+0.5)t'} \frac{1}{j2\pi (t'-n)}dt' - e^{ j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s-0.5)t'} \frac{1}{j2\pi (t'-n)}dt']\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) \cdot [e^{-j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s+0.5)(t'+n)} \frac{1}{j2\pi t'}dt' - e^{ j2\pi n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s-0.5)(t'+n)} \frac{1}{j2\pi t'}dt']\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) \cdot [e^{-j2\pi n}e^{j2\pi (f/f_s+0.5)n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s+0.5)t'} \frac{1}{j2\pi t'}dt' - e^{ j2\pi n} e^{j2\pi (f/f_s-0.5)n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s-0.5)t'} \frac{1}{j2\pi t'}dt']\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) [e^{j2\pi (f/f_s)n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s+0.5)t'} \frac{1}{j2\pi t'}dt' - e^{j2\pi (f/f_s)n} \int_{-\infty}^{\infty} e^{j2\pi (f/f_s-0.5)t'} \frac{1}{j2\pi t'}dt' ]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} [\int_{-\infty}^{\infty} e^{j2\pi (f/f_s+0.5)t'} \frac{1}{j2\pi t'}dt' - \int_{-\infty}^{\infty} e^{j2\pi (f/f_s-0.5)t'} \frac{1}{j2\pi t'}dt' ]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} \frac{1}{2}[\text{sgn}(f/f_s+0.5) - \text{sgn}(f/f_s-0.5) ]\\ &= \int_{-\infty}^{\infty} \hat{X}(f) d(f/f_s) e^{j2\pi (f/f_s)n} [\text{u}(f/f_s+0.5) - \text{u}(f/f_s-0.5) ]\\ &= \int_{-0.5}^{0.5} \hat{X}(f) e^{j2\pi (f/f_s)n} d(f/f_s)\\ \end{aligned}$$

because $\text{sgn}(t) <=> \frac{1}{j\pi f}$, then we have $\frac{1}{j\pi t} <=> \text{sgn}(-f)=-\text{sgn}(f)$,

Thus means

$$-\text{sgn}(f) = \int^{\infty}_{-\infty} \frac{1}{j\pi t} e^{-j2\pi ft} dt\\ \text{sgn}(f) =-\text{sgn}(-f) = \int^{\infty}_{-\infty} \frac{1}{j\pi t} e^{j2\pi ft} dt\\ \text{sgn}(f/f_s+0.5)= \int^{\infty}_{-\infty} \frac{1}{j\pi t} e^{j2\pi (f/f_s+0.5)t} dt = 2\text{u}(f/f_s+0.5)-1\\ \text{sgn}(f/f_s-0.5)= \int^{\infty}_{-\infty} \frac{1}{j\pi t} e^{j2\pi (f/f_s-0.5)t} dt = 2\text{u}(f/f_s-0.5)-1\\$$

To sum up:

for $\hat{x}(t) = x(t) p(t) = x(t) \cdot \sum \delta(t-nT)= \sum x(nT) \delta(t-nT)$, we have

$$\begin{aligned} \hat{X}(f) &= \int_{-\infty}^{\infty} \hat{x}(t) e^{-j2\pi ft} dt\\ &=\sum_n x(nT) e^{-j2 \pi n (f/f_s)}\\ \hat{x}^*(t) &= \int_{-\infty}^{\infty} \hat{X}(f) e^{j2\pi ft} df\\ x(nT) &= \int_{-\infty}^{\infty} \hat{x}^*(t) \frac{\sin(\pi(t-nT))}{\pi(t-nT)} dt\\ &= \int_{-0.5}^{0.5} \hat{X}(f) e^{j2\pi (f/f_s)n} d(f/f_s)\\ \end{aligned}$$

Now define $\omega = 2\pi f/f_s$, then

$$X(e^{jw}) \equiv \hat{X}(f) = \sum_n x(nT) [e^{j\omega}]^{-n}\\ x(nT)= \int_{-0.5}^{0.5} \hat{X}(f) e^{j2\pi (f/f_s)n} d(f/f_s) = \frac{1}{2\pi} \int_{2\pi} X(e^{jw}) [e^{j\omega}]^{n} d\omega$$

Formula

$$\text{FT}[\delta(t)] = \int\delta(t) e^{-j2\pi ft} dt = \int\delta(t) e^{-j2\pi f0} dt = 1\cdot \int\delta(t) dt=1(f)\\ \text{so, } \delta(t) = \text{FT}^{-1}[1(f)] = \int 1 \cdot e^{j2\pi ft} df$$

then replace $f\to t, t\to -f$, having

$$\begin{aligned} \delta(-f) &= \int1(t) \cdot e^{j2\pi t(-f)} dt = \text{FT}[1(t)]\\ &= \delta(f) \end{aligned}$$

Thus Fourier pair $1(t) <-> \delta(f)$, now we want to verify:

$$\begin{aligned} \sum_n \delta(t-nT) &= \frac{1}{T} \sum_k e^{j2\pi k(\frac{t}{T})} \quad [\alpha\delta(\alpha t') = \delta(t'), t'=\frac{t}{T}=tf_s]\\ \sum_n \delta(t'-n) &= \sum_k e^{j2\pi k t'} \end{aligned}$$

Here we have:

一个矩形脉冲的傅里叶变换是sinc函数

一个矩形脉冲离散采样之后的离散傅里叶变换就是Dirichlet函数

$$\begin{aligned} \sum_{k=-A}^{A}e^{j2\pi kt'} &= 1+ 2 \sum_{k=1}^A \cos(2\pi t' k)\\ \sin(\pi t' ) \sum_{k=-A}^{A}e^{j2\pi kt'} &= \sin(\pi t' ) + \sum_{k=1}^A[ \sin(2\pi t'(k+0.5)) - \sin(2\pi t'(k-0.5)) ]\\ &= \sin(2\pi t'(A+0.5)) = \sin(\pi t'(2A+1))\\ \sum_{k=-A}^{A}e^{j2\pi kt'} &= \frac{\sin(\pi t'(2A+1))}{\sin(\pi t' )} \end{aligned}$$

Moreover, $ \frac{\sin(\pi (t’+\Delta) (2A+1))}{\sin(\pi (t’+\Delta) )}= \frac{\sin(\pi t’(2A+1))}{\sin(\pi t’ )}, \Delta\in Z$, the period is 1:

$$\int_{-0.5}^{0.5} \frac{\sin(\pi t'(2A+1))}{\sin(\pi t' )} dt' = \sum_{k=-A}^{A} \int_{-0.5}^{0.5}e^{j2\pi kt'} dt'= \sum_{k=-A}^{A} \delta(k) = 1$$

So, $\lim_{A\to \infty} \frac{\sin(\pi t’(2A+1))}{\sin(\pi t’ )}[\text{u}(t’+0.5)-\text{u}(t’-0.5)] = \delta(t’)$, then we have

$$\begin{aligned} \lim_{A\to \infty} \sum_{k=-A}^{A}e^{j2\pi kt'} &= \lim_{A\to \infty} \frac{\sin(\pi t'(2A+1))}{\sin(\pi t' )}\\ &= \lim_{A\to \infty} \sum_n \frac{\sin(\pi t'(2A+1))}{\sin(\pi t' )} [\text{u}(t'+0.5-n)-\text{u}(t'-0.5-n)] \\ &= \lim_{A\to \infty} \sum_n \frac{\sin(\pi (t'-n)(2A+1))}{\sin(\pi (t'-n) )} [\text{u}(t'+0.5-n)-\text{u}(t'-0.5-n)] \\ &= \sum_n \lim_{A\to \infty} \frac{\sin(\pi (t'-n)(2A+1))}{\sin(\pi (t'-n) )} [\text{u}(t'+0.5-n)-\text{u}(t'-0.5-n)] \\ &= \sum_n \delta(t'-n) \\ \end{aligned}$$

Thus,

$$\sum_n \delta(t'-n) = \sum_k e^{j2\pi k t'} \quad[t=t' T]\\ T\sum_n \delta(t-nT) = \sum_n \delta(t/T-n) = \sum_k e^{j2\pi k(\frac{t}{T})}$$

DFT

consider to sample $\hat{x}(t) = x(t) p(t) = x(t) \cdot \sum \delta(t-nT)= \sum x(nT) \delta(t-nT)$ 0~NT, as a period

$$\begin{aligned} \tilde{x}(t) &= \left\{ \hat{x}(t) \left[ \text{u}(t)-\text{u}(t-NT) \right] \right\} * \sum_{n'} \delta(t-n'NT)\\ &= \left\{ \sum_{n=0}^{N-1} x(nT) \delta(t-nT) \right\} * \sum_{n'} \delta(t-n'NT)\\ &= \sum_{n'} \left\{ \sum_{n=0}^{N-1} x(nT) \delta(t-(n+n'N)T) \right\} \\ \end{aligned}$$

For freq domain:

$$\begin{aligned} \tilde{X}(f) &= \int_{-\infty}^{\infty} \tilde{x}(t) e^{-j2\pi ft} dt \\ &= \sum_{n'} \left\{ \sum_{n=0}^{N-1} x(nT) [ \int_{-\infty}^{\infty} \delta(t-(n+n'N)T) e^{-j2\pi ft} dt] \right\} \\ &= \sum_{n'} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (f/f_s)(n+n'N)} \right\} \\ &= \sum_{n'} e^{-j2\pi (\frac{f}{f_s/N})n'} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (f/f_s)n} \right\} \quad[\sum_n \delta(t'-n) = \sum_k e^{j2\pi k t'}] \\ &= \sum_{k} \delta(\frac{f}{f_s/N}-k) \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (f/f_s)n} \right\}\\ &= \frac{f_s}{N} \sum_{k} \delta(f-k\frac{f_s}{N}) \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (f/f_s)n} \right\}\\ &= \frac{f_s}{N} \sum_{k} \delta(f-k\frac{f_s}{N}) \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\}\\ &= \sum_{k} \delta(f-k\frac{f_s}{N}) \Bigg[\frac{f_s}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg] \end{aligned}$$

Reconstruction:

$$\begin{aligned} \tilde{x}^*(t) &= \int_{-\infty}^{\infty} \tilde{X}(f) e^{j2\pi ft} df\\ &=\int_{-\infty}^{\infty} e^{j2\pi ft} df \sum_{k} \delta(f-k\frac{f_s}{N}) \Bigg[\frac{f_s}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg] \\ &= \sum_{k} \Bigg[\frac{f_s}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg] \int_{-\infty}^{\infty} e^{j2\pi ft} \delta(f-k\frac{f_s}{N}) df \\ &= \sum_{k} \Bigg[\frac{f_s}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg] e^{j2\pi (k/N)f_st} \\ x(nT) &= \int_{-\infty}^{\infty} \tilde{x}^*(t) \frac{\sin(\pi(t-nT))}{\pi(t-nT)} dt\\ &= \int_{-0.5}^{0.5} \tilde{X}(f) e^{j2\pi (f/f_s)n} d(f/f_s) = \int_{0}^{1} \tilde{X}(f) e^{j2\pi (f/f_s)n} d(f/f_s)\\ &= \int_{0}^{1} e^{j2\pi (f/f_s)n} d(f/f_s) \sum_{k} \delta(f-k\frac{f_s}{N}) \Bigg[\frac{f_s}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg]\\ &= \sum_{k} \int_{0}^{1} e^{j2\pi (\frac{f}{f_s})n} \delta(\frac{f}{f_s}-\frac{k}{N}) d(\frac{f}{f_s}) \Bigg[\frac{1}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg]\\ &= \sum_{k} \int_{0}^{1} e^{j2\pi (\frac{k}{N})n} \delta(\frac{f}{f_s}-\frac{k}{N}) d(\frac{f}{f_s}) \Bigg[\frac{1}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg]\\ &= \sum_{k=0}^{N-1} e^{j2\pi (\frac{k}{N})n} \Bigg[\frac{1}{N} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (k/N)n} \right\} \Bigg]\\ &= \frac{1}{N} \sum_{k=0}^{N-1} \left\{ \sum_{n=0}^{N-1} x(nT) e^{-j2\pi (\frac{k}{N}) n} \right\} e^{j2\pi (\frac{k}{N})n} \end{aligned}$$