Fourier Transform

Statement

If $x(t) \xleftrightarrow{F} X(\omega)$ and $h(t) \xleftrightarrow{F} H(\omega)$, then

$$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau \;\xleftrightarrow{F}\; Y(\omega) = X(\omega)H(\omega)$$

Convolution in the time domain becomes multiplication in the frequency domain.

Proof

By the definition $X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$:

$$Y(\omega) = \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau\right]e^{-j\omega t}dt$$

Change the order of integration:

$$Y(\omega) = \int_{-\infty}^{\infty} x(\tau)\left[\int_{-\infty}^{\infty} h(t-\tau)e^{-j\omega t}dt\right]d\tau$$

In the inner integral put $\sigma = t - \tau$ ($dt = d\sigma$, limits unchanged):

$$\int_{-\infty}^{\infty} h(\sigma)e^{-j\omega(\sigma + \tau)}d\sigma = e^{-j\omega\tau}H(\omega)$$

This is the time-shift property. Substituting back:

$$Y(\omega) = H(\omega)\int_{-\infty}^{\infty} x(\tau)e^{-j\omega\tau}d\tau = H(\omega)X(\omega)$$ $$\boxed{x(t)*h(t) \xleftrightarrow{F} X(\omega)H(\omega)}$$

Significance

• LTI analysis: the output of an LTI system with impulse response $h(t)$ is $Y(\omega) = H(\omega)X(\omega)$. $H(\omega)$ is the frequency response: each frequency component of the input is scaled by $|H(\omega)|$ and phase-shifted by $\angle H(\omega)$.
• Filtering: an ideal low-pass filter has $H(\omega) = 1$ for $|\omega| < \omega_c$ and 0 elsewhere. It removes the high-frequency parts of $X(\omega)$ simply by multiplication.
• Cascades: systems in series give $H(\omega) = H_1(\omega)H_2(\omega)$.
• Computation: a convolution integral can be replaced by: transform, multiply, inverse transform.

Example

$x(t) = e^{-at}u(t)$ is applied to $h(t) = e^{-bt}u(t)$, with $a \ne b$ and both positive:

$$Y(\omega) = \frac{1}{(a+j\omega)(b+j\omega)} = \frac{1}{b-a}\left[\frac{1}{a+j\omega} - \frac{1}{b+j\omega}\right]$$

So $y(t) = \frac{1}{b-a}\left(e^{-at} - e^{-bt}\right)u(t)$, found without evaluating the convolution integral.

Dual (multiplication/modulation) property: $x(t)p(t) \xleftrightarrow{F} \frac{1}{2\pi}X(\omega) * P(\omega)$.

The signal $x(t) = e^{-a|t|}$, $a > 0$, is a two-sided decaying exponential. It is real and even:

$$x(t) = \begin{cases} e^{-at}, & t \ge 0 \\ e^{at}, & t < 0 \end{cases}$$

Fourier transform

$$\begin{aligned} X(\omega) &= \int_{-\infty}^{\infty} e^{-a|t|}e^{-j\omega t}dt = \int_{-\infty}^{0} e^{at}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-at}e^{-j\omega t}dt \\ &= \left[\frac{e^{(a-j\omega)t}}{a-j\omega}\right]_{-\infty}^{0} + \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} \\ &= \frac{1}{a-j\omega} + \frac{1}{a+j\omega} = \frac{(a+j\omega) + (a-j\omega)}{a^2+\omega^2} \end{aligned}$$

Both limits at $\pm\infty$ vanish because $a > 0$.

$$\boxed{X(\omega) = \frac{2a}{a^2+\omega^2}}$$

Spectrum

$X(\omega)$ is real, positive and even, so:

• Magnitude: $|X(\omega)| = \frac{2a}{a^2+\omega^2}$. The peak is $X(0) = 2/a$ (equal to the area under $x(t)$). At $\omega = \pm a$ it falls to half the peak, $1/a$, and it decays as $1/\omega^2$.
• Phase: $\angle X(\omega) = 0$ for all $\omega$.

 x(t)                       X(w)
  1        *                 2/a       *
          / \                         / \
        /     \              1/a    /     \
   __ /         \ __            __/         \__
 ---------+---------- t     ----+----+----+----- w
          0                    -a    0    a

Observation: a larger $a$ means faster decay in time and a wider spectrum. This is the inverse relation between time duration and bandwidth. The result is also the Fourier transform used for a Lorentzian, and by duality $\frac{2a}{a^2+t^2} \leftrightarrow 2\pi e^{-a|\omega|}$.

Statement

For a continuous-time aperiodic (energy) signal $x(t)$ with Fourier transform $X(\omega)$, the total energy found in the time domain equals the total energy found in the frequency domain:

$$E = \int_{-\infty}^{\infty}|x(t)|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega$$

In terms of hertz ($\omega = 2\pi f$): $\int|x(t)|^2dt = \int|X(f)|^2df$.

Proof

Write $|x(t)|^2 = x(t)x^*(t)$:

$$E = \int_{-\infty}^{\infty} x(t)x^*(t)\,dt$$

Replace $x^*(t)$ using the conjugate of the inverse Fourier transform:

$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \;\Rightarrow\; x^*(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega$$

Then

$$E = \int_{-\infty}^{\infty} x(t)\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega\right]dt$$

Interchange the order of integration:

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)\left[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\right]d\omega$$

The bracket is $X(\omega)$ by definition, so

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)X(\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega$$

Meaning

• $|X(\omega)|^2$ is the energy spectral density. The energy in a band $\omega_1$ to $\omega_2$ is $\frac{1}{2\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$.
• Energy can be computed in whichever domain is easier.
• The Fourier transform preserves energy (apart from the $1/2\pi$ factor), so the phase of $X(\omega)$ has no effect on energy.

Example

For $x(t) = e^{-at}u(t)$ with $a > 0$, $X(\omega) = \frac{1}{a+j\omega}$.

• Time domain: $E = \int_0^\infty e^{-2at}dt = \frac{1}{2a}$.
• Frequency domain: $E = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a}$.

The two results agree.

Statement

If $x[n] \xleftrightarrow{DTFT} X(e^{j\omega})$ and $h[n] \xleftrightarrow{DTFT} H(e^{j\omega})$, then

$$y[n] = x[n]*h[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \;\xleftrightarrow{DTFT}\; Y(e^{j\omega}) = X(e^{j\omega})H(e^{j\omega})$$

Convolution in time becomes multiplication in frequency.

Proof

Using $X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}$:

$$Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty}\left[\sum_{k=-\infty}^{\infty}x[k]h[n-k]\right]e^{-j\omega n}$$

Interchange the order of summation:

$$Y(e^{j\omega}) = \sum_{k=-\infty}^{\infty}x[k]\sum_{n=-\infty}^{\infty}h[n-k]e^{-j\omega n}$$

Put $m = n - k$ in the inner sum (so $n = m + k$):

$$\sum_{m=-\infty}^{\infty}h[m]e^{-j\omega(m+k)} = e^{-j\omega k}H(e^{j\omega})$$

Hence

$$Y(e^{j\omega}) = H(e^{j\omega})\sum_{k=-\infty}^{\infty}x[k]e^{-j\omega k} = X(e^{j\omega})H(e^{j\omega})$$

Significance and example

• $H(e^{j\omega})$ is the frequency response of the DT LTI system. The output spectrum is the input spectrum shaped by $H$.
• Example: with $x[n] = a^n u[n]$ and $h[n] = b^n u[n]$ ($|a|, |b| < 1$):

$$Y(e^{j\omega}) = \frac{1}{(1-ae^{-j\omega})(1-be^{-j\omega})}$$

Partial fractions give $y[n] = \frac{a^{n+1} - b^{n+1}}{a - b}u[n]$ for $a \ne b$, which matches the direct convolution sum.

The energy density spectrum (energy spectral density, ESD) of an aperiodic signal shows how the signal's total energy is spread over frequency. It is defined as $|X(\omega)|^2$.

Derivation

The energy of a CT aperiodic signal is

$$E = \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}x(t)x^*(t)\,dt$$

Express $x^*(t)$ through the inverse Fourier transform:

$$x^*(t) = \left[\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\right]^* = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega$$

Substitute and change the order of integration:

$$\begin{aligned} E &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)\left[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\right]d\omega \\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)X(\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega \end{aligned}$$

This is Parseval's relation. Write it as

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}S_{xx}(\omega)\,d\omega, \qquad S_{xx}(\omega) = |X(\omega)|^2$$

In hertz: $E = \int_{-\infty}^{\infty}|X(f)|^2df$, with ESD $|X(f)|^2$ in J/Hz.

Explanation

• $\frac{1}{2\pi}|X(\omega)|^2 d\omega$ is the energy carried by the frequency components in a small band $d\omega$ around $\omega$. The energy between $\omega_1$ and $\omega_2$ is therefore $\frac{1}{\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$ for a real signal, counting both the positive and the negative band.
• The ESD is real, non-negative, and holds no phase information. Signals with different phase but the same $|X(\omega)|$ have the same ESD.
• For real $x(t)$, $|X(-\omega)| = |X(\omega)|$, so the ESD is even.
• The ESD is the Fourier transform of the autocorrelation $R_{xx}(\tau) = \int x(t)x^*(t-\tau)dt$ (Wiener–Khinchin relation for energy signals).
• LTI systems: if $Y(\omega) = H(\omega)X(\omega)$, then $S_{yy}(\omega) = |H(\omega)|^2 S_{xx}(\omega)$.
• It applies only to energy signals (finite $E$). Power signals use the power spectral density instead.

Example

For $x(t) = e^{-at}u(t)$: $S_{xx}(\omega) = \frac{1}{a^2+\omega^2}$, and

$$E = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a}$$

Half of this energy lies in the band $|\omega| < a$:

$$\frac{1}{2\pi}\int_{-a}^{a}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{2}{a}\tan^{-1}(1) = \frac{1}{2\pi}\cdot\frac{2}{a}\cdot\frac{\pi}{4} = \frac{1}{4a}$$

The DTFT pair is derived by treating an aperiodic sequence as the limit of a periodic sequence whose period $N \to \infty$, and then using the discrete-time Fourier series.

Step 1: Build a periodic extension

Let $x[n]$ be of finite duration, non-zero only for $-N_1 \le n \le N_2$. Form $\tilde{x}[n]$ by repeating $x[n]$ with period $N$, where $N$ is large enough that the copies do not overlap. Over one period $\tilde{x}[n] = x[n]$, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.

 x[n]:        ..0 0 [###] 0 0..
 x~[n]: [###] . . . [###] . . . [###]
        |<--- N --->|

Step 2: DTFS of the periodic extension

With $\omega_0 = 2\pi/N$:

$$\tilde{x}[n] = \sum_{k=\langle N\rangle}a_k e^{jk\omega_0 n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}\tilde{x}[n]e^{-jk\omega_0 n}$$

Choose the period so that it contains $-N_1 \dots N_2$. There $\tilde{x}[n] = x[n]$, and $x[n] = 0$ outside, so the sum can run over all $n$:

$$a_k = \frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\omega_0 n}$$

Step 3: Define the envelope

Define

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$$

Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$. The DTFS coefficients are samples of this continuous envelope, spaced $\omega_0$ apart.

Step 4: Substitute back

Using $\frac{1}{N} = \frac{\omega_0}{2\pi}$:

$$\tilde{x}[n] = \sum_{k=\langle N\rangle}\frac{1}{N}X(e^{jk\omega_0})e^{jk\omega_0 n} = \frac{1}{2\pi}\sum_{k=\langle N\rangle}X(e^{jk\omega_0})e^{jk\omega_0 n}\,\omega_0$$

Step 5: Take the limit N → ∞

• $\omega_0 \to d\omega$, and $k\omega_0 \to \omega$ (a continuous variable).
• The sum over $N$ consecutive $k$ covers a frequency range of $N\omega_0 = 2\pi$, so it becomes an integral over an interval of length $2\pi$.
• $\tilde{x}[n] \to x[n]$.

The result is the DTFT pair:

$$\boxed{x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega \quad \text{(synthesis)}}$$ $$\boxed{X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \quad \text{(analysis)}}$$

Remarks

• $X(e^{j\omega})$ is continuous in $\omega$ and periodic with period $2\pi$, since $e^{-j(\omega+2\pi)n} = e^{-j\omega n}$. That is why the synthesis integral covers only $2\pi$.
• Convergence: the analysis sum converges if $\sum|x[n]| < \infty$ (absolutely summable) or $\sum|x[n]|^2 < \infty$ (finite energy, convergence in mean square).
• Example: $x[n] = a^n u[n]$ with $|a| < 1$ gives $X(e^{j\omega}) = \frac{1}{1-ae^{-j\omega}}$.

Derivation of the DTFT

1. Periodic extension: let $x[n]$ be aperiodic and of finite length ($-N_1 \le n \le N_2$). Form $\tilde{x}[n]$ by repeating $x[n]$ with a period $N$ large enough to avoid overlap. Then $\tilde{x}[n] = x[n]$ over one period, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.
2. DTFS of the extension: with $\omega_0 = 2\pi/N$,

$$a_k = \frac{1}{N}\sum_{n=\langle N\rangle}\tilde{x}[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\omega_0 n}$$

The second form holds because $x[n] = 0$ outside the chosen period. 3. Envelope: define $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$. Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$. 4. Synthesis: using $1/N = \omega_0/2\pi$,

$$\tilde{x}[n] = \sum_{k=\langle N\rangle}a_k e^{jk\omega_0 n} = \frac{1}{2\pi}\sum_{k=\langle N\rangle}X(e^{jk\omega_0})e^{jk\omega_0 n}\omega_0$$

5. Limit $N \to \infty$: $\omega_0 \to d\omega$, $k\omega_0 \to \omega$, and the sum over $N$ terms covers $N\omega_0 = 2\pi$, so it becomes an integral over $2\pi$. Also $\tilde{x}[n] \to x[n]$.

This gives

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}, \qquad x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega$$

Proof that the DTFT is periodic with period 2π

$$\begin{aligned} X(e^{j(\omega+2\pi)}) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega+2\pi)n} = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\,e^{-j2\pi n} \end{aligned}$$

Since $n$ is an integer, $e^{-j2\pi n} = \cos 2\pi n - j\sin 2\pi n = 1$. Therefore

$$X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(e^{j\omega})$$

More generally, $X(e^{j(\omega+2\pi r)}) = X(e^{j\omega})$ for any integer $r$.

Reason: discrete-time exponentials whose frequencies differ by $2\pi$ are identical sequences. Only $0 \le \omega < 2\pi$ (or $-\pi \le \omega < \pi$) needs to be considered. Low frequencies lie near $0, \pm2\pi, \dots$ and the highest frequency is $\omega = \pm\pi$.

The Fourier transform of an aperiodic signal is obtained by treating the signal as a periodic signal whose period tends to infinity, and then taking the limit of its Fourier series.

Step 1: Periodic extension

Let $x(t)$ be aperiodic and of finite duration, with $x(t) = 0$ for $|t| > T_1$. Build $\tilde{x}(t)$ by repeating $x(t)$ every $T$ seconds, with $T > 2T_1$ so the copies do not overlap. Over $-T/2 < t < T/2$, $\tilde{x}(t) = x(t)$, and as $T \to \infty$, $\tilde{x}(t) \to x(t)$ for every $t$.

 x(t):          ___/\___
 x~(t):  _/\_______/\_______/\_
           |<--T-->|

Step 2: Fourier series of the periodic extension

With $\omega_0 = 2\pi/T$:

$$\tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_{-T/2}^{T/2}\tilde{x}(t)e^{-jk\omega_0 t}dt$$

Inside $|t| < T/2$, $\tilde{x}(t) = x(t)$, and $x(t) = 0$ outside this range. So the limits can be extended to infinity:

$$a_k = \frac{1}{T}\int_{-\infty}^{\infty}x(t)e^{-jk\omega_0 t}dt$$

Step 3: Define the envelope X(ω)

$$X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad\Rightarrow\quad a_k = \frac{1}{T}X(k\omega_0)$$

The Fourier series coefficients are equally spaced samples of $X(\omega)$, scaled by $1/T$.

Step 4: Substitute in the synthesis equation

Using $\frac{1}{T} = \frac{\omega_0}{2\pi}$:

$$\tilde{x}(t) = \sum_{k=-\infty}^{\infty}\frac{1}{T}X(k\omega_0)e^{jk\omega_0 t} = \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}X(k\omega_0)e^{jk\omega_0 t}\,\omega_0$$

Step 5: Limit T → ∞

• $\omega_0 = 2\pi/T \to d\omega$ (the spectral lines merge).
• $k\omega_0 \to \omega$, a continuous variable.
• The sum $\sum(\cdot)\omega_0$ becomes the integral $\int(\cdot)d\omega$. This is the area under $X(\omega)e^{j\omega t}$, approximated by rectangles of width $\omega_0$.
• $\tilde{x}(t) \to x(t)$.

 X(w) e^{jwt}
      _|_|_
    _|     |_          area of each strip
  _|  w0    |_   =   X(k w0) e^{jk w0 t} w0
 ---+-+-+-+-+-+--- w

Therefore

$$\boxed{x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \quad \text{(inverse FT / synthesis)}}$$ $$\boxed{X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad \text{(FT / analysis)}}$$

In terms of $f$ (with $\omega = 2\pi f$): $X(f) = \int x(t)e^{-j2\pi ft}dt$ and $x(t) = \int X(f)e^{j2\pi ft}df$.

Remarks

• An aperiodic signal has a continuous spectrum, while a periodic signal has a line spectrum.
• Existence (Dirichlet conditions): $x(t)$ is absolutely integrable ($\int|x(t)|dt < \infty$), and it has a finite number of maxima, minima and finite discontinuities in any finite interval. Signals with finite energy also have a transform, converging in the mean-square sense.
• Example: for $x(t) = 1$ when $|t| < T_1$ (0 otherwise), $X(\omega) = \frac{2\sin\omega T_1}{\omega}$. Its samples $\frac{1}{T}X(k\omega_0) = \frac{\sin k\omega_0 T_1}{k\pi}$ are exactly the coefficients of the periodic square wave.

A periodic sequence is not absolutely summable, so its DTFT does not exist in the ordinary sense. It is found by allowing impulses in the frequency domain: each Fourier series term becomes a train of impulses.

Step 1: DTFT of a single complex exponential

Consider the spectrum consisting of impulses of area $2\pi$ at $\omega_0$, repeated every $2\pi$:

$$X(e^{j\omega}) = \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l)$$

Its inverse DTFT, taken over one period of $2\pi$ that contains only the impulse at $\omega_0$, is

$$x[n] = \frac{1}{2\pi}\int_{2\pi}2\pi\,\delta(\omega - \omega_0)e^{j\omega n}d\omega = e^{j\omega_0 n}$$

Hence

$$e^{j\omega_0 n} \xleftrightarrow{DTFT} \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l)$$

Step 2: Periodic sequence as a sum of exponentials

A periodic $x[n]$ with period $N$ has the DTFS

$$x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk(2\pi/N)n}$$

Step 3: Apply linearity

Each term $a_k e^{jk(2\pi/N)n}$ gives impulses of area $2\pi a_k$ at $\omega = 2\pi k/N + 2\pi l$. Since $a_k$ is periodic in $k$ with period $N$, the impulses for all $k$ and $l$ can be combined into one sum over all integers $k$:

$$\boxed{X(e^{j\omega}) = \sum_{k=-\infty}^{\infty}2\pi\,a_k\,\delta\!\left(\omega - \frac{2\pi k}{N}\right)}$$

Procedure: find the DTFS coefficients $a_k$, then place an impulse of area $2\pi a_k$ at each frequency $2\pi k/N$.

 X(e^jw)
     2pi a0   2pi a1        2pi a0 (repeat)
       ^        ^      ^      ^
       |   ^    |      |      |
 ------+---+----+------+------+---- w
       0  2pi/N 4pi/N  ...   2pi

Example: periodic impulse train

$x[n] = \sum_k\delta[n - kN]$ has $a_k = 1/N$ for all $k$, so

$$X(e^{j\omega}) = \frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta\!\left(\omega - \frac{2\pi k}{N}\right)$$

An impulse train in time gives an impulse train in frequency.

Example 2: $\cos\omega_0 n = \frac{1}{2}(e^{j\omega_0 n} + e^{-j\omega_0 n})$ gives $X(e^{j\omega}) = \sum_l \pi[\delta(\omega - \omega_0 - 2\pi l) + \delta(\omega + \omega_0 - 2\pi l)]$.

The unit step $u(t)$ is not absolutely integrable, so its Fourier transform is found as a limit, using the signum function.

Step 1: Write u(t) using the signum function

$$u(t) = \frac{1}{2} + \frac{1}{2}\mathrm{sgn}(t), \qquad \mathrm{sgn}(t) = \begin{cases} 1, & t > 0 \\ -1, & t < 0 \end{cases}$$

Step 2: FT of the constant 1/2

Since $\delta(\omega) \xleftrightarrow{F^{-1}} \frac{1}{2\pi}$, we have $1 \xleftrightarrow{F} 2\pi\delta(\omega)$, so

$$\frac{1}{2} \xleftrightarrow{F} \pi\delta(\omega)$$

Step 3: FT of sgn(t)

Write $\mathrm{sgn}(t) = \lim_{a\to 0}\left[e^{-at}u(t) - e^{at}u(-t)\right]$ with $a > 0$:

$$\begin{aligned} F\{\mathrm{sgn}(t)\} &= \lim_{a\to 0}\left[\int_0^{\infty}e^{-(a+j\omega)t}dt - \int_{-\infty}^{0}e^{(a-j\omega)t}dt\right] \\ &= \lim_{a\to 0}\left[\frac{1}{a+j\omega} - \frac{1}{a-j\omega}\right] = \lim_{a\to 0}\frac{-2j\omega}{a^2+\omega^2} = \frac{2}{j\omega} \end{aligned}$$

So $\frac{1}{2}\mathrm{sgn}(t) \xleftrightarrow{F} \frac{1}{j\omega}$.

Step 4: Combine

$$\boxed{U(\omega) = \pi\delta(\omega) + \frac{1}{j\omega}}$$

Spectrum

• The impulse $\pi\delta(\omega)$ represents the DC (average value 1/2) of $u(t)$.
• $|U(\omega)| = \frac{1}{|\omega|}$ for $\omega \ne 0$.
• $\angle U(\omega) = -\pi/2$ for $\omega > 0$ and $+\pi/2$ for $\omega < 0$.

Check: $u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau$, and the integration property gives $\frac{1}{j\omega}\cdot 1 + \pi\cdot 1\cdot\delta(\omega)$, which is the same result.

Statement

For a discrete-time aperiodic (finite-energy) signal $x[n]$ with DTFT $X(e^{j\omega})$:

$$E = \sum_{n=-\infty}^{\infty}|x[n]|^2 = \frac{1}{2\pi}\int_{2\pi}|X(e^{j\omega})|^2d\omega$$

The energy in the time domain equals the energy in the frequency domain. $|X(e^{j\omega})|^2$ is the energy density spectrum.

Proof

$$E = \sum_{n=-\infty}^{\infty}x[n]x^*[n]$$

Use the inverse DTFT for $x^*[n]$:

$$x^*[n] = \left[\frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega\right]^* = \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})e^{-j\omega n}d\omega$$

Substitute and interchange the sum and the integral:

$$\begin{aligned} E &= \sum_{n}x[n]\cdot\frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})e^{-j\omega n}d\omega \\ &= \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})\left[\sum_{n}x[n]e^{-j\omega n}\right]d\omega \\ &= \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})X(e^{j\omega})d\omega = \frac{1}{2\pi}\int_{2\pi}|X(e^{j\omega})|^2d\omega \end{aligned}$$

Example

For $x[n] = a^n u[n]$ with $|a| < 1$:

• Time domain: $E = \sum_{n=0}^{\infty}a^{2n} = \frac{1}{1-a^2}$.
• Frequency domain: $|X(e^{j\omega})|^2 = \frac{1}{1 - 2a\cos\omega + a^2}$, and the standard integral $\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{d\omega}{1-2a\cos\omega+a^2} = \frac{1}{1-a^2}$ gives the same value.

The integral covers only one period of $2\pi$ because $X(e^{j\omega})$ is periodic.

The signal $x[n] = a^n u[n]$ with $0 < a < 1$ is a right-sided decaying exponential: $x[n] = 1, a, a^2, \dots$ for $n = 0, 1, 2, \dots$

DTFT

$$\begin{aligned} X(e^{j\omega}) &= \sum_{n=-\infty}^{\infty}a^n u[n]e^{-j\omega n} = \sum_{n=0}^{\infty}\left(ae^{-j\omega}\right)^n \end{aligned}$$

This is a geometric series with ratio $r = ae^{-j\omega}$. Since $|r| = a < 1$, it converges to $\frac{1}{1-r}$:

$$\boxed{X(e^{j\omega}) = \frac{1}{1 - ae^{-j\omega}}}$$

Magnitude and phase

Write $1 - ae^{-j\omega} = (1 - a\cos\omega) + ja\sin\omega$:

$$|X(e^{j\omega})| = \frac{1}{\sqrt{1 - 2a\cos\omega + a^2}}, \qquad \angle X(e^{j\omega}) = -\tan^{-1}\!\left(\frac{a\sin\omega}{1 - a\cos\omega}\right)$$

For example, with $a = 0.5$: $|X| = 2$ at $\omega = 0$, $0.894$ at $\pi/2$, and $0.667$ at $\pi$.

 |X(e^jw)|  (a = 0.5)
  2.0            *
               *   *
  0.89       *       *
  0.67  * *             * *
       -+------+------+------+- w
      -pi      0             pi
 (repeats every 2pi)

Observations

• The magnitude is even and the phase is odd in $\omega$, because $x[n]$ is real.
• The spectrum is periodic with period $2\pi$.
• With $0 < a < 1$ the signal is low-pass: most of its energy is near $\omega = 0$. For $-1 < a < 0$ the peak moves to $\omega = \pi$ (high-pass).
• If $a \ge 1$ the sum diverges and the DTFT does not exist.