Energy spectral density (ESD)

4.4 Energy spectral density (ESD)

To understand the ESD definition, it is useful to recall the Parseval’s Theorem 2 for block transforms in Section 2.4.5. This theorem can be extended to continuous and discrete-time signals.

For continuous-time signals, the energy can be obtained in time-domain according to Eq. (1.22) or, equivalently, in frequency-domain as

E = \int_{- \infty}^{\infty} | x (t) |^{2} d t = \int_{- \infty}^{\infty} | X (f) |^{2} d f,

(4.15)

which is a version of Parseval’s Theorem.

Eq. (4.15) suggests defining

G (f) = | X (f) |^{2}

(4.16)

as the ESD in continuous-time. Therefore, $G (f)$ describes how the signal energy $E$ is spread over frequency and its integral is the energy $E$ . The adopted unit for the ESD is joules/Hz.

Note that, while $| x (t) |^{2}$ represents the instantaneous power in time-domain, in frequency-domain, $| X (f) |^{2}$ represents the ESD.

Example 4.5. ESD of a real-valued exponential. Consider the energy signal $x (t) = e^{- at} u (t)$ in volts, where $a = 0.9$ . Its Fourier transform is $X (f) = 1 ∕ (j 2 πf + a)$ and the ESD in joules/Hz is

G (f) = | X (f) |^{2} = \frac{1}{{(2 πf)}^{2} + a^{2}} .

The signal energy is $E = \int_{- \infty}^{\infty} | x (t) |^{2} d t = \int_{- \infty}^{\infty} G (f) d f \approx 0.55556$ . Listing 4.6 illustrates how $E$ can be estimated in time-domain and also using $G (f)$ . In this case, the range $] - \infty, \infty [$ is simulated as $[0, 30]$ s and $[- 1000, 1000]$ Hz in time and frequency-domain, respectively.

Listing 4.6: MatlabOctaveCodeSnippets/snip_frequency_exp_esd.m

1%% Generate time-domain signal 
2Ts=0.0001; %defines resolution in time 
3t=0:Ts:30; %time axis 
4a=0.9; %constant 
5x=exp(-a*t); %signal 
6subplot(211), plot(t,x) 
7xlabel('t (s)'); ylabel('x(t) amplitude') 
8%% Generate ESD 
9f=linspace(-2,2,1000); %frequency axis 
10Gf= 1 ./ ((2*pi*f).^2 + a^2); %ESD 
11subplot(212) 
12plot(t,x), plot(f,Gf) 
13xlabel('f (Hz)'); ylabel('ESD (J/Hz)') 
14%% Confirm Parseval theorem 
15energy_equation = 1/(2*a); %theoretical 
16energy_time = sum(abs(x).^2)*Ts; %integration 
17df=0.0001; %defines resolution in frequency 
18min_f = -1000; %range of interest 
19max_f = 1000; 
20f=min_f:df:max_f; %frequency axis 
21Gf= 1 ./ ((2*pi*f).^2 + a^2); %recalculate ESD 
22energy_frequency = sum(Gf)*df; %integration 
23disp(['Theoretical energy =' num2str(energy_equation) ' Joules']) 
24disp(['Integration in time =' num2str(energy_time) ' Joules']) 
25disp(['Integration in frequency =' num2str(energy_frequency) ' J']) 
26%% Energy in given frequency band 
27min_f = 0.2; max_f = 1; %range of interest 
28f=min_f:df:max_f; %new frequency axis 
29Gf= 1 ./ ((2*pi*f).^2 + a^2); %recalculate ESD 
30energy_band = 2*sum(Gf)*df; %integration using 2 times 
31disp(['Energy in band =' num2str(energy_band) ' Joules'])

Listing 4.6 also illustrates how $G (f)$ can be conveniently used to estimate the energy within a given frequency band. In this example, the energy within the range from 0.2 to 1 Hz, considering both negative and positive frequencies, is $E_{r} = 2 \int_{0.2}^{1} G (f) d f \approx 0.16954$ J. $□$

Example 4.6. ESD of a sinc in time-domain. Now consider the energy signal is $x (t) = 400 sinc (100 t)$ in volts. According to Eq. (A.56), one needs to consider that $2 F = 100$ , such that $F = 50$ Hz. With this assumption, Eq. (A.56) can be used together with the linearity property of the Fourier transform. The linearity takes care of a factor of 4 that multiplies Eq. (A.56), such that one has $X (f) = 4$ for $| f | \leq 50$ Hz and 0 otherwise. Hence, the ESD of $x (t)$ is $G (f) = 16$ for $| f | \leq 50$ Hz and 0 otherwise. In this case, using the ESD instead of $x (t)$ , it is easy to conclude that the total signal energy is $E = 16 \times 100 = 1600$ joules. $□$

4.4.1 ESD of discrete-time signals

The ESD in discrete-time is defined as

G (e^{j Ω}) = | X (e^{j Ω}) |^{2} .

(4.17)

Similar to Eq. (4.19), $G (e^{j Ω})$ needs to be normalized by the factor $2 π$ to be interpreted as $G (e^{j Ω}) ∕ (2 π)$ in units of joules per radians. Using this definition, one can properly interpret a version of the Parseval’s theorem in discrete-time:

E = \sum_{n = - \infty}^{\infty} | x [n] |^{2} = \frac{1}{2 π} \int_{< 2 π >} | X (e^{j Ω}) |^{2} d Ω .

(4.18)

Example 4.7. ESD of a discrete-time sinc. Consider the discrete-time, non-periodic and finite-energy signal

x [n] = 20 sinc (\frac{n}{10})

in volts. From Eq. (A.58), its DTFT is known to be:

X (e^{j Ω}) = 200 \cdot rect (\frac{5 Ω}{π}) = {\begin{matrix} 200, & | Ω | \leq 0.1 π \end{matrix}

The corresponding ESD is given by

G (e^{j Ω}) = | X (e^{j Ω}) |^{2} = {\begin{matrix} 40000, & | Ω | \leq 0.1 π \end{matrix}

Using Parseval’s Theorem for discrete-time signals, the total signal energy can be computed as

E = \frac{1}{2 π} \int_{- π}^{π} G (e^{j Ω}) d Ω = \frac{1}{2 π} \int_{- 0.1 π}^{0.1 π} 40000 d Ω = \frac{1}{2 π} \cdot 40000 \cdot 0.2 π = 4000

Thus, the total energy of the signal is $E = 4000$ joules. $□$

4.4.2 Advanced: Different definitions of ESD and their units

In continuous-time spectral analysis, the linear frequency in Hertz is more convenient and angular frequencies are seldom adopted. But this is not an option in discrete-time processing, given that $Ω$ is an angle (assumed in radians) and does not have a counterpart in Hertz. Hence, in spite of being more convenient in continuous-time signal processing to adopt $G (f)$ (using Hz), it is useful to discuss $G (ω)$ (using rad/s) in continuous-time to facilitate interpreting $G (e^{j Ω})$ (using rad) in discrete-time processing.

4.4.3 Units of ESD when angular frequencies are adopted

Similar to the discussion in Section 2.5.4 regarding the Fourier transform $X (ω)$ with $ω$ in rad/s, there is a subtle issue when using radians per second instead of Hertz for the ESD. In this case, the factor $2 π$ needs to be taken in account such that $G (ω) ∕ (2 π) = | X (ω) |^{2} ∕ (2 π)$ has the unit of joules/(rad/s). In other words, while $G (f)$ has the unit of joules/Hz, it is not strictly correct to state that $G (ω)$ has the unit of joules/(rad/s). Only after the normalization by $2 π$ , one has $G (ω) ∕ (2 π)$ with the unit of joules/(rad/s).

The energy over a band of frequencies specified in rad/s can then be calculated by integrating $| X (ω) |^{2} ∕ (2 π)$ over this band. For instance, considering a band $ω \in [- \infty, \infty]$ the energy $E$ is:

E = \int_{- \infty}^{\infty} | x (t) |^{2} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} | X (ω) |^{2} d ω = \frac{1}{2 π} \int_{- \infty}^{\infty} G (ω) d ω,

(4.19)

where $G (ω) = G (2 πf)$ and the unit of $G (ω) ∕ (2 π)$ is joules/(rad/s).

Example 4.8. The unit of $G (ω) ∕ (2 π)$ is joules/(rad/s). For instance, take the sinc signal of Example 4.6, which has a flat ESD given by $G (f) = 16$ J/Hz within the range $[- 50, 50]$ Hz and zero otherwise. The representation $G (ω)$ of this ESD in rad/s has the same constant value $G (ω) = 16$ but now over the range $[- 100 π, 100 π]$ rad/s. The value $G (ω) ∕ (2 π) = 16 ∕ (2 π)$ can be interpreted as a density in J/(rad/s). Within the range $[- 100 π, 100 π]$ , one has $(16 ∕ (2 π)) \times 200 π = 1600$ J. $□$

Table 4.2 summarizes the discussed ESD functions.

Table 4.2: ESD functions.

E

is the total energy and the column “Ind. var” (independent variable) indicates the units and symbols used for the independent variable of each function. The units of

G (f)

G (ω) ∕ (2 π)

and

G (e^{j Ω}) ∕ (2 π)

are J/Hz, J/(rad/s) and J/rad, respectively.


Time	Ind. var.	ESD definition	ESD main property

Continuous-time	$f$ (Hz)	$G (f) = \| X (f) \|^{2}$	$E = \int_{- \infty}^{\infty} G (f) d f$

Continuous-time	$ω$ (rad/s)	$G (ω) = \| X (ω) \|^{2}$	$E = \frac{1}{2 π} \int_{- \infty}^{\infty} G (ω) d ω$

Discrete-time	$Ω$ (rad)	$G (e^{j Ω}) = \| X (e^{j Ω}) \|^{2}$	$E = \frac{1}{2 π} \int_{< 2 π >} G (e^{j Ω}) d Ω$