Energy spectral density (ESD)

F.4 Energy spectral density (ESD)

To understand the ESD definition, it is useful to recall the Parseval’s Theorem 3 for block transforms in Section D.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. (C.22) or, equivalently, in frequency-domain as

E = ∫_{−∞}^{∞} |x(t) |^{2} d t = ∫_{−∞}^{∞} |X(f) |^{2} d f,

(F.15)

which is a version of Parseval’s Theorem.

Eq. (F.15) suggests defining

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

(F.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 F.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 $X(f) = 1∕(j2πf + a)$ and the ESD in joules/Hz is

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

The signal energy is $E = ∫_{−∞}^{∞} |x(t) |^{2} d t = ∫_{−∞}^{∞} G(f) d f ≈ 0.55556$ . Listing F.6 illustrates how $E$ can be estimated in time-domain and also using $G(f)$ . In this case, the range $] −∞,∞[$ is simulated as $[0,30]$ s and $[−1000,1000]$ Hz in time and frequency-domain, respectively.

Listing F.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 F.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 $E_{r} = 2 ∫_{0.2}^{1} G(f) d f ≈ 0.16954$ J. $□$

Example F.6. ESD of a sinc in time-domain. Now consider the energy signal is $x(t) = 400 sinc (100t)$ in volts. According to Eq. (A.52), one needs to consider that $2F = 100$ , such that $F = 50$ Hz. With this assumption, Eq. (A.52) 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.52), such that one has $X(f) = 4$ for $|f|≤ 50$ Hz and 0 otherwise. Hence, the ESD of $x(t)$ is $G(f) = 16$ for $|f|≤ 50$ 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 $E = 16 × 100 = 1600$ joules. $□$

F.4.1 ESD of discrete-time signals

The ESD in discrete-time is defined as

G(e^{j Ω}) = |X(e^{j Ω}) |^{2} .

(F.17)

Similar to Eq. (F.19), $G(e^{j Ω})$ needs to be normalized by the factor $2π$ to be interpreted as $G(e^{j Ω})∕(2π)$ in units of joules per radians. Using this definition, one can properly interpret a version of the Parseval’s theorem in discrete-time:

E = ∑_{n=−∞}^{∞} |x[n] |^{2} = \frac{1}{2π} ∫_{<2π>} |X(e^{j Ω}) |^{2} d Ω .

(F.18)

Example F.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.54), its DTFT is known to be:

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

The corresponding ESD is given by

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

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

E = \frac{1}{2π} ∫_{−π}^{π} G(e^{j Ω}) d Ω = \frac{1}{2π} ∫_{−0.1π}^{0.1π} 40000 d Ω = \frac{1}{2π} ⋅ 40000 ⋅ 0.2π = 4000

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

F.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.

F.4.3 Units of ESD when angular frequencies are adopted

Similar to the discussion in Section D.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 $2π$ needs to be taken in account such that $G(ω)∕(2π) = |X(ω) |^{2} ∕(2π)$ 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 $2π$ , one has $G(ω)∕(2π)$ with the unit of joules/(rad/s).

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

E = ∫_{−∞}^{∞} |x(t) |^{2} d t = \frac{1}{2π} ∫_{−∞}^{∞} |X(ω) |^{2} d ω = \frac{1}{2π} ∫_{−∞}^{∞} G(ω) d ω,

(F.19)

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

Example F.8. The unit of $G(ω)∕(2π)$ is joules/(rad/s). For instance, take the sinc signal of Example F.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 $[−100π,100π]$ rad/s. The value $G(ω)∕(2π) = 16∕(2π)$ can be interpreted as a density in J/(rad/s). Within the range $[−100π,100π]$ , one has $(16∕(2π)) × 200π = 1600$ J. $□$

Table F.2 summarizes the discussed ESD functions.

Table F.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)∕(2π)

and

G(e^{j Ω})∕(2π)

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


Time	Ind. var.	ESD definition	ESD main property

Continuous-time	$f$ (Hz)	$G(f) = \|X(f) \|^{2}$	$E = ∫_{−∞}^{∞} G(f) d f$

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

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