How to find area under the curve after converting the power spectrum from Watt to dBm

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PS=0.2; %Input Power
L=10;
h = 6.62607015e-34;
n=1.45; %Index of refraction
eps0=8.854e-12; % [F/m] Vacuum permittivity
mu0 = 4*pi*1e-7;%[H/m] Vacuum permeability
c=2.9979e8; % [m/sec] Speed of light
Z0=sqrt(mu0/eps0); %[Ohm] Vacuum impedance
dt = 2e-12; dz=dt*c/n; %Spacial and Temporal step sizes.
%dz=2.5e-4;
N=round(L/dz); % Fiber length discretization
z=0:dz:(dz*N-dz);
T=10*2*L*n/c; %time taken for 10 round trips
Nt=round(T/dt);
% Nt=2^18;
% T=Nt*dt;
%% material characteristics
lambdaS=1.064e-6; % [m] signal wavelength
lambdaP=0.915e-6; % [m] pump wavelength
WS=2*pi*c/lambdaS; %[rad/sec] signal frequency
GAMMAb=2*pi*45e6; %[1/t] phonons decay rate
A=19.6e-12; %[m^2] fiber's effective area
Va=5.9e3; %[m/s] Velocity of sound
Omega=2*pi*16e9;%[rad/sec] acoustic frequency
% Omega=2*n*Va*WS/c; %[rad/sec]
WB = WS-Omega;
fB=WB/2/pi;
lambdaB=c/fB;
% gammaE=0.8967; %electrostrictive coefficient
gammaE=1.2; %electrostrictive coefficient
kT=300*1.3806504e-23; %[Joule] at room temprature.
roh0=2201; % [kg/m^3] mean density for SiO2, bulk (http://www.virginiasemi.com/pdf/generalpropertiesSi62002.pdf) .
Q=2*kT*roh0*GAMMAb/(Va^2*A); %Noise std.
g0=(gammaE^2*WS^2)/n/c^3/Va/roh0/GAMMAb; %[m/W]
sigma=(WS*gammaE)/2/n/roh0/c;
q=2*WS*n/c;
t = (-T/2/dt:1:T/2/dt)*dt;
nu=(-1/2/dt:1/T:1/2/dt);
I1_0=PS/A; %[W/m^2] Pump intensity
vpi = 4.5;
Vsine=vpi;
fsine=0;
sine = Vsine*sin(2*pi*fsine*t);
phi=sine;
ES_0t=sqrt(I1_0/2/n/c/eps0)*exp((1i*(pi)*phi)/vpi); % Original signal in time
Power=trapz(t,2*n*c*eps0*A*abs(ES_0t).^2)/T % Area under the curve in time domain
Power = 0.2000
FFt_EL0t=fftshift(abs(fft(ES_0t/Nt))); % fourier transform of the original signal
Power_FFt=T*trapz(nu,2*n*c*eps0*A*(FFt_EL0t).^2) % Area under the curve in frequency domain
Power_FFt = 0.2000
dBm_EL0t=10*log10((2*n*c*eps0*A*abs(FFt_EL0t).^2)./1e-3);
Power_dBm=trapz(nu,dBm_EL0t)
Power_dBm = -Inf
Both area under the curve for time and frequency domain is 0.2 W but when I convert it to dBm area under it seems to be infinity.
Does anyone have any idea??

回答(1 个)

dpb
dpb 2024-9-22
编辑:dpb 2024-9-22
PS=0.2; %Input Power
L=10;
h = 6.62607015e-34;
n=1.45; %Index of refraction
eps0=8.854e-12; % [F/m] Vacuum permittivity
mu0 = 4*pi*1e-7;%[H/m] Vacuum permeability
c=2.9979e8; % [m/sec] Speed of light
Z0=sqrt(mu0/eps0); %[Ohm] Vacuum impedance
dt = 2e-12; dz=dt*c/n; %Spacial and Temporal step sizes.
%dz=2.5e-4;
N=round(L/dz); % Fiber length discretization
z=0:dz:(dz*N-dz);
T=10*2*L*n/c; %time taken for 10 round trips
Nt=round(T/dt);
% Nt=2^18;
% T=Nt*dt;
%% material characteristics
lambdaS=1.064e-6; % [m] signal wavelength
lambdaP=0.915e-6; % [m] pump wavelength
WS=2*pi*c/lambdaS; %[rad/sec] signal frequency
GAMMAb=2*pi*45e6; %[1/t] phonons decay rate
A=19.6e-12; %[m^2] fiber's effective area
Va=5.9e3; %[m/s] Velocity of sound
Omega=2*pi*16e9;%[rad/sec] acoustic frequency
% Omega=2*n*Va*WS/c; %[rad/sec]
WB = WS-Omega;
fB=WB/2/pi;
lambdaB=c/fB;
% gammaE=0.8967; %electrostrictive coefficient
gammaE=1.2; %electrostrictive coefficient
kT=300*1.3806504e-23; %[Joule] at room temprature.
roh0=2201; % [kg/m^3] mean density for SiO2, bulk (http://www.virginiasemi.com/pdf/generalpropertiesSi62002.pdf) .
Q=2*kT*roh0*GAMMAb/(Va^2*A); %Noise std.
g0=(gammaE^2*WS^2)/n/c^3/Va/roh0/GAMMAb; %[m/W]
sigma=(WS*gammaE)/2/n/roh0/c;
q=2*WS*n/c;
t = (-T/2/dt:1:T/2/dt)*dt;
nu=(-1/2/dt:1/T:1/2/dt);
I1_0=PS/A; %[W/m^2] Pump intensity
vpi = 4.5;
Vsine=vpi;
fsine=0;
sine = Vsine*sin(2*pi*fsine*t);
phi=sine;
ES_0t=sqrt(I1_0/2/n/c/eps0)*exp((1i*(pi)*phi)/vpi); % Original signal in time
Power=trapz(t,2*n*c*eps0*A*abs(ES_0t).^2)/T; % Area under the curve in time domain
FFt_EL0t=fftshift(abs(fft(ES_0t/Nt))); % fourier transform of the original signal
Power_FFt=T*trapz(nu,2*n*c*eps0*A*(FFt_EL0t).^2); % Area under the curve in frequency domain
dBm_EL0t=10*log10((2*n*c*eps0*A*abs(FFt_EL0t).^2)./1e-3);
Power_dBm=trapz(nu,dBm_EL0t)
Power_dBm = -Inf
all(abs(FFt_EL0t).^2>0)
ans = logical
0
log10(0)
ans = -Inf
Shows why -- at least one element of the spectrum is identically 0 and log10(0) --> -Inf
PSD=abs(FFt_EL0t).^2;
[numel(PSD) nnz(PSD) numel(PSD)-nnz(PSD)]
ans = 1×3
483672 483650 22
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  3 个评论
Paul
Paul 2024-9-22
"The first value is zero , even if I exclude it , the final answer is infinity."
What about the other valus of Fft_EL0t that are zero?
Also, is integrating dBm_EL0t with respect to nu physically meaningful? How would the result be interpreted?
dpb
dpb 2024-9-22
See the added to prior Answer --
PSD=abs(FFt_EL0t).^2;
[numel(PSD) nnz(PSD) numel(PSD)-nnz(PSD)]
ans = 1×3
483672 483650 22
That shows there are 22 zero elements, not just one...

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