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

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??

Answer 1

[dpb]

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

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[Yogesh]

Hey thank you for engaging ... The first value is zero , even if I exclude it , the final answer is infinity.

[Paul]

"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]

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...