what does this instruction means?

sfdr=max(ptot(nper+4:N/2))

Answer 1

[C.J. Harris]

It means sfdr will take on the maximum value contained in the array ptot between element number nper+4 and element number N/2.

If you want a better answer you really have to add more detail to your question.

[lotus]

thank you for answering.i dont actually understand the meaning of "nper".can you explain it in the simplest way?

[C.J. Harris]

nper will be an integer number, that's all I can tell, when added to the number 4 will result in an array reference, for example:

nper = 2;
ptot = [9 8 7 6 5 4 3 2 1];
nOut = ptot(nper+4)
nOut =
        4

[lotus]

so,if let say the example is like below:

                                                                                                         %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                              Example 4.11                               %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                                                         %
%                           Pipeline Converter                            %
%                                                                         %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all;

                                                                                                %-------------------------------------------------------------------------%
%                         DAC Static parameters                           %
%-------------------------------------------------------------------------%

                                                                                       %-------------------------------- Stage 1 --------------------------------%
V_DAC_L1=-0.5;
V_DAC_H1=0.5;
GAIN1=2.0;
VthL1=-0.25+eps;
VthH1=0.25+eps;

                                                                                       %-------------------------------- Stage 2 --------------------------------%
V_DAC_L2=-0.5;
V_DAC_H2=0.5;
GAIN2=2.0;
VthL2=-0.25+eps;
VthH2=0.25+eps;

                                                                                            %------------------------------- Stages 3-9 ------------------------------%
V_DAC_L3=-0.5;
V_DAC_H3=0.5;
GAIN3=2.0;
VthL3=-0.25+eps;
VthH3=0.25+eps;

Vth10=0;

                                                                                             %-------------------------------------------------------------------------%
%                             Dynamic parameters                          %
%-------------------------------------------------------------------------%

                                                                                       %-------------------------------- S&H Block-------------------------------%
srH=250e6;
f_TH=600e6;
betaH=1;
tauH=1/(2*pi*betaH*f_TH);
gainH=1;

                                                                                        %-------------------------------- Stage 1 --------------------------------%
sr1=10000e8;
f_T1=6000e6;
beta1=1/2;
tau1=1/(2*pi*beta1*f_T1);
gain1=1;

                                                                                       %-------------------------------- Stage 2 --------------------------------%
sr2=20000e8;
f_T2=6000e6;
beta2=1/2;
tau2=1/(2*pi*beta2*f_T2);
gain2=1;

                                                                                          %------------------------------- Stages 3-9 ------------------------------%
sr=25000e8;
f_T=6000e6;
beta=1/2;
tau=1/(2*pi*beta*f_T);
gain=1;

                                                                                                 %-------------------------------------------------------------------------%
%                         Simulation parameters                           %
%-------------------------------------------------------------------------%

                                                                                       Ts=1e-8;
Tmax=Ts/2;
Vfs=2;
N=2^12;
Ntransient=20;         
Tstop=Ts*(N+Ntransient);
nper=73;
Fs=1/Ts;
Fin=nper*Fs/N;                 % Input signal frequency (Fin = nper*Fs/N)
f=Fin/Fs;                      % Normalized signal frequency
bw=Fs/2;

                                                                                       Amp_dB=-0;                     % Amplitude in dB
Ampl=10^(Amp_dB/20)*Vfs/2;     % Input signal amplitude

                                                                                     finrad=Fin*2*pi;

                         %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                           Launch Simulation                             %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
options=simset('InitialStep', 1, 'RelTol', 1e-3, 'MaxStep', 1,...
    'Fixedstep', 1);
sim('Ex4_11',Tstop, options);           %Starts Simulink simulation

                                                                                               %-------------------------------------------------------------------------% 
%                             Graphic Outputs                             %
%-------------------------------------------------------------------------%
w=hann(N);
f=Fin/Fs;                             % Normalized signal frequency
fB=N*(bw/Fs);                         % Base-band frequency bins
yfft=y(1+Ntransient:N+Ntransient);
[snr,ptot]=calcSNR(yfft',f,fB,w',N);
ptot=ptot-max(ptot);                  % Normalize total spectrum

                                                                              figure(1);
clf;
plot(linspace(0,Fs/2,N/2), ptot(1:N/2), 'r');
grid on;
title('PSD of the Output')
xlabel('Frequency [Hz]')
ylabel('PSD [dB]')
axis([0 Fs/2 -140 0]);
sfdr=max(ptot(nper+4:N/2)) 
snr

i got the answer sfdr=-74.2306. but supposedly, the answer should be 70.9..can you tell me which part that i should do the correction.this is the main file

[C.J. Harris]

My guess is that the values in the array ptot are not what you expect. In your example above you are taking the maximum value in the array ptot between element 77 and 2048. In your case that maximum value is -74.2306, there is nothing you can do to 'correct' this, apart from changing the values in the array ptot.

[lotus]

thank you.i'll try my best.your explanation made me understand better about "nper"