How to calculate SNR of an exponential sine sweep?

I have an exponential sine sweep and I want to calculate its SNR. How can this be done

[Dimitris Kalogiros]

Hi Trishita.

I think you have to give some more info about your problem.

[TRISHITA BANERJEE]

%Testing using both interleaved and overlapped signal

% close all clear all; clc

      fs = 44100;
      e=3;         % no of interleaved system
      K=3;         % no of harmonic impulse response
      f_start      = 55;%553
      f_end        = fs/2;%24100;    %Possible values 20k and 24k

      T           =3; 
      A           = 0.04;%0.025; for 0.6m%0.6;%for 1m distance%0.7;%for1.5m
      L1          =9350; % for -70 db
      L2          =2250; % for -70 db
      W1=2*pi*f_start/fs;
      W2=2*pi*f_end/fs;
      tau_2=15300;
      T_1 =(((e-1)*L1)+L2).*log(W2/W1)./log(2);

      tau_k= T_1/log(W2/W1).*log(K); 

      siglen_MESM = T_1;
      x_m = A*sin(W1*siglen_MESM/log(W2/W1).*(exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1))-1)); %Multiple Sine Sweep

% t_i = ceil(L1*fs); % t_o = ceil((T_1/log(W2/W1)).*log(K)*fs); % t_i = ceil(L1); % t_o = ceil(L1+ tau_k ); imax=5; t = delay (imax, L1 ,L2, tau_k, tau_2); t_max=max(t);

    x = zeros(length(x_m)+t_max+fs,imax);
     for n = 1:imax
      x(:,n)     = [zeros(t(n)+fs,1);x_m';zeros(t_max -t(n),1)];
     end
       x1      =  x(:,1);
       x2      = x(:,2);
       x3      = x(:,3);
       x4      = x(:,4);
       x5      = x(:,5);
       y_m = x1+x2+x3+x4+x5;% Simulated Measurement
       y_m=y_m';

%Testing using both interleaved and overlapped signal

% close all clear all; clc

      fs = 44100;
      e=3;         % no of interleaved system
      K=3;         % no of harmonic impulse response
      f_start      = 55;%553
      f_end        = fs/2;%24100;    %Possible values 20k and 24k

      T           =3; 
      A           = 0.04;%0.025; for 0.6m%0.6;%for 1m distance%0.7;%for1.5m
      L1          =9350; % for -70 db
      L2          =2250; % for -70 db
      W1=2*pi*f_start/fs;
      W2=2*pi*f_end/fs;
      tau_2=15300;
      T_1 =(((e-1)*L1)+L2).*log(W2/W1)./log(2);

      tau_k= T_1/log(W2/W1).*log(K); 

      siglen_MESM = T_1;
      x_m = A*sin(W1*siglen_MESM/log(W2/W1).*(exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1))-1)); %Multiple Sine Sweep

% t_i = ceil(L1*fs); % t_o = ceil((T_1/log(W2/W1)).*log(K)*fs); % t_i = ceil(L1); % t_o = ceil(L1+ tau_k ); imax=5; t = delay (imax, L1 ,L2, tau_k, tau_2); t_max=max(t);

    x = zeros(length(x_m)+t_max+fs,imax);
     for n = 1:imax
      x(:,n)     = [zeros(t(n)+fs,1);x_m';zeros(t_max -t(n),1)];
     end
       x1      =  x(:,1);
       x2      = x(:,2);
       x3      = x(:,3);
       x4      = x(:,4);
       x5      = x(:,5);
       y_m = x1+x2+x3+x4+x5;% Simulated Measurement
       y_m=y_m';

      xinv = fliplr(x_m) .* (W2/W1).^(-(0:siglen_MESM)/siglen_MESM); %Inverse sweep

      f_t = f_start*exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1));

     [h1, h2, h1_nl, h2_nl] = farina_deconvolution2(x_m,y_m,xinv,17.6584,1 );

figure subplot(2,1,1); spectrogram(h1,1024,1000,1024,fs,'yaxis'); title('Linear Impulse Response H1');ylabel('h1 \rightarrow dB') subplot(2,1,2); spectrogram(h2,1024,1000,1024,fs,'yaxis'); title('Linear Impulse Response H2');ylabel('h2 \rightarrow dB')

Here x_m is my desired signal...And when I calculate the h1 that is the impulse response of my signal i usually take the peaks above -70db.So I can consider anything above -70db is original signal and less than -70db is noise.. So how can I approach to find SNR of this system