A plot of 'f' (frequency) versus 't' (time) in Matlab for non-linear frequency modulated signal

I have the mathematical equation for time- frequency relationship as below:

t = Tp.*(f/Bp + (1/(2*pi).* sin((2*pi*f)./Bp)));

where Bp is the Bandwidth, Tp is the pulse duration.

Can anybody help me to write this equation in terms of f. I want to use the f to compute the phase and the signal as :

phase = 2*pi* cumsum(f).* dt;

where dt is the sampling interval of the pulse duration defined as Tp/Ns, (Ns= number of samples in the pulse of duration- Tp)

I tried to use 'solve' function in Matlab, but its not successful. kindly suggest if there is a method to invert the above mentioned equation in terms of f.

thank you :)

Answer 1

[NVSL]

Hey @Gowtham_p

I can see you are trying to use “solve” function to solve a non-linear complex function. The “solve” function has its limitations when it comes to solving complex equations. You can alternately try out “fsolve” that approximates "f" from an expected initial point f0 that can be defined based on the time, bandwidth and pulse duration. I have added a sample code that can provide an idea how to use the “fsolve” function. You can then use the computed "f" to find the phase.

% Given variables 
Tp = 1; % Example value for Tp 
Bp = 2; % Example value for Bp 
t = 1;  % Example value for t 

% Define the function handle - target 'f' 
fun = @(f) Tp * (f/Bp + (1/(2*pi)) * sin((2*pi*f)/Bp)) - t; 

% Initial guess 
f0 = 0.1; % Initial guess for f, adjust based on expected range 

% Solve the equation 
options = optimoptions('fsolve', 'Display', 'iter'); % Display iteration information 
[f_solution, fval, exitflag] = fsolve(fun, f0, options); 

                                             Norm of      First-order   Trust-region
 Iteration  Func-count     ||f(x)||^2           step       optimality         radius
     0          2            0.811474                           0.879              1
     1          4            0.249989       0.923416         0.000676              1
     2          6         0.000547816              1           0.0234              1
     3          8         4.46774e-10      0.0234372         2.11e-05            2.5
     4         10         2.60817e-29     2.1137e-05         5.11e-15            2.5

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

% Display the result 

if exitflag > 0 
    fprintf('The solution is f = %.4f\n', f_solution); 
else 
    fprintf('fsolve did not converge\n'); 
end 

The solution is f = 2.0000

You can find the documentation of “fsolve” here

https://www.mathworks.com/help/optim/ug/fsolve.html