## Documentation Center |

This example shows how to perform nonlinear
fitting of complex-valued data. While most Optimization Toolbox™ solvers
and algorithms operate only on real-valued data, the `levenberg-marquardt` algorithm
works on both real-valued and complex-valued data.

Do not set the `FunValCheck` option to `'on'` when
using complex data. The solver errors.

On this page… |
---|

**Data Model**

The data model is a simple exponential:

The *x* is input data, *y* is
the response, and *v* is a complex-valued vector
of coefficients. The goal is to estimate *v* from *x* and
noisy observations *y*.

**Artificial Data with Noise**

Generate artificial data for the model. Take the complex coefficient
vector *v* as `[2;3+4i;-.5+.4i]`.
Take the observations *x* as exponentially distributed.
Add complex-valued noise to the responses *y*.

rng default % for reproducibility N = 100; % number of observations v0 = [2;3+4i;-.5+.4i]; % coefficient vector xdata = -log(rand(N,1)); % exponentially distributed noisedata = randn(N,1).*exp((1i*randn(N,1))); % complex noise cplxydata = v0(1) + v0(2).*exp(v0(3)*xdata) + noisedata;

**Fit the Model to Recover the Coefficient Vector**

The difference between the response predicted by the data model
and an observation (`xdata` for *x* and
response `cplxydata` for *y*) is:

objfcn = @(v)v(1)+v(2)*exp(v(3)*xdata) - cplxydata;

Use either `lsqnonlin` or `lsqcurvefit` to
fit the model to the data. This example first uses `lsqnonlin`.
Because the data is complex, set the `Algorithm` option
to `'levenberg-marquardt'`.

opts = optimoptions(@lsqnonlin,... 'Algorithm','levenberg-marquardt','Display','off'); x0 = (1+1i)*[1;1;1]; % arbitrary initial guess [vestimated,resnorm,residuals,exitflag,output] = lsqnonlin(objfcn,x0,[],[],opts); vestimated,resnorm,exitflag,output.firstorderopt

vestimated = 2.1581 + 0.1351i 2.7399 + 3.8012i -0.5338 + 0.4660i resnorm = 100.9933 exitflag = 3 ans = 0.0013

`lsqnonlin` recovers the complex coefficient
vector to about one significant digit. The norm of the residual is
sizable, indicating that the noise keeps the model from fitting all
the observations. The exit flag is `3`, not the preferable `1`,
because the first-order optimality measure is about `1e-3`,
not below `1e-6`.

**Alternative: Use lsqcurvefit**

To fit using `lsqcurvefit`, write the model
to give just the responses, not the responses minus the response data.

objfcn = @(v,xdata)v(1)+v(2)*exp(v(3)*xdata);

Use `lsqcurvefit` options and syntax.

```
opts = optimoptions(@lsqcurvefit,opts); % reuse the options
[vestimated,resnorm] = lsqcurvefit(objfcn,x0,xdata,cplxydata,[],[],opts)
```

vestimated = 2.1581 + 0.1351i 2.7399 + 3.8012i -0.5338 + 0.4660i resnorm = 100.9933

The results match those from `lsqnonlin`,
because the underlying algorithms are identical. Use whichever solver
you find more convenient.

**Alternative: Split Real and Imaginary Parts**

To use the `trust-region-reflective` algorithm,
such as when you want to include bounds, you must split the real and
complex parts of the coefficients into separate variables. For this
problem, split the coefficients as follows:

Write the response function for `lsqcurvefit`.

function yout = cplxreal(v,xdata) yout = zeros(length(xdata),2); % allocate yout expcoef = exp(v(5)*xdata(:)); % magnitude coscoef = cos(v(6)*xdata(:)); % real cosine term sincoef = sin(v(6)*xdata(:)); % imaginary sin term yout(:,1) = v(1) + expcoef.*(v(3)*coscoef - v(4)*sincoef); yout(:,2) = v(2) + expcoef.*(v(4)*coscoef + v(3)*sincoef);

Save this code as the file `cplxreal.m` on
your MATLAB^{®} path.

Split the response data into its real and imaginary parts.

ydata2 = [real(cplxydata),imag(cplxydata)];

The coefficient vector `v` now has six
dimensions. Initialize it as all ones, and solve the problem using `lsqcurvefit`.

```
x0 = ones(6,1);
[vestimated,resnorm,residuals,exitflag,output] = ...
lsqcurvefit(@cplxreal,x0,xdata,ydata2);
vestimated,resnorm,exitflag,output.firstorderopt
```

vestimated = 2.1582 0.1351 2.7399 3.8012 -0.5338 0.4660 resnorm = 100.9933 exitflag = 3 ans = 0.0018

Interpret the six-element vector `vestimated` as
a three-element complex vector, and you see that the solution is virtually
the same as the previous solutions.

Was this topic helpful?