adjusting swept signal equation When I do a spectrogram in matlab / octave I can create a swept signal that looks like the RED plot line below.  But how can I create a swept signal like the BLUE line in the 1st plot using the equation below.

thanks to Daniel and David for getting me this far with the code is below

startfreq=200;
fs=44100;
endfreq=20;
dursec= 10;%duration of signal in seconds
t=(0:dursec*fs)/fs; %Time vector
alpha=log(startfreq/endfreq)/dursec;
sig = exp(-j*2*pi*startfreq/alpha*exp(-alpha*t));
sig=(sig/max(abs(sig))*.8); %normalize signal
wavwrite([sig'] ,fs,32,strcat('/tmp/del.wav')); %export file
specgram(sig,150,400);


1st plot




2nd plot




How can I fix the the equation in the variable sig to get it to look like the BLUE line in the 1st plot?

3rd plot
The general formula for the exponential rise in frequency is given by

$f=A+B(1-\exp(-\alpha t))$

We need to solve for the values of $A$, $B$ and $\alpha$. Let's assume you want an initial frequency of $20$ at time $t=0$, and an asymptotic frequency of $200$ Hz at time $t=\infty$.

Substituting $t=0$ and $f=20$ into the above equation we find $A=20$. If we substitute $t=\infty$ and $f=200$ we find that $B=180$. For $\alpha$ you really have a free choice - it controls how fast the frequency rises. If you have another control point say at time $t=2$ we want the frequency to be at $95%$ of the final frequency i.e. $f=.95 \times 200=190$ then we can substitute these into the above equation (Since we know what $A$, and $B$ are).

$190=20+180(1-\exp(-2\alpha))$

Solving gives $\alpha= -0.5\ln(10/180)=-1.445$. If you want a faster rise then you'll need to choose a larger value of $\alpha$.

To find the expression for phase we simply integrate with respect to time to give

$\phi =2\pi(At+Bt+\frac{B}{\alpha}\exp(-\alpha t))$