Another thing the OP can do is use the "complex" chirp. What you really have to do is (at two different times and both times sweeping all frequencies from -Nyquist through DC to +Nyquist) pass the real part, $x_r(t)=\cos(\pi \beta t^2)$ and then pass the imaginary part $x_i(t)=\sin(\pi \beta t^2)$ of the complex chirp through the material under test and synchronously measure the two real results ($y_r(t)$ and $y_i(t)$), then assemble them as the real and imaginary parts of the response.
Let
$$ \begin{align}
x(t) & = e^{j \pi \beta t^2} \\
& = \cos(\pi \beta t^2) + j \sin(\pi \beta t^2) \\
& = x_r(t) \qquad + j \ x_i(t) \\
\end{align} $$
Then
$$ X(f) \ \triangleq \ \mathscr{F}\left\{ x(t) \right\} \ = \ \sqrt{\frac{j}{\beta}} e^{-j\frac{\pi}{\beta} f^2} $$
You will see that $ |X(f)| = \sqrt{\frac{1}{\beta}} $ for all $f$.
The respective outputs that are synchronously measured with the two driving signals are $y_r(t)$ and $y_i(t)$:
$$ \begin{align}
y_r(t) & \triangleq h(t) \circledast x_r(t) \\
& = h(t) \circledast \cos(\pi \beta t^2) \\
\\
y_i(t) & \triangleq h(t) \circledast x_i(t) \\
& = h(t) \circledast \sin(\pi \beta t^2) \\
\end{align} $$
Then this aggregate complex output, $y(t)$, can be assembled (in the mind of the computer) from the responses of the cosine sweep (the real part) and the sine sweep (the imaginary part):
$$ \begin{align}
y(t) & = y_r(t) + j \ y_i(t) \\
& = h(t) \circledast x_r(t) + j \ h(t) \circledast x_i(t) \\
& = h(t) \circledast \left(x_r(t) + j \ x_i(t)\right) \\
& = h(t) \circledast x(t) \\
& = \int_{-\infty}^{+\infty} h(u) \, x(t-u) \ du \\
& = \int_{-\infty}^{+\infty} h(u) \, e^{j \pi \beta (t-u)^2} \ du \\
& = \int_{-\infty}^{+\infty} h(u) \, e^{j \pi \beta (t^2-2tu+u^2)} \ du \\
& = e^{j \pi \beta t^2} \int_{-\infty}^{+\infty} \left(h(u) \, e^{j \pi \beta u^2}\right) \, e^{-j 2 \pi \beta t u} \ du \\
& = e^{j \pi \beta t^2} \int_{-\infty}^{+\infty} \tilde{h}(u) \, e^{-j 2 \pi (\beta t) u} \ du \\
& = e^{j \pi \beta t^2} \tilde{H}(\beta t) \\
\end{align} $$
where
$$ \tilde{H}(f) \triangleq \int_{-\infty}^{+\infty} \tilde{h}(t) e^{-j 2 \pi f t} \ dt = \mathscr{F}\left\{\tilde{h}(t) \right\} $$
and
$$ \tilde{h}(t) \triangleq h(t) e^{j \pi \beta t^2} $$
So this is what you get: first consider making this adjustment (dechirping) in the measured output:
$$ y(t) e^{-j \pi \beta t^2} = \tilde{H}(\beta t) $$
Now if the sweep rate $\beta$ is low enough, then $e^{j \pi \beta t^2} \approx 1$ for as long as $h(t)$ is not close to 0. Then $\tilde{h}(t) \approx h(t)$ and $\tilde{H}(f) \approx H(f)$ and $\tilde{H}(\beta t) \approx H(\beta t)$ and what you have measured in the output is another chirp, but with relative magnitude and phase changed by $\tilde{H}(\beta t) \approx H(\beta t)$.
Even if the sweep rate is not so low, you can compute either the true frequency response $H(f)$ by first computing
$$ \begin{align}
\tilde{h}(t) & = \mathscr{F}^{-1}\left\{ \tilde{H}(f) \right\} \\
& = \int_{-\infty}^{+\infty} \tilde{H}(f) e^{j 2 \pi f t} df \\
& = h(t) e^{j \pi \beta t^2} \\
\end{align} $$
Or from that result you can get the true impulse response $h(t)$ and then $H(f)$ if that's what you want. It should be the same $H(f)$ you would get if you divided $Y(f) = \mathscr{F}\left\{ y(t) \right\} $ by $X(f) = \mathscr{F}\left\{ x(t) \right\} = \sqrt{\frac{j}{\beta}} e^{-j\frac{\pi}{\beta} f^2}$. And there is no division by zero, because there are no ripples in $X(f)$ because the magnitude is a constant $\sqrt{\frac{1}{\beta}}$.
That is how to remove ripples (and any danger of division by zero) in the spectrum of a linearly-swept chirp.
vibroseis). So long as all relevant frequencies are contained in the chirp, and you know the chirp's initial amplitude, and post-processing you do can take account of the variations in spectral magnitude. – Peter K. Mar 20 '14 at 01:38