Expression for Frequency-shifted Gaussian pulse

I am doing some research on UWB radars which transmit frequency-shifted Gaussian pulses. These pulses are given by: $$g(t) = p(t)cos(2\pi f_{c}t) = V_{TX}exp(-\frac{t^{2}}{2\tau^2})cos(2\pi f_{c}t)$$ where $$f_{c}$$ is the center or carrier frequency, $$V_{TX}$$ is the peak amplitude and $$\tau$$ determines the Bandwidth of the pulse (you can think of $$\tau$$ as the standard deviation of $$p(t)$$. This expression in some papers is equivalent to a higher order derivative of the Gaussian function $$p(t)$$, such that: $$g(t) = \frac{d^{n}}{dt^{n}} \left( V_{TX }exp(-\frac{t^{2}}{2\tau^2}) \right)$$ where $$n$$ denotes the derivative order.

Does anyone know how these two expressions are related? Is there a way of obtaining one from the other?

• Would you mind sharing the papers you saw this expression in? – Envidia Mar 23 at 5:29

Well the expression for the $$\frac {d^n}{dt^n} p(t)$$ is $$V_n(t) p(t)$$, where $$V_n(t)$$ is a polynomial of degree $$n$$, and $$V_n(t)$$ may be determined recursivelyy

$$V_{n+1}(t)e^{-t^2/(2\tau^2)} = \frac{d}{dt}\left(V_{n}(t)e^{-t^2/(2\tau^2)}\right)$$

Solving for $$V_{n+2}(t)$$ we have

$$V_{n+1} = \frac{d}{dt} V_{n}(t) - \frac{t}{2\tau^2} V_n(t)$$

In order to link the two forms of the equation we need

$$\begin{eqnarray} g(t) &=& p(t)cos(2\pi f_{c}t) \\ &=& V_{TX}exp(-\frac{t^{2}}{2\tau^2})cos(2\pi f_{c}t) \\ &=& \frac{d^{n}}{dt^{n}} \left( V_{TX }exp(-\frac{t^{2}}{2\tau^2}) \right) \\ &=& p(t)V_n(t) \end{eqnarray}$$

Thus $$cos(2\pi f_c t) = V_n(t)$$, the right hand side is a polynomial. As $$n \to \infty$$, the are the Hermite polynomials have some asymptotic expansions in terms of cossines. So the two may assume approximate values for certain values of $$f_c$$ and $$\tau$$.

Numerically evaluating the derivaties we see some resemblance with gausiand windowed cosine.

function hermite;
tau=pi;
t = tau * linspace(-5, 5, 1000);
for order = 4:2:20
subplot(3,3,order/2-1)
a = -1/(2*tau^2);
P = V_coefs(order, a);
plot(t, polyval(P(end:-1:1), t).*exp(a*t.^2));
title(num2str(order, 'n=%d'))
end

end

function coefs = V_coefs(order, a)
coefs = 1;
for ii = 1:order
coefs = [(1:(ii-1)).*coefs(2:end), 0, 0] + [0, coefs .* (2 * a)];
end
end


• Thank you so much for this. When you are saying that the equation is not correct but just an approximation, for which of the two equations are you referring to? – Charis Hadjipanayi Mar 15 at 9:10
• sorry, I meant the equation deriving by equating the two forms of g(t) – Bob Mar 15 at 9:19
• Edited, check the post again, please – Bob Mar 15 at 9:29
• Thank you once again for this explanation. Therefore $V_n$ will be equal to $cos(2\pi f_{c} t$) as $n$ goes to $\infty$? Is this related to the Taylor series of the cosine function? – Charis Hadjipanayi Mar 15 at 9:46
• If $f_c = sqrt(n)/tau$ (maybe I am missing some constant factor), and $n$ must be a multiple of four or (multiple of 2 if you may invert the signal). Then it will be an approximation. – Bob Mar 15 at 10:02