# 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? 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? 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? 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