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We know that we have the following equation for wave.
$$g(t)=A\cos(\omega t+\theta_0)=A\cos(2\pi ft+\theta_0)$$ The equation of frequency with respect to time will be:

enter image description here $$f(t)=\frac{Bt}{\tau}+f_c-\frac{B}{2}$$ Then:

\begin{align} g(t)&=A\cos(2\pi (\frac{Bt}{\tau}+f_c-\frac{B}{2})t+\theta_0)\\ &=A\cos(2\pi(f_c-B/2)t+\mathbf{2\pi}(B/\tau)t^2+\theta_0) \end{align}

Then why that bold 2 is omitted in this answer? $$f(t)=A\cos(\theta(t))=A\cos(2\pi(f_c-B/2)t+\pi(B/\tau)t^2+\theta_0)$$

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If you have a signal

$$g(t)=\cos(2\pi \hat{f}(t)t)\tag{1}$$

then the function $\hat{f}(t)$ is not the instantaneous frequency of $g(t)$ (unless $\hat{f}(t)$ is constant).

If you want an instantaneous frequency $f(t)$, then the equation

$$\frac{\phi'(t)}{2\pi}=f(t)\tag{2}$$

must be satisfied, where $\phi(t)$ is the phase of the signal $g(t)$. So in order to obtain the phase $\phi(t)$, you have to integrate the desired instantaneous frequency $f(t)$. For

$$f(t)=\frac{Bt}{\tau}+f_c-\frac{B}{2}\tag{3}$$

you get

$$\frac{\phi(t)}{2\pi}=\frac{B}{2\tau}t^2+(f_c-\frac{B}{2})t+\frac{\theta_0}{2\pi}\tag{4}$$

So the signal with the desired instantaneous frequency is

$$g(t)=\cos(\phi(t))=\cos\left[\pi(Bt^2/\tau+(2f_c-B)t)+\theta_0\right]\tag{5}$$

Also take a look at this related question and its answer.

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