My textbook have a question p.360 problem 6.26 is like this

Consider the signal $v(t) = 8 cos (2\pi [10^5 + 4 sin(5 \times 10^3 t)]t)$. And Please Draw the amplitude spectrum of $v(t)$

So, The first step, I did, was proving the $v(t)$ is a FM signal

The the instantaneous frequency is

$\frac {d\psi} {dt} = \frac {d} {dt} 2 \pi[10^5+4sin(5 \times 10^3 t)]t = 2\pi10^5+8\pi sin(5 \times 10^3t) + 8 \pi t (5 \times 10^3) cos (5 \times 10^3 t)$

So it is linear function to $\frac {d} {dt} t[sin(5 \times 10^3 t)]$ and it is a FM signal.

But, I don't know why I cannot find out the summation of sinusoids. My methods is

$v(t) = 8 cos (2\pi [10^5 + 4 sin(5 \times 10^3 t)]t)$

$=Re \{ 8 e^{(j 2 \pi 10^5 t)} e^{j4t sin(5 \times 10^3 t)} \}$

$=Re \{ 8 e^{(j 2 \pi 10^5 t)} \sum\limits_{n=-\infty}^\infty J_n(4) e^{j 2 \pi n (\frac {5 \times 10^3}{2 \pi})t}\}$

And my problem is I dont know my last step is correct or not because it look very weird.

  • $\begingroup$ even with $\beta = 4t$?? and I miss some "t" at the sin function. $\endgroup$
    – Samuel
    Feb 7, 2013 at 16:18

1 Answer 1


The difficulty with respect to this question as well as a related one is that the authors of the text that the OP is using have chosen to adopt a definition of an FM signal that is not the one used by most other writers. This unfortunate choice muddies the study of what is already a difficult-to-understand modulation method (as compared to amplitude modulation and even phase modulation). Most people use take an FM signal to be of the form given in Wikipedia where, for a modulating signal $x(t)$ (assumed to be of magnitude at most $1$), the FM signal is of the form $$A \cos\left(2\pi f_c t + 2\pi f_\Delta \int_0^t x(\tau)\,\mathrm d\tau \right)$$ where $f_c$ is the carrier frequency and $f_\Delta$ is the maximum frequency deviation from the carrier. The instantaneous frequency is the derivative of the phase (argument of the $\cos$) with respect to time $t$, and is thus $$\begin{align*} \omega_{\scriptstyle{\text{inst}}} &= \frac{\mathrm d}{\mathrm dt}\left[2\pi f_c t + 2\pi f_\Delta \int_0^t x(\tau)\,\mathrm d\tau \right] = 2\pi f_c + 2\pi f_\Delta x(t) ~ \text{rad/sec},\\ f_{\scriptstyle{\text{inst}}} &= \frac{1}{2\pi}\frac{\mathrm d}{\mathrm dt}\left[2\pi f_c t + 2\pi f_\Delta \int_0^t x(\tau)\,\mathrm d\tau \right] = f_c + f_\Delta x(t) ~ \text{Hz}. \end{align*}$$ When $x(t)$ is a sinusoidal signal, say $\cos(2\pi f_s t)$, its integral is $$\int_0^t \cos(2\pi f_s \tau)\,\mathrm d\tau = \left.\frac{1}{2\pi f_s}\sin(2\pi f_s\tau)\right\vert_0^t = \frac{1}{2\pi f_s}\sin(2\pi f_s t)$$ is also a sinusoid at the same frequency, and so for this special case, the FM signal can be written as

$$A \cos\left( 2\pi f_c t + \frac{f_\Delta}{f_s} \sin(2\pi f_s t)\right) = A \cos\left(2\pi f_c t + k\sin(2\pi f_s t)\right).$$

Note that the instantaneous frequency of the FM signal varies (in sinusoidal fashion) between $f_c-k$ and $f_c+k$ Hz. Note also the difference between this standard result and what the OP claims is the formula in his textbook (I have not read the text myself and have no idea what it actually says):

$$v(t) = 8 cos (2\pi [10^5 + 4 sin(5 \times 10^3 t)]\,\mathbf t\,)$$

The extra $\mathbf t$ at the end changes the instantaneous frequency to $$2\pi10^5+8\pi sin(5 \times 10^3t) + 8 \pi \, \mathbf{t}\, (5 \times 10^3) cos (5 \times 10^3 t)$$ (as correctly calculated by the OP). But, instead of the instantaneous frequency fluctuating between fixed limits, the maximum frequency deviation increases with time. The claim that this is an FM signal (in the usual sense of the word) is certainly not tenable.

If OP Samuel has correctly copied the problem from his textbook, then the problem, as stated, has no solution. Perhaps it is just a typo in this particular problem. On the other hand, if the same "additional" $t$ occurs in occurs in the text in the definition of an FM signal, and this formulation is used throughout the book, then the book certainly has a major blunder in it.

  • $\begingroup$ +1 for the brief but clear explanation of FM. $\endgroup$
    – Deve
    Feb 9, 2013 at 8:37
  • $\begingroup$ +1 also, for Dilip's answer and I have double check the question. The question's equation is same as what I posted. So, may be the book have a typo. $\endgroup$
    – Samuel
    Feb 10, 2013 at 17:05
  • $\begingroup$ @Samuel As I suggested in the last paragraph of my answer, you might want to check the whole chapter as well because in the other question it looks like the same typo is there in the definition of a FM signal, and so the same mistake might be propagated elsewhere as well. $\endgroup$ Feb 10, 2013 at 20:29

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