My text book (Communication Systems analysis and design P.334, 335) have a FM signal is

$s_{FM}(t) = A cos[2\pi[f_c+ks(t)]t+\phi]$ where k is a constant and $\phi$ is the arbitrary phase angles

then the textbook suggest

while in FM, instantaneous frequency minus carrier frequency is a linear function of s(t)

So, I try to prove the $s_{FM}(t)$ is a FM signal

The instantaneous frequency is $f(t)= \frac {d\psi}{dt}=2\pi f_c+2 \pi s'(t) t +2 \pi s(t)$

and the instantaneous carrier ferquency is $f_{carrier} (t)=\frac{d\psi_{carrier}}{dt}=\frac {d}{dt} [2 \pi f_c t + \theta]=2 \pi f_c$. Assume the carrier signal is $Asin(2 \pi f_c t + \theta)$

So, $f(t)-f_{carrier}(t)=2 \pi s'(t) t +2 \pi s(t)$.

And my problem is why the text book suggest that signal is a FM signal? Clearly, I cannot prove the signal is linear.

  • 1
    $\begingroup$ I never noticed that it has the $s'(t)t$ term before. Interesting. $\endgroup$
    – Jim Clay
    Feb 4, 2013 at 15:32
  • $\begingroup$ There probably are a hundred textbooks with titles similar if not identical to Communication Systems analysis and design. Could you be a little more specific, e.g. include the authors' name(s), publisher, edition, etc? $\endgroup$ Feb 4, 2013 at 15:34
  • $\begingroup$ Harold PE stern and Samy A Mahmoud are the authors, this link have my textbook link $\endgroup$
    – Samuel
    Feb 4, 2013 at 15:35
  • $\begingroup$ What you calculate as instantaneous frequency is actually the intantaneous angular frequency $\omega(t)$ and $\omega_\mathrm{carrier}$. Besides, what you've calculated is linear in s(t) as derivation is a linear operation. $\endgroup$
    – Deve
    Feb 4, 2013 at 15:37
  • $\begingroup$ What you mean is $s'(t) t$ term is linear? $\endgroup$
    – Samuel
    Feb 4, 2013 at 15:39

1 Answer 1


As you have correctly derived, the difference of carrier and instantaneous frequency is

$$ g(s(t)) = \omega(t) - \omega_\mathrm{carrier} = 2\pi k s(t) + 2\pi k s(t)'t $$

I will omit $t$ for convenience. If $g(s)$ is a linear function of $s$ it must fulfill the following condition:

$$ g(c(a + b)) = cg(a) + cg(b) $$ So here: $$ g(c(a+b)) = 2\pi k c(a + b) + 2\pi k \left[\frac{d}{dt}(c(a + b))\right]t\\ = 2\pi k c a + 2\pi k c b + 2\pi k c a' t + 2\pi k c b' t\\ = c(2\pi k a + 2\pi k a' t) + c(2\pi k b + 2\pi k b' t)\\ = cg(a) + cg(b) $$ Thus, $g(s(t))$ is indeed a linear function of $s(t)$.

  • $\begingroup$ The awesome part is s(t) become c(a+b), then the function become linear. lol. that's cool. $\endgroup$
    – Samuel
    Feb 4, 2013 at 16:08
  • $\begingroup$ What is the function $g$? And I don't understand the first step of your "So here" proof. $\endgroup$
    – Jim Clay
    Feb 4, 2013 at 18:02
  • $\begingroup$ Samuel: We don't know whether s(t) is a linear function or not (most probably it isn't). But that's not the question. We are supposed to show that w(t) - w_c is a linear function. Namely of s(t) (and not of t). Btw, a and b are functions in t, c is a constant. That was poor notation from my side. Sorry. @JimClay g(s(t)) is w(t) - w_c as defined in the first expression and my approach is supposed to proove that g(s(t)) is a linear function of s(t) by showing that it fulfills the linearity condition. $\endgroup$
    – Deve
    Feb 4, 2013 at 18:28

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