Question

A narrowband FM does not have the following feature

• A. It has two sidebands
• B. Both sidebands are equal in amplitude
• C. It does not show amplitude variations
• D. Both sidebands have same phase difference with respect to carrier

Hint:

NBFM (Narrowband FM) has \(\beta \ll 1\).
NBFM signal \(s(t) \approx A_c \cos(\omega_c t) - A_c \beta \sin(\omega_m t) \sin(\omega_c t)\).
This can be expanded to show a carrier and two sidebands. The LSB and USB have opposite phases relative to each other (or, one is \(180^\circ\) phase shifted w.r.t the carrier's phase contribution compared to the other).
NBFM has a constant envelope (no amplitude variations).

Answer 1

The Correct Option is D

A narrowband FM (NBFM) signal, for a single-tone modulation \(m(t) = A_m \cos(\omega_m t)\), can be approximated as: \[ s_{NBFM}(t) \approx A_c \cos(\omega_c t) - A_c \beta \sin(\omega_m t) \sin(\omega_c t) \] where \(\beta = k_f A_m / \omega_m\) is the modulation index (and \(\beta \ll 1\)). Using product-to-sum identity: \(\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\). Let \(A = \omega_c t\) and \(B = \omega_m t\). (Order matters for sign if using this form) Better: \( \sin(\omega_m t) \sin(\omega_c t) = \frac{1}{2}[\cos((\omega_c - \omega_m)t) - \cos((\omega_c + \omega_m)t)] \). So, \( s_{NBFM}(t) \approx A_c \cos(\omega_c t) - \frac{A_c \beta}{2} [\cos((\omega_c - \omega_m)t) - \cos((\omega_c + \omega_m)t)] \) \[ s_{NBFM}(t) \approx A_c \cos(\omega_c t) - \frac{A_c \beta}{2} \cos((\omega_c - \omega_m)t) + \frac{A_c \beta}{2} \cos((\omega_c + \omega_m)t) \] The components are:
Carrier: \(A_c \cos(\omega_c t)\)
Lower Sideband (LSB) at \(\omega_c - \omega_m\): \(-\frac{A_c \beta}{2} \cos((\omega_c - \omega_m)t) = \frac{A_c \beta}{2} \cos((\omega_c - \omega_m)t + \pi)\) or \(\frac{A_c \beta}{2} \cos((\omega_m - \omega_c)t)\). Phase relative to carrier: \(\pi\) or \(180^\circ\) out of phase. Or, can write as \( \frac{A_c\beta}{2} \sin(\omega_c t - \pi/2) \sin(\omega_m t) \).
Upper Sideband (USB) at \(\omega_c + \omega_m\): \(+\frac{A_c \beta}{2} \cos((\omega_c + \omega_m)t)\). Phase relative to carrier: \(0^\circ\) or in phase. Let's analyze the features: (a) "It has two sidebands": True (LSB and USB). (b) "Both sidebands are equal in amplitude": True, amplitude is \(A_c \beta / 2\) for both. (c) "It does not show amplitude variations": True, NBFM (like all FM) is a constant envelope modulation, meaning its amplitude \(A_c\) remains constant. The approximation shows this. (d) "Both sidebands have same phase difference with respect to carrier": False. The LSB is \(180^\circ\) out of phase with the USB if we write them as cosine terms. Specifically, if carrier is \(A_c\cos(\omega_c t)\), LSB is \(-\frac{A_c\beta}{2}\cos((\omega_c-\omega_m)t)\) and USB is \(+\frac{A_c\beta}{2}\cos((\omega_c+\omega_m)t)\). Relative to the carrier's phase (0 at \(t=0\)), LSB has effective phase \(\pi\) and USB has phase 0. The phase difference of LSB with carrier is \(180^\circ\), while USB with carrier is \(0^\circ\). They are not the same. In AM, the two sidebands are in phase with each other, and \(90^\circ\) out of phase with the carrier if message is sine. Here, LSB and USB are \(180^\circ\) out of phase with each other. And LSB is \(180^\circ\) from the carrier's contribution to that frequency if expanded via Bessel. The LSB is \(-\pi/2\) (or \(-90^\circ\)) relative to the carrier, and the USB is \(+\pi/2\) (or \(+90^\circ\)) relative to the carrier, if we consider the standard NBFM phasor diagram or the relation to AM quadrature component. \(s_{NBFM}(t) \approx A_c \cos(\omega_c t) - A_c \beta \sin(\omega_m t) \sin(\omega_c t)\). Carrier phase reference is \(\cos(\omega_c t)\). LSB is \(+\frac{A_c\beta}{2}\cos((\omega_c+\omega_m)t - \pi/2 - \pi/2) = \frac{A_c\beta}{2}\cos((\omega_c+\omega_m)t - \pi)\). No, the term is \(-\sin(\omega_m t)\sin(\omega_c t)\). \(\sin(\omega_m t)\) is like the modulating signal. In AM it would be \(m(t)\cos(\omega_c t)\). The NBFM expression \(A_c \cos(\omega_c t) - A_c \beta \sin(\omega_m t) \sin(\omega_c t)\) shows that the carrier is \(A_c\cos(\omega_c t)\). The sidebands are from the second term. \(-A_c \beta \sin(\omega_m t) \sin(\omega_c t) = -\frac{A_c \beta}{2} [\cos((\omega_c-\omega_m)t) - \cos((\omega_c+\omega_m)t)]\). So, LSB term: \(-\frac{A_c\beta}{2}\cos((\omega_c-\omega_m)t)\). Phase relative to \(\cos(\omega_c t)\) is \(\pi\). USB term: \(+\frac{A_c\beta}{2}\cos((\omega_c+\omega_m)t)\). Phase relative to \(\cos(\omega_c t)\) is \(0\). The phase differences are \(\pi\) and \(0\). These are not the same. So (d) is false. The question asks for the feature NBFM does *not* have. \[ \boxed{\text{Both sidebands have same phase difference with respect to carrier}} \]