Frequency and Phase Sensitivity of a Modulated Signal
1. General Angle-Modulated Signal
$$ s(t) = A_c \cos[\theta(t)] $$
$$ \theta(t) = \omega_c t + \phi(t) $$
2. Frequency Sensitivity (FM)
$$ f_i(t) = \frac{1}{2\pi} \frac{d\theta(t)}{dt} $$
$$ f_i(t) = \frac{1}{2\pi} \left(\omega_c + \frac{d\phi(t)}{dt}\right) $$
For FM:
$$ \phi(t) = k_f \int m(t)\,dt $$
$$ f_i(t) = f_c + \frac{k_f}{2\pi} m(t) $$
3. Phase Sensitivity (PM)
$$ \phi(t) = k_p m(t) $$
$$ f_i(t) = f_c + \frac{k_p}{2\pi} \frac{dm(t)}{dt} $$
4. Example Signal Analysis
Given Signal
$$ x(t) = 3\cos\left[2\pi \cdot 10^6 t + 2\sin(2\pi \cdot 10^3 t)\right] $$
Step 1: Identify Components
- Carrier amplitude: \( A_c = 3 \)
- Carrier frequency: \( f_c = 10^6 \) Hz
- Modulating frequency: \( f_m = 10^3 \) Hz
Step 2: Compare with Standard FM Form
Standard FM form:
$$ x(t) = A_c \cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] $$
Comparing:
- Modulation index: \( \beta = 2 \)
Step 3: Frequency Deviation
Relation:
$$ \beta = \frac{\Delta f}{f_m} $$
$$ \Delta f = \beta f_m = 2 \times 10^3 = 2000 \text{ Hz} $$
Step 4: Frequency Sensitivity
For FM:
$$ \Delta f = k_f' A_m $$
Here \( A_m = 1 \) (since sin term has amplitude 1)
$$ k_f' = \Delta f = 2000 \text{ Hz/unit amplitude} $$
Step 5: Instantaneous Frequency
$$ f_i(t) = f_c + \Delta f \cos(2\pi f_m t) $$
$$ f_i(t) = 10^6 + 2000 \cos(2\pi \cdot 10^3 t) $$
Carrier frequency: \( 1 \text{ MHz} \)
Modulation index: \( \beta = 2 \)
Frequency deviation: \( \Delta f = 2 \text{ kHz} \)
Frequency sensitivity: \( k_f' = 2000 \text{ Hz} \)
5. Key Differences
| Quantity | FM | PM |
|---|---|---|
| Phase term | \( k_f \int m(t) dt \) | \( k_p m(t) \) |
| Frequency depends on | \( m(t) \) | \( \frac{dm(t)}{dt} \) |
6. Phase Deviation of Given Signal
Given Signal
$$ x(t) = 3\cos\left[2\pi \cdot 10^6 t + 2\sin(2\pi \cdot 10^3 t)\right] $$
Step 1: Identify Phase Term
Compare with standard form:
$$ x(t) = A_c \cos\left[2\pi f_c t + \phi(t)\right] $$
Thus, the phase term is:
$$ \phi(t) = 2\sin(2\pi \cdot 10^3 t) $$
Step 2: Definition of Phase Deviation
Phase deviation is the maximum value of \( \phi(t) \):
$$ \Delta \phi = \max |\phi(t)| $$
Step 3: Calculate Maximum Value
Since:
$$ \sin(\cdot) \in [-1, 1] $$
So:
$$ \Delta \phi = 2 \text{ radians} $$
Phase deviation: \( \Delta \phi = 2 \text{ radians} \)
Modulation index: \( \beta = 2 \)