Standard AM, FM and PM Equations

Standard Mathematical Equations for AM, FM and PM

1. Amplitude Modulation (AM)

• : Carrier amplitude, : Carrier frequency
• :  Message signal amplitude, : Message signal frequency
• : Modulation index
• : Carrier Power, : Sideband Power, : Total Power
• : Efficiency
• : Bandwidth of the AM signal

Amplitude Modulation

Q. The equation of amplitude modulated wave is given by s(t) = 40[1+0.8cos(2π ×10^3t)]cos(4π×10^5t). Find the carrier power, the total sideband power and bandwidth of the signal. The value of resistor given is 30Ω.

Solution: Here,

We are given:

$$s(t)=40[1+0.8\cos (2\pi \cdot {10}^{3}t)]\cos (4\pi \cdot {10}^{5}t)$$

And:

• Load resistance $R = 30\, \Omega$
  

Step 1: Identify the standard AM form

Standard AM wave:

$$s(t)=A_c[1+\mu \cos (2\pi f_{m}t)]\cos (2\pi f_{c}t)$$

From the given equation:

• $A_c = 40$

• $\mu =0.8$

• $f_m={10}^3=1\text{kHz}$

• $f_c=\frac{4\pi \cdot {10}^5}{2\pi }=2\times {10}^5=200\text{kHz}$

a) Carrier Power $P_c$

$$P_c=\frac{A_c^2}{2R}=\frac{{40}^2}{2\cdot 30}=\frac{1600}{60}=\boxed{{\displaystyle 26.67\text{W}}}$$

Frequency Modulation

Q. A single tone FM is represented by the voltage equation as v(t)=12cos(5x10^8t+5sin1250t). Determine following:

a) Carrier frequency

b) Modulating frequency

c) Modulation index

d) Maximum frequency deviation

Solution: Here,

We are given a single-tone FM signal:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

This equation is in the general form of an FM signal:

$$v(t)=A_c\cos (2\pi f_{c}t+\beta \sin (2\pi f_{m}t))$$

Where:

• $f_c$ = carrier frequency

• $f_m$ = modulating frequency

• $\beta$ = modulation index

• $\Delta f = \beta f_m$ = frequency deviation

Step-by-step Extraction from Given Equation

We compare:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

with:

$$v(t)=A_c\cos ({\omega }_{c}t+\beta \sin ({\omega }_{m}t))$$

From this, we get:

• $\omega_c = 5 \times 10^8 \Rightarrow f_c = \frac{\omega_c}{2\pi} = \frac{5 \times 10^8}{2\pi} \approx 79.58 \times 10^6 = 79.58 \, \text{MHz}$

• $\omega_m = 1250 \Rightarrow f_m = \frac{1250}{2\pi} \approx 199 \, \text{Hz}$
  

• $\beta = 5$

Amplitude Modulation

Q. An audio frequency signal 10sin(1000πt) is used for a single tone amplitude modulation with a carrier of 50sin(2π×10^(5)t). Calculate:

(i) Modulation index

(ii) Bandwidth requirement

(iii) Total power delivered if load = 60Ω.

Solution: Here,

We are given:

• Message signal: $m(t)=10\sin (1000\pi t)$

• Carrier signal: $c(t)=50\sin (2\pi \times {10}^{5}t)$

• Load Resistance: $R = 60\, \Omega$
  

Let’s solve the parts step-by-step.

(i) Modulation Index $\mu$

The modulation index is defined as:

$$\mu =\frac{A_m}{A_{\mathrm{c}}}$$

Where:

• $A_m =$Amplitude of message = 10

• $A_c=\; Amplitude\; of\; carrier\; =\; 50$

$$\mu =\frac{10}{50}=0.2$$

So, Modulation Index = 0.2

Analog vs Digital Communication System

Feature Analog Communication Digital Communication Signal Type Continuous-time (sine wave, etc.) Discrete-time (binary 0s and 1s) Noise Immunity Low (prone to distortion) High (noise can be detected/corrected) Bandwidth Requirement Usually, lower Usually, higher Transmission Quality Degrades with distance Maintains quality with regeneration Security Less secure More secure (encryption possible) Cost Cheaper hardware Costlier due to encoding/processing Multiplexing Frequency Division Multiplexing (FDM) Time/Frequency/Code Division Multiplexing Applications Analog radio, landline telephony Internet, mobile phones, satellite links Signal Processing Harder to analyze and process Easier (digital circuits, software tools)

Linear Block Code in Error Detection and Correction

Design a code word of a C(8, 4) block code with any suitable generation matrix.

Line Spectrum of Signal

Q. Plot the line spectrums of x(t) =12 + 6sin (140πt +30°) - 9cos(80πt-70°).

Solution:

Let’s Analyze and Plot the Line Spectrum of

x(t) = 12 + 6sin(140πt+30⁰) − 9cos(80πt−70⁰)

Step 1: Convert all terms to cosine form

Use the identity: sin(θ) = cos(θ−90⁰)

So,

6sin(140πt+30⁰) = 6cos(140πt−60⁰)

−9cos(80πt−70⁰) = 9cos(80πt+110⁰)

Thus, x(t) = 12 + 6cos(140πt−60⁰) + 9cos(80πt+110⁰)

Step 2: Frequencies, Amplitudes, Phases

Term	Frequency (Hz)	Amplitude	Phase
12 (DC)	0	12	0°
6cos(140πt−600)	f=70	6	-60°
9cos(80πt+1100)	f=40	9	+110°

 We split each cosine into two spectral lines at ±f with half the amplitude.