## Amplitude modulation math Example

Suppose we have carrier frequency ω = 2π× 10^5rad/sec; Find the frequency component of Amplitude Modulated (AM) signal s(t) of the message signal is given below:(a) m(t) = A _{0}cos(2π× 10^{3}t)(b) m(t) = A _{0}cos(2π× 10^{3}t) + A_{0}cos(4π× 10^{3}t)(c) m(t) = A _{0}cos(2π× 10^{3}t) sin(4π× 10^{3}t)(d) m(t) = A _{0}cos^{2}(2π× 10^{3}t)(e) m(t) = cos ^{2}(2π× 10^{3}t) +sin^{2}(4π× 10^{3}t)(f) m(t) = A _{0}cos^{3}(2π× 10^{3}t) |

Solutions:The AM signal is defined by ,𝑠 (𝑡) = 𝐴 _{𝑐} (1 + 𝑘_{𝑎}𝑚 (𝑡)) cos(𝜔_{𝑐}𝑡)Where 𝐴 _{𝑐}cos(𝜔_{𝑐}𝑡) is the carrier and k_{a} is a constant.We are given, ω _{c}=2π× 10^{5} rad/sec; so,𝑓_{𝑐}=100 kHz.(a) m(t) = A _{0}cos(2π× 10^{3}t)so, 𝜔 _{0 }= 2𝜋 × 10^{3}𝑡 𝑟𝑎𝑑/𝑠𝑒𝑐∴ 𝑓 _{0} = 1 𝑘𝐻𝑧The frequency components of s(t) for positive frequencies are: 𝑓 _{𝑐} = 100 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0}= 100 + 1 = 101 𝑘𝐻𝑧𝑓 _{𝑐} −𝑓_{0}= 100 − 1 = 99 𝑘𝐻𝑧 |

Solutions:(b) m(t) = A _{0}cos(2π× 10^{3}t) + A_{0}cos(4π× 10^{3}t)This message signal consists of two sinusoidal components with frequencies f _{0}=1 kHz and f_{1} = 2kHz. Hence, the frequency components of s(t) for positive frequencies are: 𝑓 _{𝑐} = 100 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0} = 100 + 1 = 101 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{0} = 100 – 1 = 99 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{1 }= 100 + 2 = 102 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{1} = 100 – 2 = 98 𝑘𝐻𝑧 |

Solutions:(c) m(t) = A _{0}cos(2π× 10^{3}t) sin(4π× 10^{3}t)First, we note that, 2cosAsinB = sin (A + B) + sin(A – B) ∴ 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 =1/2sin(𝐴 + 𝐵) +1/2sin(𝐴 – 𝐵) Hence, m(t) = A _{0}cos(2π× 10^{3}t) sin(4π× 10^{3}t)∴ 𝑚( 𝑡) =𝐴 _{0} /2 sin( 6𝜋 × 10^{3}𝑡 )+ 𝐴_{0} /2 sin(2𝜋 × 10^{3}𝑡)Which consists of two sinusoidal components with frequencies 𝑓 _{0} = 3 𝑘𝐻𝑧 and𝑓_{1} =1𝑘𝐻𝑧.The frequency components of s(t) for positive frequencies are: 𝑓𝑐 = 100𝑘𝐻𝑧 𝑓 _{𝑐} + 𝑓_{0}= 100 + 3 = 103 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{0} = 100 – 3 = 97 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{1} = 100 + 1 = 101 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{1} = 100 – 1 = 99 𝑘𝐻𝑧 |

Solutions:(d) m(t) = A _{0}cos^{2}(2π× 10^{3}t)we know, (1 + 𝑐𝑜𝑠2𝜃) = 2cos ^{2}𝜃𝑠𝑜, cos ^{2}𝜃 = 1/2 (1 + 𝑐𝑜𝑠2𝜃)Hence, m(t) = A _{0}cos^{2}(2π× 10^{3}t)∴ 𝑚 𝑡 =𝐴 _{0} /2 [1 + cos (4𝜋 × 10^{3}𝑡 )]Which consists of dc components and sinusoidal component of frequency 𝑓_{0} = 2 𝑘𝐻𝑧.The frequency components of s(t) for positive frequencies are therefore: 𝑓𝑐 = 100 𝑘𝐻𝑧 𝑓 _{𝑐} + 𝑓_{0} = 100 + 2 = 102 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{0}= 100 – 2 = 98 𝑘𝐻𝑧 |

Solutions:(e) m(t) = cos ^{2}(2π× 10^{3}t) +sin^{2}(4π× 10^{3}t)we know, (1 + 𝑐𝑜𝑠2𝜃) = 2cos ^{2}𝜃 ∴ cos^{2}𝜃 = 1/2 (1 + 𝑐𝑜𝑠2𝜃)and (1 – 𝑐𝑜𝑠2𝜃) = 2sin ^{2}𝜃 ∴ sin^{2}𝜃 = 1/2 (1 – 𝑐𝑜𝑠2𝜃)Hence, 𝑚 (𝑡) = 1/2 [1 + cos (4𝜋 x 10 ^{3}t)] + 1/2[1 – cos (8𝜋 × 10^{3}𝑡 )]∴ 𝑚 (𝑡) = 1 + cos (4𝜋 × 10^{3}𝑡 )- 1/2 [cos(8𝜋 × 10^{3}𝑡)]Which consists of dc component, and two sinusoidal components with 𝑓0 = 2 𝑘𝐻𝑧 and 𝑓1 =4 𝑘𝐻𝑧. The frequency components of s(t) for positive frequencies are:𝑓 _{𝑐 }= 100 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0} = 100 + 2 = 102 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0} = 100 – 2 = 98 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓1 = 100 + 4 = 104 𝑘𝐻𝑧𝑓 _{𝑐} – 𝑓_{1} = 100 – 4 = 96 𝑘𝐻𝑧 |

Solutions:(f) m(t) = A _{0}cos^{3}(2π× 10^{3}t)we know, cos ^{3}𝜃 = 𝑐𝑜𝑠𝜃. 1/2[1 + 𝑐𝑜𝑠2𝜃]=1/2 𝑐𝑜𝑠𝜃 +1/2 𝑐𝑜𝑠𝜃cos(2𝜃)=1/2 𝑐𝑜𝑠𝜃 +1/2 [1/2cos( 𝜃 + 2𝜃) +1/2cos(2𝜃 – 𝜃) =1/4 cos (3𝜃) +3/4 𝑐𝑜𝑠𝜃 Hence, m(t) = A _{0}cos^{3}(2π× 10^{3}t)∴ 𝑚 (𝑡) =𝐴 _{0} /4 cos (2𝜋 × 10^{3}𝑡) + 3𝐴_{0} /4 cos(2𝜋 × 10^{3}𝑡)Which consists two sinusoidal components with frequencies 𝑓 _{0} = 1𝑘𝐻𝑧and 𝑓_{1} = 3 𝑘𝐻𝑧. The frequency components of s(t) are therefore:𝑓 _{𝑐} = 100𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0}= 100 + 1 = 101 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{0} = 100 – 1 = 99 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{1}= 100 + 3 =103 𝑘𝐻𝑧𝑓 _{𝑐} + 𝑓_{1}= 100 – 3 =97 𝑘𝐻𝑧Note: For negative frequencies, the frequency components of s(t) are the negative of those for positive frequencies. |

Amplitude modulation math Example

Amplitude modulation math Example