Amplitude modulation math Example 1

 

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) = A0cos(2π× 103t) (b) m(t) = A0cos(2π× 103t) + A0cos(4π× 103t) (c) m(t) = A0cos(2π× 103t) sin(4π× 103t) (d) m(t) = A0cos2(2π× 103t) (e) m(t) = cos2(2π× 103t) +sin2(4π× 103t) (f) m(t) = A0cos3(2π× 103t)

Solutions: The AM signal is defined by ,𝑠 (𝑡) = 𝐴𝑐 (1 + 𝑘𝑎𝑚 (𝑡)) cos⁡(𝜔𝑐𝑡) Where 𝐴𝑐cos⁡(𝜔𝑐𝑡) is the carrier and ka is a constant. We are given, ωc=2π× 105 rad/sec; so,𝑓𝑐=100 kHz. (a) m(t) = A0cos(2π× 103t) so, 𝜔0 = 2𝜋 × 103𝑡 𝑟𝑎𝑑/𝑠𝑒𝑐 ∴ 𝑓0 = 1 𝑘𝐻𝑧 The frequency components of s(t) for positive frequencies are: 𝑓𝑐 = 100 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0= 100 + 1 = 101 𝑘𝐻𝑧 𝑓𝑐 −𝑓0= 100 − 1 = 99 𝑘𝐻𝑧

Solutions: (b) m(t) = A0cos(2π× 103t) + A0cos(4π× 103t) This message signal consists of two sinusoidal components with frequencies f0=1 kHz and f1 = 2 kHz. Hence, the frequency components of s(t) for positive frequencies are: 𝑓𝑐 = 100 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0 = 100 + 1 = 101 𝑘𝐻𝑧 𝑓𝑐 - 𝑓0 = 100 - 1 = 99 𝑘𝐻𝑧 𝑓𝑐 + 𝑓1 = 100 + 2 = 102 𝑘𝐻𝑧 𝑓𝑐 - 𝑓1 = 100 - 2 = 98 𝑘𝐻𝑧

Solutions: (c) m(t) = A0cos(2π× 103t) sin(4π× 103t) First, we note that, 2cosAsinB = sin (A + B) + sin⁡(A - B) ∴ 𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐵 =1/2sin(𝐴 + 𝐵) +1/2sin⁡(𝐴 - 𝐵) Hence, m(t) = A0cos(2π× 103t) sin(4π× 103t) ∴ 𝑚( 𝑡) =𝐴0 /2 sin( 6𝜋 × 103𝑡 )+ 𝐴0 /2 sin⁡(2𝜋 × 103𝑡) Which consists of two sinusoidal components with frequencies 𝑓0 = 3 𝑘𝐻𝑧 and𝑓1 =1𝑘𝐻𝑧. The frequency components of s(t) for positive frequencies are: 𝑓𝑐 = 100𝑘𝐻𝑧 𝑓𝑐 + 𝑓0= 100 + 3 = 103 𝑘𝐻𝑧 𝑓𝑐 - 𝑓0 = 100 - 3 = 97 𝑘𝐻𝑧 𝑓𝑐 + 𝑓1 = 100 + 1 = 101 𝑘𝐻𝑧 𝑓𝑐 - 𝑓1 = 100 - 1 = 99 𝑘𝐻𝑧

Solutions: (d) m(t) = A0cos2(2π× 103t) we know, (1 + 𝑐𝑜𝑠2𝜃) = 2cos2𝜃 𝑠𝑜, cos2𝜃 = 1/2 (1 + 𝑐𝑜𝑠2𝜃) Hence, m(t) = A0cos2(2π× 103t) ∴ 𝑚 𝑡 =𝐴0 /2 [1 + cos (4𝜋 × 103𝑡 )] Which consists of dc components and sinusoidal component of frequency 𝑓0 = 2 𝑘𝐻𝑧. The frequency components of s(t) for positive frequencies are therefore: 𝑓𝑐 = 100 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0 = 100 + 2 = 102 𝑘𝐻𝑧 𝑓𝑐 - 𝑓0= 100 - 2 = 98 𝑘𝐻𝑧

Solutions: (e) m(t) = cos2(2π× 103t) +sin2(4π× 103t) we know, (1 + 𝑐𝑜𝑠2𝜃) = 2cos2𝜃 ∴ cos2𝜃 = 1/2 (1 + 𝑐𝑜𝑠2𝜃) and (1 - 𝑐𝑜𝑠2𝜃) = 2sin2𝜃 ∴ sin2𝜃 = 1/2 (1 - 𝑐𝑜𝑠2𝜃) Hence, 𝑚 (𝑡) = 1/2 [1 + cos (4𝜋 x 103t)] + 1/2[1 - cos (8𝜋 × 103𝑡 )] ∴ 𝑚 (𝑡) = 1 + cos (4𝜋 × 103𝑡 )- 1/2 [cos⁡(8𝜋 × 103𝑡)] 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: 𝑓𝑐 = 100 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0 = 100 + 2 = 102 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0 = 100 - 2 = 98 𝑘𝐻𝑧 𝑓𝑐 + 𝑓1 = 100 + 4 = 104 𝑘𝐻𝑧 𝑓𝑐 - 𝑓1 = 100 - 4 = 96 𝑘𝐻𝑧

Solutions: (f) m(t) = A0cos3(2π× 103t) we know, cos3𝜃 = 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) = A0cos3(2π× 103t) ∴ 𝑚 (𝑡) =𝐴0 /4 cos (2𝜋 × 103𝑡) + 3𝐴0 /4 cos⁡(2𝜋 × 103𝑡) Which consists two sinusoidal components with frequencies 𝑓0 = 1𝑘𝐻𝑧and 𝑓1 = 3 𝑘𝐻𝑧. The frequency components of s(t) are therefore: 𝑓𝑐 = 100𝑘𝐻𝑧 𝑓𝑐 + 𝑓0= 100 + 1 = 101 𝑘𝐻𝑧 𝑓𝑐 + 𝑓0 = 100 - 1 = 99 𝑘𝐻𝑧 𝑓𝑐 + 𝑓1= 100 + 3 =103 𝑘𝐻𝑧 𝑓𝑐 + 𝑓1= 100 - 3 =97 𝑘𝐻𝑧 Note: 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