Saturday , October 23 2021

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

Check Also

what is slope overload distortion

what is slope overload distortion and granular noise?

what is slope overload distortion and granular noise Slope overload distortion When the input signal …