Assignment Solution 01
North South University
ETE 321 – Spring 2010 Instructor: Nahid Rahman
Assignment #1
Total Marks: 100 Worth: 7.5%
1. Consider the sinusoidally modulated DSB LC signal shown below. The carrier DSB-LC frequency is ωc and the message signal frequency is ωm.
(a) Determine the modulation index m. Solution: Amax = 25 Amin = 5 − 25 − 5 = = 0.67 + 25 + 5 (b) Write an expression for the modulated signal φ(t). Solution: 1 1 ) = (25 − 5) = 10 = ( − 2 2 1 1 ) = (25 + 5) = 15 = ( + 2 2 = + cos = cos + ( ) cos
Assignment 1 Sol Page: 1 of 12
= 15 cos + 10 cos cos (c) Derive time domain expressions for the upper and lower sidebands. Solution: = 15 cos + 10 cos cos = 15 cos + 5 cos( + ) + 5 cos( − ) Upper sideband: 5 cos( + ) ) Lower sideband: 5 cos( − (d) Determine the total average power of the modulated signal , the carrier power and the two sidebands. Solution: Power of carrier signal = (15 cos )2 = + −
(15)2 2
cos cos
= (cos( + ) + cos( − ))
2
1
= 112.5 W
2 (5)2 2 (5)2
Power of upper sideband = (5 cos( Power of lower sideband = (5 cos(
))2 =
= 12.5 W = 12.5 W
Power of modulated signal = 137.5 W (e) Assuming that the message signal is a voltage signal, calculate the PEP (Peak Envelop Power) across a 100Ω load. Solution: PEP =
2
))2 =
need to obtain the RMS value by dividing the peak by √2. (f) Determine the modulation efficiency η. Solution: 12.5 + 12.5 = = 18.18% 137.5
Amax is the peak value of the modulated signal. To calculate the DC power, we
=
(
)2 √2
=
(25
)2 √2 100
= 3.125 W
Assignment 1 Sol Page: 2 of 12
2. A DSB-SC modulated signal can be generated by multiplying the message signal with a periodic pulse generator and passing the resultant signal through a band-pass filter. = 2 cos 200 + cos 600 ( )= 1 2 + 2
∞
(−1) −1 cos ( =1 2 − 1
(2 − 1))
(a) Find the DSB-SC signal component in V(t). Solution: Input to the BPF: = −1 1 2 ∞ (−1) = ( ){ + cos ( (2 − 1))} =1 2 − 1 2 1 2 2 −1 2 1 =
ETE 321 – Spring 2010 Instructor: Nahid Rahman
Assignment #1
Total Marks: 100 Worth: 7.5%
1. Consider the sinusoidally modulated DSB LC signal shown below. The carrier DSB-LC frequency is ωc and the message signal frequency is ωm.
(a) Determine the modulation index m. Solution: Amax = 25 Amin = 5 − 25 − 5 = = 0.67 + 25 + 5 (b) Write an expression for the modulated signal φ(t). Solution: 1 1 ) = (25 − 5) = 10 = ( − 2 2 1 1 ) = (25 + 5) = 15 = ( + 2 2 = + cos = cos + ( ) cos
Assignment 1 Sol Page: 1 of 12
= 15 cos + 10 cos cos (c) Derive time domain expressions for the upper and lower sidebands. Solution: = 15 cos + 10 cos cos = 15 cos + 5 cos( + ) + 5 cos( − ) Upper sideband: 5 cos( + ) ) Lower sideband: 5 cos( − (d) Determine the total average power of the modulated signal , the carrier power and the two sidebands. Solution: Power of carrier signal = (15 cos )2 = + −
(15)2 2
cos cos
= (cos( + ) + cos( − ))
2
1
= 112.5 W
2 (5)2 2 (5)2
Power of upper sideband = (5 cos( Power of lower sideband = (5 cos(
))2 =
= 12.5 W = 12.5 W
Power of modulated signal = 137.5 W (e) Assuming that the message signal is a voltage signal, calculate the PEP (Peak Envelop Power) across a 100Ω load. Solution: PEP =
2
))2 =
need to obtain the RMS value by dividing the peak by √2. (f) Determine the modulation efficiency η. Solution: 12.5 + 12.5 = = 18.18% 137.5
Amax is the peak value of the modulated signal. To calculate the DC power, we
=
(
)2 √2
=
(25
)2 √2 100
= 3.125 W
Assignment 1 Sol Page: 2 of 12
2. A DSB-SC modulated signal can be generated by multiplying the message signal with a periodic pulse generator and passing the resultant signal through a band-pass filter. = 2 cos 200 + cos 600 ( )= 1 2 + 2
∞
(−1) −1 cos ( =1 2 − 1
(2 − 1))
(a) Find the DSB-SC signal component in V(t). Solution: Input to the BPF: = −1 1 2 ∞ (−1) = ( ){ + cos ( (2 − 1))} =1 2 − 1 2 1 2 2 −1 2 1 =