Assignment Solution 01

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  • Topic: Single-sideband modulation, Amplitude modulation, Phase
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  • Published : March 20, 2013
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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 = ( ) { + cos + ∙ cos 3 + ∙ cos 5 + other terms} 2 3 5 1 2 2 2 ( )+ ( ) cos ( ) cos 3 ( ) cos 5 = − + + other terms 2 3 5 Output of the BPF: 2 = cos 2

=

2 cos 200

+ cos 600

cos

(b) Specify the unwanted components in V(t) that need to be removed by a BPF of suitable design. Solution: 1 2 2 ( ), ( ) cos 3 ( ) cos 5 , ,other terms 2 3 5 (c) Assume the carrier frequency is 500 Hz. Sketch the spectral density of the resulting DSB-SC waveform. Solution: = 2 cos 200 + cos 600 =2 − 200 + 2 + 200 + − 600 + + 600

Assignment 1 Sol Page: 3 of 12

=

2 = 1

cos −

=

2 1

= 2 × 500 rad = 1000 rad See plot below. (d) In the sketch for Part (c), specify lower and upper sidebands.

+

+

1 2

+

1 2

=

1

+

1

Assignment 1 Sol Page: 4 of 12

3. Let f(t) be a real signal. The transmitter transmits the following modulated signal = cos + sin Where is the Hilbert transform of f(t). (a) Explain that the modulated signal is a lower sideband SSB signal using an example of = cos . Solution: Note that there was an error in the question. The frequency of f(t) should be ωm instead of ωc. Any students with a reasonable attempt to this question will be awarded full marks. However, the solution below refers to the corrected problem. = cos = sin cos + sin sin = cos = cos − Since, cos − = cos cos + sin sin Φ = + − + − +

For ω > 0, the impulse function is located to the left of the carrier frequency. For ω < 0, the impulse function is located to the right of –ωc. Therefore, the modulating function produces lower sideband signals. (b) Determine the frequency of the modulated signal. Solution: From the expression of , the frequency of the modulated signal is −

.

Assignment 1 Sol Page: 5 of 12

4. An SSB signal is generated by modulating an fc = 1 MHz carrier by the message signal = 2 cos 2000πt + cos 4000πt . The amplitude of the carrier signal is Ac = 1. (a) Determine the Hilbert transform of f(t). Solution: = 2 cos 2000πt + cos...
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