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CCE 401: Communication Systems Spring 2017-
Assignment 5 (Solution)
Due: 2018 Instructor: Dr. Mustafa El-Halabi
Solution 1. Instantaneous Frequency. We can write s(t) as s(t) = cos(200πt − 5 sin(2πt))
where θ(t) = 200πt − 5 sin(2πt). The instantaneous frequency is given by
fi (t) =
2 π
dθ(t) dt = 100 − 5 cos(2πt)
Solution 2. Two Tones. Let m 1 (t) = a 1 cos(2πf 1 t) and m 2 (t) = a 2 cos(2πf 2 t) be the two tones. The average of m 1 (t) and m 2 (t) are respectively P 1 = a
(^21) 2 and^ P^2 =^
a^22
- Since^ P^1 = 49P^2 , then^ a (^21) = 49a (^22) , hence
a 1 = ± 7 a 2
The modulation indices for m 1 (t) and m 2 (t) are respectively β 1 and β 2 , given by β 1 = kf f^ a 1 1 and β 2 = kf f^ a 2 2. Since β 1 = 2β 2 , then a 1 f 1
a 2 f 2 ⇒ f 2 =
f 1 (1)
Using Carson’s rule, the bandwidths of the FM modulated signal using m 1 (t) and using m 2 (t) are respectively,
Bw 1 = 2(β 1 + 1)f 1 = W ⇒ β 1 =
W
2 f 1 −^1
Bw 2 = 2(β 2 + 1)f 2 = W ⇒ β 2 =
W
2 f 2
Since β 1 = 2β 2 , then 2 W 2 f 2
W
2 f 1
Replacing Equation. (??) in Equation. (??), we get
f 1 = 3W, f 2 =
W
Solution 3.Bandwidth of Angle-Modulated Signal.
u(t) = 10 cos(2 × 108 πt + 200 cos(2000πt))
The instantaneous frequency is
fi (t) =
2 π
dθ dt = 10^8 + 2 × 105 sin(2000πt) ⇒ fi (t) − fc = 2 × 105 sin(2000πt) ⇒ ∆f = 2 × 105 Hz
Since fm = 1000 Hz, then using Carson’s rule, the bandwidth os the angle-modulated signal is
Bw = 2(∆f + fm) = 2(201000) = 402 KHz
Solution 4.Modulation Index of FM Signal. The modulation index is given by
β = max |m(t)| × k fm
We can write m(t) as m(t) = A
2 cos
( (^) π 4 − 2 πfmt
The maximum of m(t) is max |m(t)| = A
- Hence,
β =
A
2 k fm
Solution 5.Power in FM Signal. The FM modulated signal can be written as
s(t) = 100 cos(2π × 108 t + 2 sin(20, 000 πt))
where β = ∆fmf = 2010 ×^104 3 = 2. The power of the FM signal si A 22 = 5000 W. Hence, 10% of the power is 500 W. We can form the following table
n f Amplitude = AJn(β) V P ow er = A 22 J n^2 (β) (W ) − 3 100 MHz − 30 KHz AJ− 3 (2) = − 12. 9 83. 2 − 2 100 MHz − 20 KHz AJ− 2 (2) = 35. 3 623 − 1 100 MHz − 10 KHz AJ− 1 (2) = − 57. 7 1664. 645 0 100 MHz AJ 0 (2) = 22. 4 250. 88 +1 100 MHz + 10 KHz AJ 1 (2) = 57. 7 1664. 645 +2 100 MHz + 20 KHz AJ 2 (2) = 35. 5 623 +3 100 MHz + 30 KHz AJ 3 (2) = 12. 9 83. 2
From the table, the frequency components which has a power at least 10% of the unmodulated carrier power are fc ± fm and fc ± 2 fm.
Solution 6. Frequency Deviation
u(t) = 2 cos(2πfc t − 20 cos(1000πt) − 10 cos(2000πt)) = 2 cos
2 πfc t + 2πkf
∫ (^) t
0
m(τ )dτ
By inspecting, we get
2 πkf m(t) = −20(− 1000 π) sin(1000πt) − 10(− 2000 π) sin(2000πt) Since kf = 10^4 ,then we get m(t) = sin(1000πt) + sin(2000πt)
- Find the frequency deviation. Since fi (t) − fc = (^21) πdφ dt(t )= 10^4 sin(1000πt) + 10^4 sin(2000πt), hence
∆f = 10^4 max
∣ sin(1000πt) + sin(2000πt)
Let 1000πt , x. Since the maximum of sin(x) + sin(2x) is roughly 1.76, then
∆f = 10^4 × 1 .76 = 17600 Hz
