solutions hw1
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7/21/2019 Solutions HW1
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Solutions Homework 1
Problem 1
As discussed in our lectures, for a Gaussian pulse the time bandwidth product is given by:
∆∆ = 0.44
At the input the pulse duration Δ = 10 × 10−
Therefore the spectral bandwidth of the pulse is given by:
∆ = 0.44 10 × 10−
⇒ ∆ = 4.4× 10
= ⁄
=
⇒ ∆ =
∆
∆ = 3.53×10 = 0.353
We know that Δ = 10 , = 20/. and = 3
The pulse stretching is given by:
= × Δ ×
= (3×0.353×10−9 × 20) = 2.12×10−
∆ = ∆ + = √ 100+21.2
∆ = 23.44 = 2.344 × 10−
Let time difference between pulses = Δt
From our definition of acceptable time delay between pulses ∆ ≥ 2∆
⟹ ∆ = 4.688 × 10−
Maximum Transmission Frequency (Bits/Sec) = 1 4.688×10− = 21.3/
If instead a different source is used, then
∆ = 2 × 10−9
= × Δ × = 1.2 × 10−
∆ = ∆ + = 1.204×10−
∆ = 2∆ = 2.408 × 10−
Maximum Transmission Frequency (Bits/Sec) = 1 2.408×10− = 4.15/