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/ 

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Problem 2

Problem 3

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