Let’s break down the relationship between wavelength, wavelength bandwidth, pulse duration, frequency, frequency bandwidth, and time using the example of a laser with the following parameters:
- Wavelength (λ\lambdaλ) = 1550 nm
- Full Width at Half Maximum (FWHM, or wavelength bandwidth) = 0.5 nm
- Pulse Duration = 30 ns
1. Wavelength and Frequency Relationship
The relationship between wavelength (λ\lambda) and frequency (f) is given by the equation’s=cλf = \frac{c}{\lambda}f=λc
where ccc is the speed of light in a vacuum, which is approximately 3×108 m/s3 \times 10^8 \, \text{m/s}3×108m/s.
For λ=1550 nm=1.55×10−6 m\lambda = 1550 \, \text{nm} = 1.55 \times 10^{-6} \, \text{m}λ=1550nm=1.55×10−6m:f=3×1081.55×10−6=1.935×1014 Hz=193.5 THzf = \frac{3 \times 10^8}{1.55 \times 10^{-6}} = 1.935 \times 10^{14} \, \text{Hz} = 193.5 \, \text{THz}f=1.55×10−63×108=1.935×1014Hz=193.5THz
2. Wavelength Bandwidth and Frequency Bandwidth Relationship
The frequency bandwidth (Δf\Delta fΔf) can be derived from the wavelength bandwidth (Δλ\Delta \lambdaΔλ) using the following relationship:Δf=c Δλλ2\Delta f = \frac{c \, \Delta \lambda}{\lambda^2}Δf=λ2cΔλ
Given that:
- Δλ=0.5 nm=0.5×10−9 m\Delta \lambda = 0.5 \, \text{nm} = 0.5 \times 10^{-9} \, \text{m}Δλ=0.5nm=0.5×10−9m
- λ=1.55×10−6 m\lambda = 1.55 \times 10^{-6} \, \text{m}λ=1.55×10−6m
Substitute these values:Δf=3×108×0.5×10−9(1.55×10−6)2\Delta f = \frac{3 \times 10^8 \times 0.5 \times 10^{-9}}{(1.55 \times 10^{-6})^2}Δf=(1.55×10−6)23×108×0.5×10−9 Δf≈6.25×1010 Hz=62.5 GHz\Delta f \approx 6.25 \times 10^{10} \, \text{Hz} = 62.5 \, \text{GHz}Δf≈6.25×1010Hz=62.5GHz
So, the frequency bandwidth is approximately 62.5 GHz.
3. Pulse Duration and Frequency Bandwidth Relationship
The pulse duration (τ\tauτ) and frequency bandwidth (Δf\Delta fΔf) are related by the Fourier transform limit. For a Gaussian pulse, the relationship is approximately:Δf τ≈0.44\Delta f \, \tau \approx 0.44Δfτ≈0.44
Given that the pulse duration τ=30 ns\tau = 30 \, \text{ns}τ=30ns:Δf=0.44τ=0.4430×10−9=14.67 MHz\Delta f = \frac{0.44}{\tau} = \frac{0.44}{30 \times 10^{-9}} = 14.67 \, \text{MHz}Δf=τ0.44=30×10−90.44=14.67MHz
So, the frequency bandwidth corresponding to the pulse duration is approximately 14.67 MHz.
4. Relationship Summary for the Example
- Wavelength (λ\lambdaλ): 1550 nm
- Frequency (fff): 193.5 THz
- Wavelength Bandwidth (Δλ\Delta \lambdaΔλ): 0.5 nm
- Frequency Bandwidth (Δf\Delta fΔf): 62.5 GHz (from the wavelength bandwidth)
- Pulse Duration (τ\tauτ): 30 ns
- Frequency Bandwidth from Pulse Duration (Δf\Delta fΔf): 14.67 MHz