Question #319958

The Gaussian distribution also known as the Normal distribution, is given by the following


equation:


𝑦(π‘₯) = 𝑒π‘₯𝑝 βˆ’(π‘₯βˆ’πœ‡)^2/2𝜎^2



where parameter 𝝁 is the mean and 𝝈 the standard deviation.


(i) Write a MATLAB code to create a 1000 point Gaussian distribution of random numbers


having πœ‡ = 0 and 𝜎 = 1. (20)


(ii) Plot this distribution. (10)


(iii) Prove that the full width–half maximum (FWHM), of the above distribution is given by :


FWHM = 2𝜎√2ln 2 (10)

1
Expert's answer
2022-03-29T05:18:55-0400

i: >>x=normrnd(0,1,1000);

ii:

>> y=@(x)exp(-x.^2/2);

>> x=-5:0.01:5;

>> plot(x,y(x))



iii:max⁑(f(x))=max⁑(12πσ2exp⁑(βˆ’x22Οƒ2))=f(0)=12πσ2f(x)=12max⁑(f)β‡’f(x)=122πσ2β‡’12πσ2exp⁑(βˆ’x22Οƒ2)=122πσ2β‡’β‡’x22Οƒ2=ln⁑2β‡’x=Β±Οƒ2ln⁑2β‡’FWHM=x2βˆ’x1=Οƒ2ln⁑2βˆ’(βˆ’Οƒ2ln⁑2)=2Οƒ2ln⁑2iii:\\\max \left( f\left( x \right) \right) =\max \left( \frac{1}{\sqrt{2\pi \sigma ^2}}\exp \left( -\frac{x^2}{2\sigma ^2} \right) \right) =f\left( 0 \right) =\frac{1}{\sqrt{2\pi \sigma ^2}}\\f\left( x \right) =\frac{1}{2}\max \left( f \right) \Rightarrow f\left( x \right) =\frac{1}{2\sqrt{2\pi \sigma ^2}}\Rightarrow \frac{1}{\sqrt{2\pi \sigma ^2}}\exp \left( -\frac{x^2}{2\sigma ^2} \right) =\frac{1}{2\sqrt{2\pi \sigma ^2}}\Rightarrow \\\Rightarrow \frac{x^2}{2\sigma ^2}=\ln 2\Rightarrow x=\pm \sigma \sqrt{2\ln 2}\Rightarrow FWHM=x_2-x_1=\sigma \sqrt{2\ln 2}-\left( -\sigma \sqrt{2\ln 2} \right) =2\sigma \sqrt{2\ln 2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS