Finding FOURIER transform of signals involving trigonometric functions- E.X: F(w)= 2 cos(w) sin(2w) / w where w: frequency. Find the time signal for this?
In questions involving trigonometric part,we use a special function called "sinc function".It is equal to...sinc(w) = sin(w)/w. So,here we first make the numerator part as sum of two sine functions. That will be : sin(3w)-sin(w).....(using sin( A+B) and sin(A-B) formulae) So,the original question will be : 3 sinc(3w) - sinc(w). We can now easily find out the inverse as inverse fourier of sinc function is rect function. So,using some basic duality principle in FOURIER transforms,we can solve this. Here,I mixed the concepts of electronics concept(signals) with trigonometric to show the wide applications and uses of trigonometry in science fields. Hope,you got my point
Basic math is just like playing with numbers.Just have a idea about what are even numbers and what are odd numbers. One more 'simple' trick is : what do you mean by 4 3/4.....(mixed fraction)
I'd like to share a interesting fun thing with you regarding numbers. Pick any number. If that number is even, divide it by 2. If it's odd, multiply it by 3 and add 1. Now repeat the process with your new number. If you keep going, you'll eventually end up at 1. Every time. For that mixed fraction question,you can solve that directly without converting it to improper fraction. It just means 4+(3/4)---So,the answer is 4.75.
1)Well,if you are familiar with graphs ,I'd like to share something. If you were given with an algebraic equation and asked to find out the total number of real,imaginary roots,you can use graphical approach instead of solving it,which takes lot of effort. 2)If you know value of x^(n) + 1/x^(n) and asked to find out x^(n)- 1/x^(n)..You can do these kind of problems with one simple method
1) Draw the graph of the equation on Left hand side and R.H.Son the same paper.See if they are intersecting,thye number of points at which they interest are the real roots.If they dont intersect,the equation have only complex roots. 2)take x+ 1/x =k and now square it x^(2)+ 1/x^(2) = k^(2)-2..... now x - 1/x = underroot of ([(k^(2)-2)-2].. so,if x^(n)+1/x^(n) = k then x^(n) - 1/x^(n) =underrroot of [ k^(2) - 4 ]