Fourier Transform Sine

Fourier Transform Sine Fourier Transform Sine Fourier transform is used for describing the relationship in between the signal in the domain of tine and signal's representation in its domain of frequency . It is only transform the information means the information related to the signal is not created or lost in during the process of transformation , so because of that original information can be retrieved easily .Fourier transform is used to represent the complex continuous signals into the continuous time signals that are also the real valued time signals . Fourier transform is used for different waves if signals .There is one of them that is fourier transform sine wave . There are some step for evaluating the given function in an interval by using the Fourier transform sine wave as follows : Step no (I) : Fourier transform sine wave is denoted by an expression under the certain interval that is as follows : f (p) = ∑n =1 infinity b n sin (n ∏ a) / L

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Here in the above expression there is used the summation of the sin signals that is also solved by using the integration as : F [f (p)] = √ (2 / ∏) ∫ ab f (x) sin (n p ) d p . In the expression a and b are denotes the lower and upper limits respectively. Step no (II) : In this step value of the function f (p) will be put which is given by the examiner or given into the question with upper and lower limits . If we understand this function as we have the value of the function f (p) = p where lower and upper limits are (0 , ∏) respectively then according to the Fourier transformation sine wave it will be described as : F [f (p)] = √ (2 / ∏) ∫ a ∏ p sin (n p ) d p . Step no (III) : In this step after putting all the values of functions as well as values of the upper and lower limits perform the integration operation and calculate the answer for the function by Fourier transform sine wave . There is an example to understand the Fourier transform sine wave as if the given function is f (p) = 7 (∏ + p) and interval is (0 , ∏) . Solution : this problem also solved by using some steps as : step (1) : Fourier series is denoted as the above we defined : f (p) = ∑n =1 infinity b n sin (n ∏ a) / L then F [f (p)] = √ (2 / ∏) ∫ ab f (x) sin (n p ) d p .