In the original problem, parts a and b were asking for the Average power and Fou

Question

In the original problem, parts a and b were asking for the Average power and Fourier series coeff. I included their values above. Here is a link if you need it showing those first two parts being solved, (https://www.chegg.com/homework-help/questions-and-answers/question-2-consider-periodic-ct-square-wave-lecture-8-ex-35-oppenheim-book-period-t-1-dete-q24618162). Please do not re-answer parts a and b, thank you

Question 2 Consider the periodic CT square wave from Lecture 8 (Ex 3.5 in the Oppenheim book) with period T 1. 7, 0.25 7,= 0.125 · · If needed the average power is (0.5) for [T1.25] and the avg. power is (0.25) for the other value of T1 where [T1-0.125]. Given T1 = 0.25 => (a-o) = 0.5, (a-1): (a-5)-(a-9)-2/(n*pi), (a-3)=(a-7)=(a-11): (-2/(n*pi)) and b-n =0 . For T1 = 0.125 => (a-0-25, (an) =(1/(n"pi))[2'sin((n*pi)/4)] and (bo)-o c) Determine the Fourier Series coefficients (i.e., which values of k) that relate to frequency components in the range offrequences-16m 16. Hint: Determine the fundamental frequency and relate to · d) For each value of T1(i.e, T 0.25 and T 0.125), determine the power related to frequency components in the range of frequencies-16m 16. (Hint: Consider Parseval's Relation) e) For each value of T, (i.e., T-0.25 and T0.125), determine the percentage of the signal power in the range of frequencies-16 16T. Example 3.5 The periodic square wave, sketched in Figure 3.6 and defined over one period as (3.41) is a signal that we will encounter a number of times throughout this book. This signal is lo determine the Fourier series coefficienes for x(t), we use eq. (3.39). Because periodic with fundamental period T and fundamental frequency 2mm of the symmetry of x(t) about t = 0. it is ceavenient to choose-T/2 T/2 as the x(t) T I-T,T I T T I T 2 -2T 2T 2 Figure 3.6 Periodic square wave

Explanation / Answer

The DFT is the sampled Fourier Transform and therefore does not contain all frequencies forming an image, but only a set of samples which is large enough to fully describe the spatial domain image. The number of frequencies corresponds to the number of pixels in the spatial domain image, i.e. the image in the spatial and Fourier domain are of the same size.

For a square image of size N×N, the two-dimensional DFT is given by:

Eqn:eqnfour1
where f(a,b) is the image in the spatial domain and the exponential term is the basis function corresponding to each point F(k,l) in the Fourier space. The equation can be interpreted as: the value of each point F(k,l) is obtained by multiplying the spatial image with the corresponding base function and summing the result.

The basis functions are sine and cosine waves with increasing frequencies, i.e. F(0,0) represents the DC-component of the image which corresponds to the average brightness and F(N-1,N-1) represents the highest frequency.

In a similar way, the Fourier image can be re-transformed to the spatial domain. The inverse Fourier transform is given by:

Eqn:eqnfour2
Note the Eqn:oneovern2 normalization term in the inverse transformation. This normalization is sometimes applied to the forward transform instead of the inverse transform, but it should not be used for both.$$

To obtain the result for the above equations, a double sum has to be calculated for each image point. However, because the Fourier Transform is separable, it can be written as

Eqn:eqnfour3
where
Eqn:eqnfour4

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