# Discrete Fourier Transform on even function

by mathman44
Tags: discrete, fourier, function, transform
 P: 210 The DCT of an even function is comprised of just cosine coefficients, correct? I'm playing around in MATLAB and I came up with a simple even function 1.0000 0.7500 0.5000 0.2500 0 0.2500 0.5000 0.7500 1.0000 0.7500 0.5000 0.2500 0 0 0 0 0 0 0.2500 0.5000 0.7500 1.0000 0.7500 0.5000 0.2500 0 0.2500 0.5000 0.7500 1.0000 But get imaginary as well as real outputs when I use the DFT on it. Which means that my function isn't even? Can anyone help me make sense of this?
 Mentor P: 12,081 Where is x=0 in your list? And where are sin/cos coefficients in the result?
 P: 210 Hi mfb, I haven't assigned x values to the points. The DCT really just considers the index of each point, (n=0,1,2...N-1) and the total number of samples (N) right? But since it's even, let's say x ranges from -14.5 to 14.5 in steps of "1". The sin/cos coefficients are listed in the second list in my first post. The sin coefficients are the imaginary ones - I don't understand why I'm getting sin coefficients.
 Sci Advisor HW Helper PF Gold P: 3,289 Discrete Fourier Transform on even function I assume when you write DCT, you actually mean DFT. DFT = discrete Fourier transform DCT = discrete cosine transform The problem is that your input function is not even. If you remove the final element (1) then it will be even, and you will obtain a real-valued result.
 P: 210 Hi jbunniii, Yes I mean DFT. Can you explain why the final element needs to be removed? In my mind the function is even since f(x)=f(-x) if you center it on the y-axis for example.
 Sci Advisor HW Helper PF Gold P: 3,289 Yes, think about extending your function periodically in both directions. The first element in the array corresponds to x = 0. In order for the function to be even, the element to the left of this must equal the element to the right. This is not the case if we use your original function: $$\begin{array} \text{f(x)} & \ldots & 0.5 & 0.75 & 1 & 1 & 0.75 & 0.5 & \ldots \\ x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & \ldots \end{array}$$ Note that we do not have f(-1) = f(1), so the function is not even. On the other hand, if we remove the 1 at the end of the array, and again extend the function periodically, now we get an even function: $$\begin{array} \text{f(x)} & \ldots & 0.5 & 0.75 & 1 & 0.75 & 0.5 & \ldots \\ x & \ldots & -2 & -1 & 0 & 1 & 2 & \ldots \end{array}$$
 Sci Advisor HW Helper PF Gold P: 3,289 To clarify, by "extend the function periodically," I mean that the array is supposed to represent one period of the function. For example, if the array was (1, 2, 3, 4) then extending it periodically means simply repeating it over and over: (...,1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, ...) where x = 0 always corresponds to the first element in the original array.
 P: 210 Of course - thanks, that certainly clears it up. So an even function must always have an odd number of discrete samples (i.e. N is odd), right?
HW Helper
PF Gold
P: 3,289
 Quote by mathman44 Of course - thanks, that certainly clears it up. So an even function must always have an odd number of discrete samples (i.e. N is odd), right?
No, you can make an even function for any N. For example, the following are all even functions when extended periodically:

N = 1: (1)
N = 2: (1 0)
N = 3: (1 0 0)
N = 4: (1 0 0 0)
etc.
 P: 210 Sorry I wasn't being very clear. What I mean is that if I want to make a function that is symmetric with respect to the y-axis AND satisfies the condition above, then it would seem I can only use an odd N of samples. For example assume 15 points lie to the left of the y-axis, then by necessity we can only have 14 points lie to the right of the x-axis to satisfy the condition that you discussed. To use my example, if this portion lies to the left of the y-axis 1.0000 0.7500 0.5000 0.2500 0 0.2500 0.5000 0.7500 1.0000 0.7500 0.5000 0.2500 0 0 0 then the following must lie to the right of the y-axis 0 0 0 0.2500 0.5000 0.7500 1.0000 0.7500 0.5000 0.2500 0 0.2500 0.5000 0.7500 Which is a total of 29 samples.