So, if we go back to the examples that we saw before,

if we take x of n, the units signal, so equal to one for all points,

well this is the slowest possible signal in the sense that it never really changes,

it remains constant for the whole duration of its life.

And correspondingly,

it's Fourier transform only contains the lowest frequency coefficient.

It's not even a frequency in a sense,

because k equal to zero is absence of movement.

Conversely, if we take x[n] = cosine of pi n, which is equivalent to saying

-1 to the power of n, which is the fastest signal in this period of time, it's DFT

would only have one known zero coefficient exactly at the highest frequency point.

Now if you recall Parseval's theorem from module 3.3, we know that

the energy of a signal will not change if we change the underlying basis,

so conservation of energy across domains.

Now, the square magnitude of the k-th DFT coefficient gives us an indication

of the energy associated to the underlying oscillation that composes a signal.

So, the signal's energy at a given frequency

is proportional to the magnitude of the DFT coefficient at that point.

If we go back to one of the examples that we worked out before and we take

a sinusoid with a frequency that is a multiple of the fundamental frequency for

the space, we see that the magnitude is now zero, only in two points.

And that indicates that the energy of the signal is concentrated at this two

frequencies and nowhere else.

If we take the DFT of the unit step on the other hand,

we see that although most of the energy is concentrated on the low frequencies,

there will be energy all across the spectrum.

So we'll need energy at all frequencies to create a step.

Finally let's try and formulize something that you might have noticed already.

And that is the fact that the DFT of real signals is symmetric in magnitude.

Now symmetric here is in quotes because when we're dealing with finite length

vectors, symmetry depends on the parity of the vector.

So for odd length signals we have a situation, for

even length signal we have another situation.

In general,

what we say is that the magnitude of the k-th of the DFT coefficient,

will be equal to the magnitude of the coefficient computed in big N- k.

And this is valid for k that goes from one to N over two and four so

in the case of odd length signal what we have is the following.

The coefficient in zero is three, the coefficient in one will be equal to

the coefficient in big N minus one, here we have an N equal to five so

this will be equal in magnitude to the coefficient in four.

And the coefficient in 2 will be equal in magnitude to the coefficient in 3.

Now if we take N equal to 6, so

an even length space, the coefficient in 0 will be free.

The one in 1 will be equal to the one in 5 in magnitude.

The one in 2 will be equal to the one in 4 in magnitude, and

the one in k equal to 3 will be free.