Learn the fundamentals of digital signal processing theory and discover the myriad ways DSP makes everyday life more productive and fun.
3.2.c Interpreting a DFT plot
Loading...
来自 École Polytechnique Fédérale de Lausanne 的课程
数字信号处理
293 个评分
从本节课中
Module 3: Part 1 - Basics of Fourier Analysis
与讲师见面
Paolo PrandoniLecturer
School of Computer and Communication Science
Martin VetterliProfessor
School of Computer and Communication Sciences
As soon as we've seen some examples,
we can give some general guidelines on how to interpret a DFT plot.
So first of all, you will have frequency coefficients from zero to big N minus
one where N is the size of your vector space.
The first N over two coefficients correspond to frequencies less than pi.
So we're talking about counter-clockwise movement
of the point on the complex plane.
Frequencies from N over 2 to N minus 1, are frequencies are larger than Pi and
we interpret those as clockwise rotations in the plane.
Frequencies in this band, close to zero and
close to N minus 1 are low frequencies, in the sense that they are indicate a slow
rotation around a unit circle, either counterclockwise or clockwise here.
Whereas frequencies centered around N over 2 correspond to the fastest
frequencies in the vector space, either clockwise or counterclockwise.
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.
What that means is that for real valued signal, the magnitude of the DFT is
completely specified by only the floor of N over 2 plus 1 coefficients.
So for N equal to 5, for instance, we only need three coefficients.
And for N equal to 6 we only need four coefficients.
The reason why this happens is because when we take the DFT of a real signal,
although we end up with complex coefficients,
The coefficients will have to combine together to give back a real signal.
So coefficient will be complex conjugate across the N over 2 index.
