Note added February 11, 2010: We polished up the draft below and made it into a white paper called Understanding the Kramers-Kronig Relation Using A Pictorial Proof
The usual proof of the Kramers-Kronig relation involves contour integration in the complex plane of the frequency domain. Unless you’re a math genius, you don’t get much insight into what is going on. The pictorial “proof” here begins in the time domain and gives more insight. It illustrates a treatment in Advanced Signal Integrity for High-Speed Digital Designs by Stephen H. Hall and Howard L. Heck, pp.331-336 ISBN-13 978-0-470-19235-1 .
In outline, the relationship comes about because of several facts:
Let’s look at these step by step.
The Fourier integral frequency response of an arbitrary function h(t) is defined as:
Let’s think about the subset of functions that are real-valued and causal:
h(t) = 0 for t < 0, h(t) is real for t >=0
Figure 1: Example of a causal impulse response, namely a damped 30 MHz sine wave.
Let’s see how this class of functions constrains.
For reasons that you’ll see below, we’re going to build our causal impulse response out of non-causal even and odd terms. So before we do that build up, let’s assemble a mini- toolbox of relationships to do with even and odd functions.
The first tool is the relationship between an odd impulse response, defined by:
…and its Fourier integral.
Figure 2: Example of a non-causal odd impulse response, an increasing then damped 30 MHz sine wave
Tool 1: The Fourier integral of a (non-causal) odd impulse response is pure imaginary.
To see why this is so remember cosines are even, sines are odd. For an odd function, the odd-even productsintegrate out to zero because the left and right halves have equal magnitudes but opposite signs and so they always cancel each other. So the only finite terms are the odd-odd terms which are pure imaginary.
The second is the relationship between an even impulse response, defined by:
…and its Fourier integral.
Tool 2: The Fourier integral of a (non-causal) even impulse response is purely real.
The even-odd productsmust integrate to zero for the same reason as before. Only the even-even productsare finite and those are purely real.
Let’s decompose a causal function into even and odd parts, then apply these tools to each part.
You can construct any causal or non-causal h(t) out of a sum of some linear combination of even and odd components:
…but it’s not particularly useful or interesting.
However an interesting thing does happens when you construct a causal function out of even and odd components.
Let’s start with our odd function ho(t), then multiply it by the signum function, defined as:
Figure 3: The signum function is simply the sign (but not the sine!) of its argument.
The signum function gives the left hand half of ho(t) an up-down flip and this yields a new even function
Figure 4: Example of a non-causal even impulse response, created by multiplying the functions in Figures 2 and 3.
Now think about what happens when we add the odd function and the even function we derived from it.
Figure 5: Example of a non-causal impulse response, created by adding the functions in Figures 2 and 4.
Impulse responses constructed in this way are necessarily causal because the left hand half of even exactly cancels the left hand half of odd. It might seem strange to go to all this trouble to construct an impulse response, but the beauty is we can now see what the Fourier integral looks like.
Before we dive into the specifics, notice that (from Tools 1 and 2 above) the evenwill yield a real response and the odd ho(t) will yield a pure imaginary response, and that both depend on the same function ho(t).
Figure 6: The imaginary part of the frequency response comes entirely from the odd part of the time response (Figure 2 in this case). Note: For aesthetic reason I flipped the curve up-down. The imaginary part is actually negative. Physically, the peak corresponds to severe damping at the resonant frequency (30 MHz).
We can immediately see that the causality constraint means that the real and imaginary parts are related and contain the same information. But what is the exact relationship?
Multiplication in the time domain is equivalent to convolution in the frequency domain, so:
where upper case functions denotes the Fourier integral of the corresponding lower case function, and denotes convolution. The recipe for convolution in words and pictures is “flip the kernel left-right using a dummy variable, slide it over the other term, multiply, integrate over the dummy variable, rinse, repeat”
Breaking that down, take the convolution kernel (called the Hilbert kernel) in a dummy variable, flip it left-right, slide it over by to give , multiply by and integrate:
But what does the Hilbert kernel look like? Well signum(t) is odd so we know must come from the sine waves and be pure imaginary (Tool 1).
Figure 7 shows one point of the Fourier integrals we must do. Imagine Fourier integration as being a multiply and add operation on the red and green curves. We get the response at, in this case, 30 MHz. Note that only the two shaded half periods immediately to the left and right of sign change at the origin give a non-canceling products. Every other pair of half periods cancel each other out because, away from the origin, signum(t) is either constant +1 or constant -1.
Figure 7: Pictorial representation of one frequency point of the Fourier integral of the signum(t) function
The area under the non-canceling, shaded area is proportional to wavelength and so inversely proportional to period. Actually, pictures aside, it’s not too hard to do the Fourier integral because we can spilt it into one half from to 0 and one half from 0 to . You can quickly convince yourself that the Hilbert kernel is.
Here’s what it looks like:
Figure 8: The imaginary part of the Hilbert kernel. The real part is zero.
The bottom line is that the Hilbert transform in the frequency domain is equivalent to multiplication by signum(t) in the time domain.
We can re-write the real part of our frequency response as the Hilbert transform of the imaginary part:
Note thatis pure imaginary, and j times pure imaginary is purely real as expected.
Imagine convolution as running the Hilbert kernel (Figure 8) over the imaginary part (Figure 6). Here’s what you get:
Figure 9: The real part of the frequency response comes entirely from the even part of the time response (Figure 2 in this case). Physically, the switchback corresponds to a damped resonator that can respond to a stimulus below its resonant frequency (30 MHz), but that “turns off” above it, because it can no longer respond quickly enough.
Thus the real part of frequency domain response of a causal impulse response can be calculated knowing only the imaginary part.
You can apply a similar proof starting with an even function and show that the imaginary part can be calculated knowing only the real part.
In summary, you can decompose any causal impulse response as an odd function plus signum times the same function. This second term is even, and the left hand part of it exactly cancels the left hand part of the first, odd term, thus ensuring causality. The even and odd decomposition of the casual impulse response yield the real and imaginary parts of the frequency response, respectively. Using the fact that multiplication by the signum function in the time domain is equivalent to the Hilbert transform in the frequency domain we can calculate the real part solely from knowledge of the imaginary part or vise versa.
The Kramers-Kronig relations give a condition that is both necessary and sufficient, so even before applying an inverse Fourier integral, you can determine whether a given frequency response will yield a causal or a non-causal impulse response. If the real and imaginary parts are Hilbert transforms of each other, the impulse response is causal, and not otherwise.
This fact is very useful because we can test whether or not a frequency response is causal or not without leaving the frequency domain.
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