So I did rewatch the end of lecture and saw that another student asked about why we need the imaginary components for a Fourier transform and I do understand that it introduces cos(theta) + sin(theta) looking terms, but I don't understand why we don't just directly use the sum of different cos and sin terms. What is the need to introduce this imaginary part?
mpotoole
@adigrao2012 This is a fair question. Why introduce this imaginary part? Well, the short answer is that it just makes these equations a lot more elegant and easier to use (despite the fact that they involve complex numbers). If you want though, you can certainly get rid of the complex component. Here are a few options:
Let's see if we can rewrite the equation in the slide to get rid of the imaginary component.
For purely real and continuous signals $f(x)$, it turns out that the scaling parameter $F(k)$ is Hermitian symmetric---which means that $F(-k) = conj(F(k))$. The Hermitian symmetric property causes all the imaginary components of the Fourier transform to cancel out.
Let's now say $F(k) = a(k) - jb(k)$. Then, after canceling out the imaginary components, we would get the following expression: $2 \int(a(k)cos(2 \pi k x) + b(k)sin(2 \pi k x))$. So in the end, when the signal is real, the Fourier transform can be simplified to a sum of weighted sines and cosines anyways, without the imaginary part.
thuspake
A follow up question then is - wouldn't we always be dealing with real signals in this class?
mpotoole
As it happens---yes, we will generally work with real signals. But even for real signals, the standard Fourier transform is more convenient to work with. There's a relevant stackexchange thread discussing this in more detail. Let me quote the top post here:
You need to ask yourself why we use Fourier transforms. We want to transfer the signal from the space or time domain to another domain - the frequency domain. In this domain, the signal has two "properties" - magnitude and phase. If we want to get only the signal's "power" in a specific frequency bin, we indeed only need to take the absolute value of the Fourier transform, which is real. But, the Fourier transform gives the phase of each frequency as well.
While the first (magnitude's) importance is immediate, the phase is sometimes just as important. For example, for images, most of the information is contained in the phase and NOT in the amplitude. Also, frequency responses (Fourier transforms) are used in digital and analog filters, and the phase plays a major role here as well, especially for audio filters where a linear phase is required: this is what enables an audio filter to process all frequencies and output them without a different delay for each frequency (which will distort the sound - imagine a filter that makes your bass sound come a little before your treble...).
So I hope I convinced you the phase is important as well as the magnitude. And in order to get these two properties, we need something other than just real numbers, we need something with magnitude and phase. Something like a complex number.
qryy
What is the scaling parameter?
mpotoole
@qryy It is the magnitude of the Fourier coefficients, or magnitude of the complex exponential. It is the value that the corresponding point in the spectrum takes on.
So I did rewatch the end of lecture and saw that another student asked about why we need the imaginary components for a Fourier transform and I do understand that it introduces cos(theta) + sin(theta) looking terms, but I don't understand why we don't just directly use the sum of different cos and sin terms. What is the need to introduce this imaginary part?
@adigrao2012 This is a fair question. Why introduce this imaginary part? Well, the short answer is that it just makes these equations a lot more elegant and easier to use (despite the fact that they involve complex numbers). If you want though, you can certainly get rid of the complex component. Here are a few options:
There are a bunch of Fourier-related transforms to pick from. One example sine and cosine transforms, involving only a sinusoid or the cosine. Also see discrete sine transform and discrete cosine transform.
Let's see if we can rewrite the equation in the slide to get rid of the imaginary component. For purely real and continuous signals $f(x)$, it turns out that the scaling parameter $F(k)$ is Hermitian symmetric---which means that $F(-k) = conj(F(k))$. The Hermitian symmetric property causes all the imaginary components of the Fourier transform to cancel out. Let's now say $F(k) = a(k) - jb(k)$. Then, after canceling out the imaginary components, we would get the following expression: $2 \int(a(k)cos(2 \pi k x) + b(k)sin(2 \pi k x))$. So in the end, when the signal is real, the Fourier transform can be simplified to a sum of weighted sines and cosines anyways, without the imaginary part.
A follow up question then is - wouldn't we always be dealing with real signals in this class?
As it happens---yes, we will generally work with real signals. But even for real signals, the standard Fourier transform is more convenient to work with. There's a relevant stackexchange thread discussing this in more detail. Let me quote the top post here:
What is the scaling parameter?
@qryy It is the magnitude of the Fourier coefficients, or magnitude of the complex exponential. It is the value that the corresponding point in the spectrum takes on.