Sine and cosine transforms

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Sine and cosine transforms

The idea behind this can be explained somewhat like this. Say you want to examine how much of a sine wave of a particular frequency exists within some signal/function. What you can do is to multiply the signal with the sine wave in question. Let us say a component of the input is a sine wave at say 5 Hz, and say we look at the signal for a few whole periods. If we multiply that with a sine wave at, say 7 Hz, the product will sometimes be positive and sometimes negative. The sum of the products will be 0. For the moment being I leave it up to you to prove this. In the image we can see a 5 Hz sine wave at the top, then a 7 Hz sine and cos wave, and at the bottom the respective products.

Now, if we test with a 5 Hz test signal, then the product will always be positive, and the sum will be non-zero.

The reason we need a sine, and a cos transform is to catch signals with different phases.

On the following page, you can explore the above.

You can move the slider at the bottom to change the test signals frequencies. You can move it with the mouse, or by touch or by using the arrow keys.
You can see the sine and cos test signals
You can see the respective products.
You can then see the corresponding sine and cosine transform show up as you move the slider.
Integer frequencies are indicated by dotted lines.
Discrete transforms are just done at the integer frequencies, or rather at integer multiples of the fundamental frequency.
You can change the input signal by the button at the bottom.

To the sine and cos transform demo

The input for a discrete sine and cos transform consists of a number, N, of equally spaced samples, say x₀ to x_n_-1.

The discrete sine transform could be defined by

$\displaystyle {{X}_{k}}=\sum\limits_{{n=0}}^{{N-1}}{{{{x}_{n}}\cdot \sin \left( {2\pi \frac{k}{N}n} \right)}}$

Where X_k is the sum corresponding to the k^th test signal. For the cosine transform you simply replace the sin by a cos.

For the (continuous) sine and cosine transform we replace the sum with an integral and the sequence with a continuous function.

$\displaystyle X(k)=\int\limits_{a}^{b}{{f(t)\cdot }}\sin \left( {2\pi kt} \right)dt$

For a non-periodic function, we replace the bounds with minus infinity to plus infinity. For a periodic function we can, for example, put 0 to T or –T/2 to T/2 as our lower and upper bound.

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Last modified: Feb 19, 2025 @ 16:33