Add vs. multiply waveforms?

Don't know.... if I was seriously bovvered about your question then I'd type something like....

sinwt+sinwt

and

sinwt*sinwt

into Google and see what happened.

Then I would come back here and explain it to you.

NOT

DNA

"Louis Sternberg"

** Exactly what it says - the instantaneous values of two independent waveforms ( X and Y ) are electronically multiplied to get the product XY at the output. The output takes regard of the sign of X and Y when needed. ** You can draw two such waves on 5mm grid paper and do it yourself.

To make it easy, assign unity amplitude to each wave.

......... Phil

See

"Product-to-Sum Formulas"

For example, sin(u)*sin(v) = 1/2 [(cos(u-v) - cos(u+v)]

Suppose your angular frequency is w (i.e. 2pi*f), the phase difference is p, and time is t.

Then u= wt and v = wt+p.

sin(wt)*sin(wt+p) = 1/2 [cos (wt - (wt + p)) - cos (wt + (wt + p))] = 1/2 [cos (-p) - cos(2wt + p)]

The first term in the square brackets is a constant that depends only on the phase difference. The second term is another wave at twice the original frequency shifted in time by the same phase difference. The resulting wave has half the amplitude of the original and a D.C. shift depending on the phase difference. In the special case where p=0 (i.e. a squared sine wave) you get a cos wave at twice the frequency that oscillates between 0 and 1.

-- Joe Legris

Without doing a bunch of arithmetic, the simple answer is, for each point on the waveform (for each instant), multiply the value of one by the value of the other, and plot the answer.

This is the difference between "mixing" and "modulation", BTW.

Cheers! Rich

sin(A) . sin (B) = [cos (A-B) - cos(A+B)] / 2

sin(wt + phi) . sin(wt) = [cos(phi) - cos(2wt + phi)] / 2

i.e. a sine wave of twice the frequency, and the half the amplitude, with a DC offset.

The term "mix" is ambiguous. Audio people use it to mean add, and RF people usually use it to mean multiply.

Classic trigonmetric identities, in the backs of handbooks, show what happens when you multiply waveforms of the same or different frequencies, and various phases. Bottom line,

if you multiply different frequency sine waves, the output is the sum and difference frequencies

if you multiply two sines of the same frequency, you get 2F plus a DC term that depends on the phase angle of the inputs.

John

Louis Sternberg wrote:

What I am about to suggest might not seem relevant to your question, but it is.

As you fiddle with the mathematics of functions, it is very helpul to stop thinking of them as being continuous. There is no continuum - there is only the discrete, and limits toward the continuum. Instead, think of discrete points along the x-axis, with discrete f(x) values for those points, and anytime you do math, use the discrete values, for both positions on x and y, then imagine the inter-point spaces getting smaller and smaller as the points converge. It seems silly, but it helped me to begin retaining what i learned.

For example, take your last question above. Think of the two waveforms y1(x) and y2(x) being multiplied together, and think what the output y3(x)=y1(x)*y2(x) will look like. It is helpful to think of y1 and y2 as having the points on their "slanted slope" multiplied. Now imagine a y1(x) being multiplied by a y2(x) value where both their slanted slopes hover above x-axis. The result will be some number y3(x). Now imagine y3(x+dx) = y1(x + dx)*y2(x+dx), where dx is a small, positive, incremntal value. y1 might have a steep slope, while y2 might have a not-so-steep slope at x. Whatever the slopes are, each jump of dx, y3(x+dx) will change its _value_ from y3(x) at a _rate_ depending upun the _sum_ of the rates of change of y1 and y2. You should think about this for a bit until you are convinced. Both y1 and y2 are already trying to change, regardless, but if you multiply them together, the change becomes more pronounced in their product, whether their signs are equal or opposite. How pronounced? Well, naturally, according to the combined rates of change of each. (You can see the Chain Rule from calculus in this last statement, not just from two signals being multiplied, but the entire space of functional operators). This view allows intuitive appreciation of the math of modulation and Euler's formula. When multiplication of two waves are involved, the frequencies add, both in a spectrogram and in the exponents of the product wave.

Finally, if you think about it this way, you get one other benefit. When electrical engineers speak of modulation, filtering, etc...the waves are often idealistically pure sine waves with impulses in the frequency domain. In fact, the waves often have a continuum of frequencies and look weird in the time domain. If you focus on each point along the curve instead of the curve itself, it becomes clear that the frequency of a signal is actually its instantaneous rate of change of phase. Note in the preceeding sentence, the word "instantenous" is implicitly required to be inserted before the word "frequency", as you cannot speak of a waveform having a "frequency" if there are multiple frequencies in it. If it has only one frequency, then the instantaneous rate of change of phase is constant. A signal that is the product of two other signals gets its instantaneous rate of change of phase from the addition of the instantaneous rates of change of those two signals. When you work with phase detection, signal recovery, VCO's, PLL's, it is helpful to abandon the notion of frequency altogether, at least momentarily, and instead start thinking about instantaneous rates of change of phase:

There was a post about adding two wave forms whose frequencies where close, and multiplying them, which might explain better the difference between modulation and adding tones for detection by the human ear:

-Le Chaud Lapin-