# Is a voltage signal we observe on scope made up of only cosines?

Pardon me for the dumb looking question but I'm trying to comprehend a "signals and systems" subject and relate it to voltage or current signal.

A periodic signal is made up of sinusoids which is a conclusion of Fourier series. This series has the complex exponential form which means to me that a signal can be represented as sum of exponential functions.

And for a complex exponential which can also be written as cos(ὦt)+jsin(ὦt), the "Cosine" part is called the real part of a complex exponential.

So when we see any voltage waveform on a scope screen or when we view a sampled voltage data, does that actually mean the voltage signal we are dealing is made up of only cosines but no sines?

I might be confusing real part aka cosine part of the signal representation and a voltage signal which is also called a real signal. But totally confused what is meant by these..

• Signals can be made up of anything you want. The letter X, is it made up of a / and a \? Or is it made up of a > and a < ? Or a v and a ^ ? Sep 20 '18 at 23:52
• @immibis But when we write down a periodic voltage signal in terms of sum of complex functions, aren't we taking only the real part(cosine part) of the sum to define the actual voltage? Sep 20 '18 at 23:55
• Are you asking how the scope converts the complex numbers to real numbers when displaying an FFT? Sep 21 '18 at 0:06
• @immibis Almost. Is the scope's FFT showing only the cosine components of the sum of complex exponentials(signal)? Sep 21 '18 at 0:09
• Signals occur in time, not in frequency (told to me by a very wise engineer who was tutored at Bell Labs and then at Tektronix.) Those sins and cosins are result of humans using Fourier modeling of the energy, because Fourier used trigonometric functions in his 1780 thinking. Sep 21 '18 at 4:28

Both the cosine and the sine part of the complex $$\e^{j\omega t}\$$ function hold exactly the same information. Choose all cosine or all sine (and multiply with -j to get real) and you are fine. Just don't mix cosine and sine, as that results in complex gibberish.

The reason why cosine is often preferred for the representation is because it's invariant to negative time. $$\f(-t)=f(t)\$$ Because the cosine is axis-symetrical. In contrary, sine would require you to invert the result for negative time.

• I like that. Next time someone says, "Get real!", I'll remember to multiply by -j. Sep 21 '18 at 2:13
• Everytime someone told me to be rational, I eat pie out of frustration. Sep 21 '18 at 2:17
• The cosine (real) and sine (imaginary) parts of a Fourier transform, however, don't hold the same information. Sep 23 '18 at 9:21

Almost all wave forms are made up of both sine and cosine. sine and cosine are essentially the same wave form, they are just time shifted, i.e sin(x) = cos(x-pi/2) and cos(x) = sin(x+pi/2). Most wave forms are neither pure sine or pure cosine but have a time (or phase) shift somewhere in between. Hence they can be described as a mix of sine and cosine and the mix ratio determines the time shift.

Complex exponential are doing the same thing. They express the shift as a complex phase.

Treat 'j' as a vector operator, which allows signals and systems to be represented algebraically - it's a mathematical convenience that furnishes an easily manipulated algebraic representation of phase angle. Originally, naming such numbers 'imaginary' was folly, as the term creates an incorrect impression that signals prefixed by j do not really exist.

• Im just writing the same question of my comment to you as well: But when we write down a periodic voltage signal in terms of sum of complex functions, aren't we taking only the real part(cosine part) of the sum to define the actual voltage? Sep 20 '18 at 23:57
• @panic: My answer to Instantaneous vector sum of three phases may help. The OP was confused about the difference between phasors and space vectors. It's a different question to yours but it may help. Sep 21 '18 at 2:17

When you view the FFT of a signal on a scope, usually you are looking at the magnitude of each complex number.

That is, it shows $$\\sqrt{(\text{sine amount})^2 + (\text{cosine amount})^2}\$$. This works out so that the displayed value is the same no matter what the phase of the signal component is (whether it's a sine or a cosine or something in between).

• The question says "voltage waveform" however, which means a time domain display Sep 21 '18 at 7:10