It is not homework, just 73yo + quarantine.

Reading two ongoing threads in this forum I realized that, shame on me, I forgot the basics. By searching in the Web I got more doubts.

I would appreciate if answers are given in the order below with units.

a) Our AC from the mains is ideally sinusoidal, nominal 220 V - 50 Hz. Is it RMS, right? How do I express that?

b) I could set my Siglent generator to output these four sinusoidal signals. How do I express each?

Amplitude 2 V - 750 Hz - offset 0 V.
Amplitude 3 V - 650 Hz - offset 0,5 V.
Amplitude 1 V - 350 Hz - offset 3 V.
Pk to pk 3 V - 400 Hz - offset 6 V.

c) Applying a sinusoidal signal of 2 V amplitude - 1 Khz - to the both inputs of an analog multiplier gives the square of the function.
Could you by visual inspection estimate (¿?) the outcome (amplitude and frequency)?

d) What if I apply the same signal with 2 V of amplitude - 440 Hz - and 5 V offset?

e) If I wanted to multiply c) by itself and d) by itself, with my calculator, what is the right procedure?

Thanks for any help.

 

I guess for starters it might help to remember that all sinusoidal signals are expressed in terms their peak amplitude and phase. DC offsets add to the "0-level of the signal" and have the same effect on all points of the signal.

a) The signal from the mains is 220 VAC(rms). To get the peak value you multiply the RMS value by √2, or 1.414...., and you get 311.126 VAC(pk). The frequency of a sine wave is usually represented by converting the frequency from Hz (Hertz) to radians/sec. You do this by multiplying the frequency in Hz., by 2π, and you get 314.159 radians/sec. We won't worry about phase angle for the moment. What comes out of the wall is:

311.126*sin (314.159*t) VAC for all t in seconds​

​

The sine function takes values from -1 to +1 and the voltage will go from -311.126 to +311.126 as t goes from 0 to .020 seconds (1 cycle at 50 Hz.)​

b) 2nd verse - same as the first

2V@750 Hz + 0 VDC would be --> 0 VDC + 2 sin (4712.4*t) VAC​

3V@650 Hz + 0.5 VDC would be --> 0,5 VDC +3 sin (4084.1*t) VAC​

Give the other examples a try and see how it goes. Ask again if you hit a snag my friend.

​

​

​

 

In mathematics, we tend to write

y = A sin(ωt + θ)

Assuming phase angle θ is zero, in electricity and electronics we prefer to substitute
ω = 2πf

Hence we write
V(t) = Vdc + V sin(2πft)

Amplitude 3 V - 650 Hz - offset 0,5 V
becomes
0.5 + 3 x sin(650 x 2π x t)

 

I guess for starters it might help to remember that all sinusoidal signals are expressed in terms their peak amplitude and phase. DC offsets add to the "0-level of the signal" and have the same effect on all points of the signal.

a) The signal from the mains is 220 VAC(rms). To get the peak value you multiply the RMS value by √2, or 1.414...., and you get 311.126 VAC(pk). The frequency of a sine wave is usually represented by converting the frequency from Hz (Hertz) to radians/sec. You do this my multiplying the frequency in Hz., by 2π, and you get 314.159 radians/sec. We won worry about phase angle for the moment. What comes out of the wall is:

311.126*sin (314.159*t) VAC for all t in seconds​

​

The sine function takes values from -1 to +1 and the voltage will go from -311.126 to +311.126 as t goes from 0 to .020 seconds (1 cycle at 50 Hz.)​

b) 2nd verse - same as the first

2V@750 Hz + 0 VDC would be --> 0 VDC + 2 sin (4712.4*t) VAC​

3V@650 Hz + 0.5 VDC would be --> 0,5 VDC +3 sin (4084.1*t) VAC​

Give the other examples a try and see how it goes. Ask again if you hit a snag my friend.

Gracias Papabravo!! Short and to the point. I appreciate that.

 

In mathematics, we tend to write

y = A sin(ωt + θ)

Assuming phase angle θ is zero, in electricity and electronics we prefer to substitute
ω = 2πf

Hence we write
V(t) = Vdc + V sin(2πft)

Amplitude 3 V - 650 Hz - offset 0,5 V
becomes
0.5 + 3 x sin(650 x 2π x t)

Gracias MrChips! I dodged that expression for years.

 

Gracias Papabravo!! Short and to the point. I appreciate that.

de nada

 

In mathematics, we tend to write

y = A sin(ωt + θ)

Assuming phase angle θ is zero, in electricity and electronics we prefer to substitute
ω = 2πf

Hence we write
V(t) = Vdc + V sin(2πft)

Amplitude 3 V - 650 Hz - offset 0,5 V
becomes
0.5 + 3 x sin(650 x 2π x t)

Do you think it would be better to express a sinusoid in the form

y = A cos(ωt + θ)

rather than using sine function?
My rationale being that if you do a Fourier analysis on a signal the phase that's returned is relative to a cosine rather than a sine.

I know it's just a π/2 phase shift but I've always felt that it would tie in much more neatly with frequency domain analysis if cosines where the standard way of representing sinusoidal signals. Is there a particular reason that sine functions seem to be the standard in power/electrical engineering?

 

Do you think it would be better to express a sinusoid in the form

y = A cos(ωt + θ)

rather than using sine function?
My rationale being that if you do a Fourier analysis on a signal the phase that's returned is relative to a cosine rather than a sine.

I know it's just a π/2 phase shift but I've always felt that it would tie in much more neatly with frequency domain analysis if cosines where the standard way of representing sinusoidal signals. Is there a particular reason that sine functions seem to be the standard in power/electrical engineering?

Good question. I don't know the answer.

 

Do you think it would be better to express a sinusoid in the form

y = A cos(ωt + θ)

rather than using sine function?
My rationale being that if you do a Fourier analysis on a signal the phase that's returned is relative to a cosine rather than a sine.

I know it's just a π/2 phase shift but I've always felt that it would tie in much more neatly with frequency domain analysis if cosines where the standard way of representing sinusoidal signals. Is there a particular reason that sine functions seem to be the standard in power/electrical engineering?

Good qeustion but generally we like to have the origin of signals from a zero (equilibrium) rest point when looking a phase shift.
https://mathbitsnotebook.com/Algebra2/TrigGraphs/TGShift.html

 

Good qeustion but generally we like to have the origin of signals from a zero (equilibrium) rest point when looking a phase shift.
https://mathbitsnotebook.com/Algebra2/TrigGraphs/TGShift.html

I think I see what you mean. It's easier to read phase differences between sinusoids where the sinusoids cross the time axis? I still don't think it's a big deal whether they're sines or cosines though - if I understand you correctly.

 

I think I see what you mean. It's easier to read phase differences between sinusoids where the sinusoids cross the time axis? I still don't think it's a big deal whether they're sines or cosines though - if I understand you correctly.

Sometimes the cos(wt) form rather than the sin() form simplifies some circuit analysis that involve inductances. Sometimes it is better that the RESPONSE starts at zero rather than the excitation.

Also, you can convert a sin() form into a cos() form or vice versa.
You can even convert a sum like:
A*sin(w*t+ph1)+B*sin(w*t+ph2)
(or any combination of sin or cos terms)
into a single sin() or cos() form that usually includes a phase shift part also.
This means that you can convert a sum of any number of terms with the same frequency into a single sin() or cos() form.

 

Hello,

For the multiplication of two regular sines we have this identity:

A*sin(t*w1+ph1)*B*sin(t*w2+ph2)=
(cos(t*w2-t*w1+ph2-ph1)*A*B)/2-(cos(t*w2+t*w1+ph2+ph1)*A*B)/2

Notice the phases are multiplied while the frequencies are added.

 

Hello,
Or to check and have an idea you can use function plotter found on the web such as https://www.desmos.com/calculator,
you can input the functions terms in it for example as seen fig attached,
Good night
Saviour