Confused about wave

[Sonia123]

Hey there folks!

First I am extremely sorry if this is the wrong thread I am posting. Second I am having a little confusion which I really really to clarify.
Here it is, how do I sketch the output voltage versus time for sinusoidal input if output voltage = 0.98(19sin(omega)t-0.6) and input voltage = 10sin(omega)t.
I know sinusoidal wave is either sine or cosine wave but I am extremely confused on how to sketch this one. A little help would be extremely appreciated.
Thank you

 

[MrAl]

Hi,

Did you mean the output is:
0.98*(19*sin(w*t-0.6))

or did you mean the output is:
0.98*(19*sin(w*t)-0.6)

In either case though you would calculate what a good time step for t would be, then plot the output above vs t.
The first value for t would be 0, then t=tstep, then t=2*tstep, then t=3*tstep, etc.
You can first calculate the period of the sine wave to get an idea what a good value for the time step would be.

 

[Grossel]

Use Geogebra

Hi.

This is how MrAl's formula looks when drawed in Geogebra.

Just remember to replace ω with "2 Π".

Nb: space is a valid symbol for multiplication in Geogebra.

 

[JimB]

Are you trying to plot the output of a system against its input, as shown in my attachment.

Or are you trying to plot voltage against time?

Please clarify.

JimB

 

[Sonia123]

Grossel and MrAI, can you explain me one thing please. The time period of the sine wave would be t=2(pi)/(omega) and the phase shift is -0.6. Then would the time step be 2(pi)? It's been a while since I used trigonometric functions. And yes, it is 0.98*(19*sin(w*t)-0.6).
Thanks a lot for your help ^_^

 

[misterT]

yes, it is 0.98*(19*sin(w*t)-0.6).

That equation result in dc-voltage offset of (0.98 * -0.6) = -0.588

If you want the period to be: T = 2pi / w
You must have sin(t*w)

If you want phase delay of 0.6 radians, you have:
sin(t*w-0.6)

If you want time delay of 0.6 seconds, you have:
sin((t-0.6)*w)

 

[MrAl]

Grossel and MrAI, can you explain me one thing please. The time period of the sine wave would be t=2(pi)/(omega) and the phase shift is -0.6. Then would the time step be 2(pi)? It's been a while since I used trigonometric functions. And yes, it is 0.98*(19*sin(w*t)-0.6).
Thanks a lot for your help ^_^

Hi again,

First, the period Tp is 2*pi/w yes, so you know that the wave makes one complete cycle in 2*pi/w seconds. Now if you remember what a sine wave looks like, you know you can not just plot two points to show the whole wave, you'd have to plot probably at least 20 points but preferably more than than like 50 points. That means you take the whole period and divide by 50 and that gives you your time step. If you use 60 points that would mean each point would be at 5 degree increments.

Second, the way you have your function written:
0.98*(19*sin(w*t)-0.6)

the "0.6" in there is not a phase shift really, it is a DC offset. If you want it to be a phase shift then you have to write it as:
0.98*(19*sin(w*t-0.6))

which gives you a phase shift of about 34.4 degrees.

In general, in the function:
A*sin(w*t+B)+C

A is the amplitude
B is the phase shift
C is the DC offset

For your function A simplifies to 18.62 so you dont need two multiplicative constants:
A*sin(w*t-0.6)

if -0.6 is the phase shift.