Question of the week:
- Thread starter rattus
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- Lockport, IL
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- Retired Electrical Engineer
It all depends on when you throw the on/off switch to the "on" position. We are free to be as arbitrary as we like, when establishing the "initial zero" point of the variable "t." On the scope, we can move the scales around, in order to place the "time equal zero" axis at either "zero voltage" point or the "maximum voltage" point, or anywhere in between.
So it can be sine or cosine; pick your own favorite.
But we must be clear about the intended meaning of phrases like "120 volts @ 60 degrees." To use a phrase like that requires us to first establish one signal (be it current or voltage or some other strange thing) as the reference for a "zero phase angle." Then we must be clear in stating that the phase angle of a second signal is either leading or lagging behind the reference signal. When I first read the question, my impulse was to give an answer of "170 cos (wt-60)." Note the minus sign in place of your plus sign.
So it can be sine or cosine; pick your own favorite.
But we must be clear about the intended meaning of phrases like "120 volts @ 60 degrees." To use a phrase like that requires us to first establish one signal (be it current or voltage or some other strange thing) as the reference for a "zero phase angle." Then we must be clear in stating that the phase angle of a second signal is either leading or lagging behind the reference signal. When I first read the question, my impulse was to give an answer of "170 cos (wt-60)." Note the minus sign in place of your plus sign.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
It's been a long time since I was compelled to use a scope. But I vaguely recall that I would twist a particular dial one way or the other, and the on-screen image of the sinusoidal wave would shift left or right. By doing so, I could cause the peak value to take place at the "time = 0" point, or I could cause the zero value to take place at that point.
My answer remains the same: Call it cosine, or call it sine, as you will.
I don't think that an expression for voltage, as a function of time, will have an imaginary component. Voltage is the reference signal. It is generated by a source as a purely sinusoidal waveform.
It is only when you impress that voltage upon a load that has capacitive or inductive reactance that the "cos (wt) + j sin (wt)" comes into play. That expression is the solution of a differential equation of second order, with constant coefficients, and with a "forcing function" that is a pure sinusoid (i.e., the voltage source). As you well know, you get the second order differential equation from the relationships of current to voltage in capacitors and in inductors.
My answer remains the same: Call it cosine, or call it sine, as you will.
I don't think that an expression for voltage, as a function of time, will have an imaginary component. Voltage is the reference signal. It is generated by a source as a purely sinusoidal waveform.
It is only when you impress that voltage upon a load that has capacitive or inductive reactance that the "cos (wt) + j sin (wt)" comes into play. That expression is the solution of a differential equation of second order, with constant coefficients, and with a "forcing function" that is a pure sinusoid (i.e., the voltage source). As you well know, you get the second order differential equation from the relationships of current to voltage in capacitors and in inductors.
The 0, 120, 240 (-120), 60, 30 angles only are useful when they are relative another sinusoid.
Given a base sinusoid - like Charlie mentioned could be considered a sine or cosine with any angle - is the only time that you could figure the angles on the other voltages or currents.
Hopefully that helps. The following identity may be helpful:
cos(90-[some angle])=sin([some angle]).
Given a base sinusoid - like Charlie mentioned could be considered a sine or cosine with any angle - is the only time that you could figure the angles on the other voltages or currents.
Hopefully that helps. The following identity may be helpful:
cos(90-[some angle])=sin([some angle]).
pierre
Senior Member
- Location
- Westchester County, New York
I say that when the moon is in the second phase, that none of this matters!!!
In other words this is above my head :mrgreen:
But I do like Ronald's picture below his name!!
In other words this is above my head :mrgreen:
But I do like Ronald's picture below his name!!
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
I thought that also required Jupiter to align with Mars? 8)pierre said:
(Challenge of the day: Name that tune!)
kturner
Member
- Location
- East Tennessee
Age of Aquarius by the Fifth Dimension.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Sounds right to me. But wasn't the moon somewhere else (the second "house," perhaps)?
kturner
Member
- Location
- East Tennessee
When the moon is in the seventh house
And Jupitor aligns with Mars
Then peace will guide the planets
And love will steer the stars
Like wow, man.
And Jupitor aligns with Mars
Then peace will guide the planets
And love will steer the stars
Like wow, man.
kturner said:
<sigh> I have that album.
A few years ago I was in San Francisco with one of my younger engineers. We had some free time, so we decided to go sight-seeing. We both had a list of places to visit, so we took turns alternating lists. When I got down to Haight-Ashbury she asked what that was. @#$% punk rocker.
The expression burned into my mind is
E = Emax(sin wt). For this example E = Emax(sin wt + 60).
If t = 0 then E = emax(sin 60) = 170 x .87 = 148 as a starting point.
Rattus
Charlie and Pierre are trying to wreck your question AGAIN.
You are about to lose control.
E = Emax(sin wt). For this example E = Emax(sin wt + 60).
If t = 0 then E = emax(sin 60) = 170 x .87 = 148 as a starting point.
Rattus
Charlie and Pierre are trying to wreck your question AGAIN.
You are about to lose control.
eric stromberg
Senior Member
- Location
- Texas
It's just a 'phase' i'm going through
It's just a 'phase' i'm going through
No, but seriously, the voltage has to do with where the rotor of the generator is at that moment. As a matter of convention, we start the voltage sinusoid at the zero point (as it is rising). I've always used sin in all of my sinusoidal calculations. As far as the oscilloscope is concerned, it all depends on how you have set the trigger on the scope. The waveform will start (on the display) accordingly.
Eric Stromberg, P.E.
Lake Jackson, Texas
It's just a 'phase' i'm going through
No, but seriously, the voltage has to do with where the rotor of the generator is at that moment. As a matter of convention, we start the voltage sinusoid at the zero point (as it is rising). I've always used sin in all of my sinusoidal calculations. As far as the oscilloscope is concerned, it all depends on how you have set the trigger on the scope. The waveform will start (on the display) accordingly.
Eric Stromberg, P.E.
Lake Jackson, Texas
Someone is missing the point. AC voltages and currents are often represented as complex numbers (phasors). For example, the expressions,
V = Vpk(cos(phi) + jsin(phi))?trigonmetric or circular form
= Vpk(e^j(phi))?exponential form
= Vrms @ phi?polar form
are all equivalent.
Where ?phi? is the angle of lead relative to the reference axis which is the positive x axis in the xy plane. A phasor lying along this axis and pointing the right is at 0 degrees.
[Tang, AC Circuits, pp 124?126, Intl. Textbook Co., 1960]
Although we often assume zero phase angles to voltages and currents for convenience, there is no rule that says we have to. I might for example assign angles of 30, -90, and ?210 to a set of 3-phase voltages, and this would be perfectly valid.
We can make the vector rotate by inserting ?wt? in the argument.
v(t) = Vpk(cos(wt + phi) + jsin(wt + phi))
Again, ?phi? is the angle of lead relative to the reference axis. A negative value of phi indicates a phase lag relative to zero degrees.
It is my understanding that a scope synched at time zero would show Vpk(sin(wt + phi). That is, the trace would show the imaginary part of the phasor. This is contrary to what I have been thinking, and that is why I asked the question.
To do this, one would have to generate an external synch pulse at time zero, then the trace would show.
Vpk(sin(wt +phi))
One could also use delayed synch to get the proper display.
In practice, this is seldom done, but it is often done on paper to explain the relationship between a phasor and its corresponding instantaneous value.
P.S. The jokers should start their own thread.
V = Vpk(cos(phi) + jsin(phi))?trigonmetric or circular form
= Vpk(e^j(phi))?exponential form
= Vrms @ phi?polar form
are all equivalent.
Where ?phi? is the angle of lead relative to the reference axis which is the positive x axis in the xy plane. A phasor lying along this axis and pointing the right is at 0 degrees.
[Tang, AC Circuits, pp 124?126, Intl. Textbook Co., 1960]
Although we often assume zero phase angles to voltages and currents for convenience, there is no rule that says we have to. I might for example assign angles of 30, -90, and ?210 to a set of 3-phase voltages, and this would be perfectly valid.
We can make the vector rotate by inserting ?wt? in the argument.
v(t) = Vpk(cos(wt + phi) + jsin(wt + phi))
Again, ?phi? is the angle of lead relative to the reference axis. A negative value of phi indicates a phase lag relative to zero degrees.
It is my understanding that a scope synched at time zero would show Vpk(sin(wt + phi). That is, the trace would show the imaginary part of the phasor. This is contrary to what I have been thinking, and that is why I asked the question.
To do this, one would have to generate an external synch pulse at time zero, then the trace would show.
Vpk(sin(wt +phi))
One could also use delayed synch to get the proper display.
In practice, this is seldom done, but it is often done on paper to explain the relationship between a phasor and its corresponding instantaneous value.
P.S. The jokers should start their own thread.
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