Sine wave with shift formula

Thread Starter

Henry4321

Joined May 6, 2024
9
Hi everyone,
was wondering if you can help me understand something.
This diagram here shows a sinusoidal leading edge phase controlled voltage. I need to determine the RMS value as a function of the phase angle β.
The formula for RMS is sqrt(1/T * definite integral (T to 0) of mean^2)
The answers give
1715121881632.png

Now, I figure that this answer can be achieved if the formula for RMS is sqrt(Vpeak^2 * indefinite integral sin^2(pi * t + b))
My question: is this correct? and if it is, why the integral is indefinite? why it is not over a period 1/T? because every time I am doing the definite integral, all the stuff gets cancelled and I end up with just Vpeak/sqrt(2) (which is what you would expect for a sine wave, I suppose). Also, why is the angle represented as (pi * t + b)? Normally, it's omega*t + b, but in this case x-axis is already omega*t, so I guess it can be related to this... What I really need is to understand the task and the graph for any similar problems in future, as I am not sure when I am supposed to use definite integral and when indefinite and for which cases
1715121480692.png
 

WBahn

Joined Mar 31, 2012
30,290
The answer to your first question is that the integral is NOT indefinite. You need to take the definite integral over an integral number of periods and divide by the number of periods.

If the signal is aperiodic, the you take the integral over a sufficiently large amount of time and divide by the span of time used.

An indefinite integral spans over all time and if the RMS value is non-zero, then there is an infinite amount of energy divided by an infinite amount of time, which would then require taking a limiting process (L'Hospital's Rule, anyone?).

The nominal limits of integration are over one period, i.e., from 0 to 2π.

By noting and exploiting symmetry, that can be reduced to 0 to π.

Then the function can be broken up into pieces, namely 0 to ß, and ß to π.

But the function from 0 to ß is identically zero, so that is a trivial integral to evaluate.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
The answer to your first question is that the integral is NOT indefinite. You need to take the definite integral over an integral number of periods and divide by the number of periods.

If the signal is aperiodic, the you take the integral over a sufficiently large amount of time and divide by the span of time used.

An indefinite integral spans over all time and if the RMS value is non-zero, then there is an infinite amount of energy divided by an infinite amount of time, which would then require taking a limiting process (L'Hospital's Rule, anyone?).

The nominal limits of integration are over one period, i.e., from 0 to 2π.

By noting and exploiting symmetry, that can be reduced to 0 to π.

Then the function can be broken up into pieces, namely 0 to ß, and ß to π.

But the function from 0 to ß is identically zero, so that is a trivial integral to evaluate.
So when I integrate Vpeak^2*sin^2(x) from b to pi, I get the correct answer. But I wonder why it is sin(x), not sin(omega*t) or something. I feel like it’s a stupid question, but I can’t help but wonder. Is it because the x-axis is already omega*t? The task confuses me, because it says evaluate as function of phase angle b, and as far as I understand phase angles, the formula should be (omega*t+b), but the graph is not shifted along the x-axis anywhere. According to the graph, it’s just a regular sine function with a bit taken out of it. Could you maybe point out where my thinking was wrong so I don’t get confused by these tasks later?
 

WBahn

Joined Mar 31, 2012
30,290
So when I integrate Vpeak^2*sin^2(x) from b to pi, I get the correct answer. But I wonder why it is sin(x), not sin(omega*t) or something. I feel like it’s a stupid question, but I can’t help but wonder. Is it because the x-axis is already omega*t? The task confuses me, because it says evaluate as function of phase angle b, and as far as I understand phase angles, the formula should be (omega*t+b), but the graph is not shifted along the x-axis anywhere. According to the graph, it’s just a regular sine function with a bit taken out of it. Could you maybe point out where my thinking was wrong so I don’t get confused by these tasks later?
The argument to the sine function is an angle.

omega*t IS the angle.

It's just a change of variables.
 

Thread Starter

Henry4321

Joined May 6, 2024
9
The argument to the sine function is an angle.

omega*t IS the angle.

It's just a change of variables.
The thing is integrating with omega*t doesn’t give me the same answer.
Integral (b to pi) of sin^2(wt) = integral (b to pi) of 1/w * sin^2(u) = 1/2*1/w * integral (b to pi) of (1 - cos(2u)) = 1/2*1/w*(w*pi - w*b - 1/2*sin(2*w*pi) + 1/2*sin(2*w*b))

so at that stage the omega doesn’t allow to turn 1/2*sin(2*w*pi) to zero. Or maybe it does, and I just misunderstand the nature of angular frequency
 

WBahn

Joined Mar 31, 2012
30,290
When you change variables, you have to change limits accordingly.

x = wt => t = x/w

dx = w·dt

x = ß => t = ß/w
x = π => π/w

So you have

\(
U \; = \; \int_{\frac{\beta}{\omega}}^{\frac{\pi}{\omega}} \hat{u} \cdot \omega \; sin^2 \left( \omega t\right) dt
\)
 
Last edited:

Thread Starter

Henry4321

Joined May 6, 2024
9
When you change variables, you have to change limits accordingly.

x = wt => t = x/w

dx = w·dt

x = ß => t = ß/w
x = π => π/w

So you have

\(
U \; = \; \int_{\frac{\beta}{\omega}}^{\frac{\pi}{\omega}} \hat{u} \; sin^2 \left( \omega t\right) dt
\)
Ohhh, I finally see, thank you so much for explaining!
 
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