0
\$\begingroup\$

Say we have a simple rl circuit with R = 0.5 ohms, L = 0.1 H and the sinusoidal input of : $$ V_s = 13800√2sin(120πt+β)$$

Then we were tasked to:

enter image description here

I know the form of the solution of i(t) would look something like: $$ i(t) = natural + forced $$ $$ i(t) = I_ne^{-t/tau} + I_msin(ωt + β) $$

So the natural response would be the decaying exponential right? But i'm not sure how would it become zero. I was thinking of looking for beta such that the whole term inside of sin would go to zero, but then that would also make the forced response zero. What am i missing?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Why you assume amplitude of sinusoidal camponent (forced response ) and initial value of natural response equal? Thats your mistake , 1st term (natural response) will depend on B(beta ) and you can make it zero by choosing appropriate B \$\endgroup\$
    – user215805
    Oct 17, 2020 at 12:22
  • \$\begingroup\$ @user215805 I apologize, that was supposed to be n. I edited the post. And thanks. I already figured it out! I just missed something. \$\endgroup\$
    – user263783
    Oct 17, 2020 at 12:27
  • \$\begingroup\$ eeeguide.com/sinusoidal-response-of-rl-circuit/…. You can check this site and hopefully you'll understand \$\endgroup\$
    – user215805
    Oct 17, 2020 at 13:12

2 Answers 2

1
\$\begingroup\$

My interpretation: Assuming \$i(0) = 0 \space A\$ and $$ V_s(t) = V_m\sin(\omega t+\beta) $$ It's reasonable to conceive the response \$i(t)\$ as formed by the two components. The Steady state response and the Transient one:

$$ i(t) = K_1\sin(\omega t+\gamma) \space + K_2e^{-\frac{R}{L}t} $$

If there is not a transient reponse, then \$K_2=0\$

In this case

$$ i(0) = K_1\sin(\gamma) = 0 $$

Since \$K_1\$ cannot be zero, then \$\gamma = 0^\circ\$ or \$\gamma = 180^\circ \$

Choosing the first

$$ i(t) = K_1\sin(\omega t) $$

The differential equation representing the circuit is

$$ \frac{d}{dt}i(t) + \frac{R}{L}i(t) = \frac{V_m}{L}\sin(\omega t+\beta) $$

Replacing the expression by \$i(t)\$

$$ K_1\omega \cos(\omega t) + \frac{R}{L}K_1 \sin(\omega t) = \frac{V_m}{L} \sin(\omega t + \beta) $$

Expanding \$\sin(\omega t + \beta)\$ in the right side:

$$ K_1\omega \cos(\omega t) + \frac{R}{L}K_1 \sin(\omega t) = \frac{V_m}{L} \sin(\omega t)\cos(\beta) + \frac{V_m}{L} \sin(\beta)\cos(\omega t) $$

Equating the coefficients in both sides, leads to:

$$ \left\{\begin{matrix} K_1\omega = \frac{V_m}{L}\sin(\beta) & (1)\\ \frac{R}{L}K_1 = \frac{V_m}{L}\cos(\beta) & (2) \end{matrix}\right. $$

As \$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}\$, dividing (1) by (2):

$$ \tan(\beta) = \frac{\omega L}{R} $$

So, we get the condition required for the transient response to be null

$$ \begin{equation} \boxed{\beta = \arctan(\frac{\omega L}{R})} \end{equation} $$

In this case:

$$ \beta \approx 89.24^\circ $$

In other hand, squaring and summing (1) and (2):

$$ K_1^2[\omega^2 + (\frac{R}{L})^2] = (\frac{V_m}{L})^2[\sin^2(\beta) + \cos^2(\beta)]$$

As \$ \sin^2(\beta) + \cos^2(\beta) = 1 \$

$$ K_1 = \frac{V_m}{\sqrt{R^2 + \omega^2 L^2}} $$

In this case:

$$ K_1 \approx 517.63 $$

Finally

$$ i(t) = 517.63\sin(120 \pi t) $$

\$\endgroup\$
0
\$\begingroup\$

Using Laplace transform, we know that:

$$\text{I}_\text{in}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{v}_\text{in}\left(\text{s}\right)}{\text{R}+\text{sL}}\right]_{\left(t\right)}\tag1$$

With the convolution property of the Laplace transform we can write:

$$\text{I}_\text{in}\left(t\right)=\int_0^t\mathcal{L}_\text{s}^{-1}\left[\text{v}_\text{in}\left(\text{s}\right)\right]_{\left(t-\tau\right)}\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{R}+\text{sL}}\right]_{\left(\tau\right)}\space\text{d}\tau\tag2$$

Using the table of selected Laplace transforms, we can see that:

$$\text{I}_\text{in}\left(t\right)=\int_0^t\text{V}_\text{in}\left(t-\tau\right)\cdot\frac{\exp\left(-\frac{\text{R}}{\text{L}}\cdot\tau\right)}{\text{L}}\space\text{d}\tau\tag3$$

So, in your case we get:

$$\text{I}_\text{in}\left(t\right)=\int_0^t13800\sqrt{2}\sin\left(120\pi\left(t-\tau\right)+\beta\right)\cdot\frac{\exp\left(-\frac{\frac{1}{2}}{\frac{1}{10}}\cdot\tau\right)}{\frac{1}{10}}\space\text{d}\tau=$$ $$138000\sqrt{2}\int_0^t\sin\left(120\pi\left(t-\tau\right)+\beta\right)\exp\left(-5\tau\right)\space\text{d}\tau=$$ $$\displaystyle\frac{138000 \sqrt{2} \left(\sin (\beta +120 \pi t)-24 \pi \cos (\beta +120 \pi t)+e^{-5 t} (24 \pi \cos (\beta )-\sin (\beta ))\right)}{5+2880 \pi ^2}\tag4$$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.