# Find the phase angle of the of the sinusoidal input that will make the natural response zero

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:

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?

• 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 Oct 17, 2020 at 12:22
• @user215805 I apologize, that was supposed to be n. I edited the post. And thanks. I already figured it out! I just missed something.
– user263783
Oct 17, 2020 at 12:27
• eeeguide.com/sinusoidal-response-of-rl-circuit/…. You can check this site and hopefully you'll understand Oct 17, 2020 at 13:12

## 2 Answers

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

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

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

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

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:

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