Math: RLC natural response

[Gaktyt]

Hi,

I can't figure out the intermediary steps in the following calculation (example 8.11 Electric Circuits 10th Edition)


\$\alpha = 2800 Rad/Sec \quad and \quad \omega_{d} = 6800 Rad/Sec\$

The solution is given
\$i(t)=B_{1}e^{-\alpha t}cos(\omega_{d}t)+B_{2}e^{-\alpha t}sin(\omega_{d}t)\$

Since \$B_{1}=0\$ the cos part drops out and the solution becomes
\$i(t)=B_{2}e^{-\alpha t}sin(\omega_{d}t)\$

Therefore
\$\frac{d i}{dt} = 400 B_2e^{-2800t}(24cos(9600t)-7sin(9600t))\$

Herein lies my problem, how did they arrive at the "therefore" part? I can't figure the steps out, I assume the right hand side is the derivative of the solution, but when I try calculate the derivative I get:
\$\frac{d i}{dt} = B_2e^{-\alpha t}(\omega_{d} cos(\omega_{d} t)-\alpha sin(\omega_{d} t))\$
It looks a bit like above but the constants are not the same, and where did they get 400 from?

[Andy Watson]

 I suggest there was a typo. Try with \$\omega_d = 9600 Rad/Sec\$

[Gaktyt]

I suggest there was a typo. Try with \$\omega_d = 9600 Rad/Sec\$

Yes 9600 is the correct result when evaluating at t=0

\$\frac{d i}{dt} = 400 B_2e^{-2800t}(24cos(9600t)-7sin(9600t))\$
becomes
\$\frac{d i}{dt} = 9600 B_2\$

So I would say it's probably not a typo, it's the 10th edition surely someone would have noticed if it's a typo by now.

edit... wait a minute, when I evaluate my derived equation at t=0
\$\frac{d i}{dt} = B_2e^{-\alpha t}(\omega_{d} cos(\omega_{d} t)-\alpha sin(\omega_{d} t))\$
it also becomes
\$\frac{d i}{dt} = 9600 B_2\$

So I guess my question is more: what's the difference between theirs and my derivative? do they solve for complex numbers? and 24 and 7 is the imaginary part, but still where does the 400 come from?

[Andy Watson]

So I guess my question is more: what's the difference between theirs and my derivative? do they solve for complex numbers? and 24 and 7 is the imaginary part, but still where does the 400 come from?

I think you are over-thinking this part.
Starting with the derivative \$\frac{di}{dt} = B_2 e^{-\alpha t} \left[ \omega_d \cos (\omega_d t) - \alpha \sin(\omega_d t) \right]\$. Fill in \$\alpha = 2800\$ and \$\omega_d = 9600 \$, i.e.
\$\frac{di}{dt} = B_2 e^{-2800 t} \left[ 9600 \cos (9600  t) - 2800 \sin(9600 t) \right]\$. Now take 400 out.