0

I'm just having some trouble understanding a derivation in my notes. How did it go from that first line to the second?

\begin{align*} h_n&=\frac{Kj}{2r\sin(\omega_0)}\left(-(re^{j\omega_0})^{n+1}u_{n+1}+(re^{-j\omega_0})^{n+1}u_{n+1}\right) \\ &\overset{\text{how?}}{=}\begin{cases} \frac{r^nKj}{2\sin(\omega_0))}\left(e^{-j(n+1)\omega_0}-e^{j(n+1)\omega_0}\right),\quad&n\ge-1\\ 0,&n<-1 \end{cases} \end{align*}

I understand that a rule of exponents was used $a^x ⋅ b^x = (a ⋅ b)^x$ but wouldn't that mean the $r$ in the front of the second equation should also be to the power of $n+1$ and not just $n?$ Is there a mistake somewhere?

Adrian Keister
  • 10,099
  • 13
  • 30
  • 43
AlfroJang80
  • 253
  • There was formerly an $r$ on the denominator of the original expression which when cancelled will decrease the exponent of the $r$ in the numerator by one. – JMoravitz May 11 '18 at 16:05
  • 1
    I.e. you could have written an in-between step of $\dots = \dfrac{\color{red}{r^{n+1}}k_j}{2\color{red}{r}\sin(\omega_0)}(\cdots) = \dfrac{r^nk_j}{2\sin(\omega_0)}(\cdots)$ – JMoravitz May 11 '18 at 16:07
  • Ahh. That makes sense. Can't believe I didn't see that. Thank you so much!. – AlfroJang80 May 11 '18 at 16:10
  • Also, a warning: "I understand that a rule of exponents was used $a^x\cdot b^x=(a\cdot b)^x$" This "rule" only works in certain situations. Specifically when the exponent is an integer or when $a$ and $b$ are both positive reals. It can fail when both of those are false, for example with $-1=i\cdot i= (-1)^{\frac{1}{2}} \cdot(-1)^{\frac{1}{2}} \neq ((-1)\cdot (-1))^{\frac{1}{2}}=1$ – JMoravitz May 11 '18 at 16:12

0 Answers0