I am reading Proakis book "DSP using Matlab", 3rd edition.

I am reading chapter 8, section 8.3, p. 402, and I am confused regarding the equation of poles (roots of denominator of system function) eq 8.47.

How this equation has been derived from eq 8.46 especially highlighted line how the right most side of eq 8.47 $Ω_ce^{j(\pi/2N)(2k+N+1)}$ has been derived from left side $(-1)^{1/2N} jΩ_c$.

enter image description here


Assuming that you understand the left-hand side of Eq. $(8.47)$, for understanding the right-hand side you need to know that $-1=e^{j\pi}$, and that $e^{j2k\pi}=1$. So in order to obtain all $2N$ roots of $(-1)^{\frac{1}{2N}}$ you rewrite $-1$ as

$$-1=e^{j\pi}e^{j2\pi k}\tag{1}$$

from which you get

$$(-1)^{\frac{1}{2N}}=e^{j\frac{\pi}{2N}}e^{j\frac{2\pi k}{2N}}=e^{j\frac{\pi}{2N}(1+2k)},\qquad k=0,1,\ldots,2N-1\tag{2}$$

And since $j=e^{j\pi/2}$, the final result is

$$\Omega_cj(-1)^{\frac{1}{2N}}=\Omega_ce^{j\pi/2}e^{j\frac{\pi}{2N}(1+2k)}=\Omega_ce^{j\frac{\pi}{2N}(1+2k+N)},\qquad k=0,1,\ldots,2N-1\tag{3}$$

If this is all new to you, you should really read up on complex numbers. They are fundamental in understanding many concepts in signal processing.

PS: There is a typo in the equations in the book. The numerator on the right-hand of Eq. $(8.46)$ should have $\Omega_c$ instead of $\Omega$. Also, the left-hand side of Eq. $(8.47)$ should have $\Omega_c$ instead of $\Omega$.
  • $\begingroup$ Thanks alot for your answer $\endgroup$ – engr Jul 14 '20 at 6:37
  • $\begingroup$ The center side of eq 8.47,contains Ωj*(−1)^1/2N, but you wrote in last line of your answer in left most side Ωcj*(−1)^1/2N. You wrote term "Ωc",but book contains "Ω" term, please look carefully $\endgroup$ – engr Jul 14 '20 at 6:42
  • 1
    $\begingroup$ @engr: I know, that's clearly a typo in the book. $\endgroup$ – Matt L. Jul 14 '20 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.