Continuous time to Discrete time using periodic sampling

Question

From Schaum's outlines, Digital Signal Processing, 2nd Edition, 2012, page 114:

Consider the discrete time-sequence:

$$x[n] = cos(\frac{\pi}{8}n)$$

Find two different continuous time signals:

$$X_a(t) = cos(2 \pi f_0 t)$$

that would produce this sequence when sampled at a frequency of $$f_s = 10 KHz$$

So I start out with Ideal A/D converter relationship for converting from continuous to discrete time:

$$x[n]=X_a(t=nT_s)$$

$$x[n]= cos(2 \pi f_0 n T_s)$$

$$T_s = 1 / f_s $$

$$x[n]= cos(2 \pi n \frac{f_0}{f_s})$$

Since Cos is periodic we have:

$$x[n]= cos(\frac{2 \pi n f_0}{f_s} + 2 \pi k)$$

$$x[n]= cos(\frac{2 \pi n f_0}{f_s} + \frac{2 \pi k f_s}{f_s})$$

$$x[n]= cos(\frac{2 \pi n f_0 + 2 \pi k f_s}{f_s})$$

$$x[n]= cos(2 \pi \frac{n f_0 + k f_s}{f_s})$$

Here's where I have the problem. The book says at this point I should have this instead:

$$x[n]= cos(2 \pi \frac{f_0 + k f_s}{f_s} n)$$

and:

$$f = f_0 + k f_s $$

How did the book get that result instead?

Answer 1

The problem lies where you added $2\pi k$. Instead it should be $2\pi k n$. So that becomes $x[n]= cos(\frac{2 \pi n f_0}{f_s} + 2 \pi kn)$ which gives $x[n]= cos(2 \pi \frac{f_0 + k f_s}{f_s} n)$.

Answer 2

Since Cos is periodic we have:

$$x[n]= cos(\frac{2 \pi n f_0}{f_s} + 2 \pi k n)$$

$$x[n]= cos(2 \pi n \frac{f_0}{f_s} + k)$$

$$x[n]= cos(2 \pi n \frac{f_0}{f_s} + \frac{k}{f_s})$$

$$x[n]= cos(2 \pi (n \frac{1}{f_s}) (f_0 + k f_s))$$

$$x[n]= cos(2 \pi t (f_0 + k f_s))$$

$$f = f_0 + k f_s $$