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To solve this problem, we need to analyze the discrete-time signal given by

\[
x(n) = 10 \cos \left( 0.24 \pi n - \frac{\pi}{4} \right)
\]

and determine the principle alias frequency and the next two alias frequencies after sampling at a rate of \( f_s = 1000 \) samples/sec. Here’s the step-by-step solution:

### Step 1: Determine the Original Frequency
The argument of the cosine function contains a term \( 0.24 \pi n \), indicating that the angular frequency \( \omega \) of the signal is:

\[
\omega = 0.24 \pi
\]

To convert this angular frequency to a frequency in Hertz (Hz), we can use the formula:

\[
f = \frac{\omega}{2\pi} f_s
\]

Plugging in the values:

\[
f = \frac{0.24 \pi}{2 \pi} \times 1000 = 0.12 \times 1000 = 120 \text{ Hz}
\]

So, the original frequency of the signal is 120 Hz.

### Step 2: Find Aliased Frequencies
Since the signal is sampled at \( f_s = 1000 \) samples/sec, aliasing will occur at frequencies that are spaced by multiples of the sampling frequency. The aliases will be at frequencies:

\[
f_{\text{alias}} = |f \pm k f_s|
\]

where \( k \) is an integer. We want to find the first three positive aliases in ascending order.

#### Principal Alias ( \( k = 0 \) ):
The principal alias is the original frequency itself:
\[
f_0 = 120 \text{ Hz}
\]

#### Second Alias ( \( k = 1 \) ):
For \( k = 1 \):
\[
f_{\text{alias}} = |120 - 1000| = 880 \text{ Hz}
\]

#### Third Alias ( \( k = 1 \) but with positive shift ):
\[
f_{\text{alias}} = |120 + 1000| = 1120 \text{ Hz}
\]

However, 1120 Hz exceeds the Nyquist frequency (500 Hz), so we reflect it back into the range [0, 500] by subtracting it from \( f_s \):
\[
f_{\text{alias}} = |1120 - 1000| = 120 \text{ Hz}
\]

Thus, the third alias does not add a new frequency, as it is the same as the principal alias.

### Final Answer
- **Principal alias ( \( f_0 \) ):** 120 Hz
- **Second alias:** 880 Hz
- **Third alias:** 120 Hz (repeated)

Let me know if you want more details or have questions about aliasing in sampled signals.

---

**Additional Questions:**
1. What is aliasing and why does it occur in sampled signals?
2. How does the sampling rate affect the frequency components of the sampled signal?
3. What is the Nyquist frequency and how is it related to aliasing?
4. How would the aliases change if the sampling rate were lower than 1000 samples/sec?
5. How can we determine if a signal is under-sampled?

**Tip:** Always check if your sampled frequency exceeds the Nyquist rate to avoid ambiguity in identifying aliases.

A discrete-time signal is given as x(n) = 10 cos(0.24πn − π/4). This sequence was obtained by sampling a signal above the Nyquist rate at fs = 1000 samples/sec. Enter the frequency of the principal alias and the frequencies of the next two aliases in ascending order and in Hz.

Learn to determine the principal alias frequency and the next two alias frequencies for a discrete-time signal sampled at 1000 samples/sec. This problem explores key concepts in signal processing, including aliasing, the Nyquist theorem, and frequency calculation.

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