# Exercises

# 2.5. Exercises#

Imagine you are given a sampled signal \(x[n]\) of \(N\) samples, taken at \(f_s = 8000\) Hz. If you derive a new signal \(y[n]\) by taking every other sample from \(x[n]\):

What is the sampling rate \(f_s\) of \(y[n]\)?

What is the sampling period \(t_s\) of \(y[n]\)?

How many samples will \(y[n]\) have?

Hint

Be sure to consider what happens when \(N\) is even or odd.

If the sampling rate is \(f_s = 100\) Hz, use the aliasing equation (2.3) to find two aliases for each of the following frequencies:

\(f = 100\) Hz

\(f = -25\) Hz

\(f = 210\) Hz

Imagine that you want to sample a continuous wave \(x(t) = \cos(2\pi \cdot f \cdot t)\) of some arbitrary (known) frequency \(f > 0\). Which sampling rates \(f_s\) would you need to use to produce the following discrete observations by sampling \(x(t)\)? Express your answers as a fraction of \(f\).

\(x[n] = 1, 1, 1, 1, \dots\),

\(x[n] = 1, 0, -1, 0, 1, 0, -1, 0, \dots\),

\(x[n] = 1, \sqrt{1/2}, 0, -\sqrt{1/2}, -1, \dots\).

Which (if any) of these sampling rates satisfy the conditions of the Nyquist-Shannon theorem?

Using the quantize function from this chapter, apply different levels of quantization (varying `n_bits`

from 1 to 16) to a recording of your choice. For each quantized signal, subtract it from the original signal and listen to the difference:

```
# We need to normalize the signal to cover the same range
# before and after quantization if we're going to compare them.
x_span = x.max() - x.min()
x_diff = x / x_span - quantize(x, n_bits) / 2**n_bits
```

What does it sound like for each quantization level?