Nyquist Rate Sampling 63E813

1. **Problem Statement:** Solve question 1 part b from the Digital Signal Processing exam. Given the analog signal: $$x_a(t) = 3 \cos 2000\pi t + 5 \sin 6000\pi t + 10 \cos 14000\pi t$$ Tasks: - i. Find the Nyquist rate for this signal. - ii. Find the discrete-time signal obtained after sampling at sampling rate $$F_s = 9000$$ samples/s. --- 2. **Step 1: Find the Nyquist rate** - The Nyquist rate is twice the maximum frequency present in the analog signal. - The signal contains three frequency components: - $$3 \cos 2000\pi t$$ has frequency $$f_1 = \frac{2000\pi}{2\pi} = 1000$$ Hz - $$5 \sin 6000\pi t$$ has frequency $$f_2 = \frac{6000\pi}{2\pi} = 3000$$ Hz - $$10 \cos 14000\pi t$$ has frequency $$f_3 = \frac{14000\pi}{2\pi} = 7000$$ Hz - The maximum frequency is $$f_{max} = 7000$$ Hz. - Therefore, the Nyquist rate is: $$ F_N = 2 \times f_{max} = 2 \times 7000 = 14000 \text{ samples/s} $$ --- 3. **Step 2: Find the discrete-time signal after sampling at $$F_s = 9000$$ samples/s** - Sampling frequency $$F_s = 9000$$ samples/s is less than the Nyquist rate $$14000$$ samples/s. - The discrete-time signal $$x[n]$$ is obtained by sampling $$x_a(t)$$ at $$t = \frac{n}{F_s}$$: $$ x[n] = x_a\left(\frac{n}{F_s}\right) = 3 \cos \left(2000\pi \frac{n}{9000}\right) + 5 \sin \left(6000\pi \frac{n}{9000}\right) + 10 \cos \left(14000\pi \frac{n}{9000}\right) $$ - Simplify the arguments: $$ \begin{aligned} &3 \cos \left(\frac{2000\pi}{9000} n\right) = 3 \cos \left(\frac{2\pi}{9} n\right) \\ &5 \sin \left(\frac{6000\pi}{9000} n\right) = 5 \sin \left(\frac{2\pi}{3} n\right) \\ &10 \cos \left(\frac{14000\pi}{9000} n\right) = 10 \cos \left(\frac{14}{9} \pi n\right) \end{aligned} $$ - Note that $$\frac{14}{9} \pi = \pi + \frac{5}{9} \pi$$, so the last term can be rewritten using the identity: $$\cos(\pi + \theta) = -\cos \theta$$ - Thus: $$ 10 \cos \left(\pi + \frac{5}{9} \pi n\right) = -10 \cos \left(\frac{5}{9} \pi n\right) $$ - Therefore, the discrete-time signal is: $$ x[n] = 3 \cos \left(\frac{2\pi}{9} n\right) + 5 \sin \left(\frac{2\pi}{3} n\right) - 10 \cos \left(\frac{5\pi}{9} n\right) $$ --- **Final answers:** - i. Nyquist rate $$F_N = 14000$$ samples/s - ii. Discrete-time signal: $$ x[n] = 3 \cos \left(\frac{2\pi}{9} n\right) + 5 \sin \left(\frac{2\pi}{3} n\right) - 10 \cos \left(\frac{5\pi}{9} n\right) $$ --- **Explanation:** - The Nyquist rate ensures no aliasing if sampling frequency is at least twice the highest frequency. - Sampling below Nyquist rate causes aliasing, which is why the last term changes sign and frequency in the discrete-time signal.