Nyquist Rate Calculation for Modulated Signals

Understanding the Nyquist rate is crucial in digital signal processing, particularly for ensuring accurate reconstruction of a signal from its samples. The Nyquist rate is defined as twice the maximum frequency component present in a signal.

Nyquist Rate for Signal x(t)

We are given that the signal \(x(t)\) has a Nyquist rate of \(\omega_0\). Based on the definition of the Nyquist rate, if the Nyquist rate is \(2W\), where \(W\) is the maximum frequency component, then for \(x(t)\):

• Nyquist rate for \(x(t)\) = \(\omega_0\)
• This implies that the maximum frequency component present in \(x(t)\) is \(W_x = \frac{\omega_0}{2}\).
• Therefore, the spectrum of \(x(t)\), denoted by \(X(j\omega)\), extends from \(-\frac{\omega_0}{2}\) to \(\frac{\omega_0}{2}\). We can represent its bandwidth as \(B_x = \frac{\omega_0}{2}\).

Analyzing the Signal y(t)

The signal \(y(t)\) is given by the expression: $$y(t) = x(t)\cos(\omega_0t)$$ This represents a multiplication in the time domain. According to the Fourier Transform properties, multiplication in the time domain corresponds to convolution in the frequency domain. Specifically, if \(y(t) = x_1(t)x_2(t)\), then its Fourier Transform \(Y(j\omega)\) is given by: $$Y(j\omega) = \frac{1}{2\pi} [X_1(j\omega) * X_2(j\omega)]$$ where '\(*\)' denotes convolution.

Fourier Transform of Cosine Function

First, let's find the Fourier Transform of the cosine term \(\cos(\omega_0t)\). We know that: $$\cos(\omega_0t) = \frac{e^{j\omega_0t} + e^{-j\omega_0t}}{2}$$ The Fourier Transform of \(e^{j\omega_0t}\) is \(2\pi\delta(\omega - \omega_0)\), and the Fourier Transform of \(e^{-j\omega_0t}\) is \(2\pi\delta(\omega + \omega_0)\). Therefore, the Fourier Transform of \(\cos(\omega_0t)\) is: $$F[\cos(\omega_0t)] = \frac{1}{2} [2\pi\delta(\omega - \omega_0) + 2\pi\delta(\omega + \omega_0)] = \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]$$

Spectrum of y(t)

Now, let's apply the convolution property to find \(Y(j\omega)\): $$Y(j\omega) = \frac{1}{2\pi} [X(j\omega) * \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]]$$ $$Y(j\omega) = \frac{1}{2} [X(j\omega) * \delta(\omega - \omega_0) + X(j\omega) * \delta(\omega + \omega_0)]$$ When a signal's spectrum is convolved with a Dirac delta function \(\delta(\omega - a)\), the spectrum is shifted by \(a\). So, \(X(j\omega) * \delta(\omega - \omega_0) = X(j(\omega - \omega_0))\) and \(X(j\omega) * \delta(\omega + \omega_0) = X(j(\omega + \omega_0))\). Therefore: $$Y(j\omega) = \frac{1}{2} [X(j(\omega - \omega_0)) + X(j(\omega + \omega_0))]$$

Determining Maximum Frequency of y(t)

We know that the spectrum \(X(j\omega)\) of \(x(t)\) exists from \(-\frac{\omega_0}{2}\) to \(\frac{\omega_0}{2}\). Let's analyze the shifted components:

• The term \(X(j(\omega - \omega_0))\) represents \(X(j\omega)\) shifted to the right by \(\omega_0\). Its frequency range will be from \((-\frac{\omega_0}{2} + \omega_0)\) to \((\frac{\omega_0}{2} + \omega_0)\), which simplifies to \(\frac{\omega_0}{2}\) to \(\frac{3\omega_0}{2}\).
• The term \(X(j(\omega + \omega_0))\) represents \(X(j\omega)\) shifted to the left by \(\omega_0\). Its frequency range will be from \((-\frac{\omega_0}{2} - \omega_0)\) to \((\frac{\omega_0}{2} - \omega_0)\), which simplifies to \(-\frac{3\omega_0}{2}\) to \(-\frac{\omega_0}{2}\).

The spectrum \(Y(j\omega)\) is the sum of these two shifted spectra. The combined frequency range of \(Y(j\omega)\) will span from the lowest frequency of the left-shifted spectrum to the highest frequency of the right-shifted spectrum.

Thus, the overall frequency range for \(Y(j\omega)\) is from \(-\frac{3\omega_0}{2}\) to \(\frac{3\omega_0}{2}\).

The maximum frequency component in \(y(t)\), denoted as \(W_y\), is therefore \(\frac{3\omega_0}{2}\).

Calculating Nyquist Rate for y(t)

The Nyquist rate for \(y(t)\) is twice its maximum frequency component \(W_y\).

Therefore, the Nyquist rate for \(y(t)\) is \(3\omega_0\).