Consider the signal x(t) = cos(6πt) + sin(8πt), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t) = x(2t + 5) is

Understanding the Nyquist Sampling Rate for Signals

The question asks for the Nyquist sampling rate required for a transformed signal y(t), which is derived from an original signal x(t). The original signal is given as x(t) = cos(6πt) + sin(8πt), and the transformed signal is y(t) = x(2t + 5).

Analyzing the Original Signal x(t)

First, let's determine the highest frequency component present in the original signal x(t).

The signal x(t) is composed of two parts:

- cos(6πt): This is a cosine wave. The general form is cos(ωt), where ω is the angular frequency in radians per second. Here, the angular frequency is ${\omega }_1=6\pi$ rad/s. The frequency in Hertz (Hz) is calculated as $f=\frac{\omega }{2\pi }$. So, for this component, the frequency is $f_1=\frac{6\pi }{2\pi }=3$ Hz.
- sin(8πt): This is a sine wave. The general form is sin(ωt). Here, the angular frequency is ${\omega }_2=8\pi$ rad/s. The frequency in Hz is $f_2=\frac{8\pi }{2\pi }=4$ Hz.

The highest frequency present in the signal x(t) is the maximum of the frequencies of its components. Therefore, the maximum frequency for x(t) is $f_{max,x}=\max (f_1,f_2)=\max (3\text{ Hz},4\text{ Hz})=4$ Hz.

Analyzing the Transformed Signal y(t)

Next, we need to find the highest frequency component in the transformed signal y(t) = x(2t + 5).

The transformation involves time scaling (multiplying t by 2) and time shifting (adding 5 to t). A time scaling operation $t\to at$ scales the frequencies by a factor of $|a|$. A time shift $t\to t+b$ does not change the frequencies present in the signal, only the phase.

In the transformation y(t) = x(2t + 5), we have $a=2$. The time shift is 5.

The highest frequency in y(t) will be $|a|$ times the highest frequency in x(t).

So, the maximum frequency for y(t) is $f_{max,y}=|a|\times f_{max,x}=|2|\times 4\text{ Hz}=8$ Hz.

We can verify this by substituting 2t + 5 into x(t):

y(t) = cos(6π(2t + 5)) + sin(8π(2t + 5)) y(t) = cos(12πt + 30π) + sin(16πt + 40π)

Since $30\pi$ and $40\pi$ are multiples of $2\pi$, they represent full cycles and do not change the fundamental frequencies.

The frequencies in y(t) are:

- For the cosine term: ${\omega }_1^{\prime }=12\pi$ rad/s $⟹f_1^{\prime }=\frac{12\pi }{2\pi }=6$ Hz.
- For the sine term: ${\omega }_2^{\prime }=16\pi$ rad/s $⟹f_2^{\prime }=\frac{16\pi }{2\pi }=8$ Hz.

The highest frequency in y(t) is indeed $f_{max,y}=\max (6\text{ Hz},8\text{ Hz})=8$ Hz.

Calculating the Nyquist Sampling Rate for y(t)

The Nyquist sampling theorem states that to perfectly reconstruct a signal containing frequencies up to $f_{max}$, the sampling rate $f_s$ must be at least twice the maximum frequency. This minimum sampling rate is called the Nyquist rate.

Nyquist Rate $=2\times f_{max,y}$

Using the maximum frequency we found for y(t):

Nyquist Rate $=2\times 8$ Hz

Nyquist Rate $=16$ samples/second.

Therefore, the Nyquist sampling rate required for the signal y(t) is 16 samples/second.