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

Question

A continuous time periodic signal $ x(t) $ is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If $ T $ is the period of $ x(t) $, then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]

Answer 1

Step 1: Write $y(t)=x(2t+5)$ explicitly: $y(t)=\cos(12\pi t+30\pi)+\sin(16\pi t+40\pi)$, so its angular frequencies are $12\pi$ and $16\pi$ rad/s.
Step 2: Convert to Hz: $6$ Hz and $8$ Hz; the highest is $8$ Hz.
Step 3: Apply the Nyquist rate: $f_s = 2\times 8$.
\[ \boxed{f_s = 16 \text{ samples/s}} \]

Answer 2

A different way to close this out is to work backward from each candidate sampling rate and see what "highest frequency in $y(t)$" it would imply, then check that figure against the time-scaling property $y(t)=x(at)$ which scales every frequency in $x(t)$ by the factor $a$.

Here $x(t)$ has frequencies $3$ Hz and $4$ Hz, and the scaling factor from $y(t)=x(2t+5)$ is $a=2$ (the added constant $5$ inside the argument shifts phase only, it does not rescale frequency). So $y(t)$ must contain frequencies $2\times3=6$ Hz and $2\times4=8$ Hz, with $8$ Hz the highest.

8 samples/s: Implies a highest frequency of $4$ Hz, but scaling $x(t)$'s highest frequency (4 Hz) by $a=2$ gives $8$ Hz, not $4$ Hz, so this rate is too low.
12 samples/s: Implies a highest frequency of $6$ Hz, which is the scaled version of $x(t)$'s 3 Hz component, not its highest (4 Hz) component, so this rate is also too low.
16 samples/s: Implies a highest frequency of exactly $8$ Hz, matching the scaled 4 Hz component of $x(t)$, which is correct.
32 samples/s: Implies a highest frequency of $16$ Hz, double what the scaling property actually produces, over-specifying the requirement.

Therefore, the correct answer is 16.

Answer 3

A different way to close this out is to work backward from each candidate sampling rate and see what "highest frequency in $y(t)$" it would imply, then check that figure against the time-scaling property $y(t)=x(at)$ which scales every frequency in $x(t)$ by the factor $a$.

Here $x(t)$ has frequencies $3$ Hz and $4$ Hz, and the scaling factor from $y(t)=x(2t+5)$ is $a=2$ (the added constant $5$ inside the argument shifts phase only, it does not rescale frequency). So $y(t)$ must contain frequencies $2\times3=6$ Hz and $2\times4=8$ Hz, with $8$ Hz the highest.

8 samples/s: Implies a highest frequency of $4$ Hz, but scaling $x(t)$'s highest frequency (4 Hz) by $a=2$ gives $8$ Hz, not $4$ Hz, so this rate is too low.
12 samples/s: Implies a highest frequency of $6$ Hz, which is the scaled version of $x(t)$'s 3 Hz component, not its highest (4 Hz) component, so this rate is also too low.
16 samples/s: Implies a highest frequency of exactly $8$ Hz, matching the scaled 4 Hz component of $x(t)$, which is correct.
32 samples/s: Implies a highest frequency of $16$ Hz, double what the scaling property actually produces, over-specifying the requirement.

Therefore, the correct answer is 16.

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

Show Hint

The Correct Option is C

Approach Solution - 1

Approach Solution -2

Top Questions on Digital Signal Processing

Questions Asked in OJEE exam