GATE Electrical Engineering (EE) Test: Sampling Theorem and Nyquist Rate

Nyquist Sampling Theorem: 
A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.
f_s ≥ 2f_m
Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.
Calculation:
Given that,
f_m = 3.1 kHz 
⇒ f_s ≥ 2f_m
⇒ f_s ≥ 2 × 3.1 = 6.4 kHz

Concept:
The sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate f_s which is greater than twice the maximum frequency W in the modulating signal."
If the sampled frequency is less than the Nyquist frequency, overlapping of lower and upper sidebands known as aliasing takes place.
The main reason for aliasing is under-sampling, i.e. sampling the signal at a frequency less than twice the Nyquist rate.
f_s < 2f_m
f_s = Sampling frequency
f_m = Modulating frequency
Application:
The minimum sampling rate to avoid aliasing will be:
f_s = 2 × 2f_m = 2 × 4 kHz
f_s = 8 kHz
∴ For f_s ≥ 8 kHz, the sampling frequency will satisfy the Nyquist rate and no aliasing will take place.
And for both 7 kHz and 5 kHz sampling rates, the Aliasing effect will be seen in the reconstructed signal as the sampling frequency is less than the Nyquist Rate.

Aliasing is explained with the help of the spectrum as shown:

Concept:
Multiplication in the time domain is the convolution in the frequency domain and the maximum frequency of convolution of the signal is the sum of individual frequencies, i.e.
If z(t) = x(t) y(t)
The maximum frequency of Z(ω) = ω_x + ω_y
where ω_x is the maximum frequency of signal x(t)
ω_y is the maximum frequency of signal y(t)
Application:
The maximum frequency present in x(t) is:

Similarly, the maximum frequency present in y(t) is:

Now, the maximum frequency present in z(t) = x(t) y(t) will be:
f_z = f_x + f_y
f_z = 250 kHz
Nyquist sampling frequency will now be:
f_N = 2 × f_z
f_N = 500 kHz

Given,
x(t) = cos(6πt) + sin(8πt)
The bandwidth of the signal will be:

Now, y(t) = x(2t + 5) can be written as:

Taking the Fourier transform, we get:

Thus, the frequency spectrum of X(jω)  expands by 2.
This will make the highest frequency component of Y(jω) as:
2 × 4 = 8 Hz 
Hence, the Nyquist rate will be:
f_s = 2 × 8 = 16 samples/sec

Calculation:
f_s = Sampling Frequency = 56 MHz
The frequencies present at the input are:
f_m1 = 10 MHz, f_m2 = 50 MHz and, f_m3 = 70 MHz
Once sampled, the frequency content at the output will be: f_m ± f_s
For, f_m1 = 10 MHz, the output will contain frequencies of,
= f_m1, f_m1 ±, f_m1 ± 2f_s ….
= 10 MHz, 66 MHz, 122 MHz
For, f_m2 = 50 MHz, the output will contain frequencies of
= f_m2, f_m2 ± f_s, f_m2 ± 2f_s ….
= 50 MHz, 6 MHz, 106 MHz, 162 MHz
For, f_m3 = 70 MHz, the output will contain frequencies of
= f_m3, f_m3 ± f_s, f_m3 ± 2f_s ….
= 70 MHz, 14 MHz, 182 MHz ….
When these frequencies are passed through a low pass filter with a cutoff frequency of 15 MHz, the only remaining frequencies at its output will be 6MHz, 10 MHz and 14 MHz.

Period of sampling train, T_s = 20 ms

If frequency of x(t) is fx, then after sampling the signal, the sampled signal has the frequency,
f_s - f_x = 50 - f_x and f_s + f_x = 50 + f_s
Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.
f_c = 25 Hz
Now, the o/p of filter carried a single frequency component of 20 Hz
∴, only (f_s−f_x) component passes through the filter, ie.
f_s - f_x < 25
and f_s - f_x = 20
50 - f_x­ = 20
f_x = 50 - 20 = 30 Hz

Solution:

• Given:
  • Input signal: x(t) = 2 + 5sin(100πt).
  • Sampling frequency: f_s = 400 Hz.
  • Transfer function: Y(z)/X(z) = (1/N) * ((1 - z^-N) / (1 - z^-1)), where N = number of samples per cycle.
• Step 1: Determine the frequency of the input signal
  • The frequency of the sinusoidal component in x(t) is f = 50 Hz (from sin(100πt), where ω = 2πf).
• Step 2: Calculate the number of samples per cycle (N)
  • N = f_s / f.
  • Substitute values: N = 400 / 50 = 8.
• Step 3: Analyze the transfer function
  • The transfer function is a moving average system over N samples.
  • For a steady-state sinusoidal input, the moving average system passes the DC component (mean) and filters out the sinusoidal component.
• Step 4: Compute the steady-state output
  • The DC component of the input signal x(t) is 2 (the constant term).
  • The sinusoidal component is filtered out by the system.
  • Thus, the steady-state output y[n] is 2.
• Final Answer: The steady-state output y[n] is 2.

Concept:
Aliased frequency component is given by:
fa = |fs – fm|
f_s = sampling frequency
f_m = message signal frequency

Calculation:
For a sampling rate of 140 Hz, since f_s > f_m:
Aliased Frequency = f_s - f_m,
= 140 – 100 = 40 Hz
For a sampling rate of 90 Hz, since f_s < f_m:
Aliased Frequency = f_m – f_s,
= 100 – 90 = 10 Hz
For a sampling rate of 30 Hz, since f_s < f_m:
Aliased Frequency = f_m – f_s
= 100 – 30 = 70 Hz

Aliasing is explained with the help of the spectrum as shown:

Concept:
Nyquist rate: The minimum sampling rate is often called the Nyquist rate. The Nyquist sampling rate is two times the highest frequency of the input or message signal.
f_s = 2f_m
Where,
f_s is the minimum sampling frequency or Nyquist rate
f_m is the highest frequency of the input or message signal.
Calculation:
sinc(700t) + sinc(500t)

f₁ = 700π/2π = 350 Hz
f₂ = 500π/2π = 250 Hz
f_m = max (f_1, f₂) = max (350, 250) = 350 Hz
Sampling frequency,
f_s = 2f_m = 2 × 350 = 700 Hz
Sampling time period = 1/700 sec

The maximum frequency of the band-limited signal.
f_m = 5 kHz
According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist  frequency which is given as
f_N = 2 f_m = 2 × 5 = 10 kHz
So the sampling frequency f_s ≥ f_N
f_s ≥ 10 kHz
Only the option (A) does not satisfy the condition.
∴ 5 kHz is not a valid sampling frequency.