Question:
An analog speech signal contains signals from \(200\) Hz to \(2400\) Hz. It is sampled at \(6\) kHz and quantized with \(512\) levels for pulse code modulation (PCM). The bit rate will be _____ kbps.
An analog speech signal contains signals from \(200\) Hz to \(2400\) Hz. It is sampled at \(6\) kHz and quantized with \(512\) levels for pulse code modulation (PCM). The bit rate will be _____ kbps.
Show Hint
Find the number of bits needed to represent \(512\) quantization levels, then multiply by the sampling rate given in the question.
Updated On: Jul 22, 2026
- \(16\)
- \(48\)
- \(19.2\)
- \(54\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Recall the PCM bit rate formula.
For pulse code modulation, the bit rate is the sampling rate multiplied by the number of bits used to code each sample.
\[ R_b = f_s \times n \]
where \(f_s\) is the sampling frequency and \(n\) is the number of bits per sample.
Step 2: Find the number of bits per sample from the number of levels.
With \(512\) quantization levels, each sample needs \(n\) bits such that \(2^n = 512\). Since \(2^9=512\), \(n=9\) bits per sample.
Step 3: Use the sampling rate that is actually given.
The question states the sampling frequency directly as \(f_s = 6\) kHz, so we use this value as it stands rather than recomputing a Nyquist rate from the \(200\) Hz to \(2400\) Hz signal band (that band information is only there to describe the signal, the sampling rate to use is the stated \(6\) kHz).
Step 4: Compute the bit rate.
\[ R_b = 6 \times 9 = 54 \text{ kbps} \]
Step 5: Why the other options are wrong.
Option (B) \(48\) comes from miscounting the bits per sample as \(8\) instead of \(9\) (mistaking \(2^8=256\) for the level count instead of \(2^9=512\)), giving \(6\times8=48\). Option (C) \(19.2\) comes from wrongly using the upper signal frequency, \(2400\) Hz, as the sampling rate together with the same wrong \(8\)-bit count, giving \(2.4\times8=19.2\). Option (A) \(16\) comes from a similar mix-up, using a rounded \(2\) kHz bandwidth figure with \(8\) bits, giving \(2\times8=16\). None of these use the correct pairing of the stated \(6\) kHz sampling rate with the correct \(9\)-bit count.
Final Answer:
The bit rate is \(54\) kbps. \[ \boxed{R_b = 54 \text{ kbps}} \]
For pulse code modulation, the bit rate is the sampling rate multiplied by the number of bits used to code each sample.
\[ R_b = f_s \times n \]
where \(f_s\) is the sampling frequency and \(n\) is the number of bits per sample.
Step 2: Find the number of bits per sample from the number of levels.
With \(512\) quantization levels, each sample needs \(n\) bits such that \(2^n = 512\). Since \(2^9=512\), \(n=9\) bits per sample.
Step 3: Use the sampling rate that is actually given.
The question states the sampling frequency directly as \(f_s = 6\) kHz, so we use this value as it stands rather than recomputing a Nyquist rate from the \(200\) Hz to \(2400\) Hz signal band (that band information is only there to describe the signal, the sampling rate to use is the stated \(6\) kHz).
Step 4: Compute the bit rate.
\[ R_b = 6 \times 9 = 54 \text{ kbps} \]
Step 5: Why the other options are wrong.
Option (B) \(48\) comes from miscounting the bits per sample as \(8\) instead of \(9\) (mistaking \(2^8=256\) for the level count instead of \(2^9=512\)), giving \(6\times8=48\). Option (C) \(19.2\) comes from wrongly using the upper signal frequency, \(2400\) Hz, as the sampling rate together with the same wrong \(8\)-bit count, giving \(2.4\times8=19.2\). Option (A) \(16\) comes from a similar mix-up, using a rounded \(2\) kHz bandwidth figure with \(8\) bits, giving \(2\times8=16\). None of these use the correct pairing of the stated \(6\) kHz sampling rate with the correct \(9\)-bit count.
Final Answer:
The bit rate is \(54\) kbps. \[ \boxed{R_b = 54 \text{ kbps}} \]
Was this answer helpful?
0
0
Top GATE IN Signals and Systems Questions
- The sinusoidal T.F. Corresponding to the polar plot for T>0.
- The input \( x(t) \) to a system is related to its output \( y(t) \) as
\[ \frac{d y(t)}{dt} + y(t) = 3x(t - 3)u(t - 3) \] \text{Here \( u(t) \) represents a unit-step function.}
\text{The transfer function of this system is _________}
- The output of a system \( y(t) \) is related to its input \( x(t) \) according to the relation \( y(t) = x(t) \sin(2\pi t) \). This system is _________
- A signal \( x(t) \) is band-limited between 100 Hz and 200 Hz. A signal \( y(t) \) is related to \( x(t) \) as follows:
y(t) = x(2t - 5)
The statement that is always true is_________
- Consider the transfer function \[ H_c(s) = \frac{1}{(s + 1)(s + 3)} \] Bilinear transformation with a sampling period of 0.1 s is employed to obtain the discrete-time transfer function \( H_d(z) \). Then \( H_d(z) \) is _________
View More Questions
Top GATE IN Questions
- A 3- phase squirrel cage induction motor has a starting torque of 100% of full load torque and maximum torque of 300% of full load for torque neglect stator impedances the slip at maximum torque is _____%
- A 50 KVA 1- phase transformer has a full load efficiency of 95% at unity power factor and core loss is a 0 Watt. find the efficiency at 75% of full load and 0.8 power factor lag
Find dV/dt
- Consider a unity bad back system with open loop Pole zero location given in circuit. it's R.L Cut the imaginary axis at \(s=jα (α>0)\)what is the value of \(α\).
- Two Inductors are connected in series adding condition resultant is 500 MH & series opposition is 300 MH. Both inductors are equal then find coefficient of coupling?
View More Questions