A signal m(t) band-limited to 3 kHz is sampled at a rate

A signal m(t) band-limited to 3 kHz is sampled at a rate
33 1 %
3
higher than the Nyquist rate. The max acceptable error in the sample amplitude (max. quantization error) is 0.5% of peak amplitude V_p. The quantized samples are binary coded. Find out the minimum bandwidth required to transmit the encoded binary signal. If 24 such signals are TDM, determine the minimum bandwidth required to transmit the multiplexed signal—

We know that minimum bandwidth required to transmit 24 channel through TDM (Time Division Multiplexing)

= nN	f_s
	2

Where, N = no. of bits
n = no. of channel
f_s = sampling frequency
According to question,
f_s = 33% greater that the Nyquist rate

=		1 +	1		.2.f_m
			3

f_s =	4	×2×3kHz = 8kHz
	3

For the quantization step size δ the maximum quantization error is ± δ/2 and given as δ/2 = 0·5% of V_p
or

1		2V_p		=	0.5	.V_p
2		L			100

Where,

δ =	2V_p
	L

V_p = maximum amplitude
L = no. of level = 2ⁿ
N > 7 i.e. for binary coding we must take N = 8
Now,

B_{T_min} = N.n.	f_s
	2

=	8 × 24 × 8	kHz = 768 kHz
	2