Examples from chapter 3 - Stallings DCC, 8e
1. (ex3.1) T=17o C=(273+17)=290o K.
No = kT W/Hz,
k = 1.38e − 23 ≡ 10 log10 (1.38e − 23) = −228.6dBW
No = 10 log10 (k) + 10 log10 (290) = −228.6 + 24.6 = −204dBW .
2. (ex3.2) T = 294o K, B = 10 MHz, find the thermal noise at the receiver.
N = kT B, in dBW N = −228.6 + 10 log10 (294) + 10 log10 (1e7) = −228.6 + 24.7 + 70 = −133.9dBW .
3. (ex3.3) B = 3100Hz, C=Nyquist capacity=2B samples/sec=2B log2 (M ) bps with M = number of levels that
represent the signal, i.e. log2 (M ) is the number of bits needed to represent each sample.
C = 2 ∗ 3100 ∗ log2 (8) = 18600 bps.
4. (ex3.4) Channel spectrum between 4MHz and 3MHz, ⇒ B = 4 − 3 = 1M Hz and the SN RdB = 24dB ≡
1024/10 = 251.
Shannon capacity=B log2 (1 + SN R) = (1e6) ∗ log2 (1 + 251) ≈ 8e6bps
According to Nyquist, find the required voltage levels needed to represent the signal and stay within the channel
capacity calculated above?
C = 2B log2 (M ) ⇒ 8e6 = 2(1e6) log2 (M ) ⇒ log2 (M ) = 4 ⇒ M = 24 = 16 voltage levels.
5. (ex3.5) Let S is the signal power, Tb = R1 = time per bit and R is the bit rate then signal energy per bit=Eb =
S
STb = R
.
Eb
S
A better measure than S/N ratio is N
= S/R
No = kT R .
o
If Eb /No = 8.4dBW is required for some transmission criteria and if the effective noise temperature bandwidth
is 290o K (room temperature) and the data rate=2400 bps, what received signal level is required?
Eb
No
S
= NRo = kTSR
S
8.4dB = 10 log10 (x) ⇒ x = 100.84 = 6.9183 = 1.38e−23×(290)×2400
⇒ S = 6.9183 ∗ 1.38e − 23 ∗ 290 ∗ 2400 =
6.6449e − 017 and in dB, SdB = −161.7751 dBW.
6. (ex3.6) Find the minimum
Eb
No
required to achieve a spectral efficiency of 6bps/Hz.
C
Recall that C = B log2 (SN R + 1) ⇒ SN R = 2 B − 1
Eb
S×B
S
B
and N
= NoS×R = NoS×B
×B×R = N ×R = N × R
o
S
Substitute for N
and set C = R to get:
C
Eb
B
B − 1)
=
(2
No
C
For the question: B = 1, C = R = 6
6
Eb
63
1
1
No = 6 (2 − 1) = 6 = 10.5 ≡ 10 log10 (10.5) = 10.21dB
7. (P3.13)
(a) Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 X 500
picture elements (pixels), where each pixel can take on one of 32 intensity values. Assume that 30 pictures
are sent per second. (This digital source is roughly equivalent to broadcast TV standards that have been
adopted.) Find the source rate R (bps).
Pixels/sec=(480 × 500) × 30 = 7.2e6
Each pixel (sample) can take one of 32 values (voltage levels), hence it needs log2 (32) = 5 bits/pixel
(bits/sample)
R = 7.2e6 × 5 = 36e6 bps
(b) Assume that the TV picture is to be transmitted over a channel with 4.5 MHz bandwidth and a 35 dB
signal-to-noise ratio. Find the capacity of the channel (bps).
35
B = 4.5e6 Hz, SN RdB = 35 = 10 log10 (SN R) ⇒ SN R = 10 10 = 3162
C = B log2 (1 + SN R) = (4.5e6) log2 (3163) = (4.5e6) × (11.6272) = 5.2322e + 007 bps
(c) Discuss how the parameters given in part (a) could be modified to allow transmission of color TV signals
without increasing the required value for R.
Can use RGB (3 colors) and only 10 intensity levels to get 30 (voltage) levels per pixel which keeps the
overall rate within the permitted values.
8. (P3.15) What is the channel capacity for a teleprinter channel with a 300-Hz bandwidth and a signal-to-noise
ratio of 3 dB and the noise is white thermal noise?
3
B = 300Hz, SN RdB = 3dB ⇒ SN R = 10 10 = 1.995
C = B log2 (1 + SN R) = 300 × log2 (1 + 1.995) = 300 × 1.5826 = 474bps
9. (P3.22) The square wave of Figure 3.7c, with T = 1 ms, is transmitted through a low-pass filter that passes
frequencies up to 8 kHz with no attenuation.
(a) Find the power in the output waveform.
Output waveform=sin(2πf1 t) + 31 sin(2π(3f1 )t) + 15 sin(2π(5f1 )t) +
1
with f1 = T1 = 1ms
= 1kHz.
Output power= 21 (1 + ( 13 )2 + ( 15 )2 + ( 17 )2 ) = 0.586 Watt
1
7
sin(2π(7f1 )t)
(b) Assuming that at the filter input there is a thermal noise voltage with No = 0.1µ Watt/Hz find the output
signal-to-noise ratio in dB.
Noise power=B × No = 8e3Hz × 0.1e − 6 Watt/Hz=0.8e-3 Watt.
0.586
= 732.5 ≡ 10 log10 (732.5) = 28.65dB
SNR= 0.8e−3
10. (ex3.7) Signal power at input and output Si = 10mW , So = 5mW then
5e−3
the signal gain =10 log10 ( 10e−3
) = 10 log10 ( 12 ) = −10 log10 (2) = −3dB which is a loss.
11. (ex3.8) Signal power Si = 4mW , first stage gives 12 dB loss, second stage gives 35 dB gain, third stage gives 10
dB loss. Find the output power?
13
Overall dB gain of the system=−12 + 35 − 10 = 13dB ≡ 10 10 = 19.9526 (ratio) and hence the output
power=4mW*19.953=0.0798W.
12. (ex3.9) 1000W ≡ 10 log10 (1000) = 30dBW while 1mW = 1e − 3W ≡ 10 log10 (1e − 3) = −30dBW
Other measures using
dB are dBm
(milli-Watt) which uses 1mW as the reference ⇒ 0dBm ≡ 1mW and dBmV
voltage in mV
which is 20 log10
1mV