단위
Unit | Equivalent | Unit | Equivalent |
---|---|---|---|
s | 1 s | Hz | 1 Hz |
ms | 10-3 s | kHz | 103 Hz |
μs | 10-6 s | MHz | 106 Hz |
ns | 10-9 s | GHz | 109 Hz |
ps | 10-12 s | THz | 1012 Hz |
Fast means high frequency,
Slow means low frequency
Q. 만약 level이 8개라면, level 하나당 몇 개의 bit가 필요할까?
A. 3개겠지
Q. 1초에 100페이지를 다운받아야 한다. 한 페이지에 20줄, 한 줄에 80 characters, 한 character는 1 byte(8 bits). bit rate?
A. 80 * 24 * 100 bytes/s
=> 1.536Mbps
Q. 1920x1080 pixels per screen. 30frames per second. 1pixel uses 24bits. bit rate?
A. 1920 * 1080 * 24 * 30 bps
Q. baseband transmission을 이용해서 1Mbps로 보내려고 한다. What is the required bandwidth?
A. 500kHz
Q. bandwidth 100kHz. What is the maximum bit rate of this channel?
A. 200Kbps
band-pass channel을 통해 신호를 전송시키려면, 우리는 신호를 modulate해야 한다
Modulate digital signals into analog signals
dB = 10log$$_{10}$$(P2/P1)
Q. Suppose a signal travels through a transmission media and its power is reduced to one half. This means tha P2 = 0.5P1. In this case, the attenuation can be calculated as :
A.
= -3dB
3dB만큼 감소(a loss of 3dB = -3dB)하는 것은 power의 절반을 loss하는 것과 동일하다.
2배 -> P2/P1 = 2 -> 3dB
4배 -> P2/P1 = 4 -> 6dB
5dB?
10배 -> 10dB -> 이거 절반 -> 7dB
dB를 사용하게 되면 곱을 덧셈뺄셈으로 표현할 수 있다.
Q. The loss in a cable is usually defined in decibels per kilometer (dB/km). 만약 -0.3dB/km의 cable의 시작에서 signal이 2mW의 power를 갖는다면, what is the power of signals at 5km?
A. 일단 5km이므로, 총 -1.5dB의 loss 발생
-1.5dB = 10log(P2/P1)
=> P2/P1 = 10-0.15 = 0.71
=> P2 = 0.71 * P1 = 0.7 * 2mW = 1.4mW
Several types of noise, thermal(열) noise, induced noise(noise generated in a circuit by a varying magnetic or electrostatic field produced by another circuit), crosstalk(혼선), and impulse noise(원치 않는 거의 즉각적인 날카로운 사운드를 포함하는 노이즈 범주), may corrupt the signal
Signal-to-Noise Ratio(SNR)
SNR = signal power / noise power
전력의 비율을 나타낸다
단위는 W 또는 dB
Q. 신호의 세기(power)가 10mW이고, noise가 1uW이다. What are the values of SNR and SNRdB?
A. 1uW = 0.001mW
SNR = 10mW / 0.001mW = 10000
SNRdB = 10log1010000 = 40
Q. Consider a noiseless channel with a bandwidth of 3000Hz transmitting a signal with two signal levels. What is the maximum bit rate?
A. 2 x 3000 x log22 = 6000bps
Q. 265kbps를 보내야 한다. noiseless channel이고, bandwidth는 20kHz이다. 몇 개의 signal level이 필요하냐?
A. 2 x 20 x log2L = 265
=> log2L = 6.625
=> L = 98.7 levels
근데, the number of levels는 2의 거듭제곱이어야 한다.
따라서 답은 128 levels이고, 속도는 280kbps가 된다.
Maximum bit rate in a noisy channel (최대로 낼 수 있는 전송 속도)
C = B x log2(1 + SNR)
C : capacity (maximum bit rate), B : bandwidth
Q. Consider an extremely noisy channel in which SNR is zero. What is the maximum bit rate for this channel
A. SNR이 0이라는건, signal = 0이거나 noise가 ∞이다.
어쨌든,
C = B x log2(1 + SNR) = B * 0 = 0.
Q. A telephone line normally has a bandwidth of 3000Hz assigned for data communications. The signal-to-noise ratio is usually 3162. The channel capacity?
A. C = B x log2(1 + SNR)
= 3000 x log2(3163)
= 3000 x 11.62
= 34,860bps
Q. Suppose SNR is 36dB, and channel bandwidth is 2MHz. The channel capacity?
A. 먼저, SNR을 dB로 주었기 때문에 dB 아닌걸로 바꿔주어야 한다
SNRdB = 10 log10SNR
=> SNR = 10SNRdB/10 = 103.6 = 3981
C = B x log2(1 + SNR)
= 2 x 106 x log23982
= 24 Mbps
SNR이 아주 높다면, SNR과 SNR + 1을 거의 같다고 볼 수 있다. 이 경우, theoretical channel capacity는 더 간단하게 쓸 수 있다.
C = B x (SNRdB / 3)
For the previous example,
C = 2MHz x (36/3) = 24Mbps
SNR을 dB로 주면, 이게 더 편할 듯 싶다.
Bandwidth of a line is 4kHz.
This line has 56Kbps of bandwidth
Q. A network with bandwidth of 10Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
A. Throughput = (12,000 x 10,000)/60 = 2Mbps
Duration between
the time the first bit of a message leaves the source
and
the time the last bit of the message is received at the destination
Delay
= transmission delay + propagation delay + queuing delay + processing delay.
4가지의 component가 있다.
Q. What is the propagation time if the distance between the two points is 12,000km? Assume the propagation speed to be 2.4 x 108m/s in cable
A. 시 = 거 / 속
= 12 x 106 m / (2.4 x 108) s
= 50ms
Q. What are the propagation time and the transmission time for a 2.5KB message if the bandwidth of the network is 1Gbps? Assume that the distance between the sender and the receiver is 12,000km and that the signal travels at 2.4 x 108m/s
A. propagation time = (12,000 x 103) / (2.4 x 108) = 50ms
transmission time = 프레임의 크기 / 회선의 최대 전송 속도(bandwidth)
= (2500 x 8) / 109 = 0.020ms
Q. What are the propagation time and the transmission time for 5MB message if the bandwidth of the network is 1Mbps? Assume that the distance between the sender and the receiver is 12,000km and the signal travels at 2.4 x 108m/s
A. transmission time = (5,000,000 x 8) / 106 = 40s
Q. Suppose bandwidth of a link is 10Mbps, and the delay is 50ms. What is the amount of data to fill up this channel?
A. 10Mbps x 50ms = 10 x 106 x 50 x 10-3 bits = 500 x 103 = 500Kbit