Data Communication · Lecture 3

Data and Signals II

Impairments, the limits of data rate, and network performance

Md Mahbub E Noor · Assistant Professor, Department of CSE, University of Barishal

Course: Data Communication · 2nd Year, 2nd Semester

From ideal signals to real channels

Explain attenuation, distortion and noise, and where each comes from
Calculate gains and losses in decibels, including chains of stages
Compute signal-to-noise ratio, as a ratio and in dB
Find the maximum data rate using Nyquist and Shannon
Calculate throughput, latency, bandwidth-delay product and jitter

Lecture 2

Ideal signals: sine waves, spectra, bandwidth, and B ≈ N/2.

Today

Real channels: what damages signals, how fast a channel can go, and how we measure a network.

Reading: Forouzan, Data Communications and Networking, Sections 3.4 to 3.6

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

2 / 32

Signals arrive changed: three impairments

A signal at the end of a medium is never exactly the signal that entered it
Attenuation: it gets weaker
Distortion: its shape changes
Noise: unwanted energy is added
The receiver must still recover the original bits

Attenuation

Loss of energy: the signal gets weaker

Distortion

Change of shape: components arrive out of step

Noise

Unwanted energy added: random wiggles

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

3 / 32

Attenuation: the signal loses energy

Attenuation means loss of energy, so a weaker signal
Some energy is spent overcoming the resistance of the medium and turns into heat
Amplifiers boost an analog signal; repeaters regenerate a digital one
Loss grows with distance and with frequency

Point 1: original

Point 2: weakened

Amplifier

Point 3: restored

Copper loses energy as heat; fibre loses light to scattering and absorption; radio spreads out in space.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

4 / 32

Measuring gain and loss: the decibel

The decibel (dB) measures the relative strength of two signals
Negative dB means the signal was attenuated; positive dB means it was amplified
Use 10 log for power and 20 log for voltage (power is proportional to voltage squared)
Logarithms turn huge ratios into small, easy numbers

dB = 10 log₁₀(P₂ / P₁)

dB = 20 log₁₀(V₂ / V₁)

P = power (watts) · V = voltage (volts) · 1 = input, 2 = output

Power ratio P₂/P₁	Decibels	Meaning
× 2	+3 dB	power doubled
× 10	+10 dB	ten times stronger
× 100	+20 dB	a hundred times stronger
× 0.5	−3 dB	half the power lost
× 0.1	−10 dB	90% of the power lost

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

5 / 32

Worked examples: decibels

A signal's power drops to half in a cable: P₂ = 0.5 P₁
An amplifier multiplies the power by 10: P₂ = 10 P₁
A signal's voltage drops to half: V₂ = 0.5 V₁
The P₁ in the ratio cancels, so only the ratio matters

Power halved

10 log₁₀(0.5) = 10 × (−0.301)

−3 dB

Power × 10 (amplifier)

10 log₁₀(10) = 10 × 1

+10 dB

Voltage halved

20 log₁₀(0.5) = 20 × (−0.301)

−6 dB

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

6 / 32

Why engineers love decibels: they add

Through several stages, ratios multiply, but decibels simply add
Cable losses are quoted per kilometre: dB/km × length = total loss
To get back to power: P₂ = P₁ × 10^(dB/10)
dBm compares a power with 1 mW: 2 mW = +3 dBm, 0.001 mW = −30 dBm

−3 dB

−3 + 7 − 3 = +1 dB

Cable loss per km: −0.3 dB/km × 5 km = −1.5 dB. With P₁ = 2 mW: P₂ = 2 × 10^(−0.15) ≈ 1.42 mW

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

7 / 32

Try it: add up gains and losses in dB

Interactive

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

8 / 32

Checkpoint 1 · 3 minutes

Check your understanding

Q1

A signal's power drops to one quarter. What is the loss in dB?

Q2

Three stages: −2 dB, +10 dB, −5 dB. What is the total?

Q3

10 mW enters a 10 km cable losing 0.5 dB/km. What power leaves?

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

9 / 32

Distortion: the signal changes shape

Distortion means the signal's form or shape changes
It affects composite signals, made of several frequencies
Each frequency component has its own propagation speed, so they arrive with different delays
Components arrive with different phases, so the sum looks different
Equalisers at the receiver undo much of the effect

At the sender: f and 3f in step, flat-topped shape

At the receiver: 3f delayed, the shape is bent

Each frequency travels at a slightly different speed, so components drift out of phase.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

10 / 32

Noise: unwanted energy added on the way

Noise is any unwanted signal added to the one we want
Four kinds: thermal, induced, crosstalk and impulse (table)
Thermal noise is always present; it can never be removed completely
Defences: shielding, twisted pairs, filtering, error-detecting codes

Type	Cause	Everyday example
Thermal	Random motion of electrons in any conductor	The hiss on a radio between stations
Induced	Motors and appliances act as transmitters	A fan motor buzzing in speakers
Crosstalk	One wire picks up a neighbouring wire's signal	Hearing another call on a landline
Impulse	Sudden spikes: lightning, power switching	A click when a heavy load turns on

Sent + noise = received

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

11 / 32

Signal-to-noise ratio (SNR)

SNR = average signal power / average noise power
Usually quoted in decibels: SNR(dB) = 10 log₁₀ SNR
High SNR: the signal stands clearly above the noise
Low SNR: the receiver starts deciding bits wrongly
A noiseless channel would have SNR = ∞, an ideal never reached

Animation: SNR sweeps from 30 dB down to 0 dB and back

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

12 / 32

Worked examples: SNR both ways

Problem 1: signal power 10 mW, noise power 1 µW. Find SNR and SNR(dB).
Ratio: 10 × 10⁻³ / 1 × 10⁻⁶ = 10,000, so SNR(dB) = 10 log₁₀ 10,000 = 40 dB
Problem 2: a channel has SNR(dB) = 20. What is the ratio?
SNR = 10^(SNR(dB) / 10) = 10² = 100

Signal 10 mW, noise 1 µW

40 dB

SNR = 10⁻² / 10⁻⁶ = 10,000 → 10 log₁₀ 10⁴

SNR = 20 dB as a ratio

100

SNR = 10^(20/10) = 10²

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

13 / 32

Checkpoint 2 · 3 minutes

Check your understanding

Q1

Signal power 2 W, noise power 2 mW. What is SNR in dB?

Q2

A channel's SNR is 30 dB. How many times stronger is the signal than the noise?

Q3

Which noise type does twisting the wires in a cable mainly reduce?

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

14 / 32

How fast can a channel go? Two limits

Three things set the maximum data rate: bandwidth, signal levels and noise
Nyquist answers: how fast with L levels, if there were no noise?
Shannon answers: what is the absolute ceiling, given the noise?
In practice, use Shannon to find the ceiling, then Nyquist to choose the number of levels

Nyquist

Noiseless channel

N = 2 B log₂ L

Depends on bandwidth B and number of signal levels L

Shannon

Noisy channel (every real one)

C = B log₂(1 + SNR)

Depends on bandwidth B and the signal-to-noise ratio

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

15 / 32

Nyquist bit rate: the noiseless channel

BitRate = 2 × B × log₂ L (B in Hz, result in bps)
Doubling the levels adds one bit per signal element
In theory, more levels always means a higher rate
In reality, more levels sit closer together and noise makes them unreadable

Bandwidth B	Levels L	Bits per level	Max bit rate 2B log₂L
3,000 Hz	2	1	6,000 bps
3,000 Hz	4	2	12,000 bps
3,000 Hz	16	4	24,000 bps
20 kHz	64	6	240 kbps

Need 265 kbps over 20 kHz? log₂L = 265,000 / 40,000 = 6.625, so L ≈ 98.7. Levels must be a power of 2: use 128 levels (280 kbps), or 64 levels and accept 240 kbps.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

16 / 32

Shannon capacity: the noisy channel

C = B × log₂(1 + SNR), with SNR as a ratio, not in dB
C is the theoretical highest data rate with arbitrarily few errors
It does not depend on the number of levels: noise decides how many are usable
Capacity grows in proportion to bandwidth, but only logarithmically with SNR

Telephone line: B = 3,000 Hz, SNR = 35 dB = 3,162

C = 3,000 × log₂(1 + 3,162)

= 3,000 × 11.62

C ≈ 34,860 bps

Very noisy channel: SNR ≈ 0

C = B log₂(1 + 0) = B × 0 = 0

No data at all, however wide the bandwidth

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

17 / 32

Worked example: using both limits together

Problem: B = 1 MHz and SNR = 63. What rate, and how many signal levels?
Shannon gives the ceiling: 6 Mbps
Real systems stay below it for safety, so choose 4 Mbps
Nyquist then gives the levels needed: L = 4

Step 1 · Shannon ceiling

10⁶ × log₂(1 + 63) = 10⁶ × 6

6 Mbps

Step 2 · choose a safer rate

leave margin below the ceiling

4 Mbps

Step 3 · Nyquist for levels

4 Mbps = 2 × 10⁶ × log₂L → log₂L = 2

L = 4

Shortcut when SNR ≫ 1: C ≈ B × SNR(dB) / 3. For B = 2 MHz, 36 dB: C ≈ 24 Mbps.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

18 / 32

Try it: Nyquist meets Shannon

Interactive

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

19 / 32

Checkpoint 3 · 3 minutes

Check your understanding

Q1

Noiseless channel, B = 4 kHz, L = 16 levels. Maximum bit rate?

Q2

B = 3 kHz, SNR = 255. What is the Shannon capacity?

Q3

If the SNR doubles but B stays the same, does the capacity double?

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

20 / 32

Measuring a network: five performance terms

Measure	What it tells you	Unit	On a highway
Bandwidth	Capacity of the link (Hz or bps)	Hz, bps	Number of lanes
Throughput	Data actually delivered per second	bps	Cars that actually pass
Latency	Time for a message to arrive	s	Journey time
Bandwidth × delay	Bits in flight on the link at once	bits	Cars on the road at once
Jitter	Variation in delay between packets	s	Uneven arrival of buses

Bandwidth is what the link could do; throughput is what it actually does; latency is how long it takes.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

21 / 32

Throughput: what actually gets through

Throughput: how many bits per second actually pass a point
Always ≤ bandwidth; the gap comes from congestion, overhead, errors and idle time
Problem: a 10 Mbps network passes 12,000 frames per minute, each 10,000 bits
Throughput = 12,000 × 10,000 / 60 = 2 Mbps

Frames delivered per second

12,000 frames ÷ 60 s = 200 frames/s

200 frames/s

Throughput

200 × 10,000 bits

2 Mbps

The link could carry 10 Mbps, but only one fifth of that is actually used.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

22 / 32

Latency: four delays add up

Latency = propagation + transmission + queuing + processing
Propagation: time for one bit to travel the distance
Transmission: time to push all the message's bits onto the link
Queuing and processing: time spent inside routers and switches

■ Propagation

■ Transmission

■ Queuing

■ Processing

propagation = distance / speed

transmission = size / bandwidth

Queuing: waiting in router buffers behind other packets. Processing: reading headers and choosing a route.

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

23 / 32

Worked examples: which delay dominates?

Same 12,000 km cable path for both messages: propagation = 50 ms
A short email on a fast link: transmission 20 µs, so propagation dominates
A large image on a slow link: transmission 40 s, so transmission dominates
Rule: small messages care about distance; large ones care about bandwidth

Propagation · 12,000 km at 2.4 × 10⁸ m/s

12 × 10⁶ m ÷ 2.4 × 10⁸ m/s

50 ms

Email · 2.5 kB over 1 Gbps

20,000 bits ÷ 10⁹ bps

20 µs

Image · 5 MB over 1 Mbps

40 × 10⁶ bits ÷ 10⁶ bps

40 s

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

24 / 32

Try it: which delay dominates?

Interactive

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

25 / 32

Bandwidth-delay product: bits in flight

Bandwidth × delay = the number of bits that fill the link at one moment
Picture the link as a pipe: bandwidth is its cross-section, delay is its length
Example: 100 Mbps with 20 ms delay holds 2 million bits
To use a link fully, a sender must keep this many bits in flight before an answer returns

length of the pipe = delay

height = bandwidth

100 Mbps × 20 ms = 2,000,000 bits

= 250 kB in flight

A sender that waits for a reply after every packet leaves this pipe almost empty (Lecture 12).

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

26 / 32

Jitter: when delay is uneven

Jitter: variation in delay from packet to packet
Harmless for downloads, harmful for live voice and video
Caused by packets taking different paths and waiting in different queues
Fix: a jitter buffer holds packets briefly and plays them at an even pace, adding a little delay

Sent: every 20 ms · received: gaps of 25, 15 and 35 ms

Packet	Sent (ms)	Arrived (ms)	Delay (ms)
1	0	100	100
2	20	125	105
3	40	140	100
4	60	175	115

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

27 / 32

Try it yourself: Channel Lab

1 · Decibel chain: drag losses and gains; watch the power
2 · Noise and SNR: lower the SNR until bits flip
3 · Channel capacity: compare Nyquist and Shannon
4 · Latency: find which delay dominates

Scan to open the lab on your phone

Open the lab ↗

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

28 / 32

See it in action: a signal in a real channel

Watch a signal fade, get amplified, and fade again
See two frequency components drift apart (distortion)
Watch bits flip as the SNR falls, then see Shannon's limit move
When a “Pause & predict” card appears, answer before the video moves on

Click the video to play or pause · Press → for full screen

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

29 / 32

Checkpoint 4 · 3 minutes

Check your understanding

Q1

A 1,000-bit frame is sent over a 1 Mbps link that is 2,000 km long (speed 2 × 10⁸ m/s). What is the latency, ignoring queuing?

Q2

What is the bandwidth-delay product of that link?

Q3

Packets arrive 50, 52, 70 and 51 ms after being sent. Is there jitter, and which packet suffered most?

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

31 / 32

Key takeaways

Signals suffer attenuation, distortion and noise on every real channel
Decibels: 10 log for power, 20 log for voltage; stages simply add
SNR measures quality; convert dB to a ratio before using Shannon
Nyquist 2B log₂L (noiseless) and Shannon B log₂(1+SNR) (the true ceiling)
Latency = propagation + transmission + queuing + processing; BDP = bits in flight

Next lecture

Lecture 4 · Transmission Media

Twisted pair, coaxial cable, optical fibre and wireless: how each one fights attenuation and noise, and where each is used.

Reading: Forouzan, Chapter 7 · Practise in Channel Lab

Md Mahbub E Noor

Lecture 3: Data and Signals II

University of Barishal

32 / 32