LTE Data Rate Calculations

How fast can LTE actually send data? The answer comes down to counting and multiplying.

Core idea: Count how many symbols you send per second, then figure out how many bits each symbol carries.

The Building Blocks

These are fixed constants from the LTE specification:

Note: Symbols per slot depends on the Cyclic Prefix. Normal CP = 7 symbols, Extended CP = 6 symbols.

And bandwidth determines how many Resource Blocks you have:

Step 1: Count Symbols Per Second

Every RB gives you 12 subcarriers × 7 symbols per slot. You get 2 slots per subframe, and 1000 subframes per second.

Symbols/sec = RBs × 12 × 7 × 2 × 1000

Example for 20 MHz:

$\text{Symbols/sec} = 100 \times 12 \times 7 \times 2 \times 1000 = 16,800,000$

That’s 16.8 million symbols flying through the air every second.

Step 2: Bits Per Symbol (Modulation)

Each symbol carries a certain number of bits, depending on the modulation scheme:

Higher QAM packs more bits per symbol, but needs a cleaner signal to decode correctly.

Step 3: Raw Bit Rate

Multiply symbols by bits per symbol:

Raw bits/sec = Symbols/sec × bits/symbol

Example: 20 MHz with 16-QAM:

$\text{Raw rate} = 16,800,000 \times 4 = 67,200,000 \text{ bps} = 67.2 \text{ Mbps}$

But this is the theoretical maximum. Real systems have overhead.

Step 4: Apply Real-World Factors

Two things eat into your raw rate:

Coding Rate

Error correction adds redundancy. A 2/3 coding rate means for every 3 bits transmitted, only 2 are actual data.

$\text{After coding} = \text{Raw rate} \times \frac{2}{3}$

Bandwidth Efficiency

Not all spectrum carries user data. Guard bands, control channels, and reference signals take up space. Typical efficiency is 90%.

$\text{Effective rate} = \text{After coding} \times 0.90$

Step 5: MIMO Multiplication

MIMO (Multiple-Input Multiple-Output) sends multiple data streams simultaneously using multiple antennas.

Final throughput = Effective rate × MIMO layers

This is why MIMO is so powerful. It literally multiplies your capacity.

The Complete Formula

Putting it all together:

$\text{Throughput} = \underbrace{\text{RBs} \times 12 \times 7 \times 2 \times 1000}_{\text{symbols/sec}} \times \underbrace{\text{bits/symbol}}_{\text{modulation}} \times \underbrace{\text{coding rate}}_{\text{e.g. } \frac{2}{3}} \times \underbrace{\text{efficiency}}_{\text{e.g. } 0.9} \times \underbrace{\text{MIMO layers}}_{\text{e.g. } 4}$

Worked Example 1

Calculate the maximum achievable throughput of the FDD LTE system in downlink under the following assumptions:

• Normal Cyclic Prefix (CP), 10 MHz Bandwidth, 16-QAM Modulation, 4×4 MIMO

Hint: In Normal CP, there are 7 symbols and 12 subcarriers in a slot of 0.5 ms, 10 MHz contains 50 Resource Blocks, one modulation symbol carries 4 bits in 16-QAM.

Step 1: Resource Blocks for 10 MHz = 50 RBs

Step 2: Symbols per second: $50 \times 12 \times 7 \times 2 \times 1000 = 8,400,000$

Step 3: Bits per symbol for 16-QAM = 4 bits

Step 4: Raw bit rate: $8,400,000 \times 4 = 33,600,000 \text{ bps} = 33.6 \text{ Mbps}$

Step 5: With 4×4 MIMO: $33.6 \times 4 = \boxed{134.4 \text{ Mbps}}$

No coding rate or efficiency given, so we use raw throughput.

Worked Example 2

For an LTE release 8 system, consider an FDD system configured to use 20 MHz of spectrum. Consider a Type 1 frame with normal Cyclic Prefix length of seven (7) symbols, and a transmission effective bandwidth of 90%.

(i) What is the effective downlink data rate using 16-QAM and 2/3 encoding rate? (10 marks)

(ii) What is the impact on the system capacity when a 2×2 MIMO antenna configuration is used? (2 marks)

Part (i):

Step 1: Resource Blocks for 20 MHz = 100 RBs

Step 2: Symbols per second: $100 \times 12 \times 7 \times 2 \times 1000 = 16,800,000$

Step 3: Bits per symbol for 16-QAM = 4 bits

Step 4: Raw bit rate: $16,800,000 \times 4 = 67,200,000 \text{ bps} = 67.2 \text{ Mbps}$

Step 5: Apply coding rate: $67.2 \times \frac{2}{3} = 44.8 \text{ Mbps}$

Step 6: Apply bandwidth efficiency: $44.8 \times 0.9 = \boxed{40.32 \text{ Mbps}}$

Part (ii):

With 2×2 MIMO, we get 2 spatial layers, so capacity doubles: $40.32 \times 2 = \boxed{80.64 \text{ Mbps}}$

Worked Example 3

For an LTE release 8 system, consider an FDD system configured to use 20 MHz of spectrum. Consider a Type 1 frame with normal Cyclic Prefix length of seven (7) symbols, and a transmission effective bandwidth of 90%.

(i) What is the effective downlink data rate using 32-QAM and 2/3 encoding rate? (10 marks)

(ii) What is the impact on the system capacity when an 8×8 MIMO antenna configuration is used? (2 marks)

Part (i):

Step 1: Resource Blocks for 20 MHz = 100 RBs

Step 2: Symbols per second: $100 \times 12 \times 7 \times 2 \times 1000 = 16,800,000$

Step 3: Bits per symbol for 32-QAM = 5 bits

Step 4: Raw bit rate: $16,800,000 \times 5 = 84,000,000 \text{ bps} = 84 \text{ Mbps}$

Step 5: Apply coding rate: $84 \times \frac{2}{3} = 56 \text{ Mbps}$

Step 6: Apply bandwidth efficiency: $56 \times 0.9 = \boxed{50.4 \text{ Mbps}}$

Part (ii):

With 8×8 MIMO, we get 8 spatial layers: $50.4 \times 8 = \boxed{403.2 \text{ Mbps}}$

Quick Reference

The calculation is just multiplication. Count symbols, convert to bits, apply penalties, multiply by MIMO. That’s it.