Interview Question · Concept

The Baud Mismatch Budget

L1 JuniorL2 Mid-Leveluartbaud rate

The interviewer asks

Your UART and its peer run from different clocks. How much baud-rate mismatch can the link tolerate, and which bit fails first?

Reveal the Model Answer

Drift is zero at each start edge and grows with every bit. So the last sample, the stop bit, fails first. A receiver that samples exact bit centres has half a bit of slack over the nine bits before the stop. That is about ±5.6 %, shared by both ends. Real receivers do worse: the site's bench receiver first loses 8N1 frames at +4.5 % and −6 %. The 8E1 format fails earlier, because parity puts the stop bit one bit later.

Measured sweepFrames Decoded Against Rate Error, Back to Back64 seeded frames per 0.5 % step, frames back to back, the bench receiver. Positive error is a slower transmitter.

Read the plot as a table

Transmitter rate error	8N1	8E1	8N2
-8%	0	0	9
-6%	0	0	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	52	64
6%	7	4	5
8%	0	1	2

Measured, 64 frames per step, back to back. 8N1 first loses frames at −6 % and at +4.5 %, and 8E1 fails one step earlier on both sides.

Where the Budget GoesArrows read as "leads to". A slow transmitter moves the samples toward the start of each bit, a fast one toward the end.

Read the diagram as text

Rate error e: TX bit = 16 × (1 + e) ticks
Drift: 16·e·k ticks at sample k
Stop sample: largest k, drifts most
Slow side, e > 0: room to the start of the bit
Fast side, e < 0: room to the end of the bit

Rate error e → Drift
Drift → Stop sample
Stop sample → Slow side, e > 0: early sample
Stop sample → Fast side, e < 0: late sample

The budget is about half a bit spread over the frame, split by where the sample sits.

Drift
A transmitter bit longer by a fraction e ends 16·e·k ticks late at bit k. The stop bit has the largest k.
Slow side
The stop bit starts at 144·(1 + e) ticks; the centre sample is at 152. It reads data 7 once e passes 8/144, about 5.6 %.
Fast side
Data 7 ends at 144·(1 + e); its sample is at 136. Past about −5.6 % it reads the stop bit: wrong data, and with an idle line after, no flag.
Both ends
Divisor rounding plus the peer's oscillator: 115200 from 50 MHz with divisor 27 is already +0.47 %.

RUNIcarus Verilog · in your browser

Edit the code, then run it

Expected: e = +4.5 %: RX 55 ok e = +5.5 %: RX 55 ok e = +6.0 %: RX 55 FE e = -5.5 %: RX 55 ok e = -6.0 %: RX d5 ok

`timescale 1ns/1ps
// An idealized 8N1 receiver: it finds the start edge exactly and samples the centre of each
// bit by its own clock (16 ns per bit): 24, 40, ..., 152 ns after the edge. The transmitter's
// bit is 16 x (1 + e) ns. Positive e is a slower transmitter. The line idles high after each frame.
module tb;
  reg line = 1;
  real p;
  task automatic send(input [7:0] b, input real e);
    p = 16.0 * (1.0 + e);
    line = 0; #(p);                                  // start bit
    for (int i = 0; i < 8; i++) begin line = b[i]; #(p); end
    line = 1; #(p);                                  // stop bit
    #(64);                                           // idle
  endtask
  reg [7:0] got; reg stop;
  always @(negedge line) begin : rx
    #24;
    for (int i = 0; i < 8; i++) begin got[i] = line; #16; end
    stop = line;
  end
  task automatic trial(input real e);
    fork send(8'h55, e); join
    $display("e = %+0.1f %%: RX %h %s", e * 100.0, got, stop ? "ok" : "FE");
  endtask
  initial begin
    #10;
    trial(0.045); trial(0.055); trial(0.06); trial(-0.055); trial(-0.06);
    $finish;
  end
endmodule

Self-Check

Tick each key point you covered in your answer.

Key point 1 Drift is zero at each start edge and grows with every bit. Key point 2 So the last sample, the stop bit, fails first. Key point 3 A receiver that samples exact bit centres has half a bit of slack over the nine bits before the stop. Key point 4 That is about ±5.6 %, shared by both ends. Key point 5 Real receivers do worse: the site's bench receiver first loses 8N1 frames at +4.5 % and −6 %. Key point 6 The 8E1 format fails earlier, because parity puts the stop bit one bit later.

What Interviewers Listen For (Junior)

A good junior answer says the stop bit fails first because drift grows from the start edge. It says both ends share a budget of roughly 5 %.

Cross-Questions

Where the interviewer goes next. Open a node to practise that follow-up.

WhyWhat IfProve ItScale It

The Baud Mismatch Budget
WhyWhy the Stop Bit Fails FirstL1 Junior
Why is it the stop bit that fails, and not data bit 0?

Reveal the Model Answer

The start edge is the only moment both ends agree on, and drift is zero there. Each later bit adds the same timing error. So the sample furthest from the start edge, the stop sample, has drifted most. This is also why the receiver restarts its count on every start edge. The error never carries from one frame to the next.

Self-Check
Tick each key point you covered in your answer.
Key point 1 The start edge is the only moment both ends agree on, and drift is zero there. Key point 2 Each later bit adds the same timing error. Key point 3 So the sample furthest from the start edge, the stop sample, has drifted most. Key point 4 This is also why the receiver restarts its count on every start edge. Key point 5 The error never carries from one frame to the next.

What Interviewers Listen For (Junior)
A good junior answer says drift grows evenly from the start edge and resets with every new frame.
- Lesson P6.4Baud Mismatch: The Measured Budget
What IfDoes a Second Stop Bit HelpL2 Mid-Level
Would switching to two stop bits give you more tolerance?
2:30

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Reveal the Model Answer

It helps only on the fast side, and only when frames stream back to back. There, with a fast transmitter, the stop sample comes late and can land in the next start bit. It then reads 0. A second stop bit puts idle time there instead. With idle time between frames it buys nothing: the P6.4 sweep shows 8N1 and 8N2 failing at the same steps.

Measured sweepThe Same Sweep with an Idle Bit Between FramesOne idle bit plus a seeded fraction of a tick before every frame. Compare 8N1 and 8N2 on the fast side.

8N1
8E1
8N2

Read the plot as a table
Transmitter rate error 8N1 8E1 8N2
-8% 9 5 9
-6% 64 43 64
-4% 64 64 64
-2% 64 64 64
0% 64 64 64
2% 64 64 64
4% 64 48 64
6% 5 4 7
8% 2 0 2

Measured with an idle bit between frames. 8N1 and 8N2 first lose frames at the same steps, −6.5 % and +4.5 %.

Self-Check
Tick each key point you covered in your answer.
Key point 1 It helps only on the fast side, and only when frames stream back to back. Key point 2 There, with a fast transmitter, the stop sample comes late and can land in the next start bit. Key point 3 It then reads 0. Key point 4 A second stop bit puts idle time there instead. Key point 5 With idle time between frames it buys nothing: the P6.4 sweep shows 8N1 and 8N2 failing at the same steps.

What Interviewers Listen For (Mid-Level)
A good mid-level answer says a second stop bit helps only on the fast side, with frames back to back. It cites the measured sweep: with idle gaps there is no gain.
- Lesson P6.4Baud Mismatch: The Measured Budget
Scale ItSigning off a UART Link BudgetL2 Mid-LevelL3 Senior
Your SoC talks to a module clocked by an RC oscillator. How do you sign off that the UART link works across temperature and parts?

Reveal the Model Answer

Add up the error sources: this end's divisor rounding and clock tolerance, and the peer's oscillator tolerance over temperature and voltage. Compare the sum, in both directions, with the receiver's measured budget for the frame format in use, not a quoted 5 %. In simulation, sweep the peer rate in fine steps, both ways, in every supported format. Run frames back to back and with gaps. Record the first failing step as the sign-off number.

Self-Check
Tick each key point you covered in your answer.
Key point 1 Add up the error sources: this end's divisor rounding and clock tolerance, and the peer's oscillator tolerance over temperature and voltage. Key point 2 Compare the sum, in both directions, with the receiver's measured budget for the frame format in use, not a quoted 5 %. Key point 3 In simulation, sweep the peer rate in fine steps, both ways, in every supported format. Key point 4 Run frames back to back and with gaps. Key point 5 Record the first failing step as the sign-off number.

What Interviewers Listen For (Mid-Level)
A good mid-level answer adds up the errors of both ends. It compares them with the budget for the frame format really in use.
WhyWhy the Baud Budget Is LopsidedL2 Mid-LevelL3 Senior
Why is the measured 8N1 budget +4.5 % on one side and −6 % on the other, not ±5 %?

Reveal the Model Answer

A rate error moves each sample further from its bit centre, and the stop bit moves most. The total room is about half a bit over the frame. A receiver that samples a little before the centre has less room on one side and more on the other. So quote the budget with its receiver, format and pacing.

Self-Check
Tick each key point you covered in your answer.
Key point 1 A rate error moves each sample further from its bit centre, and the stop bit moves most. Key point 2 The total room is about half a bit over the frame. Key point 3 A receiver that samples a little before the centre has less room on one side and more on the other. Key point 4 So quote the budget with its receiver, format and pacing.

Misconception

"Every UART tolerates ±5 %."

The budget depends on the frame format and where the receiver samples. The lesson receiver measured +4.5 % and −6 % for 8N1.

Source ASIC Point lesson P6.4 (measured)
- Lesson P6.4Baud Mismatch: The Measured Budget
- Problem P6.4.1A Receiver That Survives the Sweep
Prove ItSolve P6.4.1: A Receiver That Survives the SweepSimulator-graded

Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
Open Problem P6.4.1

Transmitter rate error	8N1	8E1	8N2
-8%	9	5	9
-6%	64	43	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	48	64
6%	5	4	7
8%	2	0	2

What People Get Wrong

Common wrong answers and why they fail.

Missed Corner

"A rate error always shows up as a framing error."

That holds only on the slow side. A fast transmitter sends the stop bit early, so the data 7 sample reads it. If the line idles after the frame, the stop sample still reads 1. In the run, −6 % delivers D5h for 55h with no flag.

Source This question's run (Icarus Verilog 12): at −6 % it delivers D5h with no flag

Half-Truth

"Each end may be off by 5 %."

The budget belongs to the link, not to each end. Divisor rounding at one end and the peer's oscillator error add up. Real receivers also have less than the ideal ±5.6 %.

Misconception

"Adding parity makes the link more robust to clock error."

Parity makes the frame longer, so the stop sample sits one bit further from the start edge. In the P6.4 sweep, 8E1 first fails at +4 %, and 8N1 at +4.5 %.

Source ASIC Point lesson P6.4 (measured sweep): 8E1 first fails at +4 %, 8N1 at +4.5 %

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The Baud Mismatch Budget

Your Answer

Self-Check

What Interviewers Listen For (Junior)

What Interviewers Listen For (Mid-Level)

Cross-Questions

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What Interviewers Listen For (Junior)

Your Answer

Self-Check

What Interviewers Listen For (Mid-Level)

Your Answer

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What Interviewers Listen For (Mid-Level)

What Interviewers Listen For (Senior)

Your Answer

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What People Get Wrong

Learn It, Then Prove It

Transmitter rate error	8N1	8E1	8N2
-8%	0	0	9
-6%	0	0	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	52	64
6%	7	4	5
8%	0	1	2

Transmitter rate error	8N1	8E1	8N2
-8%	9	5	9
-6%	64	43	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	48	64
6%	5	4	7
8%	2	0	2

Transmitter rate error	8N1	8E1	8N2
-8%	0	0	9
-6%	0	0	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	52	64
6%	7	4	5
8%	0	1	2

Transmitter rate error	8N1	8E1	8N2
-8%	9	5	9
-6%	64	43	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	48	64
6%	5	4	7
8%	2	0	2

The Question

Your Answer

Self-Check

What Interviewers Listen For (Junior)

What Interviewers Listen For (Mid-Level)

Cross-Questions

Your Answer

Self-Check

What Interviewers Listen For (Junior)

Your Answer

Self-Check

What Interviewers Listen For (Mid-Level)

Your Answer

Self-Check

What Interviewers Listen For (Mid-Level)

What Interviewers Listen For (Senior)

Your Answer

Self-Check

What People Get Wrong

Learn It, Then Prove It

Transmitter rate error	8N1	8E1	8N2
-8%	0	0	9
-6%	0	0	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	52	64
6%	7	4	5
8%	0	1	2

Transmitter rate error	8N1	8E1	8N2
-8%	9	5	9
-6%	64	43	64
-4%	64	64	64
-2%	64	64	64
0%	64	64	64
2%	64	64	64
4%	64	48	64
6%	5	4	7
8%	2	0	2