# Baud rate for Square Wave data signal via UART if frequency is given

I was trying to solve the 2013 paper set by ISRO for electrical engineers. Question no 13 made me confused.

So since Idle in UART is usually high and start bit is low, we can find the 8 bit data to be $$\1010 1010 = (AA)_{Hex}\$$ since it ends with a high stop bit, and because we observed a square wave we know the bits are alternating.

But when I tried to find the baud rate, which is same as the bit rate in this case since we have 1 bit for 1 symbol. I am getting 19200 baud, there is no option for this.

We see the square wave has frequency 9600 Hz, so we have 2 bits being send (2 symbol changes) happening in one cycle or in one time period.

$$\text{Baud rate} = \frac{2}{\frac{1}{9600}} = 19200 \ \text{baud}$$

So I was wondering how I got this wrong. Can UART have high as the start bit? If so option B is the answer. Maybe I am counting the bits being sent incorrectly, is D the answer?

This is how I saw 2 bits being transmitted in one time period. We are sending the 8 data bits in $$\ \frac{4}{9600} secs\$$

I was also wondering why the data bits cannot be $$\ 11001100 \$$. I mean that is also a square wave right?

• I don't see how you're getting 2 bits send per cycle of the square wave. Can you maybe make a drawing of a square wave and tell us where you see these bit "boundaries"? Commented Mar 3, 2019 at 10:05
• I've added a picture which shows how I worked out 2 bits per time period. Commented Mar 3, 2019 at 10:13

Hint: did you remember that data is transmitted LSB first?

simulate this circuit – Schematic created using CircuitLab

Figure 1. 9600 Hz squarewave.

• oh ok, so the data bits can be $01010101$ because it is just that in most UART's have 0 as a start bit. There might be exceptions. While we can be sure from the calculation that the baud rate is 19200. Thanks! Commented Mar 3, 2019 at 10:40