dsp28335串口传输过程中波特率对应的中断超时时间。

Question

1. 有段代码功能是串口接受字符之间传输超时1000us判断；
2. 串口波特率是57600（1.74*e4 ns 传一位），配置为8个数据位，无奇偶校验；
3. cpu时钟频率为f=58.98MHz, SCI_B串口时钟未分频直接使用系统时钟。
   我的理解为每个字符（10位(起始位+数据位+停止位)，需要1.74*e4 10=1.74e5 ns <10^6 ns）中断一次，每获取到一个新字符然后在初始化记时函数ISB_ReceiveCommandTimeout（3）函数中初始化时间将计数变量Mruningtimeout置零。接受过程中如果对应的接受缓冲区的值为0x00(实际是不会为0x00，我理解为串口传输很慢，cpu工作频率很快),将计数变量Mruningtimeout自加一。

问题: 但却将判断值mRequiredTimeout设置为3，这个是为什么？

       if ( mNumberOfCharsInRxBuffer < SCI_RxBufferNumberOfCharsGet(SCI_B) )    // New Chars received
        {
            mNumberOfCharsInRxBuffer = SCI_RxBufferNumberOfCharsGet(SCI_B);      // Number of chars updated

            ISB_ReceiveCommandTimeoutInitialise(TIMEOUT_FOR_BAUD_RATE);                // Reset the inter-char timer
        }

        if ( mNumberOfCharsInRxBuffer != 0u )    // Chars In the RX buffer
        {
            // First we look for the CTRL_A character within the characters received
            for ( loop = 0u; loop < (mNumberOfCharsInRxBuffer - 1u); loop++ )
            {
                if ( CTRL_A == *pCtrlA )        //lint !e921 Cast to uint8_t in CTRL_A is OK
                {
                    break;
                }
                else
                {
                    pCtrlA++;
                }
            }

            // Then the message is checked.
            ISBcommandStatus = ISB_ReceivedCommandCheck(pCtrlA, mBusAddress);

            // The response to the ISB status is based on the status table in the
            // LWD / MWD / ISB Communications Opcodes document (GeMS 100553165, page 5).
            // Note that not all conditions require a response to be sent back to the master.
            switch (ISBcommandStatus)
            {
                // The packet is OK, so process the received opcode and then reset
                // the receive buffer and inter-character timer.
                case (ISB_PACKET_OK) :
                    ProcessOpcodeAndTransmitResponse(pCtrlA);
                    ReceiveBufferInitialise();
                    ISB_ReceiveCommandTimeoutInitialise(TIMEOUT_FOR_BAUD_RATE);    //  **_TIMEOUT_FOR_BAUD_RATE为3_**
                    break;

                // The reception buffer is cleared for the following cases.
                // Note that the slave does not reply to the master under these conditions.
                case ISB_NO_CTRL_A :
                case ISB_INVALID_ADDRESS :
                case ISB_COMMAND_WAIT_OPCODE_TIMEOUT :
                case ISB_CHECKSUM_FAIL :
                    ReceiveBufferInitialise();
                    break;


EIsbRxStatus_t ISB_ReceivedCommandCheck(const uint8_t * const pBuffer,
                                        const uint8_t ExpectedAddress)
{
    EIsbRxStatus_t     status             = ISB_PACKET_OK;
    uint16_t        expected_length = 6u;

    mValidOpcode = 0xffu;            // set opcode to 0xff before we use it
    mDataLength  = 0u;                // set length of data found to zero

    if (CTRL_A != pBuffer[0u])      //lint !e921 Cast to uint8_t in macro.
    {
        status = ISB_NO_CTRL_A;
    }
    else if (0x00u == pBuffer[1u])
    {
        status = ISBTimeoutCheck(ISB_WAITING_FOR_ADDRESS);
    }
    else if (ExpectedAddress != pBuffer[1u])
    {
        status = ISB_INVALID_ADDRESS;
    }
    else if ( (0x00u == pBuffer[2u]) && (0x00u == pBuffer[3u]) )
    {
        status = ISBTimeoutCheck(ISB_WAITING_FOR_MORE_COMMAND);
    }
}

// ----------------------------------------------------------------------------
/**
 * ISB_ReceiveCommandTimeoutIncrement increments the running timeout for the
 * received message.
 *
 */
// ----------------------------------------------------------------------------
void ISB_ReceiveCommandTimeoutIncrement(void)
{
    mRunningTimeout++;
}

我的解答思路和尝试过的方法

我想应该是时钟频率和波特率有差距产生的，下一个中断来以前，CPU在飞速的执行程序，预期时间IMEOUT_FOR_BAUD_RATE设为3，是不是可以理解为接受下个字符的时候我cpu给了3次机会，程序循环3次后，即计数变量mRunningTimeout自加3次之后，接受区的数据还为0x00(没更新)就属于超时了

我想要解决的问题

这个3是如何计算得出的，请大家指点迷津。