Motorola DSP56800 Manual page 254

Solution
To determine the number of instruction program words and the number of oscillator clock cycles required
for the given instruction, the user should perform the following steps:
1. Look up the number of instruction program words and the number of oscillator clock cycles
required for the opcode-operand portion of the instruction in Table A-11 on page A-18.
According to Table A-11 on page A-18, the Jcc instruction will require two instruction
program words and will execute in (4 + jx) oscillator clock cycles. The term "jx" represents
the number of additional oscillator clock cycles (if any) required for a jump-type
instruction.
2. Evaluate the "jx" term using Table A-16 on page A-20.
According to Table A-16 on page A-20, the Jcc instruction will require 2 + jx additional
oscillator clock cycles. If the "ea" condition is true, jx = 2 + 2 * ap, whereas jx = 2 * ap if
the condition is false. The term "ap" represents the number of additional oscillator clock
cycles (if any) that are required to access a P memory operand. Note that the "+ (2 * ap)"
term represents the two program memory instruction fetches executed at the end of a
one-word jump instruction to refill the instruction pipeline.
3. Evaluate the "ap" term using Table A-20 on page A-22.
According to Table A-20 on page A-22, the term "ap" depends upon where the referenced
P memory location is located in the 16-bit DSP memory space. External memory accesses
require additional oscillator clock cycles according to the number of wait states required.
Here we assume that external P memory accesses require wp = 4 wait states or additional
oscillator clock cycles. For this example the P memory reference is assumed to be an
external reference. Thus, according to Table A-20 on page A-22, the Jcc instruction will
use the value ap = wp = 4 oscillator clock cycles.
4. Compute the final results.
Thus, based upon the assumptions given for Table A-11 on page A-18, the instruction
JEQ
will require (1 + 1) = (1 + 1) = 2 instruction program word and will execute in (4 + jx) =
(4 + ea + (2 * ap)) = (4 + ea + (2 * wp)) = (4 + 2 + (2 * 4)) = 14 oscillator clock cycles.
A-24
Example A-2. Jump Instruction (Continued)
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