slider param_definition( minVal, maxVal, increment, pageIncrement, paramName )

{

statements

}

param_definition Parameter description that is printed on the slider object.

minVal An integer constant specifying the value to be passed to the function when the position of the slider is at its lowest level.

maxVal An integer constant specifying the value to be passed to the function when the position of the slider is at its highest level.

increment An integer constant specifying the increment added to the value each time the slider is moved one position.

pageIncrement An integer constant specifying the increment added to the value each time the slider is moved by one page.

paramName Parameter definition that is used inside the function.

The following example uses the slider keyword to add a volume control slider.

menuitem "My Functions";

slider VolumeControl(0, 10, 1, 1, volume)

{

/* initialize the target variable with the parameter passed

by the slider object. */

targVarVolume = volume;

}

GEL

Homework (solution)

1. Functional Units1. Functional Units

a. How many can perform an ADD? Name them.a. How many can perform an ADD? Name them.

b. Which support memory loads/stores?b. Which support memory loads/stores?

.M.M .S.S .D.D .L.L

six; .L1, .L2, .D1, .D2, .S1, .S2six; .L1, .L2, .D1, .D2, .S1, .S2

2. Memory Map2. Memory Map

a. How many a. How many externalexternal ranges exist on ‘C6201? ranges exist on ‘C6201?

FourFour

3. Conditional Code

a.a. Which registers can be used as cond’l registers?Which registers can be used as cond’l registers?

b.b. Which instructions can be conditional?Which instructions can be conditional?

4. Performance4. Performancea.a. What is the 'C6711 instruction cycle time?What is the 'C6711 instruction cycle time?

b.b. How can the 'C6711 execute 1200 MIPs?How can the 'C6711 execute 1200 MIPs?

A0, A1, A2, B0, B1, B2A0, A1, A2, B0, B1, B2

All of themAll of them

CLKOUT1CLKOUT1

1200 MIPs = 8 instructions (units) x 150 MHz1200 MIPs = 8 instructions (units) x 150 MHz

5. Coding Problems a. Move contents of A0-->A1a. Move contents of A0-->A1

MVMV .L1.L1 A0, A1A0, A1oror ADDADD .S1.S1 A0, 0, A1A0, 0, A1oror MPYMPY .M1.M1 A0, 1, A1A0, 1, A1(what’s the problem(what’s the problem

with this?)with this?)

5. Coding Problems a. Move contents of A0-->A1a. Move contents of A0-->A1

b. Move contents of CSR-->A1b. Move contents of CSR-->A1

c. Clear register A5c. Clear register A5

MVMV .L1.L1 A0, A1A0, A1oror ADDADD .S1.S1 A0, 0, A1A0, 0, A1oror MPYMPY .M1.M1 A0, 1, A1A0, 1, A1

ZEROZERO .S1.S1 A5A5oror SUBSUB .L1.L1 A5, A5, A5A5, A5, A5oror MPYMPY .M1.M1 A5, 0, A5A5, 0, A5oror CLRCLR .S1.S1 A5, 0, 31, A5A5, 0, 31, A5oror MVKMVK .S1.S1 0, A50, A5oror XORXOR .L1.L1 A5,A5,A5A5,A5,A5

(A0 can only be a (A0 can only be a 16-bit value)16-bit value)

MVCMVC CSR, A1CSR, A1

5. Coding Problems (cont’d) d. A2 = A0d. A2 = A022 + A1 + A1

e. If (B1 e. If (B1 0) then B2 = B5 * B6 0) then B2 = B5 * B6

f. A2 = A0 * A1 + 10f. A2 = A0 * A1 + 10

g. Load an g. Load an unsignedunsigned constant (19ABCh) into constant (19ABCh) into register A6. register A6.

MPY.M1MPY.M1 A0, A0, A2A0, A0, A2ADD.L1ADD.L1 A2, A1, A2A2, A1, A2

[B1] MPY.M2[B1] MPY.M2 B5, B6, B2B5, B6, B2

mvklmvkl .s1.s1 0x00019abc,a60x00019abc,a6mvkhmvkh .s1.s1 0x00019abc,a60x00019abc,a6

valuevalue .equ.equ 0x00019abc0x00019abc

mvkl.s1mvkl.s1 value,a6value,a6mvkh.s1mvkh.s1 value,a6value,a6

MPYMPY A0, A1, A2A0, A1, A2ADDADD 10, A2, A210, A2, A2

5. Coding Problems (cont’d) h.h. Load A7 with contents of mem1 and Load A7 with contents of mem1 and

post-increment the selected pointer.post-increment the selected pointer.

x16 memx16 mem

mem1mem1 10h10h

A7A7

load_mem1:load_mem1: MVKLMVKL .S1.S1 mem1,mem1, A6A6

MVKHMVKH .S1.S1 mem1,mem1, A6A6

LDHLDH .D1.D1 *A6++,*A6++, A7A7

Pointers and Double Pointers• Pointers – a variable that holds an address

Short int *Ptr, x[3];

Ptr = &x[1];

Function (ptr);

Void Function ( short *ptr)

{}

Pointers and Double Pointers• Pointers – a variable that holds an address

Short int *Ptr, x[3];

Ptr = &x[1];

Function (ptr);

Void Function ( short *ptr)

{}

Pointer is passedA-reg = &x[1]

Pointer ExampleMemory

0x100 Ptr = 0x102

0x101 X[0]

0x102 X[1]

0x103 X[2]

The value in the A-Reg = 0x102

Pointer address is passed• Pointers – a variable that holds an address

Short int *Ptr, x[3];

Ptr = &x[1];

Function (&ptr);

Void Function ( short **ptr)

{}

Address of pointer is passed• A-Reg = &ptr (not value in pointer)

Example Double Pointer

Memory

0x100 Ptr = 0x102

0x101 X[0]

0x102 X[1]

0x103 X[2]

The value in the A-Reg = 0x100

Example

Void function (short *ptr)

{

*ptr = 2;

}

// This means memory location 0x102 has a 2 in that location

Void function (short **ptr)

{

**ptr = 2;

// This means memory location 0x102 has a 2 in that location

*ptr = 2;

// this means memory location 0x100 has a 2 in its location

}

Examples

Memory

0x100 Ptr = &x[1] = 0x102

0x101 X[0]

0x102 X[1] = 2

0x103 X[2]

Memory

0x100 Ptr = &x[1] = 0x102

0x101 X[0]

0x102 X[1] = 2

Memory

0x100 2

0x101 X[0]

0x102 X[1]

Sine Wave Generation• Function Call to standard library

• Taylor Series

• Table Look-up

• Algorithm

Table LookUp

• Create a table of values for 1 cycle of a sine wave

• Create a function that takes table values and places them in an output array

• The higher the number of table entries, the higher the precision of the sine wave

Table LookUp

• A Table of 128 values from 0 to 2Pi is– Table[I] = sin (2*PI*I/128)

• Let Sampling Rate = 8000 hz, (Period=125 usec)

• If OutputArray[I] = Table[I]– 1cycle / (125usec*128) = 62.5 hz.

• If OutputArray[I] = Table[2*I]– 1 cycle / (125usec*64) = 125 hz.

Continued

• If OutputArray[I] = Table[128*I]– 1 cycle / (125usec * 1) = 8000 hz.– OutputArray[I] = 0 (DC)

• Resolution = 62 hz.

• For Higher resolution increase Table Size

Frequency as a function of step size and table size

64 128 512 1024

1 125 hz 62.5 hz 15.625 hz 7.8125 hz

2 250 hz 125 hz 31.25 hz 15.625 hz

32 4000 hz 2000 hz 500 hz 250 hz

128 x 8000 hz 2000 hz 1000 hz

stepsize

Table size

SineWave Oscillator

• H(z) = 1 / (1-a*z^-1 + z^-2)

• Y[n] = x[n] + a*y[n-1] + y[n-2]

SineWave Oscillator• Y[n] = A * sin (2PI * n * f / fs)

• Variables – frequency – f

– Sampling frequency – fs

– Amplitude - A

Z-Transform

21)cos(21

1)sin(][

211

1

2][

11

111

12

][

0 2][

/20

)/2sin(][

zz

zAzY

zzjeje

zjeje

jAzY

zjezjejAzY

nnz

j

jnwejneAzY

sffletn

nzsfnfAzX

SineWave Oscillator

• Y[n] is produced by treating Y[z] as a product of H(z)X(z), where X(z) = 1

• X[n] is an Impulse Function, x[0]=1,

• 0 elsewhere

• Now find the difference equation

][][ nnx

SineWave Oscillator

]2[]1[)cos(2]1[)sin(][

]1[)sin(]2[]1[)cos(2][

][][ 1)sin(21)cos(21

nynynxAny

nxAnynyny

zXzY zAzz

Example• Let A=1, f=1476Hz, fs=8000hz

• W= 2Pi*1476/8000=1.1592

• Sin(w)= 0.9165

• Cos(w)=0.8

Homework

• Add to “sine.gel” an Amplitude Controlled slider

• Verify it’s operation

• Page 115 – Problem 1– Problem 2