ENCM 515 W11
Midterm 1 Submitted Questions
Midterm Sample Questions
Question 1 (Part 1) – 5 Marks:
You are given the following code:
input[4] = {1, 2, 3, 5};
output = 0;
N = 4;
a0
a1
a2
a3
=
=
=
=
1;
2;
3;
5;
for (i = 0; i < 4; i++) {
output = filter(input[i]);
//Point 2
}
double filter(newvalue){
static float FIFO[N] = {0};
for (int i = 0; i < N-1; i++){
FIFO[i] = FIFO[i+1];
}
FIFO[N-1] = newvalue;
//Point 1
return a0 * FIFO[0] + a1 * FIFO[1] + a2 * FIFO[2] + a3 * FIFO[3];
}
For the above code, show the contents of FIFO at Point 1 and output at Point 2 for each of the four
iterations through the for loop.
i=0
FIFO: {____ , ____ , ____ , ____}
output: __________
i=1
FIFO: {____ , ____ , ____ , ____}
output: __________
i=2
FIFO: {____ , ____ , ____ , ____}
output: __________
i=3
FIFO: {____ , ____ , ____ , ____}
output: __________
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ENCM 515 W11
Midterm 1 Submitted Questions
Question 1 (Part 2) – 3 Marks:
Assume you are working with a moving average (low-pass) filter like the one you designed in lab 1.
Draw a rough sketch of the frequency response of the filter for N = 4 versus N = 16. Elaborate on the
differences between the two sketches.
N=4
N = 16
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Question 2 – 12 Marks:
There are several ways to design a high-pass FIR filter. Modify the following low-pass filter code to
produce a high-pass filter in C++ and assembly. If you'd like, you may use pseudo-code for your C++
implementation (not your assembly). Assume all variables used have already been properly allocated.
Some helpful skeleton code for your assembly implementation has been included.
// C++ Low Pass Filter
float sum = 0;
for (int i = 0; i < N; i++)
sum = sum + FIFO_Left[i];
float average = sum / N;
Left_Channel_Out1 = average;
// C++ High Pass Filter
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ENCM 515 W11
// Assembly
#define
#define
#define
#define
#define
#define
Midterm 1 Submitted Questions
sum_F5 F5
temp_F6 F6
count_R1 R1
FIFO_Length_R2 R2
FIFO_I4 I4
addressIndex_M4 M4
count_R1 = 0;
addressIndex_M4 = 0; // Set up even iterations
LOOP1:
CMP(count_R1, FIFO_Length_R2)
IF GE JUMP LOOP1_END (DB);
nop;
nop;
JUMP LOOP1;
LOO1_END:
LOOP2:
JUMP LOOP2;
LOO2_END:
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