Digital Filter Design, TMS320C5X Processor Instructions & Decimation

This assignment provides solutions to three distinct problems in digital signal processing. The first solution details the design of a digital notch filter using the pole-zero placement method, specifying the transfer function, difference equation, and direct-form I and canonical-form realizations based on given notch frequency, 3dB width, and sampling frequency. The second solution focuses on designing an optimum multistage decimator for down-sampling an audio band signal from 80 kHz to 5 kHz, considering passband and stopband ripple specifications, and determining the number of filter coefficients required for each stage, along with block diagrams. Lastly, the assignment includes the execution of TMS320C5X series processor instructions, tracking the changes in register contents after each instruction, assuming any missing data to facilitate the execution.

Solution 1
Given
 Notch frequency: 25 Hz
 3 dB width of notch : +/- 4 Hz
 Sampling frequency : 200 Hz
To reject the component at 25Hz we can place a complex zeros pairs at points on the unit circle
corresponding to 25Hz that is at angles of
360 ° × 25
200 =± 45°
To attain a sharp notch filter and response of better amplitude on whichever side of the notch
frequency. Complex conjugate pair of poles are there at a radius r < 1. The notch width is
strongminded by the locations of the poles. The connection amid the bandwidth and radius is
similar as given below:
Radius
r =1−
( BW
Fsample )π
r =1−( 2∗4
200 ) π
r =0.8744
The connection between the bandwidth and the radius is appropriate. Therefore the radius of the
poles is 0.8744.

Transfer Function
From the transfer function of the filter we have:
H [ z ]= [z −e− j 45 ][ z−e j 45]
[z−0.8744 e− j 45 ][ z−0.8744 e j 45 ]
H [ z ]= z2−z e j 45−z e− j 45 +1
z2−z 0.8744 e j 45−0.8744 z e− j 45+ ( 0.8744 ) 2
H [ z ]= z2− z( ( 0.525+0.85 j ) )−z ( ( 0.525−0.85 j ) )+1
z2−z 0.8744∗( ( 0.525+0.85 j ) ) −0.8744 z∗( ( 0.525−0.85 j ))+ ( 0.8744 )2
H [ z ]= z2−z +1
z2−z 0.8744+0.77
H [ z ]= 1−z−1+z−2
1−0.8744 z−1+0.77 z−2
Difference Equation
Now difference equation can be derived as
Y ( z )
X (z )= 1−z−1+ z−2
1−0.8744 z−1+0.77 z−2
Y ( z )∗( 1−0.8744 z−1 +0.77 z−2 )=X ( z )∗(1−z−1 + z−2 )
(Y ( z ) −0.8744 z−1 Y ( z ) +0.77 z−2 Y ( z ) )=( X ( z ) −z−1 X ( z ) +z−2 X ( z ) )
Using inverse z transform

( y ( n)−0.8744 y (n−1)+0.77 y ( n−2))=(x (n)− x(n−1)+ x(n−2))
y ( n ) =x ( n ) −x ( n−1 ) + x ( n−2 ) + 0.8744 y ( n−1 ) −0.77 y (n−2)
We can have coefficient as
b0=1 b1=−1 b2 =1
a0=1 a1=−0.8744 a2=0.77
Direct-Form I
From
y ( n )=x ( n )−x ( n−1 )+ x ( n−2 )+0.8744 y ( n−1 ) −0.77 y (n−2)
Canonical-Form
From
y ( n )=x ( n )−x ( n−1 )+ x ( n−2 )+0.8744 y ( n−1 ) −0.77 y (n−2)

Solution 2
A Blackman window can attain a ripple of 75dB in stop band and ripple of 0.0014dB in
passband.
This could be likened with the necessities of this antialiasing filter of
δs ≈ 0.01 , which is−20 log(0.01)=40 dB
And a passband ripple
δp ≈ 0.1 ,∨20 log (1+ 0.1)=0.827 dB
Though, giving to the low pass FIR filter design strategies, the filter coefficients number for a
Blackman window would then be:
N=−10∗log ( δs δ p )−13
14.6∗∆ f +1
N= 27
14.6∗∆ f +1
N= 1.85
∆ f +1
N= 1.85
5/80 +1

N=30
Design
Stage 1
The original sampling frequency fs = 80 kHz and the new or we can say of the decimated signal
should have a sampling frequency of f new=5 kHz .
Decimation of Multistage type with 2 stages necessitates that:
f s
f new
=D
D= 80
5
D=16=D 1× D2.
Block Diagrams
Decimation in multistage:
The values of multistage decimation could hence be D1 = 4 and D2 = 4, making a middle signal
with sampling frequency:

f s 1= f s
4 =80
4 =20 kHZ
The transition width could be lengthier with this more sampling rate. We can retain the similar
passband frequency (2 kHz). The transition width could go up to half the sampling rate:
f tw ≤ 20 kHz
2 −2 kHz=8 kHz
The number of Blackman filter coefficients for this stage is:
N 1=5.98 × f s 1
f tw
N1=5.98 × 20
8 =15( rounded up)
So N1 = 8 filter coefficients are required for the first decimation stage.
Stage 2
The intermediate signal sampled at 20 kHz is to be decimated by a factor of 4 to 5 kHz
for the second stage:
f s 2= f s 1
4 = 5
4 =1.25 kHZ
The transition width can be longer with this higher sampling rate:

f tw ≤ 1.25 kHz
2 −2kHz=−ve kHz
So N1 = 8 filter coefficients are required for the first decimation stage.
Solution 3
Given
 DP = 6
 PM=1
 CNF= 1 [311h] = 64h [02100H] = 04h
TREG0 = 22h
 PREG = 01234567h
 ACC = 76543210h
 ARCR = 2530h
ARP = 3;
 AR3 = 2350h;
 INDX = 10h; [2350h] = 132h [50h] = 4680h
a. SAMM * 0 - , AR0
Meaning

 Accept that ARP = 3 and AR3 = 2350h. The ACC content is kept to the PMST sharp at
through the 7 LSBs of AR3.
 *0+ Increment by index amount. In this the current AR Content is utilized like the data
memory address. Prior to completion of the memory access, the INDX content is
augmented to the current AR contents.
Execution
SAMM Store ACCL in memory-mapped register
 AR=2350h [132 h]+ 10h = 142 h
b. XOR # 0ABBA h, 2
Long immediate shift by 2 places
 PREG = 01234569h
 ACC = 76543212h
c. MAC 02100h,11h
 PREG = 01234569h
 ACC = 76543212h
 AR= 153 h
d. ADD 11h, 3
 PREG = 01234569h
 ACC = 76543212h
 AR= 156 h

e. CMPR 2
 PREG = 01234571h
 ACC = 76543214h
 AR= 153 h