The figure below shows the block diagram of a filter implementation compromising two delays, five multipliers with real coefficients c,C2, C3, Cg, C5, and four adder elements. For the values of c1, ... Cs, use the values given in the table on the next page. y[n] (i) Find the transfer function of the filter. and Identify the type of the filter. (ii) Comment on the type of the filter if cı and es are zero. Find the impulse response of the filter.

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The figure below shows the block diagram of a filter implementation compromising
two delays, five multipliers with real coefficients c, C2, C3, C4, C5, and four adder
elements. For the values of c1, . Cs, use the values given in the table on the next page.
x[n]
vln)
C2
(i)
Find the transfer function of the filter. and Identify the type of the filter.
(ii) Comment on the type of the filter if c4 and cs are zero.
(iii) Find the impulse response of the filter.
Note: The following table gives the values to use in this question, based on the second
most rightmost digit of your student number (URN).
Second most
fs
rightmost digit of
student number
(kHz)
C1
C2
C3
C4
Cs
(URN)
0.1
100
-1.5
-1.1
0.15
1.0
1
0.2
100
0.35
1.2
0.3
100
-1.0
0.45
1.4
3
0.4
100
-2.2
0.65
1.7
4
0.45
100
-1.4
0.75
1.9
5
0.55
200
-1.6
0.85
2.0
0.6
200
-1.7
0.95
2.4
7
0.7
200
-1.8
-0.55
2.3
8
-0.75
-0.65
0.8
200
-1.9
2.5
9.
0.9
200
-2.1
2.6
Example: if your student number (URN) is “6789012" the second most rightmost digit
is “1", so you will use: = 0.2, f= 100 kHz, c1 = -1.1, c2 = 0.35, c3 = 1.2, cC4 = 0
and cs = 0.
Transcribed Image Text:The figure below shows the block diagram of a filter implementation compromising two delays, five multipliers with real coefficients c, C2, C3, C4, C5, and four adder elements. For the values of c1, . Cs, use the values given in the table on the next page. x[n] vln) C2 (i) Find the transfer function of the filter. and Identify the type of the filter. (ii) Comment on the type of the filter if c4 and cs are zero. (iii) Find the impulse response of the filter. Note: The following table gives the values to use in this question, based on the second most rightmost digit of your student number (URN). Second most fs rightmost digit of student number (kHz) C1 C2 C3 C4 Cs (URN) 0.1 100 -1.5 -1.1 0.15 1.0 1 0.2 100 0.35 1.2 0.3 100 -1.0 0.45 1.4 3 0.4 100 -2.2 0.65 1.7 4 0.45 100 -1.4 0.75 1.9 5 0.55 200 -1.6 0.85 2.0 0.6 200 -1.7 0.95 2.4 7 0.7 200 -1.8 -0.55 2.3 8 -0.75 -0.65 0.8 200 -1.9 2.5 9. 0.9 200 -2.1 2.6 Example: if your student number (URN) is “6789012" the second most rightmost digit is “1", so you will use: = 0.2, f= 100 kHz, c1 = -1.1, c2 = 0.35, c3 = 1.2, cC4 = 0 and cs = 0.
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