IIR filter realization using direct form I & II

1. IIR Filter Realization using Direct
Form I & II structures
- SARANG JOSHI

2. IIR Filter difference equation
ak y n − k 
k=0
N
 = bk x n − k 
k=0
M

Alternative equivalent form
      ==
−=−−
M
k
k
N
k
k knxbknyany
01

3. )(.....)2()1()(
)()........3()2()1()(
210
321
Mnxbnxbnxbnxb
Nnyanyanyanyany
k
k
−++−+−+=
−−−−−−−−
      ==
−=−−
M
k
k
N
k
k knxbknyany
01

4. )2()1(
)()2()1()(
equationdifferenceorder2
21
021
−+−+
+−−−−=
nxbnxb
nxbnyanyany
nd
)2()1()()( 210 −+−+= nxbnxbnxbnv

5. h(0)
x(n)
h(0)x(n)
x(n) x(n-1)Z
-1
x(n) x(n-1)Z
-1
h(1)x(n-1)
h(1)

6. )2()1()()( 210 −+−+= nxbnxbnxbnv
V(n)
System-I

7. )2()1()()( 21 −−−−= nyanyanvny
System-II

8. )2()1(
)()2()1()(
21
021
−+−+
+−−−−=
nxbnxb
nxbnyanyany

9. DIRECT FORM -II

12. )2(2)1(3)(4)2()1(2)(3 −+−−=−+−− nxnxnxnynyny
)2(
3
1
)1(
3
2
)2(
3
2
)1()(
3
4
)( −−−+−+−−= nynynxnxnxny
Q. Implement the filter represented by following difference equation
Solution :

13. )2(
3
2
)1()(
3
4
)( −+−−= nxnxnxnv
4/3
-1
2/3
System-I

14. )2(
3
1
)1(
3
2
)()( −−−+= nynynvny
2/3
-1/3
System-II

15. -1
4/3
2/3
2/3
-1/3
System-IISystem-I
Direct Form-I

17. Direct Form-II

18. 21
21
2.02.01
6.06.33
)( −−
−−
−+
++
=
zz
zz
zH
21
21
2.02.01
6.06.33
)(
)(
)( −−
−−
−+
++
==
zz
zz
zX
zY
zH
]6.06.33)[(]2.02.01)[( 2121 −−−−
++=−+ zzzXzzzY
)(6.0)(6.3)(3)(2.0)(2.0)( 2121
zXzzXzzXzYzzYzzY −−−−
++=−+
Q. Implement the filter represented by following transfer function
Solution :

19. )(6.0)(6.3)(3)(2.0)(2.0)( 2121
zXzzXzzXzYzzYzzY −−−−
++=−+
)2(6.0)1(6.3)(3)2(2.0)1(2.0)( −+−+=−−−+ nxnxnxnynyny
)2(2.0)1(2.0)2(6.0)1(6.3)(3)( −+−−−+−+= nynynxnxnxny
)(zk)-x(n
shiftingTime
k-
z
zX

20. )2(6.0)1(6.3)(3)( −+−+= nxnxnxnv
System-I

21. )2(2.0)1(2.0)()( −+−−= nynynvny
System-II

22. System-IISystem-I
Direct Form-I

23. Direct Form-II

24. RATE, FOLLOW & SHARE
https://unacademy.com/course/digital-signal-processing-2-engineering/302E2GCR

25. THANK YOU !