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V
4
3
2
1
0					t (sec)
0	1	2	3	4	5

5.5 SYSTEM RESPONSE

• Consider that the general form for a transfer function is,

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O

--- = G

I

This can be put in the discrete domain,

------ = G B

On ( )

In

To find the output (response) of a system we can rearrange the equation,

On = G( B) In

Now, we can refer back to the sampled data form, and find a system response ‘H’,

On = G( B) F( B) Sn = H( B) Sn

•We can try an example to estimate how a system would respond,

G( B)

=

0.085B

In =

0.005B

T = 0.005 s

1-------------------------– 0.915B

-------------------Sn

( 1 – B) 2

O

=

G( B) I

=

0.085B

0.005B

S

n

n

-------------------------

-------------------

n

1 – 0.915B

( 1 – B) 2

At this point we need to calculate values. To do this we need to get the expression in usable form (note: the ‘B’ terms in the denominator can’t be directly applied to Sn).

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APPROACH 1: LONG DIVISION

On	=			0.085B		0.005B	Sn
			-------------------------	– 0.915B		-------------------
			1			( 1 – B) 2
				0.000425B2
On	=		------------------------------------------------------------------------------
			1	– 2.915B + 2.830B2			Sn
							– 0.915B3

0.000425B2 + 0.001239B3 + 0.002409B4 + 0.003904B5 +

1 – 2.915B + 2.830B2 – 0.915B3 0.000425B2

0.000425B2 – 0.001239B3 + 0.001203B4 – 0.000389B5

ETC...

On = ( 0.000425B2 + 0.001239B3 + 0.002409B4 + 0.003904B5 + … ) Sn

Now we can write down the output values at given times,

Sample #	Time (s)	Output (V)
0	0.000	0.0
1	0.005	0.0
2	0.010	0.000425
3	0.015	0.001239
4	0.020	0.002409
5	0.025	0.003904
6	0.030	etc...

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APPROACH 2: PARTIAL FRACTIONS
																				–0.085
				0.085B							0.005B									----------------	0.915					B		0.005B
On	=		-------------------------								-------------------					Sn =			----------------------									-------------------			Sn
			1 – 0.915B ( 1 – B)												2									1				( 1 – B) 2
																			B –
																							0.915------------
				D								E					F
On	=							+					+						Sn
				----------------------	1					1-----------– B					-------------------
				B –											( 1 – B)			2
				0.915
																	–0.085			B
D =				lim				B –				1					----------------0.915								0.005B
																												2
											------------					----------------------								-------------------
				1							0.915							1						( 1 – B)
			-------B → 0.915-----													B – 0.915------------
	D =			lim						–		0.085					0.005B								=
												0.915------------		B			-------------------				2
				B →	1												( 1 – B)
					------------0.915

F =

E =

On = D( ) Sn

+ E( 1 + B + B2 + … ) S	n
+ FT( B + 2B2 + 3B3 + … ) S		n

Now we can write down the output values at given times,

Sample #	Time (s)	Output (V)
0	0.000	0.0
1	0.005	0.0
2	0.010
3	0.015
4	0.020
5	0.025
6	0.030	etc...