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TRANSIENT RESPONSE AND STEP RESPONSE

Module by: Nguyen Huu Phuong

In reality we often meet the cases where the signal is suddenly applied to a system and / or turned off. In such cases we can anticipate the system would react diffirently from normal. Reaction of systems when the applying signal is suddenly turned-on or turned-off is called transient response , and the normal reaction is called stable response or, more offen, steady-state response. Transient response is another important characteristic (characterization) of a system, though not as often used as the impulse response. Transient response reveals the character and speed of reaction of a system (like human being). Most of the cases we would like the system to reach its steady-state as fast as possible whilst as smoothly as possible.
The impulse response of a system does not let us know directly its transient response. For this purpose we use a different testing signal, i.e. the unit step u(n)u(n) size 12{u \( n \) } {}rather than the unit sample δ(n)δ(n) size 12{δ \( n \) } {}. The step response, denoted s(n), is the reaction of a system when a unit step u(n) is applied to it .
As the unit step is the running sum of the unit samples :
u(n)=∑k=0∞δ(n−k)u(n)=∑k=0∞δ(n−k) size 12{u \( n \) = Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {δ \( n - k \) } } {} (1)
the step response is the running sum of the impulse responses :
s(n)=∑k=0∞h(k)s(n)=∑k=0∞h(k) size 12{s \( n \) = Sum cSub { size 8{k=0} } cSup { size 8{ infinity } } {h \( k \) } } {} (2)
Example 1 
A causal system is described by the difference equation y ( n ) = 0 . 8y ( n − 1 ) + x ( n ) y ( n ) = 0 . 8y ( n − 1 ) + x ( n ) size 12{y \( n \) =0 "." 8y \( n - 1 \) +x \( n \) } {} Find
  1. The step response
  2. Response with respect to a digital rectangular pulse consisting of five samples of amplitude 1 from n=0n=0 size 12{n=0} {} to n=4n=4 size 12{n=4} {} Figure 1a.
Solution
First we find the impulse response. This was done previously in Example. The result is shown in Figure 1b.
  1. Now we apply Equation 2 to have
s ( 0 ) = h ( 0 ) = 1 s ( 1 ) = h ( 0 ) + h ( 1 ) = 1 + 0 . 8 = 1 . 8 s ( 2 ) = s ( 1 ) + h ( 2 ) = 1 + 0 . 8 + 0 . 8 2 = 2 . 44 s ( 3 ) = s ( 2 ) + h ( 3 ) = 1 + 0 . 8 + 0 . 8 2 + 0 . 8 3 = 2 . 952 . . . s ( ∞ ) = 1 + 0 . 8 + 0 . 8 2 + 0 . 8 3 + . . . = 1 1 − 0 . 8 = 5 . 0 s ( 0 ) = h ( 0 ) = 1 s ( 1 ) = h ( 0 ) + h ( 1 ) = 1 + 0 . 8 = 1 . 8 s ( 2 ) = s ( 1 ) + h ( 2 ) = 1 + 0 . 8 + 0 . 8 2 = 2 . 44 s ( 3 ) = s ( 2 ) + h ( 3 ) = 1 + 0 . 8 + 0 . 8 2 + 0 . 8 3 = 2 . 952 . . . s ( ∞ ) = 1 + 0 . 8 + 0 . 8 2 + 0 . 8 3 + . . . = 1 1 − 0 . 8 = 5 . 0 alignl { stack { size 12{s \( 0 \) =h \( 0 \) =1} {} # size 12{s \( 1 \) =h \( 0 \) +h \( 1 \) =1+0 "." 8=1 "." 8} {} # size 12{s \( 2 \) =s \( 1 \) +h \( 2 \) =1+0 "." 8+0 "." 8 rSup { size 8{2} } =2 "." "44"} {} # s \( 3 \) =s \( 2 \) +h \( 3 \) =1+0 "." 8+0 "." 8 rSup { size 8{2} } +0 "." 8 rSup { size 8{3} } =2 "." "952" {} # "." "." "." {} # s \( infinity \) =1+0 "." 8+0 "." 8 rSup { size 8{2} } +0 "." 8 rSup { size 8{3} } + "." "." "." = { {1} over {1 - 0 "." 8} } =5 "." 0 {} } } {}
The result is shown in Figure 1c. Notice that the steady-state value (in this case the final value) is 5.0. In theory, the system needs an infinite of time to settle down. In practive the settling time may be just tens up to a hundreds of time indices, depending on the precision equired.
  2. The output signal y(n)y(n) size 12{y \( n \) } {} with respect to the given rectangular pulse can be found in two ways. One way in to consider the pulse as consiting of 5 unit samples at n=0,2,3,4n=0,2,3,4 size 12{n=0, matrix { } 2, matrix { } 3, matrix { } 4} {}. Another way is to consider the pulse as consisting of a posstive – value unit step begun at n=0n=0 size 12{n=0} {}, and a negative-value unit step begun at n=5n=5 size 12{n=5} {}. Let’s use the second way:
y ( n ) = s ( n ) − s ( n − 5 ) y ( n ) = s ( n ) − s ( n − 5 ) size 12{y \( n \) =s \( n \) - s \( n - 5 \) } {}
Thus
y ( 0 ) = s ( 0 ) − s ( − 5 ) = 1 . 0 − 0 = 1 . 0 y ( 1 ) = s ( 1 ) − s ( − 4 ) = 1 . 8 − 0 = 1 . 8 y ( 2 ) = s ( 2 ) − s ( − 3 ) = 2 . 44 − 0 = 2 . 44 y ( 3 ) = s ( 3 ) − s ( − 2 ) = 2 . 952 − 0 = 2 . 952 y ( 4 ) = s ( 4 ) − s ( − 1 ) = 3 . 362 − 0 = 3 . 362 y ( 5 ) = s ( 5 ) − s ( 0 ) = 3 . 692 − 1 = 2 . 692 . . . y ( 0 ) = s ( 0 ) − s ( − 5 ) = 1 . 0 − 0 = 1 . 0 y ( 1 ) = s ( 1 ) − s ( − 4 ) = 1 . 8 − 0 = 1 . 8 y ( 2 ) = s ( 2 ) − s ( − 3 ) = 2 . 44 − 0 = 2 . 44 y ( 3 ) = s ( 3 ) − s ( − 2 ) = 2 . 952 − 0 = 2 . 952 y ( 4 ) = s ( 4 ) − s ( − 1 ) = 3 . 362 − 0 = 3 . 362 y ( 5 ) = s ( 5 ) − s ( 0 ) = 3 . 692 − 1 = 2 . 692 . . . alignl { stack { size 12{y \( 0 \) =s \( 0 \) - s \( - 5 \) =1 "." 0 - 0=1 "." 0} {} # size 12{y \( 1 \) =s \( 1 \) - s \( - 4 \) =1 "." 8 - 0=1 "." 8} {} # size 12{y \( 2 \) =s \( 2 \) - s \( - 3 \) =2 "." "44" - 0=2 "." "44"} {} # size 12{y \( 3 \) =s \( 3 \) - s \( - 2 \) =2 "." "952" - 0=2 "." "952"} {} # size 12{y \( 4 \) =s \( 4 \) - s \( - 1 \) =3 "." "362" - 0=3 "." "362"} {} # size 12{y \( 5 \) =s \( 5 \) - s \( 0 \) =3 "." "692" - 1=2 "." "692"} {} # size 12{ "." "." "." } {} } } {}
The result is shown in Figure 1d. Notice that since the response has not reached the steady state the falling amplitude is not symmetric with respect to the rising amplitude, in other word the input – on transient and the input – off transient are not the same.
Figure 1: Example

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